\(\int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx\) [1117]

Optimal result

Integrand size = 30, antiderivative size = 217 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{4 a^2 f}+\frac {\sqrt {c+i d} \left (2 i c^2+6 c d-7 i d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 f}+\frac {(c+i d) (2 i c+5 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2} \] Output:

-1/4*I*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a^2/f+1 
/8*(c+I*d)^(1/2)*(2*I*c^2+6*c*d-7*I*d^2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c 
+I*d)^(1/2))/a^2/f+1/8*(c+I*d)*(2*I*c+5*d)*(c+d*tan(f*x+e))^(1/2)/a^2/f/(1 
+I*tan(f*x+e))+1/4*(I*c-d)*(c+d*tan(f*x+e))^(3/2)/f/(a+I*a*tan(f*x+e))^2
 

Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {\sec ^2(e+f x) \left (4 i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+2 \sqrt {c+i d} \left (2 c^2-6 i c d-7 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right ) (-i \cos (2 (e+f x))+\sin (2 (e+f x)))+2 (-i c+d) \cos (e+f x) ((4 c-5 i d) \cos (e+f x)+(2 i c+7 d) \sin (e+f x)) \sqrt {c+d \tan (e+f x)}\right )}{16 a^2 f (-i+\tan (e+f x))^2} \] Input:

Integrate[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^2,x]
 

Output:

(Sec[e + f*x]^2*((4*I)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sq 
rt[c - I*d]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) + 2*Sqrt[c + I*d]*(2* 
c^2 - (6*I)*c*d - 7*d^2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]]*( 
(-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)]) + 2*((-I)*c + d)*Cos[e + f*x]*(( 
4*c - (5*I)*d)*Cos[e + f*x] + ((2*I)*c + 7*d)*Sin[e + f*x])*Sqrt[c + d*Tan 
[e + f*x]]))/(16*a^2*f*(-I + Tan[e + f*x])^2)
 

Rubi [A] (warning: unable to verify)

Time = 1.21 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.94, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.433, Rules used = {3042, 4041, 27, 3042, 4078, 25, 3042, 4022, 3042, 4020, 25, 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.

\(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2}dx\)

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {(-d+i c) (2 c-5 i d) \sqrt {c+d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 c^3-10 i d c^2-7 d^2 c-5 i d^3\right ) a^2+d \left (2 c^2-5 i d c-9 d^2\right ) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2}}{8 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\left (4 c^3-10 i d c^2-7 d^2 c-5 i d^3\right ) a^2+d \left (2 c^2-5 i d c-9 d^2\right ) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2}+\frac {(-d+i c) (2 c-5 i d) \sqrt {c+d \tan (e+f x)}}{f (1+i \tan (e+f x))}}{8 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\left (4 c^3-10 i d c^2-7 d^2 c-5 i d^3\right ) a^2+d \left (2 c^2-5 i d c-9 d^2\right ) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2}+\frac {(-d+i c) (2 c-5 i d) \sqrt {c+d \tan (e+f x)}}{f (1+i \tan (e+f x))}}{8 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^2,x]
 

Output:

((I*c - d)*(c + d*Tan[e + f*x])^(3/2))/(4*f*(a + I*a*Tan[e + f*x])^2) + (( 
(4*a^2*(c - I*d)^(5/2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + (2*a^2*Sqrt 
[c + I*d]*(2*c^2 - (6*I)*c*d - 7*d^2)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/ 
f)/(2*a^2) + ((I*c - d)*(2*c - (5*I)*d)*Sqrt[c + d*Tan[e + f*x]])/(f*(1 + 
I*Tan[e + f*x])))/(8*a^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 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 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

rule 4041

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* 
((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m)   Int[(a + 
b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n 
 - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta 
n[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] && LtQ[m, 0] && GtQ[n, 1] && (In 
tegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4078

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), 
 x] + Simp[1/(2*a^2*m)   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* 
x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a 
*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] 
&& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.39

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\frac {i \left (i d -c \right )^{\frac {5}{2}} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 d^{3}}+\frac {i \left (\frac {-\frac {d \left (2 i c^{4}+15 i c^{2} d^{2}-7 i d^{4}+c^{3} d -19 c \,d^{3}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2 \left (2 i c d +c^{2}-d^{2}\right )}+\frac {d \left (2 i c^{5}+8 i c^{3} d^{2}-18 i c \,d^{4}-3 c^{4} d -22 c^{2} d^{3}+5 d^{5}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{4 i c d +2 c^{2}-2 d^{2}}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}+\frac {\left (5 i c^{2} d^{3}-7 i d^{5}-2 c^{5}-5 c^{3} d^{2}-15 c \,d^{4}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}\right )}{8 d^{3}}\right )}{f \,a^{2}}\)	\(302\)
default	\(\frac {2 d^{3} \left (-\frac {i \left (i d -c \right )^{\frac {5}{2}} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 d^{3}}+\frac {i \left (\frac {-\frac {d \left (2 i c^{4}+15 i c^{2} d^{2}-7 i d^{4}+c^{3} d -19 c \,d^{3}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2 \left (2 i c d +c^{2}-d^{2}\right )}+\frac {d \left (2 i c^{5}+8 i c^{3} d^{2}-18 i c \,d^{4}-3 c^{4} d -22 c^{2} d^{3}+5 d^{5}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{4 i c d +2 c^{2}-2 d^{2}}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}+\frac {\left (5 i c^{2} d^{3}-7 i d^{5}-2 c^{5}-5 c^{3} d^{2}-15 c \,d^{4}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}\right )}{8 d^{3}}\right )}{f \,a^{2}}\)	\(302\)

Input:

int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

2/f/a^2*d^3*(-1/8*I*(I*d-c)^(5/2)/d^3*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c 
)^(1/2))+1/8*I/d^3*((-1/2*d*(2*I*c^4+15*I*c^2*d^2-7*I*d^4+c^3*d-19*c*d^3)/ 
(2*I*c*d+c^2-d^2)*(c+d*tan(f*x+e))^(3/2)+1/2*d*(2*I*c^5+8*I*c^3*d^2-18*I*c 
*d^4-3*c^4*d-22*c^2*d^3+5*d^5)/(2*I*c*d+c^2-d^2)*(c+d*tan(f*x+e))^(1/2))/( 
-d*tan(f*x+e)+I*d)^2+1/2*(-5*c^3*d^2-15*c*d^4+5*I*c^2*d^3-7*I*d^5-2*c^5)/( 
2*I*c*d+c^2-d^2)/(-c-I*d)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-c-I*d)^(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. 1083 vs. \(2 (167) = 334\).

Time = 0.26 (sec) , antiderivative size = 1083, normalized size of antiderivative = 4.99 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=\text {Too large to display} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas 
")
 

Output:

1/32*(2*a^2*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 
 - I*d^5)/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(2*(c^3 - 2*I*c^2*d - c*d^2 + 
(I*a^2*f*e^(2*I*f*x + 2*I*e) + I*a^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))*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^ 
2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^4*f^2)) + (c^3 - 3*I*c^2*d - 3*c*d^ 
2 + I*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*I*c*d - d^2) 
) - 2*a^2*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - 
 I*d^5)/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(2*(c^3 - 2*I*c^2*d - c*d^2 + (- 
I*a^2*f*e^(2*I*f*x + 2*I*e) - I*a^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))*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 
 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^4*f^2)) + (c^3 - 3*I*c^2*d - 3*c*d^2 
 + I*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*I*c*d - d^2)) 
 + a^2*f*sqrt(-(4*c^5 - 20*I*c^4*d - 40*c^3*d^2 + 20*I*c^2*d^3 - 35*c*d^4 
+ 49*I*d^5)/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/8*(2*I*c^3 + 4*c^2*d - I* 
c*d^2 + 7*d^3 + (a^2*f*e^(2*I*f*x + 2*I*e) + a^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))*sqrt(-(4*c^5 - 20*I*c^ 
4*d - 40*c^3*d^2 + 20*I*c^2*d^3 - 35*c*d^4 + 49*I*d^5)/(a^4*f^2)) + (2*I*c 
^3 + 6*c^2*d - 7*I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a^2*f 
)) - a^2*f*sqrt(-(4*c^5 - 20*I*c^4*d - 40*c^3*d^2 + 20*I*c^2*d^3 - 35*c*d^ 
4 + 49*I*d^5)/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/8*(2*I*c^3 + 4*c^2*d...
 

Sympy [F]

\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {c^{2} \sqrt {c + d \tan {\left (e + f x \right )}}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \frac {d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \frac {2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \] Input:

integrate((c+d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**2,x)
 

Output:

-(Integral(c**2*sqrt(c + d*tan(e + f*x))/(tan(e + f*x)**2 - 2*I*tan(e + f* 
x) - 1), x) + Integral(d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2/(tan( 
e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(2*c*d*sqrt(c + d*tan(e 
+ f*x))*tan(e + f*x)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x))/a**2
                                                                                    
                                                                                    
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima 
")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

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. 507 vs. \(2 (167) = 334\).

Time = 0.64 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.34 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {\sqrt {2} {\left (2 i \, c^{3} + 4 \, c^{2} d - i \, c d^{2} + 7 \, d^{3}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} + i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, \sqrt {2} {\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{2} d - 2 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{3} d - 5 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c d^{2} + i \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} d^{2} + 7 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{3} - 8 \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{3} - 5 i \, \sqrt {d \tan \left (f x + e\right ) + c} d^{4}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2}}}{8 \, a^{2} f} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 6.10 (sec) , antiderivative size = 9787, normalized size of antiderivative = 45.10 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=\text {Too large to display} \] Input:

int((c + d*tan(e + f*x))^(5/2)/(a + a*tan(e + f*x)*1i)^2,x)
 

Output:

- atan((((a^2*f*(a^4*d^6*f^2*640i - 256*a^4*c*d^5*f^2 + a^4*c^2*d^4*f^2*64 
0i - 256*a^4*c^3*d^3*f^2) - 4096*a^8*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)* 
(-(d^11*45i - 15*c*d^10 + c^2*d^9*105i - 95*c^3*d^8 + c^4*d^7*20i - 72*c^5 
*d^6 - c^6*d^5*40i + 8*c^7*d^4 + a^4*f^2*(4*(256*d^6 + 256*c^2*d^4)*((((10 
5*c^3*d^13)/16 - (35*c*d^15)/8 + (295*c^5*d^11)/16 + (5*c^7*d^9)/16 - (105 
*c^9*d^7)/16 + (5*c^11*d^5)/8)*1i)/(a^8*f^4) - ((49*d^16)/64 - (149*c^2*d^ 
14)/16 - (235*c^4*d^12)/32 + (215*c^6*d^10)/16 + (505*c^8*d^8)/64 - (11*c^ 
10*d^6)/4 + (c^12*d^4)/16)/(a^8*f^4)) + (((45*d^11 + 105*c^2*d^9 + 20*c^4* 
d^7 - 40*c^6*d^5)*1i)/(a^4*f^2) - (15*c*d^10 + 95*c^3*d^8 + 72*c^5*d^6 - 8 
*c^7*d^4)/(a^4*f^2))^2)^(1/2))/(512*a^4*f^2*(d^6 + c^2*d^4)))^(1/2))*(-(d^ 
11*45i - 15*c*d^10 + c^2*d^9*105i - 95*c^3*d^8 + c^4*d^7*20i - 72*c^5*d^6 
- c^6*d^5*40i + 8*c^7*d^4 + a^4*f^2*(4*(256*d^6 + 256*c^2*d^4)*((((105*c^3 
*d^13)/16 - (35*c*d^15)/8 + (295*c^5*d^11)/16 + (5*c^7*d^9)/16 - (105*c^9* 
d^7)/16 + (5*c^11*d^5)/8)*1i)/(a^8*f^4) - ((49*d^16)/64 - (149*c^2*d^14)/1 
6 - (235*c^4*d^12)/32 + (215*c^6*d^10)/16 + (505*c^8*d^8)/64 - (11*c^10*d^ 
6)/4 + (c^12*d^4)/16)/(a^8*f^4)) + (((45*d^11 + 105*c^2*d^9 + 20*c^4*d^7 - 
 40*c^6*d^5)*1i)/(a^4*f^2) - (15*c*d^10 + 95*c^3*d^8 + 72*c^5*d^6 - 8*c^7* 
d^4)/(a^4*f^2))^2)^(1/2))/(512*a^4*f^2*(d^6 + c^2*d^4)))^(1/2) + 8*a^4*f^2 
*(c + d*tan(e + f*x))^(1/2)*(c*d^7*10i + 53*d^8 - 5*c^2*d^6 - c^3*d^5*60i 
+ 80*c^4*d^4 + c^5*d^3*40i - 8*c^6*d^2))*(-(d^11*45i - 15*c*d^10 + c^2*...
 

Reduce [F]

\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {-\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right ) c^{2}-\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right ) d^{2}-2 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right ) c d}{a^{2}} \] Input:

int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x)
 

Output:

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