\(\int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx\) [260]

Optimal result

Integrand size = 28, antiderivative size = 163 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 a^4 d}-\frac {2 e^3 \sin (c+d x)}{77 a^4 d \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{77 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )} \] Output:

-2/77*e^2*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(e*sec(d 
*x+c))^(1/2)/a^4/d-2/77*e^3*sin(d*x+c)/a^4/d/(e*sec(d*x+c))^(1/2)+4/11*I*e 
^2*(e*sec(d*x+c))^(1/2)/a/d/(a+I*a*tan(d*x+c))^3-4/77*I*e^4/d/(e*sec(d*x+c 
))^(3/2)/(a^4+I*a^4*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 1.41 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.88 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^2(c+d x) (e \sec (c+d x))^{5/2} (\cos (c+d x)+i \sin (c+d x)) \left (37 i \cos (c+d x)+11 i \cos (3 (c+d x))+3 \sin (c+d x)-4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+3 \sin (3 (c+d x))\right )}{154 a^4 d (-i+\tan (c+d x))^4} \] Input:

Integrate[(e*Sec[c + d*x])^(5/2)/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

(Sec[c + d*x]^2*(e*Sec[c + d*x])^(5/2)*(Cos[c + d*x] + I*Sin[c + d*x])*((3 
7*I)*Cos[c + d*x] + (11*I)*Cos[3*(c + d*x)] + 3*Sin[c + d*x] - 4*Sqrt[Cos[ 
c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)] 
) + 3*Sin[3*(c + d*x)]))/(154*a^4*d*(-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, 3120}

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.

\(\displaystyle \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4}dx\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \int \frac {\sqrt {e \sec (c+d x)}}{(i \tan (c+d x) a+a)^2}dx}{11 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \int \frac {\sqrt {e \sec (c+d x)}}{(i \tan (c+d x) a+a)^2}dx}{11 a^2}\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \left (\frac {3 e^2 \int \frac {1}{(e \sec (c+d x))^{3/2}}dx}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \left (\frac {3 e^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a^2}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \left (\frac {3 e^2 \left (\frac {\int \sqrt {e \sec (c+d x)}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \left (\frac {3 e^2 \left (\frac {\int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a^2}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \left (\frac {3 e^2 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \left (\frac {3 e^2 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \left (\frac {3 e^2 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a^2}\)

Input:

Int[(e*Sec[c + d*x])^(5/2)/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

(((4*I)/11)*e^2*Sqrt[e*Sec[c + d*x]])/(a*d*(a + I*a*Tan[c + d*x])^3) - (e^ 
2*((3*e^2*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + 
d*x]])/(3*d*e^2) + (2*Sin[c + d*x])/(3*d*e*Sqrt[e*Sec[c + d*x]])))/(7*a^2) 
 + (((4*I)/7)*e^2)/(d*(e*Sec[c + d*x])^(3/2)*(a^2 + I*a^2*Tan[c + d*x])))) 
/(11*a^2)
 

Defintions of rubi rules used

rule 3042

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

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 [A] (verified)

Time = 3.62 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {2 \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (56 \cos \left (d x +c \right )^{4}-16 \cos \left (d x +c \right )^{2}-1\right )+i \cos \left (d x +c \right )^{4} \left (56 \cos \left (d x +c \right )^{2}-44\right )+i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )\right ) \sqrt {e \sec \left (d x +c \right )}\, e^{2}}{77 a^{4} d}\)	\(144\)

Input:

int((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

2/77/a^4/d*(sin(d*x+c)*cos(d*x+c)*(56*cos(d*x+c)^4-16*cos(d*x+c)^2-1)+I*co 
s(d*x+c)^4*(56*cos(d*x+c)^2-44)+I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(co 
s(d*x+c)+1))^(1/2)*(cos(d*x+c)+1)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I))* 
(e*sec(d*x+c))^(1/2)*e^2
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.77 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (4 i \, \sqrt {2} e^{\frac {5}{2}} e^{\left (6 i \, d x + 6 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (4 i \, e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 17 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 20 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{154 \, a^{4} d} \] Input:

integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
 

Output:

1/154*(4*I*sqrt(2)*e^(5/2)*e^(6*I*d*x + 6*I*c)*weierstrassPInverse(-4, 0, 
e^(I*d*x + I*c)) + sqrt(2)*(4*I*e^2*e^(6*I*d*x + 6*I*c) + 17*I*e^2*e^(4*I* 
d*x + 4*I*c) + 20*I*e^2*e^(2*I*d*x + 2*I*c) + 7*I*e^2)*sqrt(e/(e^(2*I*d*x 
+ 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-6*I*d*x - 6*I*c)/(a^4*d)
 

Sympy [F]

\[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:

integrate((e*sec(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**4,x)
 

Output:

Integral((e*sec(c + d*x))**(5/2)/(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. \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((e*sec(d*x+c))^(5/2)/(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]

\[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:

integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(5/2)/(I*a*tan(d*x + c) + a)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \] Input:

int((e/cos(c + d*x))^(5/2)/(a + a*tan(c + d*x)*1i)^4,x)
 

Output:

int((e/cos(c + d*x))^(5/2)/(a + a*tan(c + d*x)*1i)^4, x)
 

Reduce [F]

\[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{4}-4 \tan \left (d x +c \right )^{3} i -6 \tan \left (d x +c \right )^{2}+4 \tan \left (d x +c \right ) i +1}d x \right ) e^{2}}{a^{4}} \] Input:

int((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^4,x)
 

Output:

(sqrt(e)*int((sqrt(sec(c + d*x))*sec(c + d*x)**2)/(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**2)/a**4