Integrand size = 40, antiderivative size = 154 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) x+\frac {a \left (3 a^2 B-8 b^2 B-9 a b C\right ) \cot (c+d x)}{3 d}-\frac {a^2 (5 b B+3 a C) \cot ^2(c+d x)}{6 d}-\frac {\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \log (\sin (c+d x))}{d}-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d} \]
(B*a^3-3*B*a*b^2-3*C*a^2*b+C*b^3)*x+1/3*a*(3*B*a^2-8*B*b^2-9*C*a*b)*cot(d* x+c)/d-1/6*a^2*(5*B*b+3*C*a)*cot(d*x+c)^2/d-(3*B*a^2*b-B*b^3+C*a^3-3*C*a*b ^2)*ln(sin(d*x+c))/d-1/3*a*B*cot(d*x+c)^3*(a+b*tan(d*x+c))^2/d
Result contains complex when optimal does not.
Time = 1.34 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.06 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {6 a \left (a^2 B-3 b^2 B-3 a b C\right ) \cot (c+d x)-3 a^2 (3 b B+a C) \cot ^2(c+d x)-2 a^3 B \cot ^3(c+d x)+3 (a+i b)^3 (-i B+C) \log (i-\tan (c+d x))-6 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \log (\tan (c+d x))+3 (a-i b)^3 (i B+C) \log (i+\tan (c+d x))}{6 d} \]
(6*a*(a^2*B - 3*b^2*B - 3*a*b*C)*Cot[c + d*x] - 3*a^2*(3*b*B + a*C)*Cot[c + d*x]^2 - 2*a^3*B*Cot[c + d*x]^3 + 3*(a + I*b)^3*((-I)*B + C)*Log[I - Tan [c + d*x]] - 6*(3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*Log[Tan[c + d*x]] + 3*(a - I*b)^3*(I*B + C)*Log[I + Tan[c + d*x]])/(6*d)
Time = 1.16 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4115, 3042, 4088, 3042, 4118, 25, 3042, 4111, 27, 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.
\(\displaystyle \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan (c+d x)^2\right )}{\tan (c+d x)^5}dx\) |
\(\Big \downarrow \) 4115 |
\(\displaystyle \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (B+C \tan (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^3 (B+C \tan (c+d x))}{\tan (c+d x)^4}dx\) |
\(\Big \downarrow \) 4088 |
\(\displaystyle \frac {1}{3} \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (-b (a B-3 b C) \tan ^2(c+d x)-3 \left (B a^2-2 b C a-b^2 B\right ) \tan (c+d x)+a (5 b B+3 a C)\right )dx-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int \frac {(a+b \tan (c+d x)) \left (-b (a B-3 b C) \tan (c+d x)^2-3 \left (B a^2-2 b C a-b^2 B\right ) \tan (c+d x)+a (5 b B+3 a C)\right )}{\tan (c+d x)^3}dx-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 4118 |
\(\displaystyle \frac {1}{3} \left (\int -\cot ^2(c+d x) \left (b^2 (a B-3 b C) \tan ^2(c+d x)+3 \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B\right ) \tan (c+d x)+a \left (3 B a^2-9 b C a-8 b^2 B\right )\right )dx-\frac {a^2 (3 a C+5 b B) \cot ^2(c+d x)}{2 d}\right )-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (-\int \cot ^2(c+d x) \left (b^2 (a B-3 b C) \tan ^2(c+d x)+3 \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B\right ) \tan (c+d x)+a \left (3 B a^2-9 b C a-8 b^2 B\right )\right )dx-\frac {a^2 (3 a C+5 b B) \cot ^2(c+d x)}{2 d}\right )-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (-\int \frac {b^2 (a B-3 b C) \tan (c+d x)^2+3 \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B\right ) \tan (c+d x)+a \left (3 B a^2-9 b C a-8 b^2 B\right )}{\tan (c+d x)^2}dx-\frac {a^2 (3 a C+5 b B) \cot ^2(c+d x)}{2 d}\right )-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 4111 |
\(\displaystyle \frac {1}{3} \left (-\int 3 \cot (c+d x) \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B-\left (B a^3-3 b C a^2-3 b^2 B a+b^3 C\right ) \tan (c+d x)\right )dx+\frac {a \left (3 a^2 B-9 a b C-8 b^2 B\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a C+5 b B) \cot ^2(c+d x)}{2 d}\right )-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (-3 \int \cot (c+d x) \left (C a^3+3 b B a^2-3 b^2 C a-b^3 B-\left (B a^3-3 b C a^2-3 b^2 B a+b^3 C\right ) \tan (c+d x)\right )dx+\frac {a \left (3 a^2 B-9 a b C-8 b^2 B\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a C+5 b B) \cot ^2(c+d x)}{2 d}\right )-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (-3 \int \frac {C a^3+3 b B a^2-3 b^2 C a-b^3 B-\left (B a^3-3 b C a^2-3 b^2 B a+b^3 C\right ) \tan (c+d x)}{\tan (c+d x)}dx+\frac {a \left (3 a^2 B-9 a b C-8 b^2 B\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a C+5 b B) \cot ^2(c+d x)}{2 d}\right )-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {1}{3} \left (-3 \left (\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \int \cot (c+d x)dx-x \left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right )\right )+\frac {a \left (3 a^2 B-9 a b C-8 b^2 B\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a C+5 b B) \cot ^2(c+d x)}{2 d}\right )-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (-3 \left (\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-x \left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right )\right )+\frac {a \left (3 a^2 B-9 a b C-8 b^2 B\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a C+5 b B) \cot ^2(c+d x)}{2 d}\right )-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (-3 \left (-\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\left (x \left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right )\right )\right )+\frac {a \left (3 a^2 B-9 a b C-8 b^2 B\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a C+5 b B) \cot ^2(c+d x)}{2 d}\right )-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {1}{3} \left (\frac {a \left (3 a^2 B-9 a b C-8 b^2 B\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a C+5 b B) \cot ^2(c+d x)}{2 d}-3 \left (\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \log (-\sin (c+d x))}{d}-x \left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right )\right )\right )-\frac {a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\) |
((a*(3*a^2*B - 8*b^2*B - 9*a*b*C)*Cot[c + d*x])/d - (a^2*(5*b*B + 3*a*C)*C ot[c + d*x]^2)/(2*d) - 3*(-((a^3*B - 3*a*b^2*B - 3*a^2*b*C + b^3*C)*x) + ( (3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*Log[-Sin[c + d*x]])/d))/3 - (a*B*C ot[c + d*x]^3*(a + b*Tan[c + d*x])^2)/(3*d)
3.1.22.3.1 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[((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 = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.12
method | result | size |
parallelrisch | \(\frac {3 \left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+6 \left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-2 B \,a^{3} \cot \left (d x +c \right )^{3}+3 \left (-3 B \,a^{2} b -C \,a^{3}\right ) \cot \left (d x +c \right )^{2}+6 a \cot \left (d x +c \right ) \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right )+6 d x \left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right )}{6 d}\) | \(172\) |
derivativedivides | \(\frac {\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {B \,a^{3}}{3 \tan \left (d x +c \right )^{3}}-\frac {a^{2} \left (3 B b +C a \right )}{2 \tan \left (d x +c \right )^{2}}+\frac {a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right )}{\tan \left (d x +c \right )}+\frac {\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(174\) |
default | \(\frac {\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {B \,a^{3}}{3 \tan \left (d x +c \right )^{3}}-\frac {a^{2} \left (3 B b +C a \right )}{2 \tan \left (d x +c \right )^{2}}+\frac {a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right )}{\tan \left (d x +c \right )}+\frac {\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(174\) |
norman | \(\frac {\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) x \tan \left (d x +c \right )^{4}+\frac {a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right ) \tan \left (d x +c \right )^{3}}{d}-\frac {B \,a^{3} \tan \left (d x +c \right )}{3 d}-\frac {a^{2} \left (3 B b +C a \right ) \tan \left (d x +c \right )^{2}}{2 d}}{\tan \left (d x +c \right )^{4}}-\frac {\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(196\) |
risch | \(B \,a^{3} x -3 B a \,b^{2} x -3 C \,a^{2} b x +C \,b^{3} x -\frac {6 i C a \,b^{2} c}{d}+\frac {6 i B \,a^{2} b c}{d}-i B \,b^{3} x +i C \,a^{3} x -\frac {2 i a \left (9 i B a b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 B \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 B \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 C a b \,{\mathrm e}^{4 i \left (d x +c \right )}-9 i B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 i C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-18 B \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-18 C a b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 B \,a^{2}+9 B \,b^{2}+9 C a b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {2 i B \,b^{3} c}{d}+\frac {2 i C \,a^{3} c}{d}-3 i C a \,b^{2} x +3 i B \,a^{2} b x -\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B \,b^{3}}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C}{d}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C a \,b^{2}}{d}\) | \(383\) |
1/6*(3*(3*B*a^2*b-B*b^3+C*a^3-3*C*a*b^2)*ln(sec(d*x+c)^2)+6*(-3*B*a^2*b+B* b^3-C*a^3+3*C*a*b^2)*ln(tan(d*x+c))-2*B*a^3*cot(d*x+c)^3+3*(-3*B*a^2*b-C*a ^3)*cot(d*x+c)^2+6*a*cot(d*x+c)*(B*a^2-3*B*b^2-3*C*a*b)+6*d*x*(B*a^3-3*B*a *b^2-3*C*a^2*b+C*b^3))/d
Time = 0.25 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.18 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {3 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} + 2 \, B a^{3} + 3 \, {\left (C a^{3} + 3 \, B a^{2} b - 2 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{3} - 6 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \]
integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")
-1/6*(3*(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)^3 + 2*B*a^3 + 3*(C*a^3 + 3*B*a^2*b - 2*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*d*x)*tan(d*x + c)^3 - 6*(B*a^3 - 3*C*a^2* b - 3*B*a*b^2)*tan(d*x + c)^2 + 3*(C*a^3 + 3*B*a^2*b)*tan(d*x + c))/(d*tan (d*x + c)^3)
Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (150) = 300\).
Time = 4.11 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.10 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{3} \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{5}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\B a^{3} x + \frac {B a^{3}}{d \tan {\left (c + d x \right )}} - \frac {B a^{3}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {3 B a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {3 B a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 B a^{2} b}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 B a b^{2} x - \frac {3 B a b^{2}}{d \tan {\left (c + d x \right )}} - \frac {B b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {C a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {C a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {C a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 C a^{2} b x - \frac {3 C a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {3 C a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 C a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + C b^{3} x & \text {otherwise} \end {cases} \]
Piecewise((nan, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))**3*(B*tan(c) + C*t an(c)**2)*cot(c)**5, 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) + 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*b**2*x - 3*B*a*b**2/(d*tan(c + d*x)) - B*b**3*log(tan(c + d*x)**2 + 1)/(2*d) + B*b**3*log(tan(c + d*x))/d + 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) - 3*C*a**2*b*x - 3*C*a**2*b/(d*tan(c + d*x)) - 3*C*a*b**2*log(tan(c + d*x )**2 + 1)/(2*d) + 3*C*a*b**2*log(tan(c + d*x))/d + C*b**3*x, True))
Time = 0.34 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.17 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {6 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} {\left (d x + c\right )} + 3 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {2 \, B a^{3} - 6 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]
integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")
1/6*(6*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*(d*x + c) + 3*(C*a^3 + 3*B* a^2*b - 3*C*a*b^2 - B*b^3)*log(tan(d*x + c)^2 + 1) - 6*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*log(tan(d*x + c)) - (2*B*a^3 - 6*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2)*tan(d*x + c)^2 + 3*(C*a^3 + 3*B*a^2*b)*tan(d*x + c))/tan(d*x + c)^3)/d
Leaf count of result is larger than twice the leaf count of optimal. 390 vs. \(2 (148) = 296\).
Time = 1.43 (sec) , antiderivative size = 390, normalized size of antiderivative = 2.53 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} {\left (d x + c\right )} + 24 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 24 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {44 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 132 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 132 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]
1/24*(B*a^3*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^3*tan(1/2*d*x + 1/2*c)^2 - 9*B* a^2*b*tan(1/2*d*x + 1/2*c)^2 - 15*B*a^3*tan(1/2*d*x + 1/2*c) + 36*C*a^2*b* tan(1/2*d*x + 1/2*c) + 36*B*a*b^2*tan(1/2*d*x + 1/2*c) + 24*(B*a^3 - 3*C*a ^2*b - 3*B*a*b^2 + C*b^3)*(d*x + c) + 24*(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) - 24*(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))) + (44*C*a^3*tan(1/2*d*x + 1/2*c)^ 3 + 132*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 132*C*a*b^2*tan(1/2*d*x + 1/2*c)^ 3 - 44*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^2 - 36 *C*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 36*B*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*C* a^3*tan(1/2*d*x + 1/2*c) - 9*B*a^2*b*tan(1/2*d*x + 1/2*c) - B*a^3)/tan(1/2 *d*x + 1/2*c)^3)/d
Time = 8.51 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.10 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-C\,a^3-3\,B\,a^2\,b+3\,C\,a\,b^2+B\,b^3\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {C\,a^3}{2}+\frac {3\,B\,b\,a^2}{2}\right )+\frac {B\,a^3}{3}+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-B\,a^3+3\,C\,a^2\,b+3\,B\,a\,b^2\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]
(log(tan(c + d*x))*(B*b^3 - C*a^3 - 3*B*a^2*b + 3*C*a*b^2))/d - (cot(c + d *x)^3*(tan(c + d*x)*((C*a^3)/2 + (3*B*a^2*b)/2) + (B*a^3)/3 + tan(c + d*x) ^2*(3*B*a*b^2 - B*a^3 + 3*C*a^2*b)))/d - (log(tan(c + d*x) - 1i)*(B + C*1i )*(a + b*1i)^3*1i)/(2*d) + (log(tan(c + d*x) + 1i)*(B - C*1i)*(a - b*1i)^3 *1i)/(2*d)