Integrand size = 30, antiderivative size = 126 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {8 i a^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}+\frac {4 a^3 (i c-4 d) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f} \]
-8*I*a^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f/(c-I*d)^(1/2)+4/3 *a^3*(I*c-4*d)*(c+d*tan(f*x+e))^(1/2)/d^2/f-2/3*(c+d*tan(f*x+e))^(1/2)*(a^ 3+I*a^3*tan(f*x+e))/d/f
Time = 2.35 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.73 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {i a^2 \left (-\frac {8 a \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {2 a (2 c+9 i d-d \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d^2}\right )}{f} \]
(I*a^2*((-8*a*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sqrt[c - I* d] + (2*a*(2*c + (9*I)*d - d*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*d^ 2)))/f
Time = 0.68 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4039, 3042, 4075, 3042, 4020, 25, 27, 73, 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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 4039 |
| \(\displaystyle \frac {2 a \int \frac {(i \tan (e+f x) a+a) (a (i c+2 d)+a (c+4 i d) \tan (e+f x))}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \int \frac {(i \tan (e+f x) a+a) (a (i c+2 d)+a (c+4 i d) \tan (e+f x))}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
| \(\Big \downarrow \) 4075 |
| \(\displaystyle \frac {2 a \left (\int \frac {6 d a^2+6 i d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (\int \frac {6 d a^2+6 i d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {2 a \left (\frac {36 i a^4 d^2 \int -\frac {1}{\sqrt {6} a^2 d \sqrt {6 c+6 d \tan (e+f x)} \left (6 a^2 d-6 i a^2 d \tan (e+f x)\right )}d\left (6 i a^2 d \tan (e+f x)\right )}{f}+\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 a \left (\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}-\frac {36 i a^4 d^2 \int \frac {1}{\sqrt {6} a^2 d \sqrt {6 c+6 d \tan (e+f x)} \left (6 a^2 d-6 i a^2 d \tan (e+f x)\right )}d\left (6 i a^2 d \tan (e+f x)\right )}{f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a \left (\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}-\frac {6 i \sqrt {6} a^2 d \int \frac {1}{\sqrt {6 c+6 d \tan (e+f x)} \left (6 a^2 d-6 i a^2 d \tan (e+f x)\right )}d\left (6 i a^2 d \tan (e+f x)\right )}{f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 a \left (\frac {12 \sqrt {6} a^4 d \int \frac {1}{36 i d^2 \tan ^2(e+f x) a^6+6 (i c+d) a^2}d\sqrt {6 c+6 d \tan (e+f x)}}{f}+\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 a \left (\frac {12 a^2 d \arctan \left (\frac {\sqrt {6} a^2 d \tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
(-2*(a^3 + I*a^3*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) + (2*a*(( 12*a^2*d*ArcTan[(Sqrt[6]*a^2*d*Tan[e + f*x])/Sqrt[c - I*d]])/(Sqrt[c - I*d ]*f) + (2*a^2*(I*c - 4*d)*Sqrt[c + d*Tan[e + f*x]])/(d*f)))/(3*d)
3.12.19.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && 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. 799 vs. \(2 (107 ) = 214\).
Time = 0.90 (sec) , antiderivative size = 800, normalized size of antiderivative = 6.35
| method | result | size |
| derivativedivides | \(\frac {2 a^{3} \left (-\frac {i \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+i c \sqrt {c +d \tan \left (f x +e \right )}-3 d \sqrt {c +d \tan \left (f x +e \right )}-4 d^{2} \left (\frac {\frac {\left (i \sqrt {c^{2}+d^{2}}+i c -d \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -\frac {\left (i \sqrt {c^{2}+d^{2}}+i c -d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {-\frac {\left (2 i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}+2 i c^{3}+2 i c \,d^{2}-c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \ln \left (\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d \tan \left (f x +e \right )-c -\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, c^{2}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, c d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}+\frac {\left (2 i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}+2 i c^{3}+2 i c \,d^{2}-c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, \left (\sqrt {c^{2}+d^{2}}\, c +c^{2}+d^{2}\right )}\right )\right )}{f \,d^{2}}\) | \(800\) |
| default | \(\frac {2 a^{3} \left (-\frac {i \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+i c \sqrt {c +d \tan \left (f x +e \right )}-3 d \sqrt {c +d \tan \left (f x +e \right )}-4 d^{2} \left (\frac {\frac {\left (i \sqrt {c^{2}+d^{2}}+i c -d \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -\frac {\left (i \sqrt {c^{2}+d^{2}}+i c -d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {-\frac {\left (2 i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}+2 i c^{3}+2 i c \,d^{2}-c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \ln \left (\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d \tan \left (f x +e \right )-c -\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, c^{2}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, c d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}+\frac {\left (2 i c^{2} \sqrt {c^{2}+d^{2}}+i d^{2} \sqrt {c^{2}+d^{2}}+2 i c^{3}+2 i c \,d^{2}-c d \sqrt {c^{2}+d^{2}}-c^{2} d -d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, \left (\sqrt {c^{2}+d^{2}}\, c +c^{2}+d^{2}\right )}\right )\right )}{f \,d^{2}}\) | \(800\) |
| parts | \(\text {Expression too large to display}\) | \(3867\) |
2/f*a^3/d^2*(-1/3*I*(c+d*tan(f*x+e))^(3/2)+I*c*(c+d*tan(f*x+e))^(1/2)-3*d* (c+d*tan(f*x+e))^(1/2)-4*d^2*(1/2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^ (1/2)*(1/2*(I*(c^2+d^2)^(1/2)+I*c-d)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1 /2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))+2*(I*(2*(c^2+d^2)^(1/2) +2*c)^(1/2)*c-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d-1/2*(I*(c^2+d^2)^(1/2)+I*c-d )*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*( c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2* c)^(1/2)))+1/2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)/((c^2+d^2)^(1 /2)*c+c^2+d^2)*(-1/2*(2*I*(c^2+d^2)^(1/2)*c^2+I*d^2*(c^2+d^2)^(1/2)+2*I*c^ 3+2*I*c*d^2-c*d*(c^2+d^2)^(1/2)-c^2*d-d^3)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c ^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))+2*(-I*(2*(c^2+d^2 )^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c^2-I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3 -I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d^2+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+ d^2)^(1/2)*c*d+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2*d+(2*(c^2+d^2)^(1/2)+2*c) ^(1/2)*d^3+1/2*(2*I*(c^2+d^2)^(1/2)*c^2+I*d^2*(c^2+d^2)^(1/2)+2*I*c^3+2*I* c*d^2-c*d*(c^2+d^2)^(1/2)-c^2*d-d^3)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^ 2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f *x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (102) = 204\).
Time = 0.25 (sec) , antiderivative size = 427, normalized size of antiderivative = 3.39 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {3 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac {{\left (8 \, a^{3} c + \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} {\left ({\left (i \, c + d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c + d\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 8 \, {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 3 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac {{\left (8 \, a^{3} c + \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} {\left ({\left (-i \, c - d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c - d\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 8 \, {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 16 \, {\left (-i \, a^{3} c + 4 \, a^{3} d + {\left (-i \, a^{3} c + 5 \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )}} \]
1/12*(3*(d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-64*I*a^6/((I*c + d)*f^2) )*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c + d)*f^2))*((I*c + d)*f*e^(2*I*f *x + 2*I*e) + (I*c + d)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/ (e^(2*I*f*x + 2*I*e) + 1)) + 8*(a^3*c - I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(- 2*I*f*x - 2*I*e)/a^3) - 3*(d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-64*I*a ^6/((I*c + d)*f^2))*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c + d)*f^2))*((- I*c - d)*f*e^(2*I*f*x + 2*I*e) + (-I*c - d)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)) + 8*(a^3*c - I*a^3*d)*e^(2* I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) - 16*(-I*a^3*c + 4*a^3*d + (-I*a ^3*c + 5*a^3*d)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)
\[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=- i a^{3} \left (\int \frac {i}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\right )\, dx\right ) \]
-I*a**3*(Integral(I/sqrt(c + d*tan(e + f*x)), x) + Integral(-3*tan(e + f*x )/sqrt(c + d*tan(e + f*x)), x) + Integral(tan(e + f*x)**3/sqrt(c + d*tan(e + f*x)), x) + Integral(-3*I*tan(e + f*x)**2/sqrt(c + d*tan(e + f*x)), x))
\[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}{\sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (102) = 204\).
Time = 0.64 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.89 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {16 i \, a^{3} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 \, {\left (i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} d^{4} f^{2} - 3 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3} c d^{4} f^{2} + 9 \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3} d^{5} f^{2}\right )}}{3 \, d^{6} f^{3}} \]
16*I*a^3*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan (f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) - I*sqrt(-2*c + 2*sqrt(c ^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/(sqrt(-2*c + 2*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) - 2/3*(I*(d*tan( f*x + e) + c)^(3/2)*a^3*d^4*f^2 - 3*I*sqrt(d*tan(f*x + e) + c)*a^3*c*d^4*f ^2 + 9*sqrt(d*tan(f*x + e) + c)*a^3*d^5*f^2)/(d^6*f^3)
Time = 7.46 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.94 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=-\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\frac {a^3\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,2{}\mathrm {i}}{3\,d^2\,f}+\frac {a^3\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {-c+d\,1{}\mathrm {i}}}\right )\,8{}\mathrm {i}}{f\,\sqrt {-c+d\,1{}\mathrm {i}}} \]