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

Optimal. Leaf size=70 \[ \frac{2 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i e^2}{a d \sqrt{e \sec (c+d x)}} \]

[Out]

((2*I)*e^2)/(a*d*Sqrt[e*Sec[c + d*x]]) + (2*e^2*EllipticE[(c + d*x)/2, 2])/(a*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[
c + d*x]])

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Rubi [A]  time = 0.0728845, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3501, 3771, 2639} \[ \frac{2 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i e^2}{a d \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*e^2)/(a*d*Sqrt[e*Sec[c + d*x]]) + (2*e^2*EllipticE[(c + d*x)/2, 2])/(a*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[
c + d*x]])

Rule 3501

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

Mathematica [C]  time = 0.383024, size = 74, normalized size = 1.06 \[ \frac{2 i e e^{-i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right ) \sqrt{e \sec (c+d x)}}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*e*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))]*Sqrt[e*Sec[c +
d*x]])/(a*d*E^(I*(c + d*x)))

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Maple [B]  time = 0.248, size = 916, normalized size = 13.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a/d*(e/cos(d*x+c))^(3/2)*(cos(d*x+c)+1)^3*(cos(d*x+c)-1)^2*(4*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*sin(
d*x+c)*cos(d*x+c)^2*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)-4*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)*cos(d*x+c)^2*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+
c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+8*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*sin(
d*x+c)*cos(d*x+c)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2)-8*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)*cos(d*x+c)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I
)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+4*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(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)*sin(d*x+c)-4*I*(1/
(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*EllipticE(I*(cos(
d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)+4*I*cos(d*x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+
c)^3*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+4*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*sin(d*x+c)+I*ln(
-(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*co
s(d*x+c)-1)/sin(d*x+c)^2)*cos(d*x+c)*sin(d*x+c)-I*ln(-2*(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-c
os(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*cos(d*x+c)*sin(d*x+c)+4*cos(d
*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)/sin(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((e*sec(d*x+c))^(3/2)/(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{{\left (a d e^{\left (i \, d x + i \, c\right )}{\rm integral}\left (-\frac{i \, \sqrt{2} e \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 )}}{a d}, x\right ) + \sqrt{2}{\left (2 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, e\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 )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*d*e^(I*d*x + I*c)*integral(-I*sqrt(2)*e*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(a*d), x)
+ sqrt(2)*(2*I*e*e^(2*I*d*x + 2*I*c) + 2*I*e)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-
I*d*x - I*c)/(a*d)

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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((e*sec(d*x+c))**(3/2)/(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{\left (e \sec \left (d x + c\right )\right )^{\frac{3}{2}}}{i \, a \tan \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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