\(\int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx\) [246]

Optimal result

Integrand size = 28, antiderivative size = 141 \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {6 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{a^3 d}-\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2} \]

[Out]

-18/5*I*e^4*(e*sec(d*x+c))^(5/2)/a^3/d+6*e^5*(e*sec(d*x+c))^(3/2)*sin(d*x+c)/a^3/d+6*e^6*(cos(1/2*d*x+1/2*c)^2
)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*sec(d*x+c))^(1/2)/a^3/d-4
*I*e^2*(e*sec(d*x+c))^(9/2)/a/d/(a+I*a*tan(d*x+c))^2

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3581, 3582, 3853, 3856, 2720} \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {6 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{a^3 d}+\frac {6 e^5 \sin (c+d x) (e \sec (c+d x))^{3/2}}{a^3 d}-\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2} \]

[In]

Int[(e*Sec[c + d*x])^(13/2)/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(6*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(a^3*d) - (((18*I)/5)*e^4*(e*Sec[c +
d*x])^(5/2))/(a^3*d) + (6*e^5*(e*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(a^3*d) - ((4*I)*e^2*(e*Sec[c + d*x])^(9/2
))/(a*d*(a + I*a*Tan[c + d*x])^2)

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3581

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] - Dist[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] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3582

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] + Dist[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] &&  !ILtQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3853

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] + Dist[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 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2}+\frac {\left (9 e^2\right ) \int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx}{a^2} \\ & = -\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2}+\frac {\left (9 e^4\right ) \int (e \sec (c+d x))^{5/2} \, dx}{a^3} \\ & = -\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2}+\frac {\left (3 e^6\right ) \int \sqrt {e \sec (c+d x)} \, dx}{a^3} \\ & = -\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2}+\frac {\left (3 e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{a^3} \\ & = \frac {6 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{a^3 d}-\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.52 \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx=\frac {e^4 (e \sec (c+d x))^{5/2} \left (-18 i-20 i \cos (2 (c+d x))+30 \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-5 \sin (2 (c+d x))\right )}{5 a^3 d} \]

[In]

Integrate[(e*Sec[c + d*x])^(13/2)/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(e^4*(e*Sec[c + d*x])^(5/2)*(-18*I - (20*I)*Cos[2*(c + d*x)] + 30*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2]
- 5*Sin[2*(c + d*x)]))/(5*a^3*d)

Maple [A] (verified)

Time = 8.77 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {2 e^{6} \sqrt {e \sec \left (d x +c \right )}\, \left (15 i \cos \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+15 i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-20 i-5 \tan \left (d x +c \right )+i \left (\sec ^{2}\left (d x +c \right )\right )\right )}{5 a^{3} d}\)	\(160\)

[In]

int((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

2/5*e^6/a^3/d*(e*sec(d*x+c))^(1/2)*(15*I*cos(d*x+c)*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)),I)*(cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)+15*I*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)-20*I-5*tan(d*x+c)+I*sec(d*x+c)^2)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.12 \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (15 i \, e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} + 36 i \, e^{6} e^{\left (2 i \, d x + 2 i \, c\right )} + 25 i \, e^{6}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 15 \, \sqrt {2} {\left (i \, e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, e^{6} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{6}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{5 \, {\left (a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

[In]

integrate((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-2/5*(sqrt(2)*(15*I*e^6*e^(4*I*d*x + 4*I*c) + 36*I*e^6*e^(2*I*d*x + 2*I*c) + 25*I*e^6)*sqrt(e/(e^(2*I*d*x + 2*
I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 15*sqrt(2)*(I*e^6*e^(4*I*d*x + 4*I*c) + 2*I*e^6*e^(2*I*d*x + 2*I*c) + I*e
^6)*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(a^3*d*e^(4*I*d*x + 4*I*c) + 2*a^3*d*e^(2*I*d*x + 2*I
*c) + a^3*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((e*sec(d*x+c))**(13/2)/(a+I*a*tan(d*x+c))**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {13}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(13/2)/(I*a*tan(d*x + c) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{13/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]

[In]

int((e/cos(c + d*x))^(13/2)/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int((e/cos(c + d*x))^(13/2)/(a + a*tan(c + d*x)*1i)^3, x)