3.253 \(\int \frac{1}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=152 \[ \frac{28 i e^2}{117 d \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}+\frac{14 e \sin (c+d x)}{117 a^3 d (e \sec (c+d x))^{3/2}}+\frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{39 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i}{13 d (a+i a \tan (c+d x))^3 \sqrt{e \sec (c+d x)}} \]

[Out]

(14*EllipticE[(c + d*x)/2, 2])/(39*a^3*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (14*e*Sin[c + d*x])/(117*a
^3*d*(e*Sec[c + d*x])^(3/2)) + ((2*I)/13)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3) + (((28*I)/117)*e^
2)/(d*(e*Sec[c + d*x])^(5/2)*(a^3 + I*a^3*Tan[c + d*x]))

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Rubi [A]  time = 0.144562, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3502, 3500, 3769, 3771, 2639} \[ \frac{28 i e^2}{117 d \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}+\frac{14 e \sin (c+d x)}{117 a^3 d (e \sec (c+d x))^{3/2}}+\frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{39 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i}{13 d (a+i a \tan (c+d x))^3 \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3),x]

[Out]

(14*EllipticE[(c + d*x)/2, 2])/(39*a^3*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (14*e*Sin[c + d*x])/(117*a
^3*d*(e*Sec[c + d*x])^(3/2)) + ((2*I)/13)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3) + (((28*I)/117)*e^
2)/(d*(e*Sec[c + d*x])^(5/2)*(a^3 + I*a^3*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 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))^3} \, dx &=\frac{2 i}{13 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac{7 \int \frac{1}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^2} \, dx}{13 a}\\ &=\frac{2 i}{13 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac{28 i e^2}{117 d (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{\left (35 e^2\right ) \int \frac{1}{(e \sec (c+d x))^{5/2}} \, dx}{117 a^3}\\ &=\frac{14 e \sin (c+d x)}{117 a^3 d (e \sec (c+d x))^{3/2}}+\frac{2 i}{13 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac{28 i e^2}{117 d (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{7 \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{39 a^3}\\ &=\frac{14 e \sin (c+d x)}{117 a^3 d (e \sec (c+d x))^{3/2}}+\frac{2 i}{13 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac{28 i e^2}{117 d (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{7 \int \sqrt{\cos (c+d x)} \, dx}{39 a^3 \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 )}{39 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{14 e \sin (c+d x)}{117 a^3 d (e \sec (c+d x))^{3/2}}+\frac{2 i}{13 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac{28 i e^2}{117 d (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 1.3221, size = 145, normalized size = 0.95 \[ \frac{\sqrt{e \sec (c+d x)} (\sin (3 (c+d x))+i \cos (3 (c+d x))) \left (-56 e^{4 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+126 i \sin (2 (c+d x))+105 i \sin (4 (c+d x))+176 \cos (2 (c+d x))+114 \cos (4 (c+d x))+62\right )}{468 a^3 d e} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3),x]

[Out]

(Sqrt[e*Sec[c + d*x]]*(I*Cos[3*(c + d*x)] + Sin[3*(c + d*x)])*(62 + 176*Cos[2*(c + d*x)] + 114*Cos[4*(c + d*x)
] - 56*E^((4*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))
] + (126*I)*Sin[2*(c + d*x)] + (105*I)*Sin[4*(c + d*x)]))/(468*a^3*d*e)

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Maple [B]  time = 0.394, size = 395, normalized size = 2.6 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{117\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}e} \left ( 36\,i \left ( \cos \left ( dx+c \right ) \right ) ^{7}\sin \left ( dx+c \right ) -36\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}-13\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}\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 ) \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 ) +31\, \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 ) \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 ) \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))^3,x)

[Out]

2/117/a^3/d*(36*I*cos(d*x+c)^7*sin(d*x+c)-36*cos(d*x+c)^8-13*I*cos(d*x+c)^5*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)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)-21*I*E
llipticE(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)+31*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)*sin(d*x+c)-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))*(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))^3,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 (-117 i \, e^{\left (9 i \, d x + 9 i \, c\right )} - 219 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 34 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 302 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 124 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 124 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 50 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 50 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 9 i \, e^{\left (i \, d x + i \, c\right )} - 9 i\right )} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )} + 936 \,{\left (a^{3} d e e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e e^{\left (7 i \, d x + 7 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 )}}{39 \,{\left (a^{3} d e e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, a^{3} d e e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d e e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{936 \,{\left (a^{3} d e e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e e^{\left (7 i \, d x + 7 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))^3,x, algorithm="fricas")

[Out]

1/936*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-117*I*e^(9*I*d*x + 9*I*c) - 219*I*e^(8*I*d*x + 8*I*c) - 34*
I*e^(7*I*d*x + 7*I*c) - 302*I*e^(6*I*d*x + 6*I*c) + 124*I*e^(5*I*d*x + 5*I*c) - 124*I*e^(4*I*d*x + 4*I*c) + 50
*I*e^(3*I*d*x + 3*I*c) - 50*I*e^(2*I*d*x + 2*I*c) + 9*I*e^(I*d*x + I*c) - 9*I)*e^(1/2*I*d*x + 1/2*I*c) + 936*(
a^3*d*e*e^(8*I*d*x + 8*I*c) - a^3*d*e*e^(7*I*d*x + 7*I*c))*integral(1/39*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^3*d*e*e^(3*I*d*x + 3*I
*c) - 2*a^3*d*e*e^(2*I*d*x + 2*I*c) + a^3*d*e*e^(I*d*x + I*c)), x))/(a^3*d*e*e^(8*I*d*x + 8*I*c) - a^3*d*e*e^(
7*I*d*x + 7*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))**3,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 )}^{3}}\,{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))^3,x, algorithm="giac")

[Out]

integrate(1/(sqrt(e*sec(d*x + c))*(I*a*tan(d*x + c) + a)^3), x)