Integrand size = 28, antiderivative size = 129 \[ \int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3 \, dx=\frac {8 (-1)^{3/4} a^3 \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {8 i a^3 \sqrt {d \tan (e+f x)}}{f}-\frac {8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}-\frac {2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f} \]
8*(-1)^(3/4)*a^3*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))*d^(1/2)/f +8*I*a^3*(d*tan(f*x+e))^(1/2)/f-8/5*a^3*(d*tan(f*x+e))^(3/2)/d/f-2/5*(d*ta n(f*x+e))^(3/2)*(a^3+I*a^3*tan(f*x+e))/d/f
Time = 1.13 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.65 \[ \int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3 \, dx=\frac {2 i a^3 \left (20 \sqrt [4]{-1} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+\sqrt {d \tan (e+f x)} \left (20+5 i \tan (e+f x)-\tan ^2(e+f x)\right )\right )}{5 f} \]
(((2*I)/5)*a^3*(20*(-1)^(1/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f* x]])/Sqrt[d]] + Sqrt[d*Tan[e + f*x]]*(20 + (5*I)*Tan[e + f*x] - Tan[e + f* x]^2)))/f
Time = 0.67 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4039, 27, 3042, 4075, 3042, 4011, 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.
\(\displaystyle \int (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4039 |
\(\displaystyle \frac {2 a \int 2 \sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a) (2 a d+3 i a \tan (e+f x) d)dx}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 a \int \sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a) (2 a d+3 i a \tan (e+f x) d)dx}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a \int \sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a) (2 a d+3 i a \tan (e+f x) d)dx}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 4075 |
\(\displaystyle \frac {4 a \left (-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{f}+\int \sqrt {d \tan (e+f x)} \left (5 d a^2+5 i d \tan (e+f x) a^2\right )dx\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a \left (-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{f}+\int \sqrt {d \tan (e+f x)} \left (5 d a^2+5 i d \tan (e+f x) a^2\right )dx\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {4 a \left (\int \frac {5 a^2 d^2 \tan (e+f x)-5 i a^2 d^2}{\sqrt {d \tan (e+f x)}}dx-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{f}+\frac {10 i a^2 d \sqrt {d \tan (e+f x)}}{f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a \left (\int \frac {5 a^2 d^2 \tan (e+f x)-5 i a^2 d^2}{\sqrt {d \tan (e+f x)}}dx-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{f}+\frac {10 i a^2 d \sqrt {d \tan (e+f x)}}{f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 4016 |
\(\displaystyle \frac {4 a \left (-\frac {50 a^4 d^4 \int \frac {1}{-5 i a^2 d^3-5 a^2 \tan (e+f x) d^3}d\sqrt {d \tan (e+f x)}}{f}+\frac {10 i a^2 d \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {4 a \left (\frac {10 (-1)^{3/4} a^2 d^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{f}+\frac {10 i a^2 d \sqrt {d \tan (e+f x)}}{f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
(-2*(d*Tan[e + f*x])^(3/2)*(a^3 + I*a^3*Tan[e + f*x]))/(5*d*f) + (4*a*((10 *(-1)^(3/4)*a^2*d^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]) /f + ((10*I)*a^2*d*Sqrt[d*Tan[e + f*x]])/f - (2*a^2*(d*Tan[e + f*x])^(3/2) )/f))/(5*d)
3.2.58.3.1 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[((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]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x ] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (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 *d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && !LeQ[m, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (107 ) = 214\).
Time = 1.02 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.53
method | result | size |
derivativedivides | \(\frac {2 a^{3} \left (-\frac {i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 i d^{2} \sqrt {d \tan \left (f x +e \right )}-4 d^{3} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\) | \(327\) |
default | \(\frac {2 a^{3} \left (-\frac {i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 i d^{2} \sqrt {d \tan \left (f x +e \right )}-4 d^{3} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\) | \(327\) |
parts | \(\frac {a^{3} d \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 \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2 i a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-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 )}{8}\right )}{f \,d^{2}}+\frac {3 i a^{3} \left (2 \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 )}{4}\right )}{f}-\frac {6 a^{3} \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 d}\) | \(625\) |
2/f*a^3/d^2*(-1/5*I*(d*tan(f*x+e))^(5/2)-d*(d*tan(f*x+e))^(3/2)+4*I*d^2*(d *tan(f*x+e))^(1/2)-4*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. 356 vs. \(2 (105) = 210\).
Time = 0.25 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.76 \[ \int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3 \, dx=\frac {5 \, \sqrt {\frac {64 i \, a^{6} d}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {64 i \, a^{6} d}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 5 \, \sqrt {\frac {64 i \, a^{6} d}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {64 i \, a^{6} d}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 16 \, {\left (-13 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 19 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{3}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{20 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
1/20*(5*sqrt(64*I*a^6*d/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I *e) + f)*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + sqrt(64*I*a^6*d/f^2)*(I *f*e^(2*I*f*x + 2*I*e) + I*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)/a^3) - 5*sqrt(64*I*a^6*d/f^2)*( f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(1/4*(-8*I*a^3*d*e ^(2*I*f*x + 2*I*e) + sqrt(64*I*a^6*d/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) - I*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)/a^3) - 16*(-13*I*a^3*e^(4*I*f*x + 4*I*e) - 19*I*a^3*e^(2*I*f *x + 2*I*e) - 8*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)))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)
\[ \int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3 \, dx=- i a^{3} \left (\int i \sqrt {d \tan {\left (e + f x \right )}}\, dx + \int \left (- 3 \sqrt {d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx + \int \sqrt {d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 3 i \sqrt {d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\right )\, dx\right ) \]
-I*a**3*(Integral(I*sqrt(d*tan(e + f*x)), x) + Integral(-3*sqrt(d*tan(e + f*x))*tan(e + f*x), x) + Integral(sqrt(d*tan(e + f*x))*tan(e + f*x)**3, x) + Integral(-3*I*sqrt(d*tan(e + f*x))*tan(e + f*x)**2, x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (105) = 210\).
Time = 0.30 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.71 \[ \int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3 \, dx=\frac {5 \, a^{3} d^{2} {\left (-\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + \frac {2 \, {\left (-i \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} - 5 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d + 20 i \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{2}\right )}}{d}}{5 \, d f} \]
1/5*(5*a^3*d^2*(-(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)) + 2*(-I*(d*tan(f*x + e))^(5/2)*a^3 - 5*(d*tan(f*x + e))^(3/2)*a^3*d + 20*I*sqrt(d*tan(f*x + e))*a^3*d^2)/d)/(d*f)
Time = 0.73 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.20 \[ \int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3 \, dx=\frac {8 \, \sqrt {2} a^{3} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 \, {\left (i \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{10} f^{4} \tan \left (f x + e\right )^{2} + 5 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{10} f^{4} \tan \left (f x + e\right ) - 20 i \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{10} f^{4}\right )}}{5 \, d^{10} f^{5}} \]
8*sqrt(2)*a^3*sqrt(d)*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)))/(f*(I*d/sqrt(d^2) + 1)) - 2/5*(I* sqrt(d*tan(f*x + e))*a^3*d^10*f^4*tan(f*x + e)^2 + 5*sqrt(d*tan(f*x + e))* a^3*d^10*f^4*tan(f*x + e) - 20*I*sqrt(d*tan(f*x + e))*a^3*d^10*f^4)/(d^10* f^5)
Time = 5.41 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.74 \[ \int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3 \, dx=\frac {a^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,8{}\mathrm {i}}{f}-\frac {2\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{d\,f}-\frac {a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}}{5\,d^2\,f}+\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,\sqrt {d}}\right )\,2{}\mathrm {i}}{f} \]