\(\int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3} \, dx\) [253]

Optimal result

Integrand size = 28, antiderivative size = 152 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3} \, dx=\frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 a^3 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 e \sin (c+d x)}{117 a^3 d (e \sec (c+d x))^{3/2}}+\frac {2 i}{13 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac {28 i e^2}{117 d (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:

14/39*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d/cos(d*x+c)^(1/2)/(e*sec( 
d*x+c))^(1/2)+14/117*e*sin(d*x+c)/a^3/d/(e*sec(d*x+c))^(3/2)+2/13*I/d/(e*s 
ec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^3+28/117*I*e^2/d/(e*sec(d*x+c))^(5/2)/ 
(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 = 2.00 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3} \, dx=\frac {\sqrt {e \sec (c+d x)} (i \cos (3 (c+d x))+\sin (3 (c+d x))) \left (62+176 \cos (2 (c+d x))+114 \cos (4 (c+d x))-56 e^{4 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+126 i \sin (2 (c+d x))+105 i \sin (4 (c+d x))\right )}{468 a^3 d e} \] Input:

Integrate[1/(Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3),x]
 

Output:

(Sqrt[e*Sec[c + d*x]]*(I*Cos[3*(c + d*x)] + Sin[3*(c + d*x)])*(62 + 176*Co 
s[2*(c + d*x)] + 114*Cos[4*(c + d*x)] - 56*E^((4*I)*(c + d*x))*Sqrt[1 + E^ 
((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] 
+ (126*I)*Sin[2*(c + d*x)] + (105*I)*Sin[4*(c + d*x)]))/(468*a^3*d*e)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3983, 3042, 3981, 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.

\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}dx\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {7 \int \frac {1}{\sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^2}dx}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7 \int \frac {1}{\sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^2}dx}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

Int[1/(Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3),x]
 

Output:

((2*I)/13)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3) + (7*((5*e^2* 
((6*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + 
d*x]]) + (2*Sin[c + d*x])/(5*d*e*(e*Sec[c + d*x])^(3/2))))/(9*a^2) + (((4* 
I)/9)*e^2)/(d*(e*Sec[c + d*x])^(5/2)*(a^2 + I*a^2*Tan[c + d*x]))))/(13*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 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 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. 278 vs. \(2 (132 ) = 264\).

Time = 4.49 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.84

method	result	size
default	\(-\frac {2 \left (\sin \left (d x +c \right ) \left (-36 \cos \left (d x +c \right )^{6}-36 \cos \left (d x +c \right )^{5}-5 \cos \left (d x +c \right )^{4}-5 \cos \left (d x +c \right )^{3}-7 \cos \left (d x +c \right )^{2}-7 \cos \left (d x +c \right )-21\right )+i \left (-36 \cos \left (d x +c \right )^{7}-36 \cos \left (d x +c \right )^{6}+13 \cos \left (d x +c \right )^{5}+13 \cos \left (d x +c \right )^{4}\right )+21 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-21 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )\right )}{117 a^{3} d \left (\cos \left (d x +c \right )+1\right ) \sqrt {e \sec \left (d x +c \right )}}\)	\(279\)

Input:

int(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

-2/117/a^3/d/(cos(d*x+c)+1)/(e*sec(d*x+c))^(1/2)*(sin(d*x+c)*(-36*cos(d*x+ 
c)^6-36*cos(d*x+c)^5-5*cos(d*x+c)^4-5*cos(d*x+c)^3-7*cos(d*x+c)^2-7*cos(d* 
x+c)-21)+I*(-36*cos(d*x+c)^7-36*cos(d*x+c)^6+13*cos(d*x+c)^5+13*cos(d*x+c) 
^4)+21*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d 
*x+c)+2+sec(d*x+c))*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)-21*I*(cos(d*x+c 
)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+2+sec(d*x+c)) 
*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (219 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 302 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 124 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 50 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 9 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 336 i \, \sqrt {2} \sqrt {e} e^{\left (7 i \, d x + 7 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{936 \, a^{3} d e} \] Input:

integrate(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas 
")
 

Output:

1/936*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(219*I*e^(8*I*d*x + 8*I*c 
) + 302*I*e^(6*I*d*x + 6*I*c) + 124*I*e^(4*I*d*x + 4*I*c) + 50*I*e^(2*I*d* 
x + 2*I*c) + 9*I)*e^(1/2*I*d*x + 1/2*I*c) + 336*I*sqrt(2)*sqrt(e)*e^(7*I*d 
*x + 7*I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I 
*c))))*e^(-7*I*d*x - 7*I*c)/(a^3*d*e)
 

Sympy [F]

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

integrate(1/(e*sec(d*x+c))**(1/2)/(a+I*a*tan(d*x+c))**3,x)
 

Output:

I*Integral(1/(sqrt(e*sec(c + d*x))*tan(c + d*x)**3 - 3*I*sqrt(e*sec(c + d* 
x))*tan(c + d*x)**2 - 3*sqrt(e*sec(c + d*x))*tan(c + d*x) + I*sqrt(e*sec(c 
 + d*x))), x)/a**3
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

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

integrate(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate(1/(sqrt(e*sec(d*x + c))*(I*a*tan(d*x + c) + a)^3), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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