Integrand size = 26, antiderivative size = 107 \[ \int (d \tan (e+f x))^{5/2} (a-i a \tan (e+f x)) \, dx=\frac {2 (-1)^{3/4} a d^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f} \] Output:
2*(-1)^(3/4)*a*d^(5/2)*arctanh((-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 = 0.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.84 \[ \int (d \tan (e+f x))^{5/2} (a-i a \tan (e+f x)) \, dx=\frac {2 a (d \tan (e+f x))^{5/2} \left (15 (-1)^{3/4} \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )+\sqrt {\tan (e+f x)} \left (15 i+5 \tan (e+f x)-3 i \tan ^2(e+f x)\right )\right )}{15 f \tan ^{\frac {5}{2}}(e+f x)} \] Input:
Integrate[(d*Tan[e + f*x])^(5/2)*(a - I*a*Tan[e + f*x]),x]
Output:
(2*a*(d*Tan[e + f*x])^(5/2)*(15*(-1)^(3/4)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[e + f*x]]] + Sqrt[Tan[e + f*x]]*(15*I + 5*Tan[e + f*x] - (3*I)*Tan[e + f*x]^2 )))/(15*f*Tan[e + f*x]^(5/2))
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, 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.
| \(\displaystyle \int (a-i a \tan (e+f x)) (d \tan (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a-i a \tan (e+f x)) (d \tan (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int (d \tan (e+f x))^{3/2} (i a d+a \tan (e+f x) d)dx-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (d \tan (e+f x))^{3/2} (i a d+a \tan (e+f x) d)dx-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \sqrt {d \tan (e+f x)} \left (i a d^2 \tan (e+f x)-a d^2\right )dx-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {d \tan (e+f x)} \left (i a d^2 \tan (e+f x)-a d^2\right )dx-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {-i a d^3-a \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}dx+\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {-i a d^3-a \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}dx+\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 4016 |
| \(\displaystyle -\frac {2 a^2 d^6 \int \frac {1}{a d^4 \tan (e+f x)-i a d^4}d\sqrt {d \tan (e+f x)}}{f}+\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 (-1)^{3/4} a d^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {2 i a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a (d \tan (e+f x))^{5/2}}{5 f}\) |
Input:
Int[(d*Tan[e + f*x])^(5/2)*(a - I*a*Tan[e + f*x]),x]
Output:
(2*(-1)^(3/4)*a*d^(5/2)*ArcTanh[((-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
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[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. 321 vs. \(2 (85 ) = 170\).
Time = 1.32 (sec) , antiderivative size = 322, normalized size of antiderivative = 3.01
| method | result | size |
| derivativedivides | \(-\frac {a \left (\frac {2 i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 i d^{2} \sqrt {d \tan \left (f x +e \right )}+2 d^{3} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\) | \(322\) |
| default | \(-\frac {a \left (\frac {2 i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 i d^{2} \sqrt {d \tan \left (f x +e \right )}+2 d^{3} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\) | \(322\) |
| parts | \(\frac {2 a d \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}-\frac {i a \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}+\frac {d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4}\right )}{f}\) | \(325\) |
Input:
int((d*tan(f*x+e))^(5/2)*(a-I*a*tan(f*x+e)),x,method=_RETURNVERBOSE)
Output:
-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*t an(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. 342 vs. \(2 (83) = 166\).
Time = 0.09 (sec) , antiderivative size = 342, normalized size of antiderivative = 3.20 \[ \int (d \tan (e+f x))^{5/2} (a-i a \tan (e+f x)) \, dx=\frac {15 \, \sqrt {-\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (2 \, a d^{3} + \sqrt {-\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{f}\right ) - 15 \, \sqrt {-\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (2 \, a d^{3} - \sqrt {-\frac {4 i \, a^{2} d^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{f}\right ) - 8 \, {\left (-13 i \, a d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 24 i \, a d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 23 i \, a d^{2}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \] Input:
integrate((d*tan(f*x+e))^(5/2)*(a-I*a*tan(f*x+e)),x, algorithm="fricas")
Output:
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*a*d^3 + sqrt(-4*I*a^2*d^5/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)/f) - 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*a*d^3 - sqrt(-4*I*a^2*d^5/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)/f) - 8*(-13*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) - 23*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)
\[ \int (d \tan (e+f x))^{5/2} (a-i a \tan (e+f x)) \, dx=- i a \left (\int i \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )}\, dx\right ) \] Input:
integrate((d*tan(f*x+e))**(5/2)*(a-I*a*tan(f*x+e)),x)
Output:
-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 = 0.11 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.95 \[ \int (d \tan (e+f x))^{5/2} (a-i a \tan (e+f x)) \, dx=\frac {15 \, a d^{4} {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - 24 i \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a d + 40 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a d^{2} + 120 i \, \sqrt {d \tan \left (f x + e\right )} a d^{3}}{60 \, d f} \] Input:
integrate((d*tan(f*x+e))^(5/2)*(a-I*a*tan(f*x+e)),x, algorithm="maxima")
Output:
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*sq rt(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)/s qrt(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 = 0.15 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.21 \[ \int (d \tan (e+f x))^{5/2} (a-i a \tan (e+f x)) \, dx=\frac {2 i \, a d^{2} {\left (\frac {15 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{-\frac {i \, d}{{\left | d \right |}} + 1} - \frac {3 \, \sqrt {d \tan \left (f x + e\right )} d^{10} \tan \left (f x + e\right )^{2} + 5 i \, \sqrt {d \tan \left (f x + e\right )} d^{10} \tan \left (f x + e\right ) - 15 \, \sqrt {d \tan \left (f x + e\right )} d^{10}}{d^{10}}\right )}}{15 \, f} \] Input:
integrate((d*tan(f*x+e))^(5/2)*(a-I*a*tan(f*x+e)),x, algorithm="giac")
Output:
2/15*I*a*d^2*(15*I*sqrt(2)*sqrt(d)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(- I*sqrt(2)*d^(3/2) + sqrt(2)*sqrt(d)*abs(d)))/(-I*d/abs(d) + 1) - (3*sqrt(d *tan(f*x + e))*d^10*tan(f*x + e)^2 + 5*I*sqrt(d*tan(f*x + e))*d^10*tan(f*x + e) - 15*sqrt(d*tan(f*x + e))*d^10)/d^10)/f
Time = 1.75 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.34 \[ \int (d \tan (e+f x))^{5/2} (a-i a \tan (e+f x)) \, dx=\frac {2\,a\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}-\frac {2\,{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}-\frac {a\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}}{5\,f}+\frac {a\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{f} \] Input:
int((d*tan(e + f*x))^(5/2)*(a - a*tan(e + f*x)*1i),x)
Output:
(2*a*d*(d*tan(e + f*x))^(3/2))/(3*f) - (a*(d*tan(e + f*x))^(5/2)*2i)/(5*f) + (a*d^2*(d*tan(e + f*x))^(1/2)*2i)/f - (2*(-1)^(1/4)*a*d^(5/2)*atan(((-1 )^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/f + ((-1)^(1/4)*a*d^(5/2)*atan(( (-1)^(1/4)*(d*tan(e + f*x))^(1/2)*1i)/d^(1/2))*1i)/f + ((-1)^(1/4)*a*d^(5/ 2)*atanh(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/f
\[ \int (d \tan (e+f x))^{5/2} (a-i a \tan (e+f x)) \, dx=\frac {\sqrt {d}\, a \,d^{2} \left (-2 \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2} i +10 \sqrt {\tan \left (f x +e \right )}\, i -5 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )}d x \right ) f i +5 \left (\int \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}d x \right ) f \right )}{5 f} \] Input:
int((d*tan(f*x+e))^(5/2)*(a-I*a*tan(f*x+e)),x)
Output:
(sqrt(d)*a*d**2*( - 2*sqrt(tan(e + f*x))*tan(e + f*x)**2*i + 10*sqrt(tan(e + f*x))*i - 5*int(sqrt(tan(e + f*x))/tan(e + f*x),x)*f*i + 5*int(sqrt(tan (e + f*x))*tan(e + f*x)**2,x)*f))/(5*f)