\(\int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx\) [247]

Optimal result

Integrand size = 28, antiderivative size = 141 \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {14 e^6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 i e^4 (e \sec (c+d x))^{3/2}}{3 a^3 d}-\frac {14 e^5 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a^3 d}+\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{a d (a+i a \tan (c+d x))^2} \] Output:

14*e^6*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)+14/3*I*e^4*(e*sec(d*x+c))^(3/2)/a^3/d-14*e^5*(e*sec(d*x+c)) 
^(1/2)*sin(d*x+c)/a^3/d+4*I*e^2*(e*sec(d*x+c))^(7/2)/a/d/(a+I*a*tan(d*x+c) 
)^2
 

Mathematica [C] (verified)

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

Time = 1.80 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.66 \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {i e^4 (e \sec (c+d x))^{3/2} \left (35+33 \cos (2 (c+d x))-7 \left (1+e^{2 i (c+d x)}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+9 i \sin (2 (c+d x))\right )}{3 a^3 d} \] Input:

Integrate[(e*Sec[c + d*x])^(11/2)/(a + I*a*Tan[c + d*x])^3,x]
 

Output:

((I/3)*e^4*(e*Sec[c + d*x])^(3/2)*(35 + 33*Cos[2*(c + d*x)] - 7*(1 + E^((2 
*I)*(c + d*x)))^(3/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x) 
)] + (9*I)*Sin[2*(c + d*x)]))/(a^3*d)
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06, 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, 3982, 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.

\(\displaystyle \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{a d (a+i a \tan (c+d x))^2}-\frac {7 e^2 \int \frac {(e \sec (c+d x))^{7/2}}{i \tan (c+d x) a+a}dx}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{a d (a+i a \tan (c+d x))^2}-\frac {7 e^2 \int \frac {(e \sec (c+d x))^{7/2}}{i \tan (c+d x) a+a}dx}{a^2}\)

\(\Big \downarrow \) 3982

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

Int[(e*Sec[c + d*x])^(11/2)/(a + I*a*Tan[c + d*x])^3,x]
 

Output:

(-7*e^2*((((-2*I)/3)*e^2*(e*Sec[c + d*x])^(3/2))/(a*d) + (e^2*((-2*e^2*Ell 
ipticE[(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))/a))/a^2 + ((4*I)*e^2*(e*Sec[c + d* 
x])^(7/2))/(a*d*(a + I*a*Tan[c + d*x])^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 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 = 4.22 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.62

method	result	size
default	\(\frac {2 \sqrt {e \sec \left (d x +c \right )}\, e^{5} \left (\sin \left (d x +c \right ) \left (12 \cos \left (d x +c \right )-9\right )+i \left (12 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+1+\sec \left (d x +c \right )\right )+i \left (21 \cos \left (d x +c \right )^{2}+42 \cos \left (d x +c \right )+21\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i \left (-21 \cos \left (d x +c \right )^{2}-42 \cos \left (d x +c \right )-21\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )\right )}{3 a^{3} d \left (\cos \left (d x +c \right )+1\right )}\)	\(229\)

Input:

int((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

2/3/a^3/d*(e*sec(d*x+c))^(1/2)*e^5/(cos(d*x+c)+1)*(sin(d*x+c)*(12*cos(d*x+ 
c)-9)+I*(12*cos(d*x+c)^2+12*cos(d*x+c)+1+sec(d*x+c))+I*(21*cos(d*x+c)^2+42 
*cos(d*x+c)+21)*(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)+I*(-21*cos(d*x+c)^2-42*cos(d*x+c)- 
21)*(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))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06 \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (-21 i \, e^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 35 i \, e^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - 12 i \, e^{5}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 21 \, \sqrt {2} {\left (-i \, e^{5} e^{\left (3 i \, d x + 3 i \, c\right )} - i \, e^{5} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{3 \, {\left (a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}\right )}} \] Input:

integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas" 
)
                                                                                    
                                                                                    
 

Output:

-2/3*(sqrt(2)*(-21*I*e^5*e^(4*I*d*x + 4*I*c) - 35*I*e^5*e^(2*I*d*x + 2*I*c 
) - 12*I*e^5)*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^(3*I*d*x + 3*I*c) - I*e^5*e^(I*d*x + I*c))*sqrt(e)*we 
ierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))/(a^3*d* 
e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d*x + I*c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((e*sec(d*x+c))**(11/2)/(a+I*a*tan(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((e*sec(d*x+c))^(11/2)/(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]

\[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(11/2)/(I*a*tan(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{11/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \] Input:

int((e/cos(c + d*x))^(11/2)/(a + a*tan(c + d*x)*1i)^3,x)
 

Output:

int((e/cos(c + d*x))^(11/2)/(a + a*tan(c + d*x)*1i)^3, x)
 

Reduce [F]

\[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{5}}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right ) e^{5}}{a^{3}} \] Input:

int((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^3,x)
 

Output:

( - sqrt(e)*int((sqrt(sec(c + d*x))*sec(c + d*x)**5)/(tan(c + d*x)**3*i + 
3*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*e**5)/a**3