\(\int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx\) [1119]

Optimal result

Integrand size = 30, antiderivative size = 126 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {8 i a^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}+\frac {4 a^3 (i c-4 d) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f} \] Output:

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

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.73 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {i a^2 \left (-\frac {8 a \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {2 a (2 c+9 i d-d \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d^2}\right )}{f} \] Input:

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

Output:

(I*a^2*((-8*a*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sqrt[c - I* 
d] + (2*a*(2*c + (9*I)*d - d*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*d^ 
2)))/f
 

Rubi [A] (warning: unable to verify)

Time = 0.71 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4039, 3042, 4075, 3042, 4020, 25, 27, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4075

\(\displaystyle \frac {2 a \left (\int \frac {6 d a^2+6 i d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (\int \frac {6 d a^2+6 i d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 a \left (\frac {36 i a^4 d^2 \int -\frac {1}{\sqrt {6} a^2 d \sqrt {6 c+6 d \tan (e+f x)} \left (6 a^2 d-6 i a^2 d \tan (e+f x)\right )}d\left (6 i a^2 d \tan (e+f x)\right )}{f}+\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 a \left (\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}-\frac {36 i a^4 d^2 \int \frac {1}{\sqrt {6} a^2 d \sqrt {6 c+6 d \tan (e+f x)} \left (6 a^2 d-6 i a^2 d \tan (e+f x)\right )}d\left (6 i a^2 d \tan (e+f x)\right )}{f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 a \left (\frac {12 \sqrt {6} a^4 d \int \frac {1}{36 i d^2 \tan ^2(e+f x) a^6+6 (i c+d) a^2}d\sqrt {6 c+6 d \tan (e+f x)}}{f}+\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a \left (\frac {12 a^2 d \arctan \left (\frac {\sqrt {6} a^2 d \tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {2 a^2 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{d f}\right )}{3 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\)

Input:

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

Output:

(-2*(a^3 + I*a^3*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) + (2*a*(( 
12*a^2*d*ArcTan[(Sqrt[6]*a^2*d*Tan[e + f*x])/Sqrt[c - I*d]])/(Sqrt[c - I*d 
]*f) + (2*a^2*(I*c - 4*d)*Sqrt[c + d*Tan[e + f*x]])/(d*f)))/(3*d)
 

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 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. 792 vs. \(2 (107 ) = 214\).

Time = 0.51 (sec) , antiderivative size = 793, normalized size of antiderivative = 6.29

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {i \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+i \sqrt {c +d \tan \left (f x +e \right )}\, c -3 \sqrt {c +d \tan \left (f x +e \right )}\, d -4 d^{2} \left (\frac {\frac {\left (-2 i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}-2 i c^{3}-2 i c \,d^{2}+c d \sqrt {c^{2}+d^{2}}+c^{2} d +d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c d -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}+\frac {\left (-2 i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}-2 i c^{3}-2 i c \,d^{2}+c d \sqrt {c^{2}+d^{2}}+c^{2} d +d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, \left (\sqrt {c^{2}+d^{2}}\, c +c^{2}+d^{2}\right )}+\frac {\frac {\left (i \sqrt {c^{2}+d^{2}}+i c -d \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -\frac {\left (i \sqrt {c^{2}+d^{2}}+i c -d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}\right )\right )}{f \,d^{2}}\)	\(793\)
default	\(\frac {2 a^{3} \left (-\frac {i \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+i \sqrt {c +d \tan \left (f x +e \right )}\, c -3 \sqrt {c +d \tan \left (f x +e \right )}\, d -4 d^{2} \left (\frac {\frac {\left (-2 i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}-2 i c^{3}-2 i c \,d^{2}+c d \sqrt {c^{2}+d^{2}}+c^{2} d +d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{2}-\sqrt {c^{2}+d^{2}}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c d -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{2} d -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{3}+\frac {\left (-2 i c^{2} \sqrt {c^{2}+d^{2}}-i d^{2} \sqrt {c^{2}+d^{2}}-2 i c^{3}-2 i c \,d^{2}+c d \sqrt {c^{2}+d^{2}}+c^{2} d +d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, \left (\sqrt {c^{2}+d^{2}}\, c +c^{2}+d^{2}\right )}+\frac {\frac {\left (i \sqrt {c^{2}+d^{2}}+i c -d \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d -\frac {\left (i \sqrt {c^{2}+d^{2}}+i c -d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}\right )\right )}{f \,d^{2}}\)	\(793\)
parts	\(\text {Expression too large to display}\)	\(3867\)

Input:

int((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/f*a^3/d^2*(-1/3*I*(c+d*tan(f*x+e))^(3/2)+I*(c+d*tan(f*x+e))^(1/2)*c-3*(c 
+d*tan(f*x+e))^(1/2)*d-4*d^2*(1/2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^ 
(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^2)*(1/2*(-2*I*(c^2+d^2)^(1/2)*c^2-I*d^2*(c^ 
2+d^2)^(1/2)-2*I*c^3-2*I*c*d^2+c*d*(c^2+d^2)^(1/2)+c^2*d+d^3)*ln(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*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+I*(2*(c^2+d^2)^(1 
/2)+2*c)^(1/2)*c^3+I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d^2-(c^2+d^2)^(1/2)*( 
2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2*d-(2*(c 
^2+d^2)^(1/2)+2*c)^(1/2)*d^3+1/2*(-2*I*(c^2+d^2)^(1/2)*c^2-I*d^2*(c^2+d^2) 
^(1/2)-2*I*c^3-2*I*c*d^2+c*d*(c^2+d^2)^(1/2)+c^2*d+d^3)*(2*(c^2+d^2)^(1/2) 
+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+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+d^2)^(1/2)+I*c-d)*ln 
(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+ 
d^2)^(1/2))+2*(I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-(2*(c^2+d^2)^(1/2)+2*c)^( 
1/2)*d-1/2*(I*(c^2+d^2)^(1/2)+I*c-d)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^ 
2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2 
)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)))))
                                                                                    
                                                                                    
 

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. 427 vs. \(2 (104) = 208\).

Time = 0.12 (sec) , antiderivative size = 427, normalized size of antiderivative = 3.39 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {3 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac {{\left (8 \, a^{3} c + \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} {\left ({\left (i \, c + d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c + d\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 8 \, {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 3 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac {{\left (8 \, a^{3} c + \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} {\left ({\left (-i \, c - d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c - d\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 8 \, {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 16 \, {\left (-i \, a^{3} c + 4 \, a^{3} d + {\left (-i \, a^{3} c + 5 \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )}} \] Input:

integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x, algorithm="fricas 
")
 

Output:

1/12*(3*(d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-64*I*a^6/((I*c + d)*f^2) 
)*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c + d)*f^2))*((I*c + d)*f*e^(2*I*f 
*x + 2*I*e) + (I*c + d)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/ 
(e^(2*I*f*x + 2*I*e) + 1)) + 8*(a^3*c - I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(- 
2*I*f*x - 2*I*e)/a^3) - 3*(d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-64*I*a 
^6/((I*c + d)*f^2))*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c + d)*f^2))*((- 
I*c - d)*f*e^(2*I*f*x + 2*I*e) + (-I*c - d)*f)*sqrt(((c - I*d)*e^(2*I*f*x 
+ 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)) + 8*(a^3*c - I*a^3*d)*e^(2* 
I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) - 16*(-I*a^3*c + 4*a^3*d + (-I*a 
^3*c + 5*a^3*d)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + 
 c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)
 

Sympy [F]

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

integrate((a+I*a*tan(f*x+e))**3/(c+d*tan(f*x+e))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima 
")
 

Output:

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (104) = 208\).

Time = 0.50 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.76 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {2 \, a^{3} {\left (-\frac {12 i \, \sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{4} - 3 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{4} + 9 \, \sqrt {d \tan \left (f x + e\right ) + c} d^{5}}{d^{6}}\right )}}{3 \, f} \] Input:

integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")
 

Output:

-2/3*a^3*(-12*I*sqrt(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*s 
qrt(2)*sqrt(-c + sqrt(c^2 + d^2))*d - sqrt(2)*sqrt(c^2 + d^2)*sqrt(-c + sq 
rt(c^2 + d^2))))/(sqrt(-c + sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 
 1)) + (I*(d*tan(f*x + e) + c)^(3/2)*d^4 - 3*I*sqrt(d*tan(f*x + e) + c)*c* 
d^4 + 9*sqrt(d*tan(f*x + e) + c)*d^5)/d^6)/f
 

Mupad [B] (verification not implemented)

Time = 3.58 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.94 \[ \int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=-\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\frac {a^3\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,2{}\mathrm {i}}{3\,d^2\,f}+\frac {a^3\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {-c+d\,1{}\mathrm {i}}}\right )\,8{}\mathrm {i}}{f\,\sqrt {-c+d\,1{}\mathrm {i}}} \] Input:

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

Output:

(a^3*atan((c + d*tan(e + f*x))^(1/2)/(d*1i - c)^(1/2))*8i)/(f*(d*1i - c)^( 
1/2)) - (a^3*(c + d*tan(e + f*x))^(3/2)*2i)/(3*d^2*f) - ((a^3*(c - d*1i)*2 
i)/(d^2*f) - (a^3*(c + d*1i)*4i)/(d^2*f))*(c + d*tan(e + f*x))^(1/2)
 

Reduce [F]

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

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

Output:

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