\(\int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^3} \, dx\) [248]

Optimal result

Integrand size = 28, antiderivative size = 116 \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {10 i e^4 \sqrt {e \sec (c+d x)}}{3 a^3 d}-\frac {10 e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 a^3 d}+\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{3 a d (a+i a \tan (c+d x))^2} \] Output:

10/3*I*e^4*(e*sec(d*x+c))^(1/2)/a^3/d-10/3*e^4*cos(d*x+c)^(1/2)*InverseJac 
obiAM(1/2*d*x+1/2*c,2^(1/2))*(e*sec(d*x+c))^(1/2)/a^3/d+4/3*I*e^2*(e*sec(d 
*x+c))^(5/2)/a/d/(a+I*a*tan(d*x+c))^2
 

Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.08 \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 e^4 \sec ^3(c+d x) \sqrt {e \sec (c+d x)} \left (-7 i \cos (c+d x)+5 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (c+d x)+i \sin (c+d x))+3 \sin (c+d x)\right ) (-i \cos (2 (c+d x))+\sin (2 (c+d x)))}{3 a^3 d (-i+\tan (c+d x))^3} \] Input:

Integrate[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x])^3,x]
 

Output:

(2*e^4*Sec[c + d*x]^3*Sqrt[e*Sec[c + d*x]]*((-7*I)*Cos[c + d*x] + 5*Sqrt[C 
os[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[c + d*x] + I*Sin[c + d*x]) + 3 
*Sin[c + d*x])*((-I)*Cos[2*(c + d*x)] + Sin[2*(c + d*x)]))/(3*a^3*d*(-I + 
Tan[c + d*x])^3)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3981, 3042, 3982, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3982

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

Int[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x])^3,x]
 

Output:

(-5*e^2*(((-2*I)*e^2*Sqrt[e*Sec[c + d*x]])/(a*d) + (2*e^2*Sqrt[Cos[c + d*x 
]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(a*d)))/(3*a^2) + (((4* 
I)/3)*e^2*(e*Sec[c + d*x])^(5/2))/(a*d*(a + I*a*Tan[c + d*x])^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 3982

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + 
f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Simp[d^2*((m - 2)/(a*(m + n - 1))) 
 Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ 
[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && GtQ[m, 1] &&  !IL 
tQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 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 = 4.01 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.01

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

Input:

int((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

-2/3/a^3/d*(I*(5*cos(d*x+c)+5)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d 
*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)+I*(-4*cos(d*x+c)^2- 
3)-4*cos(d*x+c)*sin(d*x+c))*(e*sec(d*x+c))^(1/2)*e^4
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^3} \, dx=-\frac {2 \, {\left (-5 i \, \sqrt {2} e^{\frac {9}{2}} e^{\left (2 i \, d x + 2 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-5 i \, e^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, e^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{3 \, a^{3} d} \] Input:

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

-2/3*(-5*I*sqrt(2)*e^(9/2)*e^(2*I*d*x + 2*I*c)*weierstrassPInverse(-4, 0, 
e^(I*d*x + I*c)) + sqrt(2)*(-5*I*e^4*e^(2*I*d*x + 2*I*c) - 2*I*e^4)*sqrt(e 
/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-2*I*d*x - 2*I*c)/ 
(a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((e*sec(d*x+c))**(9/2)/(a+I*a*tan(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^3} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(9/2)/(I*a*tan(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^3} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{9/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \] Input:

int((e/cos(c + d*x))^(9/2)/(a + a*tan(c + d*x)*1i)^3,x)
 

Output:

int((e/cos(c + d*x))^(9/2)/(a + a*tan(c + d*x)*1i)^3, x)
 

Reduce [F]

\[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right ) e^{4}}{a^{3}} \] Input:

int((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^3,x)
 

Output:

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