\(\int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx\) [261]

Optimal result

Integrand size = 28, antiderivative size = 163 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{13 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac {4 i e^4}{117 d (e \sec (c+d x))^{5/2} \left (a^4+i a^4 \tan (c+d x)\right )} \] Output:

2/39*e^2*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d/cos(d*x+c)^(1/2)/(e*s 
ec(d*x+c))^(1/2)+2/117*e^3*sin(d*x+c)/a^4/d/(e*sec(d*x+c))^(3/2)+4/13*I*e^ 
2/a/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^3+4/117*I*e^4/d/(e*sec(d*x+c 
))^(5/2)/(a^4+I*a^4*tan(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.90 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.87 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {i e^{-i d x} \sec ^2(c+d x) (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x)) \left (28+40 \cos (2 (c+d x))+\frac {24 e^{4 i (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+22 i \sin (2 (c+d x))\right )}{234 a^4 d (-i+\tan (c+d x))^4} \] Input:

Integrate[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

((I/234)*Sec[c + d*x]^2*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I*Sin[d*x])*(28 
 + 40*Cos[2*(c + d*x)] + (24*E^((4*I)*(c + d*x))*Hypergeometric2F1[-1/4, 1 
/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + (22*I)*Sin 
[2*(c + d*x)]))/(a^4*d*E^(I*d*x)*(-I + Tan[c + d*x])^4)
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3981, 3042, 3981, 3042, 4256, 3042, 4258, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

\(\displaystyle \frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^2}dx}{13 a^2}+\frac {4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^2}dx}{13 a^2}+\frac {4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {e^2 \left (\frac {5 e^2 \int \frac {1}{(e \sec (c+d x))^{5/2}}dx}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (\frac {5 e^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {e^2 \left (\frac {5 e^2 \left (\frac {3 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (\frac {5 e^2 \left (\frac {3 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {e^2 \left (\frac {5 e^2 \left (\frac {3 \int \sqrt {\cos (c+d x)}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (\frac {5 e^2 \left (\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {e^2 \left (\frac {5 e^2 \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

Input:

Int[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

(((4*I)/13)*e^2)/(a*d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3) + (e^ 
2*((5*e^2*((6*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[ 
e*Sec[c + d*x]]) + (2*Sin[c + d*x])/(5*d*e*(e*Sec[c + d*x])^(3/2))))/(9*a^ 
2) + (((4*I)/9)*e^2)/(d*(e*Sec[c + d*x])^(5/2)*(a^2 + I*a^2*Tan[c + d*x])) 
))/(13*a^2)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3981

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[2*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + 
 f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Simp[d^2*((m - 2)/(b^2*(m + 2*n))) 
Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[ 
{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] 
 && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (IntegersQ[n, m + 
1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
 

rule 4256

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* 
n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 290 vs. \(2 (143 ) = 286\).

Time = 3.86 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.79

method	result	size
default	\(\frac {2 \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (72 \cos \left (d x +c \right )^{6}+72 \cos \left (d x +c \right )^{5}-16 \cos \left (d x +c \right )^{4}-16 \cos \left (d x +c \right )^{3}+\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right )+i \cos \left (d x +c \right )^{5} \left (72 \cos \left (d x +c \right )^{3}+72 \cos \left (d x +c \right )^{2}-52 \cos \left (d x +c \right )-52\right )+i \left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+i \left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right ) \sqrt {e \sec \left (d x +c \right )}\, e}{117 a^{4} d \left (\cos \left (d x +c \right )+1\right )}\)	\(291\)

Input:

int((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

2/117/a^4/d*(sin(d*x+c)*cos(d*x+c)*(72*cos(d*x+c)^6+72*cos(d*x+c)^5-16*cos 
(d*x+c)^4-16*cos(d*x+c)^3+cos(d*x+c)^2+cos(d*x+c)+3)+I*cos(d*x+c)^5*(72*co 
s(d*x+c)^3+72*cos(d*x+c)^2-52*cos(d*x+c)-52)+I*(3*cos(d*x+c)^2+6*cos(d*x+c 
)+3)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1 
/2)*(1/(cos(d*x+c)+1))^(1/2)+I*(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*EllipticF( 
I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x 
+c)+1))^(1/2))*(e*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)*e
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.81 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (24 i \, \sqrt {2} e^{\frac {3}{2}} e^{\left (7 i \, d x + 7 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (24 i \, e e^{\left (8 i \, d x + 8 i \, c\right )} + 55 i \, e e^{\left (6 i \, d x + 6 i \, c\right )} + 59 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 37 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 9 i \, e\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{468 \, a^{4} d} \] Input:

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
 

Output:

1/468*(24*I*sqrt(2)*e^(3/2)*e^(7*I*d*x + 7*I*c)*weierstrassZeta(-4, 0, wei 
erstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(24*I*e*e^(8*I*d*x + 8 
*I*c) + 55*I*e*e^(6*I*d*x + 6*I*c) + 59*I*e*e^(4*I*d*x + 4*I*c) + 37*I*e*e 
^(2*I*d*x + 2*I*c) + 9*I*e)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x 
 + 1/2*I*c))*e^(-7*I*d*x - 7*I*c)/(a^4*d)
 

Sympy [F]

\[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:

integrate((e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**4,x)
 

Output:

Integral((e*sec(c + d*x))**(3/2)/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 
6*tan(c + d*x)**2 + 4*I*tan(c + d*x) + 1), x)/a**4
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(3/2)/(I*a*tan(d*x + c) + a)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \] Input:

int((e/cos(c + d*x))^(3/2)/(a + a*tan(c + d*x)*1i)^4,x)
 

Output:

int((e/cos(c + d*x))^(3/2)/(a + a*tan(c + d*x)*1i)^4, x)
 

Reduce [F]

\[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )}{\tan \left (d x +c \right )^{4}-4 \tan \left (d x +c \right )^{3} i -6 \tan \left (d x +c \right )^{2}+4 \tan \left (d x +c \right ) i +1}d x \right ) e}{a^{4}} \] Input:

int((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^4,x)
 

Output:

(sqrt(e)*int((sqrt(sec(c + d*x))*sec(c + d*x))/(tan(c + d*x)**4 - 4*tan(c 
+ d*x)**3*i - 6*tan(c + d*x)**2 + 4*tan(c + d*x)*i + 1),x)*e)/a**4