3.229 \(\int \frac{1}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))} \, dx\)

Optimal. Leaf size=80 \[ \frac{6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i}{5 d (a+i a \tan (c+d x)) \sqrt{e \sec (c+d x)}} \]

[Out]

(6*EllipticE[(c + d*x)/2, 2])/(5*a*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + ((2*I)/5)/(d*Sqrt[e*Sec[c + d*
x]]*(a + I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.0696769, antiderivative size = 80, 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 = {3502, 3771, 2639} \[ \frac{6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i}{5 d (a+i a \tan (c+d x)) \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*EllipticE[(c + d*x)/2, 2])/(5*a*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + ((2*I)/5)/(d*Sqrt[e*Sec[c + d*
x]]*(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 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}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))} \, dx &=\frac{2 i}{5 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))}+\frac{3 \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{5 a}\\ &=\frac{2 i}{5 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))}+\frac{3 \int \sqrt{\cos (c+d x)} \, dx}{5 a \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i}{5 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.799048, size = 109, normalized size = 1.36 \[ \frac{(\tan (c+d x)+i) \left (-2 e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+3 i \sin (2 (c+d x))+4 \cos (2 (c+d x))+4\right )}{5 a d \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4 + 4*Cos[2*(c + d*x)] - 2*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4
, -E^((2*I)*(c + d*x))] + (3*I)*Sin[2*(c + d*x)])*(I + Tan[c + d*x]))/(5*a*d*Sqrt[e*Sec[c + d*x]])

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Maple [B]  time = 0.444, size = 358, normalized size = 4.5 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{5\,ad \left ( \sin \left ( dx+c \right ) \right ) ^{5}e} \left ( 3\,i{\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 ) \cos \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}}}-3\,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 ) +i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3\,i{\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 ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-3\,i\sin \left ( dx+c \right ){\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}- \left ( \cos \left ( dx+c \right ) \right ) ^{4}-2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,\cos \left ( dx+c \right ) \right ) \sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/a/d*(3*I*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+
c)/(cos(d*x+c)+1))^(1/2)-3*I*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)*(1/(cos(d*x+c)+1))
^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*sin(d*x+c)*cos(d*x+c)^3+3*I*EllipticF(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)-3*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)-cos(d*x+c)^4-2*cos(d*x+c)^2+3*cos
(d*x+c))*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(e/cos(d*x+c))^(1/2)/sin(d*x+c)^5/e

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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))^(1/2)/(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 (-5 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 7 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (i \, d x + i \, c\right )} - i\right )} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )} + 10 \,{\left (a d e e^{\left (4 i \, d x + 4 i \, c\right )} - a d e e^{\left (3 i \, d x + 3 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 (-3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, e^{\left (i \, d x + i \, c\right )} - 3 i\right )} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}}{5 \,{\left (a d e e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, a d e e^{\left (2 i \, d x + 2 i \, c\right )} + a d e e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{10 \,{\left (a d e e^{\left (4 i \, d x + 4 i \, c\right )} - a d e e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-5*I*e^(5*I*d*x + 5*I*c) - 7*I*e^(4*I*d*x + 4*I*c) - 4*I*e^(3
*I*d*x + 3*I*c) - 8*I*e^(2*I*d*x + 2*I*c) + I*e^(I*d*x + I*c) - I)*e^(1/2*I*d*x + 1/2*I*c) + 10*(a*d*e*e^(4*I*
d*x + 4*I*c) - a*d*e*e^(3*I*d*x + 3*I*c))*integral(1/5*sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-3*I*e^(2*I*
d*x + 2*I*c) - 6*I*e^(I*d*x + I*c) - 3*I)*e^(1/2*I*d*x + 1/2*I*c)/(a*d*e*e^(3*I*d*x + 3*I*c) - 2*a*d*e*e^(2*I*
d*x + 2*I*c) + a*d*e*e^(I*d*x + I*c)), x))/(a*d*e*e^(4*I*d*x + 4*I*c) - a*d*e*e^(3*I*d*x + 3*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))**(1/2)/(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}{\sqrt{e \sec \left (d x + c\right )}{\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))^(1/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

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