Integrand size = 28, antiderivative size = 80 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx=-\frac {8 (-1)^{3/4} a^3 \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{3/2} f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}} \] Output:
-8*(-1)^(3/4)*a^3*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(3/2)/ f-2*(a^3+I*a^3*tan(f*x+e))/d/f/(d*tan(f*x+e))^(1/2)
Time = 0.70 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx=-\frac {2 a^3 \left (4 (-1)^{3/4} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+(i+\cot (e+f x)) \sqrt {d \tan (e+f x)}\right )}{d^2 f} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(3/2),x]
Output:
(-2*a^3*(4*(-1)^(3/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqr t[d]] + (I + Cot[e + f*x])*Sqrt[d*Tan[e + f*x]]))/(d^2*f)
Time = 0.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 4036, 27, 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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4036 |
| \(\displaystyle -\frac {2 \int -\frac {2 i d \left (i \tan (e+f x) a^3+a^3\right )}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 i \int \frac {i \tan (e+f x) a^3+a^3}{\sqrt {d \tan (e+f x)}}dx}{d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 i \int \frac {i \tan (e+f x) a^3+a^3}{\sqrt {d \tan (e+f x)}}dx}{d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4016 |
| \(\displaystyle \frac {8 i a^6 \int \frac {1}{a^3 d-i a^3 d \tan (e+f x)}d\sqrt {d \tan (e+f x)}}{d f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {8 (-1)^{3/4} a^3 \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{3/2} f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{d f \sqrt {d \tan (e+f x)}}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(3/2),x]
Output:
(-8*(-1)^(3/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^( 3/2)*f) - (2*(a^3 + I*a^3*Tan[e + f*x]))/(d*f*Sqrt[d*Tan[e + f*x]])
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/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] )^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si mp[a/(d*(b*c + a*d)*(n + 1)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[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[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (67 ) = 134\).
Time = 1.61 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.86
| method | result | size |
| derivativedivides | \(\frac {2 a^{3} \left (-i \sqrt {d \tan \left (f x +e \right )}-\frac {d}{\sqrt {d \tan \left (f x +e \right )}}+4 d \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\) | \(309\) |
| default | \(\frac {2 a^{3} \left (-i \sqrt {d \tan \left (f x +e \right )}-\frac {d}{\sqrt {d \tan \left (f x +e \right )}}+4 d \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\) | \(309\) |
| parts | \(\frac {2 a^{3} d \left (-\frac {1}{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 )}{8 d^{2} \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}-\frac {2 i a^{3} \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 \,d^{2}}+\frac {3 i a^{3} \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 f \,d^{2}}-\frac {3 a^{3} \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 f d \left (d^{2}\right )^{\frac {1}{4}}}\) | \(596\) |
Input:
int((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/f*a^3/d^2*(-I*(d*tan(f*x+e))^(1/2)-d/(d*tan(f*x+e))^(1/2)+4*d*(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. 341 vs. \(2 (68) = 136\).
Time = 0.09 (sec) , antiderivative size = 341, normalized size of antiderivative = 4.26 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx=\frac {-16 i \, a^{3} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{3} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d^{2} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{3} f^{2}}} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) + {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{3} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, d^{2} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{3} f^{2}}} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right )}{4 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{2} f\right )}} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(3/2),x, algorithm="fricas")
Output:
1/4*(-16*I*a^3*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) - (d^2*f*e^(2*I*f*x + 2*I*e) - d^2*f)*sqrt(64*I* a^6/(d^3*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (I*d^2*f*e^(2*I*f *x + 2*I*e) + I*d^2*f)*sqrt(64*I*a^6/(d^3*f^2))*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^3) + (d^2*f *e^(2*I*f*x + 2*I*e) - d^2*f)*sqrt(64*I*a^6/(d^3*f^2))*log(1/4*(-8*I*a^3*d *e^(2*I*f*x + 2*I*e) + (-I*d^2*f*e^(2*I*f*x + 2*I*e) - I*d^2*f)*sqrt(64*I* a^6/(d^3*f^2))*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^3))/(d^2*f*e^(2*I*f*x + 2*I*e) - d^2*f)
\[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx=- i a^{3} \left (\int \frac {i}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\right )\, dx\right ) \] Input:
integrate((a+I*a*tan(f*x+e))**3/(d*tan(f*x+e))**(3/2),x)
Output:
-I*a**3*(Integral(I/(d*tan(e + f*x))**(3/2), x) + Integral(-3*tan(e + f*x) /(d*tan(e + f*x))**(3/2), x) + Integral(tan(e + f*x)**3/(d*tan(e + f*x))** (3/2), x) + Integral(-3*I*tan(e + f*x)**2/(d*tan(e + f*x))**(3/2), x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (68) = 136\).
Time = 0.13 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.42 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx=-\frac {a^{3} {\left (-\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + \frac {2 \, a^{3}}{\sqrt {d \tan \left (f x + e\right )}} + \frac {2 i \, \sqrt {d \tan \left (f x + e\right )} a^{3}}{d}}{d f} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(3/2),x, algorithm="maxima")
Output:
-(a^3*(-(2*I - 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*t an(f*x + e)))/sqrt(d))/sqrt(d) - (2*I - 2)*sqrt(2)*arctan(-1/2*sqrt(2)*(sq rt(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)) + 2*a^3/sqrt(d*tan(f*x + e)) + 2*I*sqrt(d*tan(f*x + e))*a^3/ d)/(d*f)
Time = 0.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.14 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx=-\frac {2 \, a^{3} {\left (\frac {4 \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\sqrt {d} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {1}{\sqrt {d \tan \left (f x + e\right )}} + \frac {i \, \sqrt {d \tan \left (f x + e\right )}}{d}\right )}}{d f} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(3/2),x, algorithm="giac")
Output:
-2*a^3*(4*sqrt(2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*sqrt(2)*d^(3/2) + sqrt(2)*sqrt(d)*abs(d)))/(sqrt(d)*(I*d/abs(d) + 1)) + 1/sqrt(d*tan(f*x + e)) + I*sqrt(d*tan(f*x + e))/d)/(d*f)
Time = 1.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx=-\frac {2\,a^3}{d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {a^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{d^2\,f}+\frac {2\,\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atanh}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,\sqrt {d}}\right )}{d^{3/2}\,f} \] Input:
int((a + a*tan(e + f*x)*1i)^3/(d*tan(e + f*x))^(3/2),x)
Output:
(2*16i^(1/2)*a^3*atanh((16i^(1/2)*(d*tan(e + f*x))^(1/2))/(4*d^(1/2))))/(d ^(3/2)*f) - (a^3*(d*tan(e + f*x))^(1/2)*2i)/(d^2*f) - (2*a^3)/(d*f*(d*tan( e + f*x))^(1/2))
\[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {d}\, a^{3} \left (-2 \sqrt {\tan \left (f x +e \right )}\, i +\left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{2}}d x \right ) f +4 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )}d x \right ) f i -3 \left (\int \sqrt {\tan \left (f x +e \right )}d x \right ) f \right )}{d^{2} f} \] Input:
int((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(3/2),x)
Output:
(sqrt(d)*a**3*( - 2*sqrt(tan(e + f*x))*i + int(sqrt(tan(e + f*x))/tan(e + f*x)**2,x)*f + 4*int(sqrt(tan(e + f*x))/tan(e + f*x),x)*f*i - 3*int(sqrt(t an(e + f*x)),x)*f))/(d**2*f)