3.231 \(\int \frac{1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx\)

Optimal. Leaf size=114 \[ \frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 a d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac{2 i}{9 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}} \]

[Out]

(14*EllipticE[(c + d*x)/2, 2])/(15*a*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (14*Sin[c + d*x])/(45*a*
d*e*(e*Sec[c + d*x])^(3/2)) + ((2*I)/9)/(d*(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.0919698, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3502, 3769, 3771, 2639} \[ \frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 a d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac{2 i}{9 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])),x]

[Out]

(14*EllipticE[(c + d*x)/2, 2])/(15*a*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (14*Sin[c + d*x])/(45*a*
d*e*(e*Sec[c + d*x])^(3/2)) + ((2*I)/9)/(d*(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x]))

Rule 3502

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] + Dist[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 3769

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
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

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]

Rule 2639

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

Rubi steps

\begin{align*} \int \frac{1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx &=\frac{2 i}{9 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))}+\frac{7 \int \frac{1}{(e \sec (c+d x))^{5/2}} \, dx}{9 a}\\ &=\frac{14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac{2 i}{9 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))}+\frac{7 \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{15 a e^2}\\ &=\frac{14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac{2 i}{9 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))}+\frac{7 \int \sqrt{\cos (c+d x)} \, dx}{15 a e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 a d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac{2 i}{9 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.982436, size = 134, normalized size = 1.18 \[ \frac{(\tan (c+d x)+i) \left (-56 e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+70 i \sin (2 (c+d x))-7 i \sin (4 (c+d x))+104 \cos (2 (c+d x))-2 \cos (4 (c+d x))+106\right )}{180 a d e^2 \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])),x]

[Out]

((106 + 104*Cos[2*(c + d*x)] - 2*Cos[4*(c + d*x)] - 56*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hyper
geometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + (70*I)*Sin[2*(c + d*x)] - (7*I)*Sin[4*(c + d*x)])*(I + Tan
[c + d*x]))/(180*a*d*e^2*Sqrt[e*Sec[c + d*x]])

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Maple [B]  time = 0.408, size = 376, normalized size = 3.3 \begin{align*} -{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{45\,ad{e}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{5}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}} \left ( -5\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +5\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}-21\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +21\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -21\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) +21\,i{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+2\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+14\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-21\,\cos \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c)),x)

[Out]

-2/45/a/d*cos(d*x+c)^2*(e/cos(d*x+c))^(5/2)*(cos(d*x+c)-1)^2*(cos(d*x+c)+1)^2*(-5*I*cos(d*x+c)^5*sin(d*x+c)+5*
cos(d*x+c)^6-21*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*
x+c),I)*cos(d*x+c)*sin(d*x+c)+21*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos
(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)-21*I*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*(1/(co
s(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+21*I*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*
(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*cos(d*x+c)^4+14*cos(d*x+c)^2-21*cos(d*x+c))/e^5/s
in(d*x+c)^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-9 i \, e^{\left (9 i \, d x + 9 i \, c\right )} + 9 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 162 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 174 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 124 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 212 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 34 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 34 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{\left (i \, d x + i \, c\right )} - 5 i\right )} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )} + 360 \,{\left (a d e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - a d e^{3} e^{\left (5 i \, d x + 5 i \, c\right )}\right )}{\rm integral}\left (\frac{\sqrt{2} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 14 i \, e^{\left (i \, d x + i \, c\right )} - 7 i\right )} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}}{15 \,{\left (a d e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{360 \,{\left (a d e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - a d e^{3} e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/360*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-9*I*e^(9*I*d*x + 9*I*c) + 9*I*e^(8*I*d*x + 8*I*c) - 162*I*e
^(7*I*d*x + 7*I*c) - 174*I*e^(6*I*d*x + 6*I*c) - 124*I*e^(5*I*d*x + 5*I*c) - 212*I*e^(4*I*d*x + 4*I*c) + 34*I*
e^(3*I*d*x + 3*I*c) - 34*I*e^(2*I*d*x + 2*I*c) + 5*I*e^(I*d*x + I*c) - 5*I)*e^(1/2*I*d*x + 1/2*I*c) + 360*(a*d
*e^3*e^(6*I*d*x + 6*I*c) - a*d*e^3*e^(5*I*d*x + 5*I*c))*integral(1/15*sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1)
)*(-7*I*e^(2*I*d*x + 2*I*c) - 14*I*e^(I*d*x + I*c) - 7*I)*e^(1/2*I*d*x + 1/2*I*c)/(a*d*e^3*e^(3*I*d*x + 3*I*c)
- 2*a*d*e^3*e^(2*I*d*x + 2*I*c) + a*d*e^3*e^(I*d*x + I*c)), x))/(a*d*e^3*e^(6*I*d*x + 6*I*c) - a*d*e^3*e^(5*I
*d*x + 5*I*c))

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))**(5/2)/(a+I*a*tan(d*x+c)),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(5/2)*(I*a*tan(d*x + c) + a)), x)