Integrand size = 33, antiderivative size = 219 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {\left (8 a^2 A-3 A b^2-12 a b B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{4 \sqrt {a} d}-\frac {(a-i b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {(5 A b+4 a B) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d} \]
-(a-I*b)^(3/2)*(A-I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d-(a+ I*b)^(3/2)*(A+I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d+1/4*(8* A*a^2-3*A*b^2-12*B*a*b)*arctanh((a+b*tan(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)- 1/4*(5*A*b+4*B*a)*cot(d*x+c)*(a+b*tan(d*x+c))^(1/2)/d-1/2*a*A*cot(d*x+c)^2 *(a+b*tan(d*x+c))^(1/2)/d
Time = 2.70 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.89 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {\left (8 a^2 A-3 A b^2-12 a b B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )-\sqrt {a} \left (4 (a-i b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+4 (a+i b)^{3/2} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\cot (c+d x) (5 A b+4 a B+2 a A \cot (c+d x)) \sqrt {a+b \tan (c+d x)}\right )}{4 \sqrt {a} d} \]
((8*a^2*A - 3*A*b^2 - 12*a*b*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]] - Sqrt[a]*(4*(a - I*b)^(3/2)*(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sq rt[a - I*b]] + 4*(a + I*b)^(3/2)*(A + I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x] ]/Sqrt[a + I*b]] + Cot[c + d*x]*(5*A*b + 4*a*B + 2*a*A*Cot[c + d*x])*Sqrt[ a + b*Tan[c + d*x]]))/(4*Sqrt[a]*d)
Time = 1.85 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 4088, 27, 3042, 4132, 27, 3042, 4136, 27, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 73, 221}
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.
| \(\displaystyle \int \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan (c+d x)^3}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {1}{2} \int \frac {\cot ^2(c+d x) \left (-b (3 a A-4 b B) \tan ^2(c+d x)-4 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (5 A b+4 a B)\right )}{2 \sqrt {a+b \tan (c+d x)}}dx-\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \frac {\cot ^2(c+d x) \left (-b (3 a A-4 b B) \tan ^2(c+d x)-4 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (5 A b+4 a B)\right )}{\sqrt {a+b \tan (c+d x)}}dx-\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {-b (3 a A-4 b B) \tan (c+d x)^2-4 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (5 A b+4 a B)}{\tan (c+d x)^2 \sqrt {a+b \tan (c+d x)}}dx-\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {\cot (c+d x) \left (a b (5 A b+4 a B) \tan ^2(c+d x)+8 a \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (8 A a^2-12 b B a-3 A b^2\right )\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {\cot (c+d x) \left (a b (5 A b+4 a B) \tan ^2(c+d x)+8 a \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (8 A a^2-12 b B a-3 A b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx}{2 a}-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {a b (5 A b+4 a B) \tan (c+d x)^2+8 a \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (8 A a^2-12 b B a-3 A b^2\right )}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx}{2 a}-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\) |
| \(\Big \downarrow \) 4136 |
| \(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {8 \left (a \left (B a^2+2 A b a-b^2 B\right )-a \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)}}dx+a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {\cot (c+d x) \left (\tan ^2(c+d x)+1\right )}{\sqrt {a+b \tan (c+d x)}}dx}{2 a}-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (-\frac {8 \int \frac {a \left (B a^2+2 A b a-b^2 B\right )-a \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {\cot (c+d x) \left (\tan ^2(c+d x)+1\right )}{\sqrt {a+b \tan (c+d x)}}dx}{2 a}-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (-\frac {8 \int \frac {a \left (B a^2+2 A b a-b^2 B\right )-a \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx}{2 a}-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle -\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+8 \left (\frac {1}{2} a (a-i b)^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a (a+i b)^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )}{2 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+8 \left (\frac {1}{2} a (a-i b)^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a (a+i b)^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )}{2 a}\right )\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle -\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+8 \left (\frac {i a (a-i b)^2 (B+i A) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i a (a+i b)^2 (-B+i A) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\right )}{2 a}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+8 \left (-\frac {i a (a-i b)^2 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i a (a+i b)^2 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\right )}{2 a}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+8 \left (\frac {a (a-i b)^2 (B+i A) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}-\frac {a (a+i b)^2 (-B+i A) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}\right )}{2 a}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+8 \left (\frac {a (a-i b)^{3/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {a (a+i b)^{3/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )}{2 a}\right )\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle -\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {\frac {a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {\cot (c+d x)}{\sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}+8 \left (\frac {a (a-i b)^{3/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {a (a+i b)^{3/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )}{2 a}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {\frac {2 a \left (8 a^2 A-12 a b B-3 A b^2\right ) \int \frac {1}{\frac {a+b \tan (c+d x)}{b}-\frac {a}{b}}d\sqrt {a+b \tan (c+d x)}}{b d}+8 \left (\frac {a (a-i b)^{3/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {a (a+i b)^{3/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )}{2 a}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {a A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (-\frac {(4 a B+5 A b) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {-\frac {2 \sqrt {a} \left (8 a^2 A-12 a b B-3 A b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{d}+8 \left (\frac {a (a-i b)^{3/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {a (a+i b)^{3/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )}{2 a}\right )\) |
-1/2*(a*A*Cot[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]])/d + (-1/2*(8*((a*(a - I *b)^(3/2)*(I*A + B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d - (a*(a + I*b)^( 3/2)*(I*A - B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d) - (2*Sqrt[a]*(8*a^2* A - 3*A*b^2 - 12*a*b*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/d)/a - ((5*A*b + 4*a*B)*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/d)/4
3.4.30.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
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_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
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[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* (m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d )*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ [b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^ n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ (A*b^2 - a*b*B + a^2*C)/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & & !GtQ[n, 0] && !LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(1804\) vs. \(2(185)=370\).
Time = 0.26 (sec) , antiderivative size = 1805, normalized size of antiderivative = 8.24
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1805\) |
| default | \(\text {Expression too large to display}\) | \(1805\) |
-5/4/d/tan(d*x+c)^2*A*(a+b*tan(d*x+c))^(3/2)+2*a^(3/2)*A*arctanh((a+b*tan( d*x+c))^(1/2)/a^(1/2))/d+1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+ b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a) ^(1/2))*B*(a^2+b^2)^(1/2)+1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^ 2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^ (1/2))*A*(a^2+b^2)^(1/2)*a-1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+ b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a) ^(1/2))*A*(a^2+b^2)^(1/2)*a+1/4/d/b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2) ^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a )^(1/2)*a^2-1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2) +2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*(a^ 2+b^2)^(1/2)+2/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2 )+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*a- 2/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a ^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*a-1/4/d/b*ln(b* tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2 )^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-1/4/d/b*ln((a+b*tan(d*x+c))^( 1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*B*(2*(a ^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a+1/4/d/b*ln(b*tan(d*x+c)+a+(a+b* tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*B*(2*(...
Leaf count of result is larger than twice the leaf count of optimal. 3170 vs. \(2 (179) = 358\).
Time = 6.20 (sec) , antiderivative size = 6357, normalized size of antiderivative = 29.03 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
\[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \cot ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Time = 9.95 (sec) , antiderivative size = 23016, normalized size of antiderivative = 105.10 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
atan(((((932*A^3*a^3*b^12*d^2 + 1344*A^3*a^5*b^10*d^2 - 192*A^3*a^7*b^8*d^ 2 + 1600*B^3*a^2*b^13*d^2 + 704*B^3*a^4*b^11*d^2 - 896*B^3*a^6*b^9*d^2 + 3 48*A^2*B*b^15*d^2 - 604*A^3*a*b^14*d^2 + 1760*A*B^2*a*b^14*d^2 - 3232*A*B^ 2*a^3*b^12*d^2 - 4416*A*B^2*a^5*b^10*d^2 + 576*A*B^2*a^7*b^8*d^2 - 3780*A^ 2*B*a^2*b^13*d^2 - 1440*A^2*B*a^4*b^11*d^2 + 2688*A^2*B*a^6*b^9*d^2)/(2*d^ 5) - (((384*A*b^12*d^4 + 1280*B*a*b^11*d^4 - 384*A*a^2*b^10*d^4 - 768*A*a^ 4*b^8*d^4 + 1280*B*a^3*b^9*d^4)/(2*d^5) + ((512*b^10*d^4 + 768*a^2*b^8*d^4 )*(a + b*tan(c + d*x))^(1/2)*((((8*A^2*a^3*d^2 - 8*B^2*a^3*d^2 + 16*A*B*b^ 3*d^2 - 24*A^2*a*b^2*d^2 + 24*B^2*a*b^2*d^2 - 48*A*B*a^2*b*d^2)^2/64 - d^4 *(A^4*a^6 + A^4*b^6 + B^4*a^6 + B^4*b^6 + 2*A^2*B^2*a^6 + 2*A^2*B^2*b^6 + 3*A^4*a^2*b^4 + 3*A^4*a^4*b^2 + 3*B^4*a^2*b^4 + 3*B^4*a^4*b^2 + 6*A^2*B^2* a^2*b^4 + 6*A^2*B^2*a^4*b^2))^(1/2) + A^2*a^3*d^2 - B^2*a^3*d^2 + 2*A*B*b^ 3*d^2 - 3*A^2*a*b^2*d^2 + 3*B^2*a*b^2*d^2 - 6*A*B*a^2*b*d^2)/(4*d^4))^(1/2 ))/d^4)*((((8*A^2*a^3*d^2 - 8*B^2*a^3*d^2 + 16*A*B*b^3*d^2 - 24*A^2*a*b^2* d^2 + 24*B^2*a*b^2*d^2 - 48*A*B*a^2*b*d^2)^2/64 - d^4*(A^4*a^6 + A^4*b^6 + B^4*a^6 + B^4*b^6 + 2*A^2*B^2*a^6 + 2*A^2*B^2*b^6 + 3*A^4*a^2*b^4 + 3*A^4 *a^4*b^2 + 3*B^4*a^2*b^4 + 3*B^4*a^4*b^2 + 6*A^2*B^2*a^2*b^4 + 6*A^2*B^2*a ^4*b^2))^(1/2) + A^2*a^3*d^2 - B^2*a^3*d^2 + 2*A*B*b^3*d^2 - 3*A^2*a*b^2*d ^2 + 3*B^2*a*b^2*d^2 - 6*A*B*a^2*b*d^2)/(4*d^4))^(1/2) + ((a + b*tan(c + d *x))^(1/2)*(1088*A^2*a^3*b^10*d^2 - 576*A^2*a^5*b^8*d^2 - 1472*B^2*a^3*...