\(\int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx\) [1107]

Optimal result

Integrand size = 30, antiderivative size = 181 \[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=-\frac {8 i a^3 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {8 a^3 (i c+d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {8 i a^3 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a^3 (i c-8 d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f} \] Output:

-8*I*a^3*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+8*a 
^3*(I*c+d)*(c+d*tan(f*x+e))^(1/2)/f+8/3*I*a^3*(c+d*tan(f*x+e))^(3/2)/f+4/3 
5*a^3*(I*c-8*d)*(c+d*tan(f*x+e))^(5/2)/d^2/f-2/7*(a^3+I*a^3*tan(f*x+e))*(c 
+d*tan(f*x+e))^(5/2)/d/f
 

Mathematica [A] (verified)

Time = 2.15 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.88 \[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=-\frac {2 i a^3 \left (420 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )-\frac {\sqrt {c+d \tan (e+f x)} \left (6 c^3+63 i c^2 d+560 c d^2-420 i d^3+d \left (-3 c^2+126 i c d+140 d^2\right ) \tan (e+f x)-3 (8 c-21 i d) d^2 \tan ^2(e+f x)-15 d^3 \tan ^3(e+f x)\right )}{d^2}\right )}{105 f} \] Input:

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

Output:

(((-2*I)/105)*a^3*(420*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sq 
rt[c - I*d]] - (Sqrt[c + d*Tan[e + f*x]]*(6*c^3 + (63*I)*c^2*d + 560*c*d^2 
 - (420*I)*d^3 + d*(-3*c^2 + (126*I)*c*d + 140*d^2)*Tan[e + f*x] - 3*(8*c 
- (21*I)*d)*d^2*Tan[e + f*x]^2 - 15*d^3*Tan[e + f*x]^3))/d^2))/f
 

Rubi [A] (warning: unable to verify)

Time = 1.19 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 4039, 3042, 4075, 3042, 4011, 3042, 4011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4075

\(\displaystyle \frac {2 a \left (\int (c+d \tan (e+f x))^{3/2} \left (14 d a^2+14 i d \tan (e+f x) a^2\right )dx+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (\int (c+d \tan (e+f x))^{3/2} \left (14 d a^2+14 i d \tan (e+f x) a^2\right )dx+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 a \left (\int \sqrt {c+d \tan (e+f x)} \left (14 (c-i d) d a^2+14 d (i c+d) \tan (e+f x) a^2\right )dx+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {28 i a^2 d (c+d \tan (e+f x))^{3/2}}{3 f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (\int \sqrt {c+d \tan (e+f x)} \left (14 (c-i d) d a^2+14 d (i c+d) \tan (e+f x) a^2\right )dx+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {28 i a^2 d (c+d \tan (e+f x))^{3/2}}{3 f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 a \left (\int \frac {14 a^2 d (c-i d)^2+14 i a^2 d \tan (e+f x) (c-i d)^2}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {28 i a^2 d (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {28 a^2 d (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (\int \frac {14 a^2 d (c-i d)^2+14 i a^2 d \tan (e+f x) (c-i d)^2}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {28 i a^2 d (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {28 a^2 d (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 a \left (\frac {196 i a^4 d^2 (c-i d)^4 \int -\frac {1}{\sqrt {14} a^2 (c-i d)^2 d \sqrt {14 c+14 d \tan (e+f x)} \left (14 a^2 (c-i d)^2 d-14 i a^2 (c-i d)^2 d \tan (e+f x)\right )}d\left (14 i a^2 (c-i d)^2 d \tan (e+f x)\right )}{f}+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {28 i a^2 d (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {28 a^2 d (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 a \left (-\frac {196 i a^4 d^2 (c-i d)^4 \int \frac {1}{\sqrt {14} a^2 (c-i d)^2 d \sqrt {14 c+14 d \tan (e+f x)} \left (14 a^2 (c-i d)^2 d-14 i a^2 (c-i d)^2 d \tan (e+f x)\right )}d\left (14 i a^2 (c-i d)^2 d \tan (e+f x)\right )}{f}+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {28 i a^2 d (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {28 a^2 d (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a \left (-\frac {14 i \sqrt {14} a^2 d (c-i d)^2 \int \frac {1}{\sqrt {14 c+14 d \tan (e+f x)} \left (14 a^2 (c-i d)^2 d-14 i a^2 (c-i d)^2 d \tan (e+f x)\right )}d\left (14 i a^2 (c-i d)^2 d \tan (e+f x)\right )}{f}+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {28 i a^2 d (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {28 a^2 d (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 a \left (\frac {28 \sqrt {14} a^4 d (c-i d)^4 \int \frac {1}{196 i a^6 (c-i d)^6 d^2 \tan ^2(e+f x)-14 a^2 (i c+d)^3}d\sqrt {14 c+14 d \tan (e+f x)}}{f}+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {28 i a^2 d (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {28 a^2 d (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a \left (\frac {28 a^2 d (c-i d)^{3/2} \arctan \left (\sqrt {14} a^2 d (c-i d)^{3/2} \tan (e+f x)\right )}{f}+\frac {2 a^2 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {28 i a^2 d (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {28 a^2 d (d+i c) \sqrt {c+d \tan (e+f x)}}{f}\right )}{7 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

Input:

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

Output:

(-2*(a^3 + I*a^3*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2))/(7*d*f) + (2*a* 
((28*a^2*(c - I*d)^(3/2)*d*ArcTan[Sqrt[14]*a^2*(c - I*d)^(3/2)*d*Tan[e + f 
*x]])/f + (28*a^2*d*(I*c + d)*Sqrt[c + d*Tan[e + f*x]])/f + (((28*I)/3)*a^ 
2*d*(c + d*Tan[e + f*x])^(3/2))/f + (2*a^2*(I*c - 8*d)*(c + d*Tan[e + f*x] 
)^(5/2))/(5*d*f)))/(7*d)
 

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 73

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]
 

rule 221

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]
 

rule 3042

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

rule 4011

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]
 

rule 4020

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]
 

rule 4039

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])
 

rule 4075

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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 907 vs. \(2 (154 ) = 308\).

Time = 0.48 (sec) , antiderivative size = 908, normalized size of antiderivative = 5.02

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {i \left (c +d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {i c \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {4 i d^{2} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {3 d \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+4 i \sqrt {c +d \tan \left (f x +e \right )}\, d^{2} c +4 \sqrt {c +d \tan \left (f x +e \right )}\, d^{3}-4 d^{2} \left (\frac {\frac {\left (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}+i c \,d^{2}+2 c d \sqrt {c^{2}+d^{2}}+c^{2} d +d^{3}\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^{3}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}+\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 (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}+i c \,d^{2}+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 {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 (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}+i c \,d^{2}+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}\, c^{3}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}-\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 (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}+i c \,d^{2}+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}}}\right )\right )}{f \,d^{2}}\)	\(908\)
default	\(\frac {2 a^{3} \left (-\frac {i \left (c +d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {i c \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {4 i d^{2} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {3 d \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+4 i \sqrt {c +d \tan \left (f x +e \right )}\, d^{2} c +4 \sqrt {c +d \tan \left (f x +e \right )}\, d^{3}-4 d^{2} \left (\frac {\frac {\left (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}+i c \,d^{2}+2 c d \sqrt {c^{2}+d^{2}}+c^{2} d +d^{3}\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^{3}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}+\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 (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}+i c \,d^{2}+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 {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 (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}+i c \,d^{2}+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}\, c^{3}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}-\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 (i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}+i c^{3}+i c \,d^{2}+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}}}\right )\right )}{f \,d^{2}}\)	\(908\)
parts	\(\text {Expression too large to display}\)	\(3358\)

Input:

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

Output:

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

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. 786 vs. \(2 (149) = 298\).

Time = 0.23 (sec) , antiderivative size = 786, normalized size of antiderivative = 4.34 \[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

2/105*(105*(d^2*f*e^(6*I*f*x + 6*I*e) + 3*d^2*f*e^(4*I*f*x + 4*I*e) + 3*d^ 
2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-(a^6*c^3 - 3*I*a^6*c^2*d - 3*a^6*c* 
d^2 + I*a^6*d^3)/f^2)*log(2*(-I*a^3*c^2 - a^3*c*d + (f*e^(2*I*f*x + 2*I*e) 
 + 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))*sqrt(-(a^6*c^3 - 3*I*a^6*c^2*d - 3*a^6*c*d^2 + I*a^6*d^3)/f^2) + (-I 
*a^3*c^2 - 2*a^3*c*d + I*a^3*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e 
)/(-I*a^3*c - a^3*d)) - 105*(d^2*f*e^(6*I*f*x + 6*I*e) + 3*d^2*f*e^(4*I*f* 
x + 4*I*e) + 3*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-(a^6*c^3 - 3*I*a^6 
*c^2*d - 3*a^6*c*d^2 + I*a^6*d^3)/f^2)*log(2*(-I*a^3*c^2 - a^3*c*d - (f*e^ 
(2*I*f*x + 2*I*e) + 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))*sqrt(-(a^6*c^3 - 3*I*a^6*c^2*d - 3*a^6*c*d^2 + I*a^ 
6*d^3)/f^2) + (-I*a^3*c^2 - 2*a^3*c*d + I*a^3*d^2)*e^(2*I*f*x + 2*I*e))*e^ 
(-2*I*f*x - 2*I*e)/(-I*a^3*c - a^3*d)) - 2*(-3*I*a^3*c^3 + 30*a^3*c^2*d - 
229*I*a^3*c*d^2 - 164*a^3*d^3 + (-3*I*a^3*c^3 + 33*a^3*c^2*d - 355*I*a^3*c 
*d^2 - 319*a^3*d^3)*e^(6*I*f*x + 6*I*e) + (-9*I*a^3*c^3 + 96*a^3*c^2*d - 8 
91*I*a^3*c*d^2 - 646*a^3*d^3)*e^(4*I*f*x + 4*I*e) + (-9*I*a^3*c^3 + 93*a^3 
*c^2*d - 765*I*a^3*c*d^2 - 551*a^3*d^3)*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^(6* 
I*f*x + 6*I*e) + 3*d^2*f*e^(4*I*f*x + 4*I*e) + 3*d^2*f*e^(2*I*f*x + 2*I*e) 
 + d^2*f)
 

Sympy [F]

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

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

Output:

-I*a**3*(Integral(I*c*sqrt(c + d*tan(e + f*x)), x) + Integral(-3*c*sqrt(c 
+ d*tan(e + f*x))*tan(e + f*x), x) + Integral(c*sqrt(c + d*tan(e + f*x))*t 
an(e + f*x)**3, x) + Integral(-3*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)** 
2, x) + Integral(d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**4, x) + Integral 
(-3*I*c*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(I*d*sqrt(c 
 + d*tan(e + f*x))*tan(e + f*x), x) + Integral(-3*I*d*sqrt(c + d*tan(e + f 
*x))*tan(e + f*x)**3, x))
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.60 \[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=-\frac {2 \, a^{3} {\left (\frac {420 \, \sqrt {2} {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {15 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} d^{12} - 21 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c d^{12} + 63 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} d^{13} - 140 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{14} - 420 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{14} - 420 \, \sqrt {d \tan \left (f x + e\right ) + c} d^{15}}{d^{14}}\right )}}{105 \, f} \] Input:

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

Output:

-2/105*a^3*(420*sqrt(2)*(-I*c^2 - 2*c*d + I*d^2)*arctan(2*(sqrt(d*tan(f*x 
+ e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(sqrt(2)*c*sqrt(-c 
 + sqrt(c^2 + d^2)) - I*sqrt(2)*sqrt(-c + sqrt(c^2 + d^2))*d - sqrt(2)*sqr 
t(c^2 + d^2)*sqrt(-c + sqrt(c^2 + d^2))))/(sqrt(-c + sqrt(c^2 + d^2))*(-I* 
d/(c - sqrt(c^2 + d^2)) + 1)) + (15*I*(d*tan(f*x + e) + c)^(7/2)*d^12 - 21 
*I*(d*tan(f*x + e) + c)^(5/2)*c*d^12 + 63*(d*tan(f*x + e) + c)^(5/2)*d^13 
- 140*I*(d*tan(f*x + e) + c)^(3/2)*d^14 - 420*I*sqrt(d*tan(f*x + e) + c)*c 
*d^14 - 420*sqrt(d*tan(f*x + e) + c)*d^15)/d^14)/f
 

Mupad [B] (verification not implemented)

Time = 11.32 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.71 \[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=-\left (\frac {\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )}{3}+\frac {a^3\,{\left (c+d\,1{}\mathrm {i}\right )}^2\,2{}\mathrm {i}}{3\,d^2\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}-\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{5\,d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{5\,d^2\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}-\left (c-d\,1{}\mathrm {i}\right )\,\left (\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )+\frac {a^3\,{\left (c+d\,1{}\mathrm {i}\right )}^2\,2{}\mathrm {i}}{d^2\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\frac {a^3\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}\,2{}\mathrm {i}}{7\,d^2\,f}+\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,{\left (-d-c\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,\left (c^2\,1{}\mathrm {i}+2\,c\,d-d^2\,1{}\mathrm {i}\right )}\right )\,{\left (-d-c\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{f} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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