Integrand size = 28, antiderivative size = 132 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=\frac {8 (-1)^{3/4} a^3 \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}-\frac {8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}+\frac {8 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}} \] Output:
8*(-1)^(3/4)*a^3*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(7/2)/f -8/5*I*a^3/d^2/f/(d*tan(f*x+e))^(3/2)+8*a^3/d^3/f/(d*tan(f*x+e))^(1/2)-2/5 *(a^3+I*a^3*tan(f*x+e))/d/f/(d*tan(f*x+e))^(5/2)
Time = 1.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.64 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=\frac {2 a^3 \left (20 (-1)^{3/4} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-\frac {d \left (-20+5 i \cot (e+f x)+\cot ^2(e+f x)\right )}{\sqrt {d \tan (e+f x)}}\right )}{5 d^4 f} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]
Output:
(2*a^3*(20*(-1)^(3/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqr t[d]] - (d*(-20 + (5*I)*Cot[e + f*x] + Cot[e + f*x]^2))/Sqrt[d*Tan[e + f*x ]]))/(5*d^4*f)
Time = 0.76 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 4036, 27, 3042, 4074, 27, 3042, 4012, 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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}}dx\) |
\(\Big \downarrow \) 4036 |
\(\displaystyle -\frac {2 \int -\frac {2 (i \tan (e+f x) a+a) \left (3 i a^2 d-2 a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{5/2}}dx}{5 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \int \frac {(i \tan (e+f x) a+a) \left (3 i a^2 d-2 a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{5/2}}dx}{5 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \int \frac {(i \tan (e+f x) a+a) \left (3 i a^2 d-2 a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{5/2}}dx}{5 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 4074 |
\(\displaystyle \frac {4 \left (\frac {\int -\frac {5 \left (d^2 a^3+i d^2 \tan (e+f x) a^3\right )}{(d \tan (e+f x))^{3/2}}dx}{d^2}-\frac {2 i a^3}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \left (-\frac {5 \int \frac {d^2 a^3+i d^2 \tan (e+f x) a^3}{(d \tan (e+f x))^{3/2}}dx}{d^2}-\frac {2 i a^3}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \left (-\frac {5 \int \frac {d^2 a^3+i d^2 \tan (e+f x) a^3}{(d \tan (e+f x))^{3/2}}dx}{d^2}-\frac {2 i a^3}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {4 \left (-\frac {5 \left (-\frac {2 a^3 d}{f \sqrt {d \tan (e+f x)}}+\frac {\int \frac {i a^3 d^3-a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{d^2}\right )}{d^2}-\frac {2 i a^3}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \left (-\frac {5 \left (-\frac {2 a^3 d}{f \sqrt {d \tan (e+f x)}}+\frac {\int \frac {i a^3 d^3-a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{d^2}\right )}{d^2}-\frac {2 i a^3}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 4016 |
\(\displaystyle \frac {4 \left (-\frac {5 \left (-\frac {2 a^3 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^6 d^4 \int \frac {1}{i a^3 d^4+a^3 \tan (e+f x) d^4}d\sqrt {d \tan (e+f x)}}{f}\right )}{d^2}-\frac {2 i a^3}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {4 \left (-\frac {5 \left (-\frac {2 (-1)^{3/4} a^3 \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {2 a^3 d}{f \sqrt {d \tan (e+f x)}}\right )}{d^2}-\frac {2 i a^3}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]
Output:
(-2*(a^3 + I*a^3*Tan[e + f*x]))/(5*d*f*(d*Tan[e + f*x])^(5/2)) + (4*(((-2* I)*a^3)/(f*(d*Tan[e + f*x])^(3/2)) - (5*((-2*(-1)^(3/4)*a^3*Sqrt[d]*ArcTan [((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - (2*a^3*d)/(f*Sqrt[d*Tan[e + f*x]])))/d^2))/(5*d^2)
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[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (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*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
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. 324 vs. \(2 (110 ) = 220\).
Time = 1.65 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.46
method | result | size |
derivativedivides | \(\frac {2 a^{3} \left (-\frac {i}{\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {d}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {4}{d \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (d^{2}\right )^{\frac {1}{4}}}}{d}\right )}{f \,d^{2}}\) | \(325\) |
default | \(\frac {2 a^{3} \left (-\frac {i}{\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {d}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {4}{d \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (d^{2}\right )^{\frac {1}{4}}}}{d}\right )}{f \,d^{2}}\) | \(325\) |
parts | \(\frac {2 a^{3} d \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 )}{8 d^{4} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {1}{5 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}\right )}{f}-\frac {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^{4}}+\frac {3 i a^{3} \left (-\frac {2}{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 )}{4 d^{4}}\right )}{f}-\frac {6 a^{3} \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 d}\) | \(636\) |
Input:
int((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x,method=_RETURNVERBOSE)
Output:
2/f*a^3/d^2*(-I/(d*tan(f*x+e))^(3/2)-1/5*d/(d*tan(f*x+e))^(5/2)+4/d/(d*tan (f*x+e))^(1/2)+1/d*(-1/2*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) )+1/2/(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. 468 vs. \(2 (110) = 220\).
Time = 0.10 (sec) , antiderivative size = 468, normalized size of antiderivative = 3.55 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=\frac {5 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{7} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d^{4} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {64 i \, a^{6}}{d^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 5 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt {\frac {64 i \, a^{6}}{d^{7} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, d^{4} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {64 i \, a^{6}}{d^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 16 \, {\left (-13 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 6 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 11 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{3}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{20 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )}} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="fricas")
Output:
1/20*(5*(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4*f *e^(2*I*f*x + 2*I*e) - d^4*f)*sqrt(64*I*a^6/(d^7*f^2))*log(1/4*(-8*I*a^3*d *e^(2*I*f*x + 2*I*e) + (I*d^4*f*e^(2*I*f*x + 2*I*e) + I*d^4*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^7*f ^2)))*e^(-2*I*f*x - 2*I*e)/a^3) - 5*(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e ^(4*I*f*x + 4*I*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*f)*sqrt(64*I*a^6/(d ^7*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (-I*d^4*f*e^(2*I*f*x + 2*I*e) - I*d^4*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^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) - 16*(-13*I*a ^3*e^(6*I*f*x + 6*I*e) + 6*I*a^3*e^(4*I*f*x + 4*I*e) + 11*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)))/(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3* d^4*f*e^(2*I*f*x + 2*I*e) - d^4*f)
\[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=- i a^{3} \left (\int \frac {i}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\right )\, dx\right ) \] Input:
integrate((a+I*a*tan(f*x+e))**3/(d*tan(f*x+e))**(7/2),x)
Output:
-I*a**3*(Integral(I/(d*tan(e + f*x))**(7/2), x) + Integral(-3*tan(e + f*x) /(d*tan(e + f*x))**(7/2), x) + Integral(tan(e + f*x)**3/(d*tan(e + f*x))** (7/2), x) + Integral(-3*I*tan(e + f*x)**2/(d*tan(e + f*x))**(7/2), x))
Time = 0.12 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.66 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=\frac {\frac {5 \, a^{3} {\left (-\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d^{2}} + \frac {2 \, {\left (20 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} - 5 i \, a^{3} d^{2} \tan \left (f x + e\right ) - a^{3} d^{2}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} d^{2}}}{5 \, d f} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="maxima")
Output:
1/5*(5*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 + 2*(20*a^3*d^2*tan(f*x + e)^2 - 5*I*a^3*d^2*tan(f* x + e) - a^3*d^2)/((d*tan(f*x + e))^(5/2)*d^2))/(d*f)
Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.87 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=\frac {2 \, a^{3} {\left (\frac {20 \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{d^{\frac {7}{2}} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {20 \, d^{2} \tan \left (f x + e\right )^{2} - 5 i \, d^{2} \tan \left (f x + e\right ) - d^{2}}{\sqrt {d \tan \left (f x + e\right )} d^{5} \tan \left (f x + e\right )^{2}}\right )}}{5 \, f} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="giac")
Output:
2/5*a^3*(20*sqrt(2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*sqrt(2)*d^(3/2 ) + sqrt(2)*sqrt(d)*abs(d)))/(d^(7/2)*(I*d/abs(d) + 1)) + (20*d^2*tan(f*x + e)^2 - 5*I*d^2*tan(f*x + e) - d^2)/(sqrt(d*tan(f*x + e))*d^5*tan(f*x + e )^2))/f
Time = 1.53 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.72 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=-\frac {\frac {2\,a^3}{5\,d\,f}-\frac {8\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{d\,f}+\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}}{d\,f}}{{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}-\frac {2\,\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atanh}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,\sqrt {d}}\right )}{d^{7/2}\,f} \] Input:
int((a + a*tan(e + f*x)*1i)^3/(d*tan(e + f*x))^(7/2),x)
Output:
- ((2*a^3)/(5*d*f) + (a^3*tan(e + f*x)*2i)/(d*f) - (8*a^3*tan(e + f*x)^2)/ (d*f))/(d*tan(e + f*x))^(5/2) - (2*16i^(1/2)*a^3*atanh((16i^(1/2)*(d*tan(e + f*x))^(1/2))/(4*d^(1/2))))/(d^(7/2)*f)
\[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=\frac {\sqrt {d}\, a^{3} \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{4}}d x +3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{3}}d x \right ) i -3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{2}}d x \right )-\left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )}d x \right ) i \right )}{d^{4}} \] Input:
int((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x)
Output:
(sqrt(d)*a**3*(int(sqrt(tan(e + f*x))/tan(e + f*x)**4,x) + 3*int(sqrt(tan( e + f*x))/tan(e + f*x)**3,x)*i - 3*int(sqrt(tan(e + f*x))/tan(e + f*x)**2, x) - int(sqrt(tan(e + f*x))/tan(e + f*x),x)*i))/d**4