Integrand size = 28, antiderivative size = 260 \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=-\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}+\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}-\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a f}+\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))} \]
(-1/8+3/8*I)*d^(3/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a/f*2^ (1/2)+(1/8-3/8*I)*d^(3/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a /f*2^(1/2)-(1/16+3/16*I)*d^(3/2)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d ^(1/2)*tan(f*x+e))/a/f*2^(1/2)+(1/16+3/16*I)*d^(3/2)*ln(d^(1/2)+2^(1/2)*(d *tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a/f*2^(1/2)-1/2*d*(d*tan(f*x+e))^(1 /2)/f/(a+I*a*tan(f*x+e))
Time = 0.81 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.43 \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\frac {d \left (\sqrt [4]{-1} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-2 \sqrt [4]{-1} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+\frac {i \sqrt {d \tan (e+f x)}}{-i+\tan (e+f x)}\right )}{2 a f} \]
(d*((-1)^(1/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] - 2*(-1)^(1/4)*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] + (I*Sqrt[d*Tan[e + f*x]])/(-I + Tan[e + f*x])))/(2*a*f)
Time = 0.56 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.93, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4033, 27, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {(d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 4033 |
| \(\displaystyle \frac {\int \frac {a d^2-3 i a d^2 \tan (e+f x)}{2 \sqrt {d \tan (e+f x)}}dx}{2 a^2}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a d^2-3 i a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{4 a^2}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a d^2-3 i a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{4 a^2}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {\int \frac {a d^2 (d-3 i d \tan (e+f x))}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{2 a^2 f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 \int \frac {d-3 i d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{2 a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {d^2 \left (\left (\frac {1}{2}+\frac {3 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {1}{2}-\frac {3 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{2 a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {d^2 \left (\left (\frac {1}{2}+\frac {3 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {1}{2}-\frac {3 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )\right )}{2 a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {d^2 \left (\left (\frac {1}{2}+\frac {3 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {1}{2}-\frac {3 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {d^2 \left (\left (\frac {1}{2}+\frac {3 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {1}{2}-\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {d^2 \left (\left (\frac {1}{2}+\frac {3 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {1}{2}-\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {d^2 \left (\left (\frac {1}{2}+\frac {3 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {1}{2}-\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 \left (\left (\frac {1}{2}+\frac {3 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\left (\frac {1}{2}-\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {d^2 \left (\left (\frac {1}{2}-\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\left (\frac {1}{2}+\frac {3 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{2 a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\) |
(d^2*((1/2 - (3*I)/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d] ]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/ (Sqrt[2]*Sqrt[d])) + (1/2 + (3*I)/2)*(-1/2*Log[d + d*Tan[e + f*x] - Sqrt[2 ]*Sqrt[d]*Sqrt[d*Tan[e + f*x]]]/(Sqrt[2]*Sqrt[d]) + Log[d + d*Tan[e + f*x] + Sqrt[2]*Sqrt[d]*Sqrt[d*Tan[e + f*x]]]/(2*Sqrt[2]*Sqrt[d]))))/(2*a*f) - (d*Sqrt[d*Tan[e + f*x]])/(2*f*(a + I*a*Tan[e + f*x]))
3.2.66.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a*c, 2]}, Simp[(d*q + a*e)/(2*a*c) Int[(q + c*x^2)/(a + c*x^4), x], x] + Simp[(d*q - a*e)/(2*a*c) Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a , c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- a)*c]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2/f Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & & NeQ[c^2 + d^2, 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((c + d*Tan[e + f*x])^(n - 1)/( 2*a*f*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a^2) Int[(c + d*Tan[e + f*x] )^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1]
Time = 0.73 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.36
| method | result | size |
| derivativedivides | \(\frac {2 d^{2} \left (-\frac {\sqrt {d \tan \left (f x +e \right )}}{4 \left (i d \tan \left (f x +e \right )+d \right )}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 \sqrt {i d}}\right )}{f a}\) | \(93\) |
| default | \(\frac {2 d^{2} \left (-\frac {\sqrt {d \tan \left (f x +e \right )}}{4 \left (i d \tan \left (f x +e \right )+d \right )}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 \sqrt {i d}}\right )}{f a}\) | \(93\) |
2/f/a*d^2*(-1/4*(d*tan(f*x+e))^(1/2)/(I*d*tan(f*x+e)+d)-1/2*I/(-I*d)^(1/2) *arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))-1/4*I/(I*d)^(1/2)*arctan((d*tan (f*x+e))^(1/2)/(I*d)^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 524 vs. \(2 (186) = 372\).
Time = 0.25 (sec) , antiderivative size = 524, normalized size of antiderivative = 2.02 \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\frac {{\left (a f \sqrt {-\frac {i \, d^{3}}{4 \, a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 {i \, d^{3}}{4 \, a^{2} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d}\right ) - a f \sqrt {-\frac {i \, d^{3}}{4 \, a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 \, {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 {i \, d^{3}}{4 \, a^{2} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d}\right ) + a f \sqrt {\frac {i \, d^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (i \, d^{2} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 {i \, d^{3}}{a^{2} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) - a f \sqrt {\frac {i \, d^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (i \, d^{2} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 {i \, d^{3}}{a^{2} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) - {\left (d e^{\left (2 i \, f x + 2 i \, e\right )} + 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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]
1/4*(a*f*sqrt(-1/4*I*d^3/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(-2*(I*d^2*e^(2 *I*f*x + 2*I*e) + 2*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/4*I*d^3/(a^2*f^2)))*e^( -2*I*f*x - 2*I*e)/d) - a*f*sqrt(-1/4*I*d^3/(a^2*f^2))*e^(2*I*f*x + 2*I*e)* log(-2*(I*d^2*e^(2*I*f*x + 2*I*e) - 2*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt ((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/4*I*d ^3/(a^2*f^2)))*e^(-2*I*f*x - 2*I*e)/d) + a*f*sqrt(I*d^3/(a^2*f^2))*e^(2*I* f*x + 2*I*e)*log((I*d^2 + (a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt((-I*d*e^(2* I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(I*d^3/(a^2*f^2)))*e^ (-2*I*f*x - 2*I*e)/(a*f)) - a*f*sqrt(I*d^3/(a^2*f^2))*e^(2*I*f*x + 2*I*e)* log((I*d^2 - (a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e ) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(I*d^3/(a^2*f^2)))*e^(-2*I*f*x - 2 *I*e)/(a*f)) - (d*e^(2*I*f*x + 2*I*e) + d)*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*f)
\[ \int \frac {(d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=- \frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\tan {\left (e + f x \right )} - i}\, dx}{a} \]
Exception generated. \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.48 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.67 \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=-\frac {1}{2} \, d^{2} {\left (\frac {i \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a \sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {2 i \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a \sqrt {d} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {i \, \sqrt {d \tan \left (f x + e\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a f}\right )} \]
-1/2*d^2*(I*sqrt(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)))/(a*sqrt(d)*f*(I*d/sqrt(d^2) + 1)) + 2*I*sqrt(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)))/(a*sqrt(d)*f*(-I*d/sqrt(d^2) + 1)) - I*sq rt(d*tan(f*x + e))/((d*tan(f*x + e) - I*d)*a*f))
Time = 6.83 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.53 \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=-\mathrm {atan}\left (\frac {2\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^3\,1{}\mathrm {i}}{4\,a^2\,f^2}}}{d^2}\right )\,\sqrt {\frac {d^3\,1{}\mathrm {i}}{4\,a^2\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {4\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^3\,1{}\mathrm {i}}{16\,a^2\,f^2}}}{d^2}\right )\,\sqrt {-\frac {d^3\,1{}\mathrm {i}}{16\,a^2\,f^2}}\,2{}\mathrm {i}-\frac {d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{2\,a\,f\,\left (-d\,\mathrm {tan}\left (e+f\,x\right )+d\,1{}\mathrm {i}\right )} \]