Integrand size = 28, antiderivative size = 113 \[ \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2 \, dx=\frac {4 \sqrt [4]{-1} a^2 d^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {4 a^2 d \sqrt {d \tan (e+f x)}}{f}+\frac {4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f} \]
4*(-1)^(1/4)*a^2*d^(3/2)*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f +4*a^2*d*(d*tan(f*x+e))^(1/2)/f+4/3*I*a^2*(d*tan(f*x+e))^(3/2)/f-2/5*a^2*( d*tan(f*x+e))^(5/2)/d/f
Time = 0.99 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.78 \[ \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2 \, dx=\frac {2 a^2 d \left ((-15+15 i) \sqrt {2} \sqrt {d} \text {arctanh}\left (\frac {(1+i) \sqrt {d \tan (e+f x)}}{\sqrt {2} \sqrt {d}}\right )+\sqrt {d \tan (e+f x)} \left (30+10 i \tan (e+f x)-3 \tan ^2(e+f x)\right )\right )}{15 f} \]
(2*a^2*d*((-15 + 15*I)*Sqrt[2]*Sqrt[d]*ArcTanh[((1 + I)*Sqrt[d*Tan[e + f*x ]])/(Sqrt[2]*Sqrt[d])] + Sqrt[d*Tan[e + f*x]]*(30 + (10*I)*Tan[e + f*x] - 3*Tan[e + f*x]^2)))/(15*f)
Time = 0.58 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4026, 3042, 4011, 3042, 4011, 3042, 4016, 218}
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.
\(\displaystyle \int (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}dx\) |
\(\Big \downarrow \) 4026 |
\(\displaystyle -\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\int (d \tan (e+f x))^{3/2} \left (2 i \tan (e+f x) a^2+2 a^2\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\int (d \tan (e+f x))^{3/2} \left (2 i \tan (e+f x) a^2+2 a^2\right )dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \sqrt {d \tan (e+f x)} \left (2 a^2 d \tan (e+f x)-2 i a^2 d\right )dx-\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac {4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {d \tan (e+f x)} \left (2 a^2 d \tan (e+f x)-2 i a^2 d\right )dx-\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac {4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \frac {-2 a^2 d^2-2 i a^2 \tan (e+f x) d^2}{\sqrt {d \tan (e+f x)}}dx-\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac {4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a^2 d \sqrt {d \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {-2 a^2 d^2-2 i a^2 \tan (e+f x) d^2}{\sqrt {d \tan (e+f x)}}dx-\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac {4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a^2 d \sqrt {d \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 4016 |
\(\displaystyle \frac {8 a^4 d^4 \int \frac {1}{2 i a^2 d^3 \tan (e+f x)-2 a^2 d^3}d\sqrt {d \tan (e+f x)}}{f}+\frac {4 a^2 d \sqrt {d \tan (e+f x)}}{f}+\frac {4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {4 \sqrt [4]{-1} a^2 d^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac {4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a^2 d \sqrt {d \tan (e+f x)}}{f}\) |
(4*(-1)^(1/4)*a^2*d^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d] ])/f + (4*a^2*d*Sqrt[d*Tan[e + f*x]])/f + (((4*I)/3)*a^2*(d*Tan[e + f*x])^ (3/2))/f - (2*a^2*(d*Tan[e + f*x])^(5/2))/(5*d*f)
3.2.50.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2*(c^2/f) Subst[Int[1/(b*c - d*x^2), x], x, Sqrt[b *Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ [m, -1] && !(EqQ[m, 2] && EqQ[a, 0])
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (92 ) = 184\).
Time = 0.99 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.88
method | result | size |
derivativedivides | \(\frac {2 a^{2} \left (-\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 d^{2} \sqrt {d \tan \left (f x +e \right )}-2 d^{3} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f d}\) | \(326\) |
default | \(\frac {2 a^{2} \left (-\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 d^{2} \sqrt {d \tan \left (f x +e \right )}-2 d^{3} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f d}\) | \(326\) |
parts | \(\frac {2 a^{2} d \left (\sqrt {d \tan \left (f x +e \right )}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{f}+\frac {2 i a^{2} \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}-\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-d^{2} \sqrt {d \tan \left (f x +e \right )}+\frac {d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{f d}\) | \(482\) |
2/f*a^2/d*(-1/5*(d*tan(f*x+e))^(5/2)+2/3*I*d*(d*tan(f*x+e))^(3/2)+2*d^2*(d *tan(f*x+e))^(1/2)-2*d^3*(1/8/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2 )^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4 )*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)* (d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2) +1))+1/8*I/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e) )^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2 )*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+ 1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (91) = 182\).
Time = 0.25 (sec) , antiderivative size = 372, normalized size of antiderivative = 3.29 \[ \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2 \, dx=-\frac {15 \, \sqrt {-\frac {16 i \, a^{4} d^{3}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-4 i \, a^{2} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {-\frac {16 i \, a^{4} d^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2} d}\right ) - 15 \, \sqrt {-\frac {16 i \, a^{4} d^{3}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-4 i \, a^{2} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - \sqrt {-\frac {16 i \, a^{4} d^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2} d}\right ) - 8 \, {\left (43 \, a^{2} d e^{\left (4 i \, f x + 4 i \, e\right )} + 54 \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} + 23 \, a^{2} d\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
-1/60*(15*sqrt(-16*I*a^4*d^3/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(1/2*(-4*I*a^2*d^2*e^(2*I*f*x + 2*I*e) + sqrt(-16*I*a^4*d ^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/ (e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a^2*d)) - 15*sqrt(-16*I* a^4*d^3/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(1/2 *(-4*I*a^2*d^2*e^(2*I*f*x + 2*I*e) - sqrt(-16*I*a^4*d^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a^2*d)) - 8*(43*a^2*d*e^(4*I*f*x + 4*I*e) + 5 4*a^2*d*e^(2*I*f*x + 2*I*e) + 23*a^2*d)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I *d)/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)
\[ \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2 \, dx=- a^{2} \left (\int \left (- \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\right )\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- 2 i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}\right )\, dx\right ) \]
-a**2*(Integral(-(d*tan(e + f*x))**(3/2), x) + Integral((d*tan(e + f*x))** (3/2)*tan(e + f*x)**2, x) + Integral(-2*I*(d*tan(e + f*x))**(3/2)*tan(e + f*x), x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (91) = 182\).
Time = 0.29 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.89 \[ \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2 \, dx=\frac {15 \, a^{2} d^{3} {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - 12 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{2} + 40 i \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{2} d + 120 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{2}}{30 \, d f} \]
1/30*(15*a^2*d^3*(-(2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - (2*I + 2)*sqrt(2)*arctan(-1/2* sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + (I - 1)*sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d) /sqrt(d) - (I - 1)*sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e ))*sqrt(d) + d)/sqrt(d)) - 12*(d*tan(f*x + e))^(5/2)*a^2 + 40*I*(d*tan(f*x + e))^(3/2)*a^2*d + 120*sqrt(d*tan(f*x + e))*a^2*d^2)/(d*f)
Time = 0.75 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.39 \[ \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2 \, dx=-\frac {2}{15} \, {\left (\frac {30 i \, \sqrt {2} a^{2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{10} f^{4} \tan \left (f x + e\right )^{2} - 10 i \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{10} f^{4} \tan \left (f x + e\right ) - 30 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{10} f^{4}}{d^{10} f^{5}}\right )} d \]
-2/15*(30*I*sqrt(2)*a^2*sqrt(d)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4 *I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(f*(I*d/sqrt(d^2) + 1)) + (3*sqrt(d*tan(f*x + e))*a^2*d^10*f^4*tan(f*x + e)^2 - 10*I*sqrt(d*tan(f *x + e))*a^2*d^10*f^4*tan(f*x + e) - 30*sqrt(d*tan(f*x + e))*a^2*d^10*f^4) /(d^10*f^5))*d
Time = 5.69 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.86 \[ \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2 \, dx=\frac {a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,4{}\mathrm {i}}{3\,f}-\frac {2\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,d\,f}+\frac {4\,a^2\,d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}-\frac {\sqrt {4{}\mathrm {i}}\,a^2\,{\left (-d\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {4{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {-d}}\right )\,2{}\mathrm {i}}{f} \]