\(\int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx\) [1153]

Optimal result

Integrand size = 32, antiderivative size = 257 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {2 \sqrt [4]{-1} d^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{a^{3/2} f}-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(c+i d) (i c+3 d) \sqrt {c+d \tan (e+f x)}}{2 a f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 3.79 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.56 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {3 a (c-i d)^2 \sqrt {-2 c+2 i d} \arctan \left (\frac {\sqrt {-c+i d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}\right ) (1+i \tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}+24 \sqrt [4]{-1} \sqrt {i a} \sqrt {i a (c+i d)} d^{5/2} \text {arcsinh}\left (\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a} \sqrt {i a (c+i d)}}\right ) (1+i \tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}+2 a^{3/2} (c+i d) \left (c (5 c-9 i d)+\left (3 i c^2+16 c d-9 i d^2\right ) \tan (e+f x)+d (3 i c+11 d) \tan ^2(e+f x)\right )}{12 a^{5/2} f (-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \] Input:

Integrate[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(3/2),x]
 

Output:

(3*a*(c - I*d)^2*Sqrt[-2*c + (2*I)*d]*ArcTan[(Sqrt[-c + I*d]*Sqrt[a + I*a* 
Tan[e + f*x]])/(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])]*(1 + I*Tan[e + 
f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] + 24*(-1)^(1/4)* 
Sqrt[I*a]*Sqrt[I*a*(c + I*d)]*d^(5/2)*ArcSinh[((-1)^(1/4)*Sqrt[a]*Sqrt[d]* 
Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[I*a]*Sqrt[I*a*(c + I*d)])]*(1 + I*Tan[e 
+ f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)] + 
2*a^(3/2)*(c + I*d)*(c*(5*c - (9*I)*d) + ((3*I)*c^2 + 16*c*d - (9*I)*d^2)* 
Tan[e + f*x] + d*((3*I)*c + 11*d)*Tan[e + f*x]^2))/(12*a^(5/2)*f*(-I + Tan 
[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 1.61 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4084, 3042, 4027, 221, 4082, 66, 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.

\(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4084

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

Int[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(3/2),x]
 

Output:

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

Defintions of rubi rules used

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 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

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 4027

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

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]
 

rule 4082

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[b*(B/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], 
x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

rule 4084

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)/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] 
 - Simp[B/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ 
e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - 
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3160 vs. \(2 (201 ) = 402\).

Time = 0.39 (sec) , antiderivative size = 3161, normalized size of antiderivative = 12.30

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(3161\)
default	\(\text {Expression too large to display}\)	\(3161\)

Input:

int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(3/2),x,method=_RETURNVERBOS 
E)
 

Output:

1/24*I/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^2*(3*I*ln((3* 
a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*( 
a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*(I*a*d)^(1/2)* 
2^(1/2)*(-a*(I*d-c))^(1/2)*c^3*tan(f*x+e)^3-9*I*ln((3*a*c+I*a*tan(f*x+e)*c 
-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*( 
1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c)) 
^(1/2)*d^3*tan(f*x+e)^2-9*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x 
+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/ 
2))/(I+tan(f*x+e)))*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^3*tan(f*x+e 
)+9*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c 
))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*(I*a 
*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^3*tan(f*x+e)^2-9*ln((3*a*c+I*a*tan( 
f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f 
*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*(I*a*d)^(1/2)*2^(1/2)*(-a* 
(I*d-c))^(1/2)*d^3*tan(f*x+e)-3*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan 
(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e))) 
^(1/2))/(I+tan(f*x+e)))*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^2+24* 
I*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e))) 
^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^3*tan(f*x+e)^3+3*I*ln((3*a* 
c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-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. 1071 vs. \(2 (189) = 378\).

Time = 0.21 (sec) , antiderivative size = 1071, normalized size of antiderivative = 4.17 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="fr 
icas")
 

Output:

-1/12*(3*sqrt(1/2)*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^3*f^2))*e^(3*I*f*x + 3*I*e)*log(-(2*I*sqrt(1/2)*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^3*f^2))*e^(I*f*x + I*e) - sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 - 2*I*c*d 
 - 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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))/(c^2 - 2*I 
*c*d - d^2)) - 3*sqrt(1/2)*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^3*f^2))*e^(3*I*f*x + 3*I*e)*log(-(-2*I*sqr 
t(1/2)*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^3*f^2))*e^(I*f*x + I*e) - sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 
- 2*I*c*d - 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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))/( 
c^2 - 2*I*c*d - d^2)) + 3*a^2*f*sqrt(4*I*d^5/(a^3*f^2))*e^(3*I*f*x + 3*I*e 
)*log(2*(4*sqrt(2)*(d^3*e^(3*I*f*x + 3*I*e) + d^3*e^(I*f*x + 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))*sqrt(a/ 
(e^(2*I*f*x + 2*I*e) + 1)) + ((a^2*c - 3*I*a^2*d)*f*e^(2*I*f*x + 2*I*e) + 
(a^2*c + I*a^2*d)*f)*sqrt(4*I*d^5/(a^3*f^2)))/(I*c^3 + c^2*d + I*c*d^2 + d 
^3 + (I*c^3 + c^2*d + I*c*d^2 + d^3)*e^(2*I*f*x + 2*I*e))) - 3*a^2*f*sqrt( 
4*I*d^5/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(2*(4*sqrt(2)*(d^3*e^(3*I*f*x + 
3*I*e) + d^3*e^(I*f*x + I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c +...
 

Sympy [F]

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

integrate((c+d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(3/2),x)
 

Output:

Integral((c + d*tan(e + f*x))**(5/2)/(I*a*(tan(e + f*x) - I))**(3/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="ma 
xima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="gi 
ac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeError: Bad Argument TypeError: Bad Argument TypeRecursive ass 
umption s
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \] Input:

int((c + d*tan(e + f*x))^(5/2)/(a + a*tan(e + f*x)*1i)^(3/2),x)
 

Output:

int((c + d*tan(e + f*x))^(5/2)/(a + a*tan(e + f*x)*1i)^(3/2), x)
 

Reduce [F]

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

int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(3/2),x)
 

Output:

int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(3/2),x)