Integrand size = 40, antiderivative size = 233 \[ \int \cot ^7(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\left (\left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) x\right )-\frac {\left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) \cot (c+d x)}{d}+\frac {\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \cot ^2(c+d x)}{2 d}+\frac {a \left (5 a^2 B-12 b^2 B-15 a b C\right ) \cot ^3(c+d x)}{15 d}-\frac {a^2 (7 b B+5 a C) \cot ^4(c+d x)}{20 d}+\frac {\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \log (\sin (c+d x))}{d}-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d} \]
-(B*a^3-3*B*a*b^2-3*C*a^2*b+C*b^3)*x-(B*a^3-3*B*a*b^2-3*C*a^2*b+C*b^3)*cot (d*x+c)/d+1/2*(3*B*a^2*b-B*b^3+C*a^3-3*C*a*b^2)*cot(d*x+c)^2/d+1/15*a*(5*B *a^2-12*B*b^2-15*C*a*b)*cot(d*x+c)^3/d-1/20*a^2*(7*B*b+5*C*a)*cot(d*x+c)^4 /d+(3*B*a^2*b-B*b^3+C*a^3-3*C*a*b^2)*ln(sin(d*x+c))/d-1/5*a*B*cot(d*x+c)^5 *(a+b*tan(d*x+c))^2/d
Result contains complex when optimal does not.
Time = 1.24 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.02 \[ \int \cot ^7(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {-60 \left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) \cot (c+d x)+30 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \cot ^2(c+d x)+20 a \left (a^2 B-3 b^2 B-3 a b C\right ) \cot ^3(c+d x)-15 a^2 (3 b B+a C) \cot ^4(c+d x)-12 a^3 B \cot ^5(c+d x)+30 i (a+i b)^3 (B+i C) \log (i-\tan (c+d x))+60 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \log (\tan (c+d x))+30 (i a+b)^3 (B-i C) \log (i+\tan (c+d x))}{60 d} \]
(-60*(a^3*B - 3*a*b^2*B - 3*a^2*b*C + b^3*C)*Cot[c + d*x] + 30*(3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*Cot[c + d*x]^2 + 20*a*(a^2*B - 3*b^2*B - 3*a* b*C)*Cot[c + d*x]^3 - 15*a^2*(3*b*B + a*C)*Cot[c + d*x]^4 - 12*a^3*B*Cot[c + d*x]^5 + (30*I)*(a + I*b)^3*(B + I*C)*Log[I - Tan[c + d*x]] + 60*(3*a^2 *b*B - b^3*B + a^3*C - 3*a*b^2*C)*Log[Tan[c + d*x]] + 30*(I*a + b)^3*(B - I*C)*Log[I + Tan[c + d*x]])/(60*d)
Time = 1.72 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.04, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4115, 3042, 4088, 3042, 4118, 25, 3042, 4111, 27, 3042, 4012, 25, 3042, 4012, 3042, 4014, 3042, 25, 3956}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \cot ^7(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan (c+d x)^2\right )}{\tan (c+d x)^7}dx\) |
| \(\Big \downarrow \) 4115 |
| \(\displaystyle \int \cot ^6(c+d x) (a+b \tan (c+d x))^3 (B+C \tan (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^3 (B+C \tan (c+d x))}{\tan (c+d x)^6}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {1}{5} \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (-b (3 a B-5 b C) \tan ^2(c+d x)-5 \left (B a^2-2 b C a-b^2 B\right ) \tan (c+d x)+a (7 b B+5 a C)\right )dx-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {(a+b \tan (c+d x)) \left (-b (3 a B-5 b C) \tan (c+d x)^2-5 \left (B a^2-2 b C a-b^2 B\right ) \tan (c+d x)+a (7 b B+5 a C)\right )}{\tan (c+d x)^5}dx-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4118 |
| \(\displaystyle \frac {1}{5} \left (\int -\cot ^4(c+d x) \left (b^2 (3 a B-5 b C) \tan ^2(c+d x)+5 \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B\right ) \tan (c+d x)+a \left (5 B a^2-15 b C a-12 b^2 B\right )\right )dx-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (-\int \cot ^4(c+d x) \left (b^2 (3 a B-5 b C) \tan ^2(c+d x)+5 \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B\right ) \tan (c+d x)+a \left (5 B a^2-15 b C a-12 b^2 B\right )\right )dx-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-\int \frac {b^2 (3 a B-5 b C) \tan (c+d x)^2+5 \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B\right ) \tan (c+d x)+a \left (5 B a^2-15 b C a-12 b^2 B\right )}{\tan (c+d x)^4}dx-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4111 |
| \(\displaystyle \frac {1}{5} \left (-\int 5 \cot ^3(c+d x) \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B-\left (B a^3-3 b C a^2-3 b^2 B a+b^3 C\right ) \tan (c+d x)\right )dx+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-5 \int \cot ^3(c+d x) \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B-\left (B a^3-3 b C a^2-3 b^2 B a+b^3 C\right ) \tan (c+d x)\right )dx+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-5 \int \frac {C a^3+3 b B a^2-3 b^2 C a-b^3 B-\left (B a^3-3 b C a^2-3 b^2 B a+b^3 C\right ) \tan (c+d x)}{\tan (c+d x)^3}dx+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\int -\cot ^2(c+d x) \left (B a^3-3 b C a^2-3 b^2 B a+b^3 C+\left (C a^3+3 b B a^2-3 b^2 C a-b^3 B\right ) \tan (c+d x)\right )dx-\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot ^2(c+d x)}{2 d}\right )+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (-\int \cot ^2(c+d x) \left (B a^3-3 b C a^2-3 b^2 B a+b^3 C+\left (C a^3+3 b B a^2-3 b^2 C a-b^3 B\right ) \tan (c+d x)\right )dx-\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot ^2(c+d x)}{2 d}\right )+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (-\int \frac {B a^3-3 b C a^2-3 b^2 B a+b^3 C+\left (C a^3+3 b B a^2-3 b^2 C a-b^3 B\right ) \tan (c+d x)}{\tan (c+d x)^2}dx-\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot ^2(c+d x)}{2 d}\right )+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (-\int \cot (c+d x) \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B-\left (B a^3-3 b C a^2-3 b^2 B a+b^3 C\right ) \tan (c+d x)\right )dx-\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot ^2(c+d x)}{2 d}+\frac {\left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right ) \cot (c+d x)}{d}\right )+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (-\int \frac {C a^3+3 b B a^2-3 b^2 C a-b^3 B-\left (B a^3-3 b C a^2-3 b^2 B a+b^3 C\right ) \tan (c+d x)}{\tan (c+d x)}dx-\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot ^2(c+d x)}{2 d}+\frac {\left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right ) \cot (c+d x)}{d}\right )+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4014 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (-\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \int \cot (c+d x)dx-\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot ^2(c+d x)}{2 d}+\frac {\left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right ) \cot (c+d x)}{d}+x \left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right )\right )+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (-\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot ^2(c+d x)}{2 d}+\frac {\left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right ) \cot (c+d x)}{d}+x \left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right )\right )+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot ^2(c+d x)}{2 d}+\frac {\left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right ) \cot (c+d x)}{d}+x \left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right )\right )+\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle \frac {1}{5} \left (\frac {a \left (5 a^2 B-15 a b C-12 b^2 B\right ) \cot ^3(c+d x)}{3 d}-\frac {a^2 (5 a C+7 b B) \cot ^4(c+d x)}{4 d}-5 \left (-\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot ^2(c+d x)}{2 d}+\frac {\left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right ) \cot (c+d x)}{d}-\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \log (-\sin (c+d x))}{d}+x \left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right )\right )\right )-\frac {a B \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\) |
((a*(5*a^2*B - 12*b^2*B - 15*a*b*C)*Cot[c + d*x]^3)/(3*d) - (a^2*(7*b*B + 5*a*C)*Cot[c + d*x]^4)/(4*d) - 5*((a^3*B - 3*a*b^2*B - 3*a^2*b*C + b^3*C)* x + ((a^3*B - 3*a*b^2*B - 3*a^2*b*C + b^3*C)*Cot[c + d*x])/d - ((3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*Cot[c + d*x]^2)/(2*d) - ((3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*Log[-Sin[c + d*x]])/d))/5 - (a*B*Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2)/(5*d)
3.1.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N eQ[a*c + b*d, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x ])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d* (b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1) + a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[ e + f*x] - b*(d*(A*b*c + a*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] & & LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x ] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B , C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0 ]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*(b*B - a*C + b*C*Tan[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_. )*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(c^2*C - B*c*d + A*d^2)*((c + d*Tan[e + f*x])^(n + 1)/(d^2*f*(n + 1)*(c^2 + d^2))), x] + Simp[1/(d*(c^2 + d^2)) Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b* (c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d + a*C*d) *Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n , -1]
Time = 0.36 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.04
| method | result | size |
| parallelrisch | \(\frac {\left (-90 B \,a^{2} b +30 B \,b^{3}-30 C \,a^{3}+90 C a \,b^{2}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (180 B \,a^{2} b -60 B \,b^{3}+60 C \,a^{3}-180 C a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-12 B \,a^{3} \cot \left (d x +c \right )^{5}+\left (-45 B \,a^{2} b -15 C \,a^{3}\right ) \cot \left (d x +c \right )^{4}+20 a \cot \left (d x +c \right )^{3} \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right )+\left (90 B \,a^{2} b -30 B \,b^{3}+30 C \,a^{3}-90 C a \,b^{2}\right ) \cot \left (d x +c \right )^{2}+\left (-60 B \,a^{3}+180 B a \,b^{2}+180 C \,a^{2} b -60 C \,b^{3}\right ) \cot \left (d x +c \right )-60 d x \left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right )}{60 d}\) | \(243\) |
| derivativedivides | \(\frac {-\frac {-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}}{2 \tan \left (d x +c \right )^{2}}+\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}}{\tan \left (d x +c \right )}-\frac {B \,a^{3}}{5 \tan \left (d x +c \right )^{5}}-\frac {a^{2} \left (3 B b +C a \right )}{4 \tan \left (d x +c \right )^{4}}+\frac {a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right )}{3 \tan \left (d x +c \right )^{3}}+\frac {\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-B \,a^{3}+3 B a \,b^{2}+3 C \,a^{2} b -C \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(248\) |
| default | \(\frac {-\frac {-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}}{2 \tan \left (d x +c \right )^{2}}+\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}}{\tan \left (d x +c \right )}-\frac {B \,a^{3}}{5 \tan \left (d x +c \right )^{5}}-\frac {a^{2} \left (3 B b +C a \right )}{4 \tan \left (d x +c \right )^{4}}+\frac {a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right )}{3 \tan \left (d x +c \right )^{3}}+\frac {\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-B \,a^{3}+3 B a \,b^{2}+3 C \,a^{2} b -C \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(248\) |
| norman | \(\frac {\left (-B \,a^{3}+3 B a \,b^{2}+3 C \,a^{2} b -C \,b^{3}\right ) x \tan \left (d x +c \right )^{6}+\frac {\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \tan \left (d x +c \right )^{4}}{2 d}-\frac {\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \tan \left (d x +c \right )^{5}}{d}-\frac {B \,a^{3} \tan \left (d x +c \right )}{5 d}+\frac {a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right ) \tan \left (d x +c \right )^{3}}{3 d}-\frac {a^{2} \left (3 B b +C a \right ) \tan \left (d x +c \right )^{2}}{4 d}}{\tan \left (d x +c \right )^{6}}+\frac {\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(275\) |
| risch | \(-B \,a^{3} x +3 B a \,b^{2} x +3 C \,a^{2} b x -C \,b^{3} x -\frac {6 i B \,a^{2} b c}{d}-\frac {2 i \left (-60 C \,a^{2} b -60 B a \,b^{2}+15 C \,b^{3}+23 B \,a^{3}-70 B \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-60 C \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-45 i C a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+180 i B \,a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-135 i C a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-180 i B \,a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+135 i C a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+90 i B \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-90 i B \,a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+45 i C a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+90 C \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-90 B \,a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+45 B \,a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+15 C \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-60 C \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+140 B \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-90 C \,a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+270 B a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+270 C \,a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-330 B a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-330 C \,a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+210 B a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+210 C \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-60 i C \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-15 i B \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+30 i C \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-90 B a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+15 i B \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-30 i C \,a^{3} {\mathrm e}^{8 i \left (d x +c \right )}-45 i B \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+60 i C \,a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+45 i B \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+i B \,b^{3} x -i C \,a^{3} x +\frac {6 i C a \,b^{2} c}{d}+\frac {2 i B \,b^{3} c}{d}-\frac {2 i C \,a^{3} c}{d}+3 i C a \,b^{2} x -3 i B \,a^{2} b x +\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B \,b^{3}}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C}{d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C a \,b^{2}}{d}\) | \(755\) |
1/60*((-90*B*a^2*b+30*B*b^3-30*C*a^3+90*C*a*b^2)*ln(sec(d*x+c)^2)+(180*B*a ^2*b-60*B*b^3+60*C*a^3-180*C*a*b^2)*ln(tan(d*x+c))-12*B*a^3*cot(d*x+c)^5+( -45*B*a^2*b-15*C*a^3)*cot(d*x+c)^4+20*a*cot(d*x+c)^3*(B*a^2-3*B*b^2-3*C*a* b)+(90*B*a^2*b-30*B*b^3+30*C*a^3-90*C*a*b^2)*cot(d*x+c)^2+(-60*B*a^3+180*B *a*b^2+180*C*a^2*b-60*C*b^3)*cot(d*x+c)-60*d*x*(B*a^3-3*B*a*b^2-3*C*a^2*b+ C*b^3))/d
Time = 0.26 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.14 \[ \int \cot ^7(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {30 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{5} + 15 \, {\left (3 \, C a^{3} + 9 \, B a^{2} b - 6 \, C a b^{2} - 2 \, B b^{3} - 4 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{5} - 60 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \tan \left (d x + c\right )^{4} - 12 \, B a^{3} + 30 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )^{3} + 20 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \tan \left (d x + c\right )^{2} - 15 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{60 \, d \tan \left (d x + c\right )^{5}} \]
integrate(cot(d*x+c)^7*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")
1/60*(30*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*log(tan(d*x + c)^2/(tan(d *x + c)^2 + 1))*tan(d*x + c)^5 + 15*(3*C*a^3 + 9*B*a^2*b - 6*C*a*b^2 - 2*B *b^3 - 4*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*d*x)*tan(d*x + c)^5 - 60* (B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*tan(d*x + c)^4 - 12*B*a^3 + 30*(C* a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*tan(d*x + c)^3 + 20*(B*a^3 - 3*C*a^2* b - 3*B*a*b^2)*tan(d*x + c)^2 - 15*(C*a^3 + 3*B*a^2*b)*tan(d*x + c))/(d*ta n(d*x + c)^5)
Time = 12.36 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.98 \[ \int \cot ^7(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{3} \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{7}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\- B a^{3} x - \frac {B a^{3}}{d \tan {\left (c + d x \right )}} + \frac {B a^{3}}{3 d \tan ^{3}{\left (c + d x \right )}} - \frac {B a^{3}}{5 d \tan ^{5}{\left (c + d x \right )}} - \frac {3 B a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 B a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {3 B a^{2} b}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {3 B a^{2} b}{4 d \tan ^{4}{\left (c + d x \right )}} + 3 B a b^{2} x + \frac {3 B a b^{2}}{d \tan {\left (c + d x \right )}} - \frac {B a b^{2}}{d \tan ^{3}{\left (c + d x \right )}} + \frac {B b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B b^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {C a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {C a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {C a^{3}}{4 d \tan ^{4}{\left (c + d x \right )}} + 3 C a^{2} b x + \frac {3 C a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {C a^{2} b}{d \tan ^{3}{\left (c + d x \right )}} + \frac {3 C a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {3 C a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 C a b^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - C b^{3} x - \frac {C b^{3}}{d \tan {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]
Piecewise((nan, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))**3*(B*tan(c) + C*t an(c)**2)*cot(c)**7, Eq(d, 0)), (nan, Eq(c, -d*x)), (-B*a**3*x - B*a**3/(d *tan(c + d*x)) + B*a**3/(3*d*tan(c + d*x)**3) - B*a**3/(5*d*tan(c + d*x)** 5) - 3*B*a**2*b*log(tan(c + d*x)**2 + 1)/(2*d) + 3*B*a**2*b*log(tan(c + d* x))/d + 3*B*a**2*b/(2*d*tan(c + d*x)**2) - 3*B*a**2*b/(4*d*tan(c + d*x)**4 ) + 3*B*a*b**2*x + 3*B*a*b**2/(d*tan(c + d*x)) - B*a*b**2/(d*tan(c + d*x)* *3) + B*b**3*log(tan(c + d*x)**2 + 1)/(2*d) - B*b**3*log(tan(c + d*x))/d - B*b**3/(2*d*tan(c + d*x)**2) - C*a**3*log(tan(c + d*x)**2 + 1)/(2*d) + C* a**3*log(tan(c + d*x))/d + C*a**3/(2*d*tan(c + d*x)**2) - C*a**3/(4*d*tan( c + d*x)**4) + 3*C*a**2*b*x + 3*C*a**2*b/(d*tan(c + d*x)) - C*a**2*b/(d*ta n(c + d*x)**3) + 3*C*a*b**2*log(tan(c + d*x)**2 + 1)/(2*d) - 3*C*a*b**2*lo g(tan(c + d*x))/d - 3*C*a*b**2/(2*d*tan(c + d*x)**2) - C*b**3*x - C*b**3/( d*tan(c + d*x)), True))
Time = 0.38 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.07 \[ \int \cot ^7(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {60 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} {\left (d x + c\right )} + 30 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {60 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \tan \left (d x + c\right )^{4} + 12 \, B a^{3} - 30 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )^{3} - 20 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \tan \left (d x + c\right )^{2} + 15 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]
integrate(cot(d*x+c)^7*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")
-1/60*(60*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*(d*x + c) + 30*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*log(tan(d*x + c)^2 + 1) - 60*(C*a^3 + 3*B*a ^2*b - 3*C*a*b^2 - B*b^3)*log(tan(d*x + c)) + (60*(B*a^3 - 3*C*a^2*b - 3*B *a*b^2 + C*b^3)*tan(d*x + c)^4 + 12*B*a^3 - 30*(C*a^3 + 3*B*a^2*b - 3*C*a* b^2 - B*b^3)*tan(d*x + c)^3 - 20*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2)*tan(d*x + c)^2 + 15*(C*a^3 + 3*B*a^2*b)*tan(d*x + c))/tan(d*x + c)^5)/d
Leaf count of result is larger than twice the leaf count of optimal. 670 vs. \(2 (225) = 450\).
Time = 1.57 (sec) , antiderivative size = 670, normalized size of antiderivative = 2.88 \[ \int \cot ^7(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {6 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 45 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 70 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 180 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 540 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 360 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 660 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1800 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1800 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 480 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 960 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} {\left (d x + c\right )} - 960 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 960 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2192 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6576 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6576 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2192 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 660 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1800 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1800 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 480 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 180 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 540 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 70 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 45 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]
1/960*(6*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 15*C*a^3*tan(1/2*d*x + 1/2*c)^4 - 45*B*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 70*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 120* C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 120*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 180* C*a^3*tan(1/2*d*x + 1/2*c)^2 + 540*B*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 360*C* a*b^2*tan(1/2*d*x + 1/2*c)^2 - 120*B*b^3*tan(1/2*d*x + 1/2*c)^2 + 660*B*a^ 3*tan(1/2*d*x + 1/2*c) - 1800*C*a^2*b*tan(1/2*d*x + 1/2*c) - 1800*B*a*b^2* tan(1/2*d*x + 1/2*c) + 480*C*b^3*tan(1/2*d*x + 1/2*c) - 960*(B*a^3 - 3*C*a ^2*b - 3*B*a*b^2 + C*b^3)*(d*x + c) - 960*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*log(tan(1/2*d*x + 1/2*c)^2 + 1) + 960*(C*a^3 + 3*B*a^2*b - 3*C*a*b ^2 - B*b^3)*log(abs(tan(1/2*d*x + 1/2*c))) - (2192*C*a^3*tan(1/2*d*x + 1/2 *c)^5 + 6576*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 6576*C*a*b^2*tan(1/2*d*x + 1 /2*c)^5 - 2192*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 660*B*a^3*tan(1/2*d*x + 1/2* c)^4 - 1800*C*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 1800*B*a*b^2*tan(1/2*d*x + 1/ 2*c)^4 + 480*C*b^3*tan(1/2*d*x + 1/2*c)^4 - 180*C*a^3*tan(1/2*d*x + 1/2*c) ^3 - 540*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 360*C*a*b^2*tan(1/2*d*x + 1/2*c) ^3 + 120*B*b^3*tan(1/2*d*x + 1/2*c)^3 - 70*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 120*C*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 120*B*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 15*C*a^3*tan(1/2*d*x + 1/2*c) + 45*B*a^2*b*tan(1/2*d*x + 1/2*c) + 6*B*a^3) /tan(1/2*d*x + 1/2*c)^5)/d
Time = 8.43 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.02 \[ \int \cot ^7(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {{\mathrm {cot}\left (c+d\,x\right )}^5\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {C\,a^3}{4}+\frac {3\,B\,b\,a^2}{4}\right )+\frac {B\,a^3}{5}+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {B\,a^3}{3}+C\,a^2\,b+B\,a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (B\,a^3-3\,C\,a^2\,b-3\,B\,a\,b^2+C\,b^3\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-\frac {C\,a^3}{2}-\frac {3\,B\,a^2\,b}{2}+\frac {3\,C\,a\,b^2}{2}+\frac {B\,b^3}{2}\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-C\,a^3-3\,B\,a^2\,b+3\,C\,a\,b^2+B\,b^3\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]
(log(tan(c + d*x) - 1i)*(B + C*1i)*(a + b*1i)^3*1i)/(2*d) - (log(tan(c + d *x))*(B*b^3 - C*a^3 - 3*B*a^2*b + 3*C*a*b^2))/d - (cot(c + d*x)^5*(tan(c + d*x)*((C*a^3)/4 + (3*B*a^2*b)/4) + (B*a^3)/5 + tan(c + d*x)^2*(B*a*b^2 - (B*a^3)/3 + C*a^2*b) + tan(c + d*x)^4*(B*a^3 + C*b^3 - 3*B*a*b^2 - 3*C*a^2 *b) + tan(c + d*x)^3*((B*b^3)/2 - (C*a^3)/2 - (3*B*a^2*b)/2 + (3*C*a*b^2)/ 2)))/d - (log(tan(c + d*x) + 1i)*(B - C*1i)*(a - b*1i)^3*1i)/(2*d)