\(\int (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)} \, dx\) [1137]

Optimal result

Integrand size = 32, antiderivative size = 263 \[ \int (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)} \, dx=-\frac {\sqrt [4]{-1} a^{5/2} \left (c^2+10 i c d+23 d^2\right ) \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}+\frac {a^2 (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f} \] Output:

-1/4*(-1)^(1/4)*a^(5/2)*(c^2+10*I*c*d+23*d^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))/d^(3/2)/f-4*I*2^(1 
/2)*a^(5/2)*(c-I*d)^(1/2)*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))/f+1/4*a^2*(c+9*I*d)*(a+I*a*tan(f*x+ 
e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/d/f-1/2*a^2*(a+I*a*tan(f*x+e))^(1/2)*(c+d 
*tan(f*x+e))^(3/2)/d/f
 

Mathematica [A] (verified)

Time = 6.13 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.12 \[ \int (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)} \, dx=\frac {a^2 \left (-(-1)^{3/4} \sqrt {a} \sqrt {c+i d} \left (c^2+10 i c d+23 d^2\right ) \arcsin \left (\frac {\sqrt [4]{-1} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+i d}}\right ) \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}+\sqrt {d} \left (16 i \sqrt {2} \sqrt {-a (c-i d)} d \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right ) \sqrt {c+d \tan (e+f x)}-\sqrt {a+i a \tan (e+f x)} \left (c (c-9 i d)+3 (c-3 i d) d \tan (e+f x)+2 d^2 \tan ^2(e+f x)\right )\right )\right )}{4 d^{3/2} f \sqrt {c+d \tan (e+f x)}} \] Input:

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

Output:

(a^2*(-((-1)^(3/4)*Sqrt[a]*Sqrt[c + I*d]*(c^2 + (10*I)*c*d + 23*d^2)*ArcSi 
n[((-1)^(1/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + I*d])] 
*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)]) + Sqrt[d]*((16*I)*Sqrt[2]*Sqrt[-(a* 
(c - I*d))]*d*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a + I*a*Tan[e + f*x]])/(Sq 
rt[2]*a*Sqrt[c + d*Tan[e + f*x]])]*Sqrt[c + d*Tan[e + f*x]] - Sqrt[a + I*a 
*Tan[e + f*x]]*(c*(c - (9*I)*d) + 3*(c - (3*I)*d)*d*Tan[e + f*x] + 2*d^2*T 
an[e + f*x]^2))))/(4*d^(3/2)*f*Sqrt[c + d*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 1.64 (sec) , antiderivative size = 274, 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, 4039, 27, 3042, 4080, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4080

\(\displaystyle \frac {a \left (\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (22 c d+i \left (c^2-9 d^2\right )\right ) a^2+\left (c^2+10 i d c+23 d^2\right ) \tan (e+f x) a^2\right )}{2 \sqrt {c+d \tan (e+f x)}}dx}{a}+\frac {a (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \left (\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (22 c d+i \left (c^2-9 d^2\right )\right ) a^2+\left (c^2+10 i d c+23 d^2\right ) \tan (e+f x) a^2\right )}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (22 c d+i \left (c^2-9 d^2\right )\right ) a^2+\left (c^2+10 i d c+23 d^2\right ) \tan (e+f x) a^2\right )}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

\(\Big \downarrow \) 4084

\(\displaystyle \frac {a \left (\frac {32 a^2 d (c-i d) \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-a \left (10 c d-i \left (c^2+23 d^2\right )\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (\frac {32 a^2 d (c-i d) \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-a \left (10 c d-i \left (c^2+23 d^2\right )\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {a \left (\frac {-\frac {64 i a^4 d (c-i d) \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f}-a \left (10 c d-i \left (c^2+23 d^2\right )\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {a \left (\frac {-a \left (10 c d-i \left (c^2+23 d^2\right )\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-\frac {32 i \sqrt {2} a^{5/2} d \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {a \left (\frac {-\frac {a^3 \left (10 c d-i \left (c^2+23 d^2\right )\right ) \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f}-\frac {32 i \sqrt {2} a^{5/2} d \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {a \left (\frac {-\frac {2 a^3 \left (10 c d-i \left (c^2+23 d^2\right )\right ) \int \frac {1}{i a-\frac {d (i \tan (e+f x) a+a)}{c+d \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}}{f}-\frac {32 i \sqrt {2} a^{5/2} d \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {a \left (\frac {-\frac {2 (-1)^{3/4} a^{5/2} \left (10 c d-i \left (c^2+23 d^2\right )\right ) \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f}-\frac {32 i \sqrt {2} a^{5/2} d \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a (c+9 i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

Input:

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

Output:

-1/2*(a^2*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(d*f) + ( 
a*(((-2*(-1)^(3/4)*a^(5/2)*(10*c*d - I*(c^2 + 23*d^2))*ArcTanh[((-1)^(3/4) 
*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/ 
(Sqrt[d]*f) - ((32*I)*Sqrt[2]*a^(5/2)*Sqrt[c - I*d]*d*ArcTanh[(Sqrt[2]*Sqr 
t[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])] 
)/f)/(2*a) + (a*(c + (9*I)*d)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e 
+ f*x]])/f))/(4*d)
 

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 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 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. 1078 vs. \(2 (208 ) = 416\).

Time = 0.52 (sec) , antiderivative size = 1079, normalized size of antiderivative = 4.10

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1079\)
default	\(\text {Expression too large to display}\)	\(1079\)

Input:

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

Output:

1/16/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a^2*(-10*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d+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))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c^2+32*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d+16*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)*a*d^2+32*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*t 
an(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^( 
1/2)*a*d^2-4*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))* 
(1+I*tan(f*x+e)))^(1/2)*d*tan(f*x+e)-2*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^ 
(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c+23*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))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*d^2+16*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*t 
an(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*(I*a*d)^(1/2)*a*c*d+18 
*I*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f 
*x+e)))^(1/2)*d-16*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+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. 1039 vs. \(2 (199) = 398\).

Time = 0.13 (sec) , antiderivative size = 1039, normalized size of antiderivative = 3.95 \[ \int (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/8*(16*sqrt(2)*(d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(a^5*c - I*a^5*d)/f 
^2)*log(-(I*sqrt(2)*f*sqrt(-(a^5*c - I*a^5*d)/f^2)*e^(I*f*x + I*e) - sqrt( 
2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*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)))*e^(- 
I*f*x - I*e)/a^2) - 16*sqrt(2)*(d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(a^5* 
c - I*a^5*d)/f^2)*log(-(-I*sqrt(2)*f*sqrt(-(a^5*c - I*a^5*d)/f^2)*e^(I*f*x 
 + I*e) - sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*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)))*e^(-I*f*x - I*e)/a^2) + 2*sqrt(2)*((a^2*c - 11*I*a^2*d)*e^(3*I* 
f*x + 3*I*e) + (a^2*c - 7*I*a^2*d)*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)) + (d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt((I*a^5*c^4 - 20*a^5*c^3 
*d - 54*I*a^5*c^2*d^2 - 460*a^5*c*d^3 + 529*I*a^5*d^4)/(d^3*f^2))*log((2*I 
*d^2*f*sqrt((I*a^5*c^4 - 20*a^5*c^3*d - 54*I*a^5*c^2*d^2 - 460*a^5*c*d^3 + 
 529*I*a^5*d^4)/(d^3*f^2))*e^(I*f*x + I*e) + sqrt(2)*(a^2*c^2 + 10*I*a^2*c 
*d + 23*a^2*d^2 + (a^2*c^2 + 10*I*a^2*c*d + 23*a^2*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)))*e^(-I*f*x - I*e)/(a^2*c^2 + 10*I*a^2 
*c*d + 23*a^2*d^2)) - (d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt((I*a^5*c^4 - 20 
*a^5*c^3*d - 54*I*a^5*c^2*d^2 - 460*a^5*c*d^3 + 529*I*a^5*d^4)/(d^3*f^2...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(3*d-c>0)', see `assume?` for mor 
e details)
 

Giac [F(-2)]

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

integrate((a+I*a*tan(f*x+e))^(5/2)*(c+d*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 TypeRecursive ass 
umption s
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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