\(\int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx\) [163]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 159 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx=-\frac {8 \sqrt [4]{-1} a^3 \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{9/2} f}-\frac {32 i a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}+\frac {8 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac {8 i a^3}{d^4 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}} \] Output:

-8*(-1)^(1/4)*a^3*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(9/2)/ 
f-32/35*I*a^3/d^2/f/(d*tan(f*x+e))^(5/2)+8/3*a^3/d^3/f/(d*tan(f*x+e))^(3/2 
)+8*I*a^3/d^4/f/(d*tan(f*x+e))^(1/2)-2/7*(a^3+I*a^3*tan(f*x+e))/d/f/(d*tan 
(f*x+e))^(7/2)
 

Mathematica [A] (verified)

Time = 1.91 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.62 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx=-\frac {2 a^3 \left (420 \sqrt [4]{-1} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+\frac {d \left (-420 i-140 \cot (e+f x)+63 i \cot ^2(e+f x)+15 \cot ^3(e+f x)\right )}{\sqrt {d \tan (e+f x)}}\right )}{105 d^5 f} \] Input:

Integrate[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(9/2),x]
 

Output:

(-2*a^3*(420*(-1)^(1/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/S 
qrt[d]] + (d*(-420*I - 140*Cot[e + f*x] + (63*I)*Cot[e + f*x]^2 + 15*Cot[e 
 + f*x]^3))/Sqrt[d*Tan[e + f*x]]))/(105*d^5*f)
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.09, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4036, 27, 3042, 4074, 27, 3042, 4012, 3042, 4012, 25, 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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}}dx\)

\(\Big \downarrow \) 4036

\(\displaystyle -\frac {2 \int -\frac {2 (i \tan (e+f x) a+a) \left (4 i a^2 d-3 a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{7/2}}dx}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 \int \frac {(i \tan (e+f x) a+a) \left (4 i a^2 d-3 a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{7/2}}dx}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 \int \frac {(i \tan (e+f x) a+a) \left (4 i a^2 d-3 a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{7/2}}dx}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 4074

\(\displaystyle \frac {4 \left (\frac {\int -\frac {7 \left (d^2 a^3+i d^2 \tan (e+f x) a^3\right )}{(d \tan (e+f x))^{5/2}}dx}{d^2}-\frac {8 i a^3}{5 f (d \tan (e+f x))^{5/2}}\right )}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 \left (-\frac {7 \int \frac {d^2 a^3+i d^2 \tan (e+f x) a^3}{(d \tan (e+f x))^{5/2}}dx}{d^2}-\frac {8 i a^3}{5 f (d \tan (e+f x))^{5/2}}\right )}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 \left (-\frac {7 \int \frac {d^2 a^3+i d^2 \tan (e+f x) a^3}{(d \tan (e+f x))^{5/2}}dx}{d^2}-\frac {8 i a^3}{5 f (d \tan (e+f x))^{5/2}}\right )}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {4 \left (-\frac {7 \left (-\frac {2 a^3 d}{3 f (d \tan (e+f x))^{3/2}}+\frac {\int \frac {i a^3 d^3-a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{d^2}\right )}{d^2}-\frac {8 i a^3}{5 f (d \tan (e+f x))^{5/2}}\right )}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 \left (-\frac {7 \left (-\frac {2 a^3 d}{3 f (d \tan (e+f x))^{3/2}}+\frac {\int \frac {i a^3 d^3-a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{d^2}\right )}{d^2}-\frac {8 i a^3}{5 f (d \tan (e+f x))^{5/2}}\right )}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {4 \left (-\frac {7 \left (-\frac {2 a^3 d}{3 f (d \tan (e+f x))^{3/2}}+\frac {\frac {\int -\frac {a^3 d^4+i a^3 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {2 i a^3 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}\right )}{d^2}-\frac {8 i a^3}{5 f (d \tan (e+f x))^{5/2}}\right )}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4 \left (-\frac {7 \left (-\frac {2 a^3 d}{3 f (d \tan (e+f x))^{3/2}}+\frac {-\frac {\int \frac {a^3 d^4+i a^3 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {2 i a^3 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}\right )}{d^2}-\frac {8 i a^3}{5 f (d \tan (e+f x))^{5/2}}\right )}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 \left (-\frac {7 \left (-\frac {2 a^3 d}{3 f (d \tan (e+f x))^{3/2}}+\frac {-\frac {\int \frac {a^3 d^4+i a^3 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {2 i a^3 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}\right )}{d^2}-\frac {8 i a^3}{5 f (d \tan (e+f x))^{5/2}}\right )}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 4016

\(\displaystyle \frac {4 \left (-\frac {7 \left (-\frac {2 a^3 d}{3 f (d \tan (e+f x))^{3/2}}+\frac {-\frac {2 a^6 d^6 \int \frac {1}{a^3 d^5-i a^3 d^5 \tan (e+f x)}d\sqrt {d \tan (e+f x)}}{f}-\frac {2 i a^3 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}\right )}{d^2}-\frac {8 i a^3}{5 f (d \tan (e+f x))^{5/2}}\right )}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {4 \left (-\frac {7 \left (-\frac {2 a^3 d}{3 f (d \tan (e+f x))^{3/2}}+\frac {\frac {2 \sqrt [4]{-1} a^3 d^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {2 i a^3 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}\right )}{d^2}-\frac {8 i a^3}{5 f (d \tan (e+f x))^{5/2}}\right )}{7 d^2}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\)

Input:

Int[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(9/2),x]
 

Output:

(-2*(a^3 + I*a^3*Tan[e + f*x]))/(7*d*f*(d*Tan[e + f*x])^(7/2)) + (4*((((-8 
*I)/5)*a^3)/(f*(d*Tan[e + f*x])^(5/2)) - (7*((-2*a^3*d)/(3*f*(d*Tan[e + f* 
x])^(3/2)) + ((2*(-1)^(1/4)*a^3*d^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + 
f*x]])/Sqrt[d]])/f - ((2*I)*a^3*d^2)/(f*Sqrt[d*Tan[e + f*x]]))/d^2))/d^2)) 
/(7*d^2)
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218
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]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4012
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 
]
 

rule 4016
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]
 

rule 4036
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])
 

rule 4074
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]
 
Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (132 ) = 264\).

Time = 1.66 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.14

method result size
derivativedivides \(\frac {2 a^{3} \left (-\frac {3 i}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {d}{7 \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {4 i}{d^{2} \sqrt {d \tan \left (f x +e \right )}}+\frac {4}{3 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 d}+\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{2}}\right )}{f \,d^{2}}\) \(341\)
default \(\frac {2 a^{3} \left (-\frac {3 i}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {d}{7 \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {4 i}{d^{2} \sqrt {d \tan \left (f x +e \right )}}+\frac {4}{3 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 d}+\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{2}}\right )}{f \,d^{2}}\) \(341\)
parts \(\frac {2 a^{3} d \left (\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^{6}}-\frac {1}{7 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {1}{3 d^{4} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f}-\frac {2 i 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^{2}}+\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^{4} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2}{5 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {2}{d^{4} \sqrt {d \tan \left (f x +e \right )}}\right )}{f}-\frac {6 a^{3} \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 d}\) \(673\)

Input:

int((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x,method=_RETURNVERBOSE)
 

Output:

2/f*a^3/d^2*(-3/5*I/(d*tan(f*x+e))^(5/2)-1/7*d/(d*tan(f*x+e))^(7/2)+4*I/d^ 
2/(d*tan(f*x+e))^(1/2)+4/3/d/(d*tan(f*x+e))^(3/2)+1/d^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*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))))
                                                                                    
                                                                                    
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 521 vs. \(2 (131) = 262\).

Time = 0.10 (sec) , antiderivative size = 521, normalized size of antiderivative = 3.28 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx=\frac {105 \, {\left (d^{5} f e^{\left (8 i \, f x + 8 i \, e\right )} - 4 \, d^{5} f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, d^{5} f e^{\left (4 i \, f x + 4 i \, e\right )} - 4 \, d^{5} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{5} f\right )} \sqrt {-\frac {64 i \, a^{6}}{d^{9} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (d^{5} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{5} 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^{9} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 105 \, {\left (d^{5} f e^{\left (8 i \, f x + 8 i \, e\right )} - 4 \, d^{5} f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, d^{5} f e^{\left (4 i \, f x + 4 i \, e\right )} - 4 \, d^{5} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{5} f\right )} \sqrt {-\frac {64 i \, a^{6}}{d^{9} f^{2}}} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (d^{5} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{5} 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^{9} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 16 \, {\left (319 \, a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} - 327 \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} - 95 \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 387 \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 164 \, 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}}}{420 \, {\left (d^{5} f e^{\left (8 i \, f x + 8 i \, e\right )} - 4 \, d^{5} f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, d^{5} f e^{\left (4 i \, f x + 4 i \, e\right )} - 4 \, d^{5} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{5} f\right )}} \] Input:

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x, algorithm="fricas")
 

Output:

1/420*(105*(d^5*f*e^(8*I*f*x + 8*I*e) - 4*d^5*f*e^(6*I*f*x + 6*I*e) + 6*d^ 
5*f*e^(4*I*f*x + 4*I*e) - 4*d^5*f*e^(2*I*f*x + 2*I*e) + d^5*f)*sqrt(-64*I* 
a^6/(d^9*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (d^5*f*e^(2*I*f*x 
 + 2*I*e) + d^5*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^9*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) - 105*(d^5* 
f*e^(8*I*f*x + 8*I*e) - 4*d^5*f*e^(6*I*f*x + 6*I*e) + 6*d^5*f*e^(4*I*f*x + 
 4*I*e) - 4*d^5*f*e^(2*I*f*x + 2*I*e) + d^5*f)*sqrt(-64*I*a^6/(d^9*f^2))*l 
og(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) - (d^5*f*e^(2*I*f*x + 2*I*e) + d^5* 
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^9*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) - 16*(319*a^3*e^(8*I*f*x + 
8*I*e) - 327*a^3*e^(6*I*f*x + 6*I*e) - 95*a^3*e^(4*I*f*x + 4*I*e) + 387*a^ 
3*e^(2*I*f*x + 2*I*e) - 164*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^5*f*e^(8*I*f*x + 8*I*e) - 4*d^5*f*e^(6*I*f*x + 
 6*I*e) + 6*d^5*f*e^(4*I*f*x + 4*I*e) - 4*d^5*f*e^(2*I*f*x + 2*I*e) + d^5* 
f)
 

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx=- i a^{3} \left (\int \frac {i}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\right )\, dx\right ) \] Input:

integrate((a+I*a*tan(f*x+e))**3/(d*tan(f*x+e))**(9/2),x)
 

Output:

-I*a**3*(Integral(I/(d*tan(e + f*x))**(9/2), x) + Integral(-3*tan(e + f*x) 
/(d*tan(e + f*x))**(9/2), x) + Integral(tan(e + f*x)**3/(d*tan(e + f*x))** 
(9/2), x) + Integral(-3*I*tan(e + f*x)**2/(d*tan(e + f*x))**(9/2), x))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.48 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx=\frac {\frac {105 \, a^{3} {\left (\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d^{3}} - \frac {2 \, {\left (-420 i \, a^{3} d^{3} \tan \left (f x + e\right )^{3} - 140 \, a^{3} d^{3} \tan \left (f x + e\right )^{2} + 63 i \, a^{3} d^{3} \tan \left (f x + e\right ) + 15 \, a^{3} d^{3}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} d^{3}}}{105 \, d f} \] Input:

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x, algorithm="maxima")
 

Output:

1/105*(105*a^3*((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*sqr 
t(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)/sq 
rt(d) + (I - 1)*sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))* 
sqrt(d) + d)/sqrt(d))/d^3 - 2*(-420*I*a^3*d^3*tan(f*x + e)^3 - 140*a^3*d^3 
*tan(f*x + e)^2 + 63*I*a^3*d^3*tan(f*x + e) + 15*a^3*d^3)/((d*tan(f*x + e) 
)^(7/2)*d^3))/(d*f)
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.81 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx=-\frac {2 \, a^{3} {\left (-\frac {420 i \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{d^{\frac {9}{2}} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {-420 i \, d^{3} \tan \left (f x + e\right )^{3} - 140 \, d^{3} \tan \left (f x + e\right )^{2} + 63 i \, d^{3} \tan \left (f x + e\right ) + 15 \, d^{3}}{\sqrt {d \tan \left (f x + e\right )} d^{7} \tan \left (f x + e\right )^{3}}\right )}}{105 \, f} \] Input:

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x, algorithm="giac")
 

Output:

-2/105*a^3*(-420*I*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^(9/2)*(I*d/abs(d) + 1)) + (-420*I*d 
^3*tan(f*x + e)^3 - 140*d^3*tan(f*x + e)^2 + 63*I*d^3*tan(f*x + e) + 15*d^ 
3)/(sqrt(d*tan(f*x + e))*d^7*tan(f*x + e)^3))/f
 

Mupad [B] (verification not implemented)

Time = 2.15 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.75 \[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx=-\frac {\frac {2\,a^3}{7\,d\,f}+\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )\,6{}\mathrm {i}}{5\,d\,f}-\frac {8\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{3\,d\,f}-\frac {a^3\,{\mathrm {tan}\left (e+f\,x\right )}^3\,8{}\mathrm {i}}{d\,f}}{{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}+\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,\sqrt {-d}}\right )\,2{}\mathrm {i}}{{\left (-d\right )}^{9/2}\,f} \] Input:

int((a + a*tan(e + f*x)*1i)^3/(d*tan(e + f*x))^(9/2),x)
 

Output:

(16i^(1/2)*a^3*atan((16i^(1/2)*(d*tan(e + f*x))^(1/2))/(4*(-d)^(1/2)))*2i) 
/((-d)^(9/2)*f) - ((2*a^3)/(7*d*f) + (a^3*tan(e + f*x)*6i)/(5*d*f) - (8*a^ 
3*tan(e + f*x)^2)/(3*d*f) - (a^3*tan(e + f*x)^3*8i)/(d*f))/(d*tan(e + f*x) 
)^(7/2)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx=\frac {\sqrt {d}\, a^{3} \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{5}}d x +3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{4}}d x \right ) i -3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{3}}d x \right )-\left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{2}}d x \right ) i \right )}{d^{5}} \] Input:

int((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x)
 

Output:

(sqrt(d)*a**3*(int(sqrt(tan(e + f*x))/tan(e + f*x)**5,x) + 3*int(sqrt(tan( 
e + f*x))/tan(e + f*x)**4,x)*i - 3*int(sqrt(tan(e + f*x))/tan(e + f*x)**3, 
x) - int(sqrt(tan(e + f*x))/tan(e + f*x)**2,x)*i))/d**5