\(\int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx\) [256]

Optimal result

Integrand size = 28, antiderivative size = 157 \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {10 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{a^4 d}-\frac {10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )} \] Output:

-10*e^6*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-10*e^5*(e*sec(d*x+c))^(3/2)*sin(d*x+c)/a^4/d+4/3*I*e^2*(e 
*sec(d*x+c))^(9/2)/a/d/(a+I*a*tan(d*x+c))^3+12*I*e^4*(e*sec(d*x+c))^(5/2)/ 
d/(a^4+I*a^4*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 1.49 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.85 \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {i e^6 \sec ^5(c+d x) \sqrt {e \sec (c+d x)} \left (21+19 \cos (2 (c+d x))+30 i \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (c+d x)+i \sin (c+d x))+11 i \sin (2 (c+d x))\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))}{3 a^4 d (-i+\tan (c+d x))^4} \] Input:

Integrate[(e*Sec[c + d*x])^(13/2)/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

((I/3)*e^6*Sec[c + d*x]^5*Sqrt[e*Sec[c + d*x]]*(21 + 19*Cos[2*(c + d*x)] + 
 (30*I)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2]*(Cos[c + d*x] + I*Sin 
[c + d*x]) + (11*I)*Sin[2*(c + d*x)])*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x 
)]))/(a^4*d*(-I + Tan[c + d*x])^4)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.09, 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, 4255, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}-\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )}{a^2}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\)

Input:

Int[(e*Sec[c + d*x])^(13/2)/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

(((4*I)/3)*e^2*(e*Sec[c + d*x])^(9/2))/(a*d*(a + I*a*Tan[c + d*x])^3) - (3 
*e^2*((5*e^2*((2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*S 
ec[c + d*x]])/(3*d) + (2*e*(e*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)))/a^ 
2 - ((4*I)*e^2*(e*Sec[c + d*x])^(5/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))))/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 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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 = 4.75 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {2 \sqrt {e \sec \left (d x +c \right )}\, \left (\tan \left (d x +c \right ) \left (-8 \cos \left (d x +c \right )^{2}-1\right )+i \left (15 \cos \left (d x +c \right )+15\right ) \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \left (-8 \cos \left (d x +c \right )^{2}-12\right )\right ) e^{6}}{3 a^{4} d}\)	\(122\)

Input:

int((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

-2/3/a^4/d*(e*sec(d*x+c))^(1/2)*(tan(d*x+c)*(-8*cos(d*x+c)^2-1)+I*(15*cos( 
d*x+c)+15)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2) 
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*(-8*cos(d*x+c)^2-12))*e^6
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.94 \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (-15 i \, e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} - 21 i \, e^{6} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, e^{6}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 15 \, \sqrt {2} {\left (-i \, e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} - i \, e^{6} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{3 \, {\left (a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \] Input:

integrate((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas" 
)
 

Output:

-2/3*(sqrt(2)*(-15*I*e^6*e^(4*I*d*x + 4*I*c) - 21*I*e^6*e^(2*I*d*x + 2*I*c 
) - 4*I*e^6)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 1 
5*sqrt(2)*(-I*e^6*e^(4*I*d*x + 4*I*c) - I*e^6*e^(2*I*d*x + 2*I*c))*sqrt(e) 
*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(a^4*d*e^(4*I*d*x + 4*I*c) + 
 a^4*d*e^(2*I*d*x + 2*I*c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate((e*sec(d*x+c))**(13/2)/(a+I*a*tan(d*x+c))**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((e*sec(d*x+c))^(13/2)/(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]

\[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {13}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:

integrate((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(13/2)/(I*a*tan(d*x + c) + a)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{13/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \] Input:

int((e/cos(c + d*x))^(13/2)/(a + a*tan(c + d*x)*1i)^4,x)
 

Output:

int((e/cos(c + d*x))^(13/2)/(a + a*tan(c + d*x)*1i)^4, x)
 

Reduce [F]

\[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{6}}{\tan \left (d x +c \right )^{4}-4 \tan \left (d x +c \right )^{3} i -6 \tan \left (d x +c \right )^{2}+4 \tan \left (d x +c \right ) i +1}d x \right ) e^{6}}{a^{4}} \] Input:

int((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^4,x)
 

Output:

(sqrt(e)*int((sqrt(sec(c + d*x))*sec(c + d*x)**6)/(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**6)/a**4