\(\int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx\) [1152]

Optimal result

Integrand size = 32, antiderivative size = 250 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\frac {\sqrt [4]{-1} (5 i c-d) d^{3/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a} f}-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f}-\frac {(c+2 i d) d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{a f}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 3.63 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.34 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\frac {-\frac {i \sqrt {2} (c-i d)^3 \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right ) \sqrt {c+d \tan (e+f x)}}{\sqrt {-a (c-i d)}}+\frac {2 (-1)^{3/4} (i a)^{3/2} \sqrt {i a (c+i d)} (5 c+i d) d^{3/2} \text {arcsinh}\left (\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a} \sqrt {i a (c+i d)}}\right ) \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{a^{5/2}}+\frac {2 i c \left (c^2+2 i c d-2 d^2\right )+2 i d \left (c^2+i c d-2 d^2\right ) \tan (e+f x)+2 d^3 \tan ^2(e+f x)}{\sqrt {a+i a \tan (e+f x)}}}{2 f \sqrt {c+d \tan (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.63 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.406, Rules used = {3042, 4041, 27, 3042, 4080, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4080

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4084

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

((I*c - d)*(c + d*Tan[e + f*x])^(3/2))/(f*Sqrt[a + I*a*Tan[e + f*x]]) + (( 
(2*(-1)^(3/4)*a^(5/2)*(5*c + I*d)*d^(3/2)*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)/a - (2*a*(c + (2*I)*d)* 
d*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/f)/(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 4080

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*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 
1/(a*(m + n))   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim 
p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T 
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 + b^2, 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. 2646 vs. \(2 (200 ) = 400\).

Time = 0.42 (sec) , antiderivative size = 2647, normalized size of antiderivative = 10.59

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2647\)
default	\(\text {Expression too large to display}\)	\(2647\)

Input:

int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x,method=_RETURNVERBOS 
E)
 

Output:

-1/4/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a*(-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+2*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)+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^2-10*I*ln(1/2*(2*I*a*d*t 
an(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^2*d^2*tan(f*x+e)^2-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^3-24*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)+2 
0*I*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d^2*tan(f* 
x+e)+8*I*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^3+4*l 
n(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*d^4*tan(f*x+e)-12*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)^...
 

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. 1163 vs. \(2 (190) = 380\).

Time = 0.21 (sec) , antiderivative size = 1163, normalized size of antiderivative = 4.65 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="fr 
icas")
 

Output:

-1/4*(sqrt(2)*a*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*f^2))*e^(I*f*x + I*e)*log(-(I*sqrt(2)*a*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*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)) - sqrt(2 
)*a*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*f^2))*e^(I*f*x + I*e)*log(-(-I*sqrt(2)*a*f*sqrt(-(c^5 - 5*I*c^4*d - 1 
0*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a*f^2))*e^(I*f*x + I*e) - sqr 
t(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))*sqr 
t(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqr 
t(a/(e^(2*I*f*x + 2*I*e) + 1)))/(c^2 - 2*I*c*d - d^2)) + a*f*sqrt((-25*I*c 
^2*d^3 + 10*c*d^4 + I*d^5)/(a*f^2))*e^(I*f*x + I*e)*log(-4*(2*sqrt(2)*((5* 
I*c*d^3 - d^4)*e^(3*I*f*x + 3*I*e) + (5*I*c*d^3 - d^4)*e^(I*f*x + I*e))*sq 
rt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sq 
rt(a/(e^(2*I*f*x + 2*I*e) + 1)) + ((a*c*d - 3*I*a*d^2)*f*e^(2*I*f*x + 2*I* 
e) + (a*c*d + I*a*d^2)*f)*sqrt((-25*I*c^2*d^3 + 10*c*d^4 + I*d^5)/(a*f^2)) 
)/(5*c^4 - 4*I*c^3*d + 6*c^2*d^2 - 4*I*c*d^3 + d^4 + (5*c^4 - 4*I*c^3*d + 
6*c^2*d^2 - 4*I*c*d^3 + d^4)*e^(2*I*f*x + 2*I*e))) - a*f*sqrt((-25*I*c^2*d 
^3 + 10*c*d^4 + I*d^5)/(a*f^2))*e^(I*f*x + I*e)*log(-4*(2*sqrt(2)*((5*I...
 

Sympy [F]

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

integrate((c+d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="ma 
xima")
 

Output:

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

Giac [F(-2)]

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

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),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 TypeError: Bad Ar 
gument Ty
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\text {too large to display} \] Input:

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

Output:

(sqrt(a)*(6*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x) 
*c**5*d*i + 10*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f 
*x)*c**4*d**2 + 2*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e 
+ f*x)*c**3*d**3*i + 10*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)* 
tan(e + f*x)*c**2*d**4 - 4*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + 
c)*tan(e + f*x)*c*d**5*i - 2*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d 
+ c)*c**6*i - 10*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c**5*d 
+ 6*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c**4*d**2*i - 10*sqr 
t(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c**3*d**3 + 8*sqrt(tan(e + 
f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c**2*d**4*i - int((sqrt(tan(e + f*x)* 
i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**3)/(tan(e + f*x)**2 + 1),x)* 
tan(e + f*x)**2*c**4*d**2*f*i - 2*int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e 
 + f*x)*d + c)*tan(e + f*x)**3)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*c 
**2*d**4*f*i - int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan( 
e + f*x)**3)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*d**6*f*i - int((sqrt 
(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**3)/(tan(e + f* 
x)**2 + 1),x)*c**4*d**2*f*i - 2*int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + 
 f*x)*d + c)*tan(e + f*x)**3)/(tan(e + f*x)**2 + 1),x)*c**2*d**4*f*i - int 
((sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**3)/(tan( 
e + f*x)**2 + 1),x)*d**6*f*i - 2*int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan...