\(\int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^3} \, dx\) [251]

Optimal result

Integrand size = 28, antiderivative size = 132 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{9 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}+\frac {2 i e^2}{45 d \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:

2/15*e^2*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d/cos(d*x+c)^(1/2)/(e*s 
ec(d*x+c))^(1/2)+4/9*I*e^2/a/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^2+2 
/45*I*e^2/d/(e*sec(d*x+c))^(1/2)/(a^3+I*a^3*tan(d*x+c))
 

Mathematica [C] (verified)

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

Time = 1.59 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.06 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^3} \, dx=-\frac {e^{-i d x} \sec ^2(c+d x) (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x)) \left (8+8 \cos (2 (c+d x))+6 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+3 i \sin (2 (c+d x))\right )}{45 a^3 d (-i+\tan (c+d x))^3} \] Input:

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

Output:

-1/45*(Sec[c + d*x]^2*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I*Sin[d*x])*(8 + 
8*Cos[2*(c + d*x)] + 6*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*H 
ypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + (3*I)*Sin[2*(c + 
d*x)]))/(a^3*d*E^(I*d*x)*(-I + Tan[c + d*x])^3)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.01, 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, 3983, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3983

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

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

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 3983

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ 
(b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n))   Int[(d*Sec[e + 
f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x 
] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && NeQ[m + 2*n, 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 [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 496 vs. \(2 (116 ) = 232\).

Time = 3.57 (sec) , antiderivative size = 497, normalized size of antiderivative = 3.77

method	result	size
default	\(-\frac {2 \left (-6 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \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}}\, \sqrt {\frac {1}{\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 )+3 i \left (2 \cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+\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}}\, \sqrt {\frac {1}{\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 )-6 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \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}}\, \sqrt {\frac {1}{\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 )+3 i \left (-2 \cos \left (d x +c \right )^{4}-4 \cos \left (d x +c \right )^{3}-\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}}\, \sqrt {\frac {1}{\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 )-\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-5 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right )+i \cos \left (d x +c \right )^{2} \left (5 \cos \left (d x +c \right )^{2}+11 \cos \left (d x +c \right )+6\right )\right ) \sqrt {e \sec \left (d x +c \right )}\, e}{45 a^{3} d \left (2 i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-\cos \left (d x +c \right )-1\right )-2 \cos \left (d x +c \right )^{3}-2 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+1\right )}\)	\(497\)

Input:

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

Output:

-2/45/a^3/d*(-6*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)^2+2*cos(d*x+c)+1)*(cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+ 
c)-cot(d*x+c)),I)+3*I*(2*cos(d*x+c)^4+4*cos(d*x+c)^3+cos(d*x+c)^2-2*cos(d* 
x+c)-1)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*Ellipti 
cE(I*(csc(d*x+c)-cot(d*x+c)),I)-6*sin(d*x+c)*cos(d*x+c)*(-cos(d*x+c)^2-2*c 
os(d*x+c)-1)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*El 
lipticF(I*(csc(d*x+c)-cot(d*x+c)),I)+3*I*(-2*cos(d*x+c)^4-4*cos(d*x+c)^3-c 
os(d*x+c)^2+2*cos(d*x+c)+1)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+ 
c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)-sin(d*x+c)*cos(d*x+c)* 
(-5*cos(d*x+c)^2+cos(d*x+c)+3)+I*cos(d*x+c)^2*(5*cos(d*x+c)^2+11*cos(d*x+c 
)+6))*(e*sec(d*x+c))^(1/2)/(2*I*sin(d*x+c)*cos(d*x+c)*(-cos(d*x+c)-1)-2*co 
s(d*x+c)^3-2*cos(d*x+c)^2+cos(d*x+c)+1)*e
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.91 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (12 i \, \sqrt {2} e^{\frac {3}{2}} e^{\left (5 i \, d x + 5 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (12 i \, e e^{\left (6 i \, d x + 6 i \, c\right )} + 23 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{90 \, a^{3} d} \] Input:

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

Output:

1/90*(12*I*sqrt(2)*e^(3/2)*e^(5*I*d*x + 5*I*c)*weierstrassZeta(-4, 0, weie 
rstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(12*I*e*e^(6*I*d*x + 6* 
I*c) + 23*I*e*e^(4*I*d*x + 4*I*c) + 16*I*e*e^(2*I*d*x + 2*I*c) + 5*I*e)*sq 
rt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-5*I*d*x - 5*I 
*c)/(a^3*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((e*sec(d*x+c))^(3/2)/(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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