\(\int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2} \, dx\) [243]

Optimal result

Integrand size = 28, antiderivative size = 150 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2} \, dx=\frac {42 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{65 a^2 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 e \sin (c+d x)}{13 a^2 d (e \sec (c+d x))^{7/2}}+\frac {14 \sin (c+d x)}{65 a^2 d e (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{13 d (e \sec (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )} \] Output:

42/65*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d/e^2/cos(d*x+c)^(1/2)/(e* 
sec(d*x+c))^(1/2)+2/13*e*sin(d*x+c)/a^2/d/(e*sec(d*x+c))^(7/2)+14/65*sin(d 
*x+c)/a^2/d/e/(e*sec(d*x+c))^(3/2)+4/13*I*e^2/d/(e*sec(d*x+c))^(9/2)/(a^2+ 
I*a^2*tan(d*x+c))
 

Mathematica [C] (verified)

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

Time = 2.66 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2} \, dx=\frac {(\cos (2 (c+d x))-i \sin (2 (c+d x))) \left (88 i+416 i \cos (2 (c+d x))-8 i \cos (4 (c+d x))-\frac {224 i e^{4 i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-356 \sin (2 (c+d x))+18 \sin (4 (c+d x))\right )}{520 a^2 d e^2 \sqrt {e \sec (c+d x)}} \] Input:

Integrate[1/((e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^2),x]
 

Output:

((Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)])*(88*I + (416*I)*Cos[2*(c + d*x)] 
- (8*I)*Cos[4*(c + d*x)] - ((224*I)*E^((4*I)*(c + d*x))*Hypergeometric2F1[ 
1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] - 356* 
Sin[2*(c + d*x)] + 18*Sin[4*(c + d*x)]))/(520*a^2*d*e^2*Sqrt[e*Sec[c + d*x 
]])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.08, 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, 4256, 3042, 4256, 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.

\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {9 e^2 \int \frac {1}{(e \sec (c+d x))^{9/2}}dx}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 e^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}}dx}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {9 e^2 \left (\frac {7 \int \frac {1}{(e \sec (c+d x))^{5/2}}dx}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 e^2 \left (\frac {7 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \sqrt {\cos (c+d x)}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {9 e^2 \left (\frac {7 \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\)

Input:

Int[1/((e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^2),x]
 

Output:

(9*e^2*((2*Sin[c + d*x])/(9*d*e*(e*Sec[c + d*x])^(7/2)) + (7*((6*EllipticE 
[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*S 
in[c + d*x])/(5*d*e*(e*Sec[c + d*x])^(3/2))))/(9*e^2)))/(13*a^2) + (((4*I) 
/13)*e^2)/(d*(e*Sec[c + d*x])^(9/2)*(a^2 + I*a^2*Tan[c + d*x]))
 

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 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 [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 519 vs. \(2 (132 ) = 264\).

Time = 4.53 (sec) , antiderivative size = 520, normalized size of antiderivative = 3.47

method	result	size
default	\(\frac {\frac {84 i \sin \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}{65}+\frac {42 \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (2 \cos \left (d x +c \right )^{3}+4 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )-\sec \left (d x +c \right )-2\right )}{65}+\frac {84 i \sin \left (d x +c \right ) \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}{65}+\frac {42 \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (-2 \cos \left (d x +c \right )^{3}-4 \cos \left (d x +c \right )^{2}+\sec \left (d x +c \right )-\cos \left (d x +c \right )+2\right )}{65}+\frac {2 i \sin \left (d x +c \right ) \left (9 \cos \left (d x +c \right )^{4}+9 \cos \left (d x +c \right )^{3}+35 \cos \left (d x +c \right )^{2}-7 \cos \left (d x +c \right )-21\right )}{65}+\frac {8 \cos \left (d x +c \right )^{5}}{65}+\frac {8 \cos \left (d x +c \right )^{4}}{65}+\frac {56 \cos \left (d x +c \right )^{3}}{65}-\frac {28 \cos \left (d x +c \right )^{2}}{65}-\frac {84 \cos \left (d x +c \right )}{65}}{a^{2} d \left (2 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-\cos \left (d x +c \right )-1\right )+2 i \cos \left (d x +c \right )^{3}+2 i \cos \left (d x +c \right )^{2}-i \cos \left (d x +c \right )-i\right ) \sqrt {e \sec \left (d x +c \right )}\, e^{2}}\)	\(520\)

Input:

int(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

2/65/a^2/d/(2*sin(d*x+c)*cos(d*x+c)*(-cos(d*x+c)-1)+2*I*cos(d*x+c)^3+2*I*c 
os(d*x+c)^2-I*cos(d*x+c)-I)/(e*sec(d*x+c))^(1/2)*(42*I*sin(d*x+c)*(cos(d*x 
+c)^2+2*cos(d*x+c)+1)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)*(cos(d*x+c)/( 
cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)+21*EllipticE(I*(cot(d*x+c)-c 
sc(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*( 
2*cos(d*x+c)^3+4*cos(d*x+c)^2+cos(d*x+c)-sec(d*x+c)-2)+42*I*sin(d*x+c)*(-c 
os(d*x+c)^2-2*cos(d*x+c)-1)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*(cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)+21*EllipticF(I*(cot(d* 
x+c)-csc(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^( 
1/2)*(-2*cos(d*x+c)^3-4*cos(d*x+c)^2+sec(d*x+c)-cos(d*x+c)+2)+I*sin(d*x+c) 
*(9*cos(d*x+c)^4+9*cos(d*x+c)^3+35*cos(d*x+c)^2-7*cos(d*x+c)-21)+4*cos(d*x 
+c)^5+4*cos(d*x+c)^4+28*cos(d*x+c)^3-14*cos(d*x+c)^2-42*cos(d*x+c))/e^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-13 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 373 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 474 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 118 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 35 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 672 i \, \sqrt {2} \sqrt {e} e^{\left (7 i \, d x + 7 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{1040 \, a^{2} d e^{3}} \] Input:

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas 
")
 

Output:

1/1040*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-13*I*e^(10*I*d*x + 10* 
I*c) + 373*I*e^(8*I*d*x + 8*I*c) + 474*I*e^(6*I*d*x + 6*I*c) + 118*I*e^(4* 
I*d*x + 4*I*c) + 35*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(1/2*I*d*x + 1/2*I*c) + 
 672*I*sqrt(2)*sqrt(e)*e^(7*I*d*x + 7*I*c)*weierstrassZeta(-4, 0, weierstr 
assPInverse(-4, 0, e^(I*d*x + I*c))))*e^(-7*I*d*x - 7*I*c)/(a^2*d*e^3)
 

Sympy [F]

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

integrate(1/(e*sec(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**2,x)
 

Output:

-Integral(1/((e*sec(c + d*x))**(5/2)*tan(c + d*x)**2 - 2*I*(e*sec(c + d*x) 
)**(5/2)*tan(c + d*x) - (e*sec(c + d*x))**(5/2)), x)/a**2
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(e*sec(d*x+c))^(5/2)/(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]

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

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate(1/((e*sec(d*x + c))^(5/2)*(I*a*tan(d*x + c) + a)^2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^2,x)
 

Output:

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