Integrand size = 28, antiderivative size = 109 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=\frac {8 \sqrt [4]{-1} a^3 \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {16 i a^3}{3 d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}} \]
8*(-1)^(1/4)*a^3*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(5/2)/f -16/3*I*a^3/d^2/f/(d*tan(f*x+e))^(1/2)-2/3*(a^3+I*a^3*tan(f*x+e))/d/f/(d*t an(f*x+e))^(3/2)
Time = 1.49 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.69 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=\frac {2 a^3 \left (12 \sqrt [4]{-1} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-\frac {d (9 i+\cot (e+f x))}{\sqrt {d \tan (e+f x)}}\right )}{3 d^3 f} \]
(2*a^3*(12*(-1)^(1/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqr t[d]] - (d*(9*I + Cot[e + f*x]))/Sqrt[d*Tan[e + f*x]]))/(3*d^3*f)
Time = 0.60 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4036, 27, 3042, 4074, 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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4036 |
| \(\displaystyle -\frac {2 \int -\frac {2 (i \tan (e+f x) a+a) \left (2 i a^2 d-a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{3/2}}dx}{3 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 \int \frac {(i \tan (e+f x) a+a) \left (2 i a^2 d-a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{3/2}}dx}{3 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \int \frac {(i \tan (e+f x) a+a) \left (2 i a^2 d-a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{3/2}}dx}{3 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4074 |
| \(\displaystyle \frac {4 \left (\frac {\int -\frac {3 \left (d^2 a^3+i d^2 \tan (e+f x) a^3\right )}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {4 i a^3}{f \sqrt {d \tan (e+f x)}}\right )}{3 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 \left (-\frac {3 \int \frac {d^2 a^3+i d^2 \tan (e+f x) a^3}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {4 i a^3}{f \sqrt {d \tan (e+f x)}}\right )}{3 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (-\frac {3 \int \frac {d^2 a^3+i d^2 \tan (e+f x) a^3}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {4 i a^3}{f \sqrt {d \tan (e+f x)}}\right )}{3 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4016 |
| \(\displaystyle \frac {4 \left (-\frac {6 a^6 d^2 \int \frac {1}{a^3 d^3-i a^3 d^3 \tan (e+f x)}d\sqrt {d \tan (e+f x)}}{f}-\frac {4 i a^3}{f \sqrt {d \tan (e+f x)}}\right )}{3 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {4 \left (\frac {6 \sqrt [4]{-1} a^3 \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {4 i a^3}{f \sqrt {d \tan (e+f x)}}\right )}{3 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
(-2*(a^3 + I*a^3*Tan[e + f*x]))/(3*d*f*(d*Tan[e + f*x])^(3/2)) + (4*((6*(- 1)^(1/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[d]*f ) - ((4*I)*a^3)/(f*Sqrt[d*Tan[e + f*x]])))/(3*d^2)
3.2.61.3.1 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/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])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b *c - a*d)*(A*b - a*B)*((a + b*Tan[e + f*x])^(m + 1)/(b*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*A*c + b*B*c + A*b*d - a*B*d - (A*b*c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && LtQ[m , -1] && NeQ[a^2 + b^2, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (89 ) = 178\).
Time = 0.81 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.80
| method | result | size |
| derivativedivides | \(\frac {2 a^{3} \left (-\frac {3 i}{\sqrt {d \tan \left (f x +e \right )}}-\frac {d}{3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 d}-\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f \,d^{2}}\) | \(305\) |
| default | \(\frac {2 a^{3} \left (-\frac {3 i}{\sqrt {d \tan \left (f x +e \right )}}-\frac {d}{3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 d}-\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f \,d^{2}}\) | \(305\) |
| parts | \(\frac {2 a^{3} 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^{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^{2} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {3 i a^{3} \left (-\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}}}-\frac {2}{d^{2} \sqrt {d \tan \left (f x +e \right )}}\right )}{f}-\frac {3 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^{3}}\) | \(601\) |
2/f*a^3/d^2*(-3*I/(d*tan(f*x+e))^(1/2)-1/3*d/(d*tan(f*x+e))^(3/2)-1/2/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))-1/2*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*ta n(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. 402 vs. \(2 (89) = 178\).
Time = 0.25 (sec) , antiderivative size = 402, normalized size of antiderivative = 3.69 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {3 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {-\frac {64 i \, a^{6}}{d^{5} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {64 i \, a^{6}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 3 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {-\frac {64 i \, a^{6}}{d^{5} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {64 i \, a^{6}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 16 \, {\left (5 \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, a^{3}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )}} \]
-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(-64*I*a^6/(d^5*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (d^3* f*e^(2*I*f*x + 2*I*e) + d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2 *I*f*x + 2*I*e) + 1))*sqrt(-64*I*a^6/(d^5*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) - 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)*sqr t(-64*I*a^6/(d^5*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) - (d^3*f*e^ (2*I*f*x + 2*I*e) + d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f *x + 2*I*e) + 1))*sqrt(-64*I*a^6/(d^5*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) - 1 6*(5*a^3*e^(4*I*f*x + 4*I*e) + a^3*e^(2*I*f*x + 2*I*e) - 4*a^3)*sqrt((-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)
\[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=- i a^{3} \left (\int \frac {i}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\right )\, dx\right ) \]
-I*a**3*(Integral(I/(d*tan(e + f*x))**(5/2), x) + Integral(-3*tan(e + f*x) /(d*tan(e + f*x))**(5/2), x) + Integral(tan(e + f*x)**3/(d*tan(e + f*x))** (5/2), x) + Integral(-3*I*tan(e + f*x)**2/(d*tan(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. 198 vs. \(2 (89) = 178\).
Time = 0.29 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.82 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {\frac {3 \, a^{3} {\left (\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d} + \frac {2 \, {\left (9 i \, a^{3} d \tan \left (f x + e\right ) + a^{3} d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d}}{3 \, d f} \]
-1/3*(3*a^3*((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 + 2*(9*I*a^3*d*tan(f*x + e) + a^3*d)/((d*tan(f*x + e) )^(3/2)*d))/(d*f)
Time = 0.91 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.04 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {8 i \, \sqrt {2} a^{3} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{d^{\frac {5}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 \, {\left (9 i \, a^{3} d \tan \left (f x + e\right ) + a^{3} d\right )}}{3 \, \sqrt {d \tan \left (f x + e\right )} d^{3} f \tan \left (f x + e\right )} \]
-8*I*sqrt(2)*a^3*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)))/(d^(5/2)*f*(I*d/sqrt(d^2) + 1)) - 2/3* (9*I*a^3*d*tan(f*x + e) + a^3*d)/(sqrt(d*tan(f*x + e))*d^3*f*tan(f*x + e))
Time = 5.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.73 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {\frac {2\,a^3}{3\,d\,f}+\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )\,6{}\mathrm {i}}{d\,f}}{{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}-\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,\sqrt {-d}}\right )\,2{}\mathrm {i}}{{\left (-d\right )}^{5/2}\,f} \]