Integrand size = 26, antiderivative size = 87 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx=\frac {2 \sqrt [4]{-1} a \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {2 a}{3 d f (d \tan (e+f x))^{3/2}}+\frac {2 i a}{d^2 f \sqrt {d \tan (e+f x)}} \] Output:
2*(-1)^(1/4)*a*arctanh((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(5/2)/f- 2/3*a/d/f/(d*tan(f*x+e))^(3/2)+2*I*a/d^2/f/(d*tan(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.47 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {2 a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-i \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}} \] Input:
Integrate[(a - I*a*Tan[e + f*x])/(d*Tan[e + f*x])^(5/2),x]
Output:
(-2*a*Hypergeometric2F1[-3/2, 1, -1/2, (-I)*Tan[e + f*x]])/(3*d*f*(d*Tan[e + f*x])^(3/2))
Time = 0.50 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 4012, 25, 3042, 4012, 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.
| \(\displaystyle \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle -\frac {2 a}{3 d f (d \tan (e+f x))^{3/2}}+\frac {\int -\frac {i a d+a \tan (e+f x) d}{(d \tan (e+f x))^{3/2}}dx}{d^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 a}{3 d f (d \tan (e+f x))^{3/2}}-\frac {\int \frac {i a d+a \tan (e+f x) d}{(d \tan (e+f x))^{3/2}}dx}{d^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 a}{3 d f (d \tan (e+f x))^{3/2}}-\frac {\int \frac {i a d+a \tan (e+f x) d}{(d \tan (e+f x))^{3/2}}dx}{d^2}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle -\frac {2 a}{3 d f (d \tan (e+f x))^{3/2}}-\frac {\frac {\int \frac {a d^2-i a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {2 i a}{f \sqrt {d \tan (e+f x)}}}{d^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 a}{3 d f (d \tan (e+f x))^{3/2}}-\frac {\frac {\int \frac {a d^2-i a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {2 i a}{f \sqrt {d \tan (e+f x)}}}{d^2}\) |
| \(\Big \downarrow \) 4016 |
| \(\displaystyle -\frac {2 a}{3 d f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \int \frac {1}{a d^3+i a \tan (e+f x) d^3}d\sqrt {d \tan (e+f x)}}{f}-\frac {2 i a}{f \sqrt {d \tan (e+f x)}}}{d^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {2 a}{3 d f (d \tan (e+f x))^{3/2}}-\frac {-\frac {2 \sqrt [4]{-1} a \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {2 i a}{f \sqrt {d \tan (e+f x)}}}{d^2}\) |
Input:
Int[(a - I*a*Tan[e + f*x])/(d*Tan[e + f*x])^(5/2),x]
Output:
(-2*a)/(3*d*f*(d*Tan[e + f*x])^(3/2)) - ((-2*(-1)^(1/4)*a*ArcTanh[((-1)^(3 /4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[d]*f) - ((2*I)*a)/(f*Sqrt[d*Tan[ e + f*x]]))/d^2
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[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
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. 310 vs. \(2 (70 ) = 140\).
Time = 1.32 (sec) , antiderivative size = 311, normalized size of antiderivative = 3.57
| method | result | size |
| derivativedivides | \(-\frac {a \left (-\frac {2 i}{d^{2} \sqrt {d \tan \left (f x +e \right )}}+\frac {2}{3 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{2}}\right )}{f}\) | \(311\) |
| default | \(-\frac {a \left (-\frac {2 i}{d^{2} \sqrt {d \tan \left (f x +e \right )}}+\frac {2}{3 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{2}}\right )}{f}\) | \(311\) |
| parts | \(\frac {2 a d \left (-\frac {1}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{4}}\right )}{f}-\frac {i a \left (-\frac {2}{d^{2} \sqrt {d \tan \left (f x +e \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d^{2} \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}\) | \(316\) |
Input:
int((a-I*a*tan(f*x+e))/(d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/f*a*(-2*I/d^2/(d*tan(f*x+e))^(1/2)+2/3/d/(d*tan(f*x+e))^(3/2)+2/d^2*(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*ar ctan(-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*ar ctan(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. 376 vs. \(2 (69) = 138\).
Time = 0.09 (sec) , antiderivative size = 376, normalized size of antiderivative = 4.32 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {3 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {4 i \, a^{2}}{d^{5} f^{2}}} \log \left (-\frac {{\left ({\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {4 i \, a^{2}}{d^{5} f^{2}}} + 2 i \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2} f}\right ) - 3 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {4 i \, a^{2}}{d^{5} f^{2}}} \log \left (\frac {{\left ({\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {4 i \, a^{2}}{d^{5} f^{2}}} - 2 i \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2} f}\right ) + 16 \, {\left (a e^{\left (4 i \, f x + 4 i \, e\right )} - a e^{\left (2 i \, f x + 2 i \, e\right )} - 2 \, a\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )}} \] Input:
integrate((a-I*a*tan(f*x+e))/(d*tan(f*x+e))^(5/2),x, algorithm="fricas")
Output:
-1/12*(3*(d^3*f*e^(4*I*f*x + 4*I*e) - 2*d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f) *sqrt(4*I*a^2/(d^5*f^2))*log(-((d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt((- I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(4*I*a^2/(d^ 5*f^2)) + 2*I*a)*e^(-2*I*f*x - 2*I*e)/(d^2*f)) - 3*(d^3*f*e^(4*I*f*x + 4*I *e) - 2*d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f)*sqrt(4*I*a^2/(d^5*f^2))*log(((d ^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e ^(2*I*f*x + 2*I*e) + 1))*sqrt(4*I*a^2/(d^5*f^2)) - 2*I*a)*e^(-2*I*f*x - 2* I*e)/(d^2*f)) + 16*(a*e^(4*I*f*x + 4*I*e) - a*e^(2*I*f*x + 2*I*e) - 2*a)*s qrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^3*f*e^ (4*I*f*x + 4*I*e) - 2*d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f)
\[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx=- i a \left (\int \frac {i}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {\tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx\right ) \] Input:
integrate((a-I*a*tan(f*x+e))/(d*tan(f*x+e))**(5/2),x)
Output:
-I*a*(Integral(I/(d*tan(e + f*x))**(5/2), x) + Integral(tan(e + f*x)/(d*ta n(e + f*x))**(5/2), x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (69) = 138\).
Time = 0.15 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.22 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {\frac {3 \, a {\left (-\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d} - \frac {8 \, {\left (3 i \, a d \tan \left (f x + e\right ) - a d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d}}{12 \, d f} \] Input:
integrate((a-I*a*tan(f*x+e))/(d*tan(f*x+e))^(5/2),x, algorithm="maxima")
Output:
-1/12*(3*a*(-(2*I - 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqr t(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)*sq rt(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))*sqr t(d) + d)/sqrt(d))/d - 8*(3*I*a*d*tan(f*x + e) - a*d)/((d*tan(f*x + e))^(3 /2)*d))/(d*f)
Time = 0.22 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.10 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx=\frac {2 i \, a {\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{d^{\frac {5}{2}} {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {3 \, d \tan \left (f x + e\right ) + i \, d}{\sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )}\right )}}{3 \, f} \] Input:
integrate((a-I*a*tan(f*x+e))/(d*tan(f*x+e))^(5/2),x, algorithm="giac")
Output:
2/3*I*a*(3*sqrt(2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(-I*sqrt(2)*d^(3/2 ) + sqrt(2)*sqrt(d)*abs(d)))/(d^(5/2)*(-I*d/abs(d) + 1)) + (3*d*tan(f*x + e) + I*d)/(sqrt(d*tan(f*x + e))*d^3*tan(f*x + e)))/f
Time = 1.48 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx=\frac {a\,2{}\mathrm {i}}{d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {2\,a}{3\,d\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{d^{5/2}\,f} \] Input:
int((a - a*tan(e + f*x)*1i)/(d*tan(e + f*x))^(5/2),x)
Output:
(a*2i)/(d^2*f*(d*tan(e + f*x))^(1/2)) - (2*a)/(3*d*f*(d*tan(e + f*x))^(3/2 )) + ((-1)^(1/4)*a*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2))*2i)/( d^(5/2)*f)
\[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {d}\, a \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{3}}d x -\left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{2}}d x \right ) i \right )}{d^{3}} \] Input:
int((a-I*a*tan(f*x+e))/(d*tan(f*x+e))^(5/2),x)
Output:
(sqrt(d)*a*(int(sqrt(tan(e + f*x))/tan(e + f*x)**3,x) - int(sqrt(tan(e + f *x))/tan(e + f*x)**2,x)*i))/d**3