\(\int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx\) [680]

Optimal result

Integrand size = 28, antiderivative size = 126 \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\frac {10 \cos ^{\frac {9}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d (e \cos (c+d x))^{9/2}}+\frac {10 \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (e \cos (c+d x))^{9/2}}-\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )} \] Output:

10/3*cos(d*x+c)^(9/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^2/d/(e*cos( 
d*x+c))^(9/2)+10/3*cos(d*x+c)^3*sin(d*x+c)/a^2/d/(e*cos(d*x+c))^(9/2)-4*I* 
cos(d*x+c)^2/d/(e*cos(d*x+c))^(9/2)/(a^2+I*a^2*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\frac {2 \left (-6 i \cos (c+d x)+5 \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\sin (c+d x)\right )}{3 a^2 d e^3 (e \cos (c+d x))^{3/2}} \] Input:

Integrate[1/((e*Cos[c + d*x])^(9/2)*(a + I*a*Tan[c + d*x])^2),x]
 

Output:

(2*((-6*I)*Cos[c + d*x] + 5*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] - 
 Sin[c + d*x]))/(3*a^2*d*e^3*(e*Cos[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3998, 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.

\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \cos (c+d x))^{9/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \cos (c+d x))^{9/2}}dx\)

\(\Big \downarrow \) 3998

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

Int[1/((e*Cos[c + d*x])^(9/2)*(a + I*a*Tan[c + d*x])^2),x]
 

Output:

((5*e^2*((2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[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])))/((e*Cos[ 
c + d*x])^(9/2)*(e*Sec[c + d*x])^(9/2))
 

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 3998

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m   Int[( 
a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m 
, n}, x] &&  !IntegerQ[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 = 4.16 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.65

method	result	size
default	\(\frac {-\frac {20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}-8 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\frac {4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{3}+\frac {10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3}+4 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, e^{4} d}\)	\(208\)

Input:

int(1/(e*cos(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

2/3/(2*sin(1/2*d*x+1/2*c)^2-1)/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2* 
c)^2*e+e)^(1/2)/e^4*(-10*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d* 
x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-12 
*I*sin(1/2*d*x+1/2*c)^3+2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2* 
d*x+1/2*c),2^(1/2))+6*I*sin(1/2*d*x+1/2*c))/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {4 \, {\left (\sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (5 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 7 i \, e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 5 \, \sqrt {\frac {1}{2}} \sqrt {e} {\left (i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{3 \, {\left (a^{2} d e^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} d e^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{5}\right )}} \] Input:

integrate(1/(e*cos(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas 
")
                                                                                    
                                                                                    
 

Output:

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

Sympy [F(-1)]

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

integrate(1/(e*cos(d*x+c))**(9/2)/(a+I*a*tan(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(e*cos(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima 
")
 

Output:

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

Giac [F]

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

integrate(1/(e*cos(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate(1/((e*cos(d*x + c))^(9/2)*(I*a*tan(d*x + c) + a)^2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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