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

Optimal. Leaf size=116 \[ \frac{4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}+\frac{2 e \sin (c+d x)}{9 a^2 d (e \sec (c+d x))^{3/2}}+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0843109, antiderivative size = 116, 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 = {3500, 3769, 3771, 2639} \[ \frac{4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}+\frac{2 e \sin (c+d x)}{9 a^2 d (e \sec (c+d x))^{3/2}}+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3500

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 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}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^2} \, dx &=\frac{4 i e^2}{9 d (e \sec (c+d x))^{5/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (5 e^2\right ) \int \frac{1}{(e \sec (c+d x))^{5/2}} \, dx}{9 a^2}\\ &=\frac{2 e \sin (c+d x)}{9 a^2 d (e \sec (c+d x))^{3/2}}+\frac{4 i e^2}{9 d (e \sec (c+d x))^{5/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{3 a^2}\\ &=\frac{2 e \sin (c+d x)}{9 a^2 d (e \sec (c+d x))^{3/2}}+\frac{4 i e^2}{9 d (e \sec (c+d x))^{5/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\int \sqrt{\cos (c+d x)} \, dx}{3 a^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 e \sin (c+d x)}{9 a^2 d (e \sec (c+d x))^{3/2}}+\frac{4 i e^2}{9 d (e \sec (c+d x))^{5/2} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 1.26681, size = 123, normalized size = 1.06 \[ \frac{(\sin (2 (c+d x))+i \cos (2 (c+d x))) \left (2 (7 i \sin (2 (c+d x))+8 \cos (2 (c+d x))+2)-\frac{8 e^{4 i (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{18 a^2 d \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-8*E^((4*I)*(c + d*x))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))
] + 2*(2 + 8*Cos[2*(c + d*x)] + (7*I)*Sin[2*(c + d*x)]))*(I*Cos[2*(c + d*x)] + Sin[2*(c + d*x)]))/(18*a^2*d*Sq
rt[e*Sec[c + d*x]])

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Maple [B]  time = 0.386, size = 366, normalized size = 3.2 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{9\,{a}^{2}d \left ( \sin \left ( dx+c \right ) \right ) ^{5}e} \left ( 2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +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 ) -2\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}+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))^2,x)

[Out]

2/9/a^2/d*(2*I*cos(d*x+c)^5*sin(d*x+c)+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)-2*cos(d*x+c)^6+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*(co
s(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))^2,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 (7 i \, d x + 7 i \, c\right )} - 15 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 5 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 19 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 5 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 )} + 36 \,{\left (a^{2} d e e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e 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 (-i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 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 )}}{3 \,{\left (a^{2} d e e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, a^{2} d e e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{36 \,{\left (a^{2} d e e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e 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))^(1/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/36*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-9*I*e^(7*I*d*x + 7*I*c) - 15*I*e^(6*I*d*x + 6*I*c) - 5*I*e^(
5*I*d*x + 5*I*c) - 19*I*e^(4*I*d*x + 4*I*c) + 5*I*e^(3*I*d*x + 3*I*c) - 5*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) + 36*(a^2*d*e*e^(6*I*d*x + 6*I*c) - a^2*d*e*e^(5*I*d*x + 5*I*c))*integral(1
/3*sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-I*e^(2*I*d*x + 2*I*c) - 2*I*e^(I*d*x + I*c) - I)*e^(1/2*I*d*x +
1/2*I*c)/(a^2*d*e*e^(3*I*d*x + 3*I*c) - 2*a^2*d*e*e^(2*I*d*x + 2*I*c) + a^2*d*e*e^(I*d*x + I*c)), x))/(a^2*d*
e*e^(6*I*d*x + 6*I*c) - a^2*d*e*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))**(1/2)/(a+I*a*tan(d*x+c))**2,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 )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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