Integrand size = 28, antiderivative size = 179 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^3 \, dx=-\frac {8 (-1)^{3/4} a^3 d^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {8 i a^3 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {8 a^3 d (d \tan (e+f x))^{3/2}}{3 f}+\frac {8 i a^3 (d \tan (e+f x))^{5/2}}{5 f}-\frac {40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}-\frac {2 (d \tan (e+f x))^{7/2} \left (a^3+i a^3 \tan (e+f x)\right )}{9 d f} \] Output:
-8*(-1)^(3/4)*a^3*d^(5/2)*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/ f-8*I*a^3*d^2*(d*tan(f*x+e))^(1/2)/f+8/3*a^3*d*(d*tan(f*x+e))^(3/2)/f+8/5* I*a^3*(d*tan(f*x+e))^(5/2)/f-40/63*a^3*(d*tan(f*x+e))^(7/2)/d/f-2/9*(d*tan (f*x+e))^(7/2)*(a^3+I*a^3*tan(f*x+e))/d/f
Time = 1.25 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.61 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^3 \, dx=-\frac {2 a^3 d^2 \left (1260 (-1)^{3/4} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+\sqrt {d \tan (e+f x)} \left (1260 i-420 \tan (e+f x)-252 i \tan ^2(e+f x)+135 \tan ^3(e+f x)+35 i \tan ^4(e+f x)\right )\right )}{315 f} \] Input:
Integrate[(d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x])^3,x]
Output:
(-2*a^3*d^2*(1260*(-1)^(3/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x ]])/Sqrt[d]] + Sqrt[d*Tan[e + f*x]]*(1260*I - 420*Tan[e + f*x] - (252*I)*T an[e + f*x]^2 + 135*Tan[e + f*x]^3 + (35*I)*Tan[e + f*x]^4)))/(315*f)
Time = 1.08 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4039, 27, 3042, 4075, 3042, 4011, 3042, 4011, 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.
| \(\displaystyle \int (a+i a \tan (e+f x))^3 (d \tan (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (e+f x))^3 (d \tan (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 4039 |
| \(\displaystyle \frac {2 a \int 2 (d \tan (e+f x))^{5/2} (i \tan (e+f x) a+a) (4 a d+5 i a \tan (e+f x) d)dx}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 a \int (d \tan (e+f x))^{5/2} (i \tan (e+f x) a+a) (4 a d+5 i a \tan (e+f x) d)dx}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \int (d \tan (e+f x))^{5/2} (i \tan (e+f x) a+a) (4 a d+5 i a \tan (e+f x) d)dx}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 4075 |
| \(\displaystyle \frac {4 a \left (-\frac {10 a^2 (d \tan (e+f x))^{7/2}}{7 f}+\int (d \tan (e+f x))^{5/2} \left (9 d a^2+9 i d \tan (e+f x) a^2\right )dx\right )}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \left (-\frac {10 a^2 (d \tan (e+f x))^{7/2}}{7 f}+\int (d \tan (e+f x))^{5/2} \left (9 d a^2+9 i d \tan (e+f x) a^2\right )dx\right )}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {4 a \left (\int (d \tan (e+f x))^{3/2} \left (9 a^2 d^2 \tan (e+f x)-9 i a^2 d^2\right )dx-\frac {10 a^2 (d \tan (e+f x))^{7/2}}{7 f}+\frac {18 i a^2 d (d \tan (e+f x))^{5/2}}{5 f}\right )}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \left (\int (d \tan (e+f x))^{3/2} \left (9 a^2 d^2 \tan (e+f x)-9 i a^2 d^2\right )dx-\frac {10 a^2 (d \tan (e+f x))^{7/2}}{7 f}+\frac {18 i a^2 d (d \tan (e+f x))^{5/2}}{5 f}\right )}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {4 a \left (\int \sqrt {d \tan (e+f x)} \left (-9 a^2 d^3-9 i a^2 \tan (e+f x) d^3\right )dx+\frac {6 a^2 d^2 (d \tan (e+f x))^{3/2}}{f}-\frac {10 a^2 (d \tan (e+f x))^{7/2}}{7 f}+\frac {18 i a^2 d (d \tan (e+f x))^{5/2}}{5 f}\right )}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \left (\int \sqrt {d \tan (e+f x)} \left (-9 a^2 d^3-9 i a^2 \tan (e+f x) d^3\right )dx+\frac {6 a^2 d^2 (d \tan (e+f x))^{3/2}}{f}-\frac {10 a^2 (d \tan (e+f x))^{7/2}}{7 f}+\frac {18 i a^2 d (d \tan (e+f x))^{5/2}}{5 f}\right )}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {4 a \left (\int \frac {9 i a^2 d^4-9 a^2 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx-\frac {18 i a^2 d^3 \sqrt {d \tan (e+f x)}}{f}+\frac {6 a^2 d^2 (d \tan (e+f x))^{3/2}}{f}-\frac {10 a^2 (d \tan (e+f x))^{7/2}}{7 f}+\frac {18 i a^2 d (d \tan (e+f x))^{5/2}}{5 f}\right )}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \left (\int \frac {9 i a^2 d^4-9 a^2 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx-\frac {18 i a^2 d^3 \sqrt {d \tan (e+f x)}}{f}+\frac {6 a^2 d^2 (d \tan (e+f x))^{3/2}}{f}-\frac {10 a^2 (d \tan (e+f x))^{7/2}}{7 f}+\frac {18 i a^2 d (d \tan (e+f x))^{5/2}}{5 f}\right )}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 4016 |
| \(\displaystyle \frac {4 a \left (-\frac {162 a^4 d^8 \int \frac {1}{9 i a^2 d^5+9 a^2 \tan (e+f x) d^5}d\sqrt {d \tan (e+f x)}}{f}-\frac {18 i a^2 d^3 \sqrt {d \tan (e+f x)}}{f}+\frac {6 a^2 d^2 (d \tan (e+f x))^{3/2}}{f}+\frac {18 i a^2 d (d \tan (e+f x))^{5/2}}{5 f}-\frac {10 a^2 (d \tan (e+f x))^{7/2}}{7 f}\right )}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {4 a \left (-\frac {18 (-1)^{3/4} a^2 d^{7/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {18 i a^2 d^3 \sqrt {d \tan (e+f x)}}{f}+\frac {6 a^2 d^2 (d \tan (e+f x))^{3/2}}{f}+\frac {18 i a^2 d (d \tan (e+f x))^{5/2}}{5 f}-\frac {10 a^2 (d \tan (e+f x))^{7/2}}{7 f}\right )}{9 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{7/2}}{9 d f}\) |
Input:
Int[(d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x])^3,x]
Output:
(-2*(d*Tan[e + f*x])^(7/2)*(a^3 + I*a^3*Tan[e + f*x]))/(9*d*f) + (4*a*((-1 8*(-1)^(3/4)*a^2*d^(7/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] )/f - ((18*I)*a^2*d^3*Sqrt[d*Tan[e + f*x]])/f + (6*a^2*d^2*(d*Tan[e + f*x] )^(3/2))/f + (((18*I)/5)*a^2*d*(d*Tan[e + f*x])^(5/2))/f - (10*a^2*(d*Tan[ e + f*x])^(7/2))/(7*f)))/(9*d)
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. 357 vs. \(2 (148 ) = 296\).
Time = 1.71 (sec) , antiderivative size = 358, normalized size of antiderivative = 2.00
| method | result | size |
| derivativedivides | \(\frac {2 a^{3} \left (-\frac {i \left (d \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {3 d \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {4 i d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {4 d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-4 i d^{4} \sqrt {d \tan \left (f x +e \right )}+4 d^{5} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\) | \(358\) |
| default | \(\frac {2 a^{3} \left (-\frac {i \left (d \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {3 d \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {4 i d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {4 d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-4 i d^{4} \sqrt {d \tan \left (f x +e \right )}+4 d^{5} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\) | \(358\) |
| parts | \(\frac {2 a^{3} d \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}-\frac {2 i a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+d^{4} \sqrt {d \tan \left (f x +e \right )}-\frac {d^{4} \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 (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-2 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 )}{4}\right )}{f}-\frac {6 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}-\frac {d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+\frac {d^{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 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\) | \(690\) |
Input:
int((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
Output:
2/f*a^3/d^2*(-1/9*I*(d*tan(f*x+e))^(9/2)-3/7*d*(d*tan(f*x+e))^(7/2)+4/5*I* d^2*(d*tan(f*x+e))^(5/2)+4/3*d^3*(d*tan(f*x+e))^(3/2)-4*I*d^4*(d*tan(f*x+e ))^(1/2)+4*d^5*(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*arcta n(-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. 489 vs. \(2 (145) = 290\).
Time = 0.11 (sec) , antiderivative size = 489, normalized size of antiderivative = 2.73 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^3 \, dx=-\frac {315 \, \sqrt {\frac {64 i \, a^{6} d^{5}}{f^{2}}} {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-8 i \, a^{3} d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {64 i \, a^{6} d^{5}}{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} d^{2}}\right ) - 315 \, \sqrt {\frac {64 i \, a^{6} d^{5}}{f^{2}}} {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-8 i \, a^{3} d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {64 i \, a^{6} d^{5}}{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} d^{2}}\right ) + 16 \, {\left (1051 i \, a^{3} d^{2} e^{\left (8 i \, f x + 8 i \, e\right )} + 2735 i \, a^{3} d^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 3633 i \, a^{3} d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 2165 i \, a^{3} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 496 i \, a^{3} d^{2}\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}}}{1260 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \] Input:
integrate((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")
Output:
-1/1260*(315*sqrt(64*I*a^6*d^5/f^2)*(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f* x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)*log(1/ 4*(-8*I*a^3*d^3*e^(2*I*f*x + 2*I*e) + sqrt(64*I*a^6*d^5/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*d^2)) - 315*sqrt(64*I*a^6*d^5/f^2)*(f *e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)*log(1/4*(-8*I*a^3*d^3*e^(2*I*f*x + 2*I*e) + sqrt(64*I*a^6*d^5/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* d^2)) + 16*(1051*I*a^3*d^2*e^(8*I*f*x + 8*I*e) + 2735*I*a^3*d^2*e^(6*I*f*x + 6*I*e) + 3633*I*a^3*d^2*e^(4*I*f*x + 4*I*e) + 2165*I*a^3*d^2*e^(2*I*f*x + 2*I*e) + 496*I*a^3*d^2)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f *x + 2*I*e) + 1)))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f* e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)
\[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^3 \, dx=- i a^{3} \left (\int i \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx + \int \left (- 3 \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )}\right )\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 3 i \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan ^{2}{\left (e + f x \right )}\right )\, dx\right ) \] Input:
integrate((d*tan(f*x+e))**(5/2)*(a+I*a*tan(f*x+e))**3,x)
Output:
-I*a**3*(Integral(I*(d*tan(e + f*x))**(5/2), x) + Integral(-3*(d*tan(e + f *x))**(5/2)*tan(e + f*x), x) + Integral((d*tan(e + f*x))**(5/2)*tan(e + f* x)**3, x) + Integral(-3*I*(d*tan(e + f*x))**(5/2)*tan(e + f*x)**2, x))
Time = 0.12 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.43 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^3 \, dx=-\frac {315 \, a^{3} d^{4} {\left (-\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - \frac {2 \, {\left (-35 i \, \left (d \tan \left (f x + e\right )\right )^{\frac {9}{2}} a^{3} - 135 \, \left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} a^{3} d + 252 i \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} d^{2} + 420 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d^{3} - 1260 i \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{4}\right )}}{d}}{315 \, d f} \] Input:
integrate((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")
Output:
-1/315*(315*a^3*d^4*(-(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*sqrt(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)/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*(-35*I*(d*tan(f*x + e))^(9/2)*a^3 - 135*(d *tan(f*x + e))^(7/2)*a^3*d + 252*I*(d*tan(f*x + e))^(5/2)*a^3*d^2 + 420*(d *tan(f*x + e))^(3/2)*a^3*d^3 - 1260*I*sqrt(d*tan(f*x + e))*a^3*d^4)/d)/(d* f)
Time = 0.32 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.97 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^3 \, dx=-\frac {2 \, {\left (\frac {1260 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\frac {i \, d}{{\left | d \right |}} + 1} + \frac {35 i \, \sqrt {d \tan \left (f x + e\right )} d^{20} \tan \left (f x + e\right )^{4} + 135 \, \sqrt {d \tan \left (f x + e\right )} d^{20} \tan \left (f x + e\right )^{3} - 252 i \, \sqrt {d \tan \left (f x + e\right )} d^{20} \tan \left (f x + e\right )^{2} - 420 \, \sqrt {d \tan \left (f x + e\right )} d^{20} \tan \left (f x + e\right ) + 1260 i \, \sqrt {d \tan \left (f x + e\right )} d^{20}}{d^{18}}\right )} a^{3}}{315 \, f} \] Input:
integrate((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e))^3,x, algorithm="giac")
Output:
-2/315*(1260*sqrt(2)*d^(5/2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*sqrt( 2)*d^(3/2) + sqrt(2)*sqrt(d)*abs(d)))/(I*d/abs(d) + 1) + (35*I*sqrt(d*tan( f*x + e))*d^20*tan(f*x + e)^4 + 135*sqrt(d*tan(f*x + e))*d^20*tan(f*x + e) ^3 - 252*I*sqrt(d*tan(f*x + e))*d^20*tan(f*x + e)^2 - 420*sqrt(d*tan(f*x + e))*d^20*tan(f*x + e) + 1260*I*sqrt(d*tan(f*x + e))*d^20)/d^18)*a^3/f
Time = 1.89 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.77 \[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^3 \, dx=\frac {a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,8{}\mathrm {i}}{5\,f}-\frac {a^3\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,8{}\mathrm {i}}{f}-\frac {6\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}{7\,d\,f}-\frac {a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{9/2}\,2{}\mathrm {i}}{9\,d^2\,f}+\frac {8\,a^3\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}-\frac {\sqrt {16{}\mathrm {i}}\,a^3\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,\sqrt {d}}\right )\,2{}\mathrm {i}}{f} \] Input:
int((d*tan(e + f*x))^(5/2)*(a + a*tan(e + f*x)*1i)^3,x)
Output:
(a^3*(d*tan(e + f*x))^(5/2)*8i)/(5*f) - (a^3*d^2*(d*tan(e + f*x))^(1/2)*8i )/f - (6*a^3*(d*tan(e + f*x))^(7/2))/(7*d*f) - (a^3*(d*tan(e + f*x))^(9/2) *2i)/(9*d^2*f) + (8*a^3*d*(d*tan(e + f*x))^(3/2))/(3*f) - (16i^(1/2)*a^3*d ^(5/2)*atan((16i^(1/2)*(d*tan(e + f*x))^(1/2)*1i)/(4*d^(1/2)))*2i)/f
\[ \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^3 \, dx=\frac {\sqrt {d}\, a^{3} d^{2} \left (-10 \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{4} i +72 \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2} i -360 \sqrt {\tan \left (f x +e \right )}\, i +180 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )}d x \right ) f i -135 \left (\int \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{4}d x \right ) f +45 \left (\int \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}d x \right ) f \right )}{45 f} \] Input:
int((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e))^3,x)
Output:
(sqrt(d)*a**3*d**2*( - 10*sqrt(tan(e + f*x))*tan(e + f*x)**4*i + 72*sqrt(t an(e + f*x))*tan(e + f*x)**2*i - 360*sqrt(tan(e + f*x))*i + 180*int(sqrt(t an(e + f*x))/tan(e + f*x),x)*f*i - 135*int(sqrt(tan(e + f*x))*tan(e + f*x) **4,x)*f + 45*int(sqrt(tan(e + f*x))*tan(e + f*x)**2,x)*f))/(45*f)