\(\int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1125]

Optimal result

Integrand size = 30, antiderivative size = 139 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {8 i a^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}+\frac {4 a^3 c \sqrt {c+d \tan (e+f x)}}{d^2 (i c+d) f} \] Output:

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

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.82 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 a^3 \left (-\frac {4 i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2}}+\frac {-2 i c^2+c d+i d^2-i (c-i d) d \tan (e+f x)}{(c-i d) d^2 \sqrt {c+d \tan (e+f x)}}\right )}{f} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.75 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4036, 25, 3042, 4075, 3042, 4020, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4036

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4075

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 \left (\frac {2 a^3 c \sqrt {c+d \tan (e+f x)}}{d f}+\frac {4 i \sqrt {2} a^6 d \int \frac {1}{2 a^3 (i c+d)-4 i a^9 d^2 \tan ^2(e+f x)}d\sqrt {2 c+2 d \tan (e+f x)}}{f}\right )}{d (d+i c)}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*(c + I*d)*(a^3 + I*a^3*Tan[e + f*x]))/((c - I*d)*d*f*Sqrt[c + d*Tan[e + 
 f*x]]) + (2*((-4*a^3*d*ArcTanh[(Sqrt[2]*a^3*d*Tan[e + f*x])/Sqrt[c - I*d] 
])/(Sqrt[c - I*d]*f) + (2*a^3*c*Sqrt[c + d*Tan[e + f*x]])/(d*f)))/(d*(I*c 
+ 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 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 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. 2587 vs. \(2 (122 ) = 244\).

Time = 0.50 (sec) , antiderivative size = 2588, normalized size of antiderivative = 18.62

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2588\)
default	\(\text {Expression too large to display}\)	\(2588\)
parts	\(\text {Expression too large to display}\)	\(7395\)

Input:

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

Output:

4*I/f*a^3*d^2/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)/(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))*c+4*I/f*a^3*d^2/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2 
)+c)/(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))*c-2/f*a^3*d/(c^2+d 
^2)/(c+d*tan(f*x+e))^(1/2)-2*I/f*a^3/d^2*(c+d*tan(f*x+e))^(1/2)+6*I/f*a^3/ 
(c^2+d^2)/(c+d*tan(f*x+e))^(1/2)*c+6/f*a^3/d/(c^2+d^2)/(c+d*tan(f*x+e))^(1 
/2)*c^2+2/f*a^3*d/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)*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*(c^ 
2+d^2)^(1/2)+2*c)^(1/2)*c+8/f*a^3*d/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)/(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))*c^2-2/f*a^3*d/(c^2+d^2)^( 
3/2)/((c^2+d^2)^(1/2)+c)*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*(c^2+d^2)^(1/2)+2*c)^(1/2)*c+I/f 
*a^3*d^2/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)*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*(c^2+d^2)^(1 
/2)+2*c)^(1/2)+I/f*a^3/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)*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*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+8/f*a^3*d/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2 
)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(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. 604 vs. \(2 (115) = 230\).

Time = 0.12 (sec) , antiderivative size = 604, normalized size of antiderivative = 4.35 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

\[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] 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 [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (115) = 230\).

Time = 0.61 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.67 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 \, a^{3} {\left (\frac {-i \, c^{2} + 2 \, c d + i \, d^{2}}{{\left (c d^{2} - i \, d^{3}\right )} \sqrt {d \tan \left (f x + e\right ) + c}} + \frac {4 \, \sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{{\left (-i \, c - d\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {i \, \sqrt {d \tan \left (f x + e\right ) + c}}{d^{2}}\right )}}{f} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.70 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.31 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {a^3\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{d^2\,f}+\frac {a^3\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\left (2\,c^4\,f^2+4\,c^2\,d^2\,f^2+2\,d^4\,f^2\right )}{2\,f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}\,\left (f\,c^3+1{}\mathrm {i}\,f\,c^2\,d+f\,c\,d^2+1{}\mathrm {i}\,f\,d^3\right )}\right )\,8{}\mathrm {i}}{f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\left (a^3\,c^2+a^3\,c\,d\,2{}\mathrm {i}-a^3\,d^2\right )\,2{}\mathrm {i}}{d^2\,f\,\left (c-d\,1{}\mathrm {i}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \] Input:

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

Output:

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

Reduce [F]

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

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