\(\int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx\) [224]

Optimal result

Integrand size = 28, antiderivative size = 105 \[ \int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx=\frac {2 e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 a d}-\frac {2 i e^2 (e \sec (c+d x))^{5/2}}{5 a d}+\frac {2 e^3 (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 a d} \]

[Out]

-2/5*I*e^2*(e*sec(d*x+c))^(5/2)/a/d+2/3*e^3*(e*sec(d*x+c))^(3/2)*sin(d*x+c)/a/d+2/3*e^4*(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/d

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3582, 3853, 3856, 2720} \[ \int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx=\frac {2 e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 a d}+\frac {2 e^3 \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 a d}-\frac {2 i e^2 (e \sec (c+d x))^{5/2}}{5 a d} \]

[In]

Int[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x]),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.59 \[ \int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx=\frac {e^2 (e \sec (c+d x))^{5/2} \left (-6 i+10 \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+5 \sin (2 (c+d x))\right )}{15 a d} \]

[In]

Integrate[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x]),x]

[Out]

(e^2*(e*Sec[c + d*x])^(5/2)*(-6*I + 10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d*x)]))/(15
*a*d)

Maple [A] (verified)

Time = 7.63 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {2 e^{4} \sqrt {e \sec \left (d x +c \right )}\, \left (5 i \cos \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+5 i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+5 \tan \left (d x +c \right )-3 i \left (\sec ^{2}\left (d x +c \right )\right )\right )}{15 a d}\)	\(158\)

[In]

int((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/15*e^4/a/d*(e*sec(d*x+c))^(1/2)*(5*I*cos(d*x+c)*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(
1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+5*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-csc(d*x+c)+cot(d*x+
c)),I)*(1/(cos(d*x+c)+1))^(1/2)+5*tan(d*x+c)-3*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 = 0.07 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.45 \[ \int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (5 i \, e^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 12 i \, e^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, e^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 5 \, \sqrt {2} {\left (i \, e^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, e^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{15 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

[In]

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/15*(sqrt(2)*(5*I*e^4*e^(4*I*d*x + 4*I*c) + 12*I*e^4*e^(2*I*d*x + 2*I*c) - 5*I*e^4)*sqrt(e/(e^(2*I*d*x + 2*I
*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 5*sqrt(2)*(I*e^4*e^(4*I*d*x + 4*I*c) + 2*I*e^4*e^(2*I*d*x + 2*I*c) + I*e^4
)*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(a*d*e^(4*I*d*x + 4*I*c) + 2*a*d*e^(2*I*d*x + 2*I*c) +
a*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((e*sec(d*x+c))**(9/2)/(a+I*a*tan(d*x+c)),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {9}{2}}}{i \, a \tan \left (d x + c\right ) + a} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(9/2)/(I*a*tan(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{9/2}}{a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

[In]

int((e/cos(c + d*x))^(9/2)/(a + a*tan(c + d*x)*1i),x)

[Out]

int((e/cos(c + d*x))^(9/2)/(a + a*tan(c + d*x)*1i), x)