\(\int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx\) [223]

Optimal result

Integrand size = 28, antiderivative size = 136 \[ \int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx=-\frac {6 e^6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {2 i e^2 (e \sec (c+d x))^{7/2}}{7 a d}+\frac {6 e^5 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a d}+\frac {2 e^3 (e \sec (c+d x))^{5/2} \sin (c+d x)}{5 a d} \] Output:

-6/5*e^6*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d/cos(d*x+c)^(1/2)/(e*sec 
(d*x+c))^(1/2)-2/7*I*e^2*(e*sec(d*x+c))^(7/2)/a/d+6/5*e^5*(e*sec(d*x+c))^( 
1/2)*sin(d*x+c)/a/d+2/5*e^3*(e*sec(d*x+c))^(5/2)*sin(d*x+c)/a/d
 

Mathematica [C] (verified)

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

Time = 2.15 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.94 \[ \int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx=\frac {e^4 (e \sec (c+d x))^{3/2} \left (76+28 \cos (2 (c+d x))-7 e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+7 i \sec (c+d x) \sin (3 (c+d x))-13 i \tan (c+d x)\right ) (-i+\tan (c+d x))}{70 a d} \] Input:

Integrate[(e*Sec[c + d*x])^(11/2)/(a + I*a*Tan[c + d*x]),x]
 

Output:

(e^4*(e*Sec[c + d*x])^(3/2)*(76 + 28*Cos[2*(c + d*x)] - (7*(1 + E^((2*I)*( 
c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E 
^((2*I)*(c + d*x)) + (7*I)*Sec[c + d*x]*Sin[3*(c + d*x)] - (13*I)*Tan[c + 
d*x])*(-I + Tan[c + d*x]))/(70*a*d)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3982, 3042, 4255, 3042, 4255, 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.

\(\displaystyle \int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3982

\(\displaystyle \frac {e^2 \int (e \sec (c+d x))^{7/2}dx}{a}-\frac {2 i e^2 (e \sec (c+d x))^{7/2}}{7 a d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

\(\displaystyle \frac {e^2 \left (\frac {3}{5} e^2 \int (e \sec (c+d x))^{3/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{a}-\frac {2 i e^2 (e \sec (c+d x))^{7/2}}{7 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (\frac {3}{5} e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{a}-\frac {2 i e^2 (e \sec (c+d x))^{7/2}}{7 a d}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{a}-\frac {2 i e^2 (e \sec (c+d x))^{7/2}}{7 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{a}-\frac {2 i e^2 (e \sec (c+d x))^{7/2}}{7 a d}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{a}-\frac {2 i e^2 (e \sec (c+d x))^{7/2}}{7 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{a}-\frac {2 i e^2 (e \sec (c+d x))^{7/2}}{7 a d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{a}-\frac {2 i e^2 (e \sec (c+d x))^{7/2}}{7 a d}\)

Input:

Int[(e*Sec[c + d*x])^(11/2)/(a + I*a*Tan[c + d*x]),x]
 

Output:

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

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 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 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 = 4.26 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.73

method	result	size
default	\(\frac {2 e^{5} \sqrt {e \sec \left (d x +c \right )}\, \left (21 \sin \left (d x +c \right )+7 \tan \left (d x +c \right )+7 \sec \left (d x +c \right ) \tan \left (d x +c \right )+5 i \left (-\sec \left (d x +c \right )^{3}-\sec \left (d x +c \right )^{2}\right )+21 i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+21 i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right )}{35 a d \left (\cos \left (d x +c \right )+1\right )}\)	\(235\)

Input:

int((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

2/35*e^5/a/d*(e*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)*(21*sin(d*x+c)+7*tan(d*x+ 
c)+7*sec(d*x+c)*tan(d*x+c)+5*I*(-sec(d*x+c)^3-sec(d*x+c)^2)+21*I*(cos(d*x+ 
c)^2+2*cos(d*x+c)+1)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^ 
(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)+21*I*(-cos(d*x+c)^2-2*cos(d*x 
+c)-1)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elliptic 
E(I*(csc(d*x+c)-cot(d*x+c)),I))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.51 \[ \int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (21 i \, e^{5} e^{\left (7 i \, d x + 7 i \, c\right )} + 77 i \, e^{5} e^{\left (5 i \, d x + 5 i \, c\right )} + 103 i \, e^{5} e^{\left (3 i \, d x + 3 i \, c\right )} + 7 i \, e^{5} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 21 \, \sqrt {2} {\left (i \, e^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, e^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, e^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{5}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{35 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \] Input:

integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
 

Output:

-2/35*(sqrt(2)*(21*I*e^5*e^(7*I*d*x + 7*I*c) + 77*I*e^5*e^(5*I*d*x + 5*I*c 
) + 103*I*e^5*e^(3*I*d*x + 3*I*c) + 7*I*e^5*e^(I*d*x + I*c))*sqrt(e/(e^(2* 
I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 21*sqrt(2)*(I*e^5*e^(6*I*d* 
x + 6*I*c) + 3*I*e^5*e^(4*I*d*x + 4*I*c) + 3*I*e^5*e^(2*I*d*x + 2*I*c) + I 
*e^5)*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + 
 I*c))))/(a*d*e^(6*I*d*x + 6*I*c) + 3*a*d*e^(4*I*d*x + 4*I*c) + 3*a*d*e^(2 
*I*d*x + 2*I*c) + a*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((e*sec(d*x+c))**(11/2)/(a+I*a*tan(d*x+c)),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}}{i \, a \tan \left (d x + c\right ) + a} \,d x } \] Input:

integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(11/2)/(I*a*tan(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{11/2}}{a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \] Input:

int((e/cos(c + d*x))^(11/2)/(a + a*tan(c + d*x)*1i),x)
 

Output:

int((e/cos(c + d*x))^(11/2)/(a + a*tan(c + d*x)*1i), x)
 

Reduce [F]

\[ \int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{5}}{\tan \left (d x +c \right ) i +1}d x \right ) e^{5}}{a} \] Input:

int((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c)),x)
 

Output:

(sqrt(e)*int((sqrt(sec(c + d*x))*sec(c + d*x)**5)/(tan(c + d*x)*i + 1),x)* 
e**5)/a