Integrand size = 26, antiderivative size = 107 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) \, dx=-\frac {2 (-1)^{3/4} a d^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f} \]
-2*(-1)^(3/4)*a*d^(5/2)*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f- 2*I*a*d^2*(d*tan(f*x+e))^(1/2)/f+2/3*a*d*(d*tan(f*x+e))^(3/2)/f+2/5*I*a*(d *tan(f*x+e))^(5/2)/f
Time = 0.84 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.79 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) \, dx=\frac {2 a \left (-15 (-1)^{3/4} d^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+d^2 \sqrt {d \tan (e+f x)} \left (-15 i+5 \tan (e+f x)+3 i \tan ^2(e+f x)\right )\right )}{15 f} \]
(2*a*(-15*(-1)^(3/4)*d^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt [d]] + d^2*Sqrt[d*Tan[e + f*x]]*(-15*I + 5*Tan[e + f*x] + (3*I)*Tan[e + f* x]^2)))/(15*f)
Time = 0.57 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 4011, 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.
| \(\displaystyle \int (a+i a \tan (e+f x)) (d \tan (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (e+f x)) (d \tan (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int (d \tan (e+f x))^{3/2} (a d \tan (e+f x)-i a d)dx+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (d \tan (e+f x))^{3/2} (a d \tan (e+f x)-i a d)dx+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \sqrt {d \tan (e+f x)} \left (-a d^2-i a \tan (e+f x) d^2\right )dx+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {d \tan (e+f x)} \left (-a d^2-i a \tan (e+f x) d^2\right )dx+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {i a d^3-a d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx-\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {i a d^3-a d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx-\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 4016 |
| \(\displaystyle -\frac {2 a^2 d^6 \int \frac {1}{i a d^4+a \tan (e+f x) d^4}d\sqrt {d \tan (e+f x)}}{f}-\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {2 (-1)^{3/4} a d^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}\) |
(-2*(-1)^(3/4)*a*d^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] )/f - ((2*I)*a*d^2*Sqrt[d*Tan[e + f*x]])/f + (2*a*d*(d*Tan[e + f*x])^(3/2) )/(3*f) + (((2*I)/5)*a*(d*Tan[e + f*x])^(5/2))/f
3.2.35.3.1 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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (85 ) = 170\).
Time = 0.79 (sec) , antiderivative size = 321, normalized size of antiderivative = 3.00
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {2 i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 i d^{2} \sqrt {d \tan \left (f x +e \right )}+2 d^{3} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\) | \(321\) |
| default | \(\frac {a \left (\frac {2 i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 i d^{2} \sqrt {d \tan \left (f x +e \right )}+2 d^{3} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\) | \(321\) |
| parts | \(\frac {2 a d \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}+\frac {i a \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}+\frac {d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4}\right )}{f}\) | \(325\) |
1/f*a*(2/5*I*(d*tan(f*x+e))^(5/2)+2/3*d*(d*tan(f*x+e))^(3/2)-2*I*d^2*(d*ta n(f*x+e))^(1/2)+2*d^3*(1/8*I/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/(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. 371 vs. \(2 (83) = 166\).
Time = 0.25 (sec) , antiderivative size = 371, normalized size of antiderivative = 3.47 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) \, dx=-\frac {15 \, \sqrt {\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-2 i \, a d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a d^{2}}\right ) - 15 \, \sqrt {\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-2 i \, a d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a d^{2}}\right ) + 8 \, {\left (23 i \, a d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 24 i \, a d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 13 i \, a d^{2}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
-1/60*(15*sqrt(4*I*a^2*d^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log((-2*I*a*d^3*e^(2*I*f*x + 2*I*e) + sqrt(4*I*a^2*d^5/f^2)*(I *f*e^(2*I*f*x + 2*I*e) + I*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*d^2)) - 15*sqrt(4*I*a^2*d^5/ f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log((-2*I*a*d^3 *e^(2*I*f*x + 2*I*e) + sqrt(4*I*a^2*d^5/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) - I *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*d^2)) + 8*(23*I*a*d^2*e^(4*I*f*x + 4*I*e) + 24*I*a*d^2 *e^(2*I*f*x + 2*I*e) + 13*I*a*d^2)*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)
\[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) \, dx=i a \left (\int \left (- i \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\right )\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )}\, dx\right ) \]
I*a*(Integral(-I*(d*tan(e + f*x))**(5/2), x) + Integral((d*tan(e + f*x))** (5/2)*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. 209 vs. \(2 (83) = 166\).
Time = 0.47 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.95 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) \, dx=\frac {15 \, a d^{4} {\left (\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + 24 i \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a d + 40 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a d^{2} - 120 i \, \sqrt {d \tan \left (f x + e\right )} a d^{3}}{60 \, d f} \]
1/60*(15*a*d^4*((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*sqr t(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)/sq rt(d) - (I + 1)*sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))* sqrt(d) + d)/sqrt(d)) + 24*I*(d*tan(f*x + e))^(5/2)*a*d + 40*(d*tan(f*x + e))^(3/2)*a*d^2 - 120*I*sqrt(d*tan(f*x + e))*a*d^3)/(d*f)
Time = 0.61 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.36 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) \, dx=-\frac {2}{15} \, a d {\left (\frac {15 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {-3 i \, \sqrt {d \tan \left (f x + e\right )} d^{6} f^{4} \tan \left (f x + e\right )^{2} - 5 \, \sqrt {d \tan \left (f x + e\right )} d^{6} f^{4} \tan \left (f x + e\right ) + 15 i \, \sqrt {d \tan \left (f x + e\right )} d^{6} f^{4}}{d^{5} f^{5}}\right )} \]
-2/15*a*d*(15*sqrt(2)*d^(3/2)*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*I*sqrt(d*tan(f*x + e))*d^6*f^4*tan(f*x + e)^2 - 5*sqrt(d*tan(f*x + e) )*d^6*f^4*tan(f*x + e) + 15*I*sqrt(d*tan(f*x + e))*d^6*f^4)/(d^5*f^5))
Time = 5.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.07 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) \, dx=\frac {2\,a\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}+\frac {a\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}}{5\,f}-\frac {a\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{f}-\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{f} \]