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]

14/117*e*sin(d*x+c)/a^3/d/(e*sec(d*x+c))^(3/2)+14/39*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE
(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)+2/13*I/d/(e*sec(d*x+c))^(1/2)/(a+I*a*
tan(d*x+c))^3+28/117*I*e^2/d/(e*sec(d*x+c))^(5/2)/(a^3+I*a^3*tan(d*x+c))

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Rubi [A]  time = 0.14, antiderivative size = 152, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [C]  time = 1.45, size = 145, normalized size = 0.95 \[ \frac {\sqrt {e \sec (c+d x)} (\sin (3 (c+d x))+i \cos (3 (c+d x))) \left (-56 e^{4 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+126 i \sin (2 (c+d x))+105 i \sin (4 (c+d x))+176 \cos (2 (c+d x))+114 \cos (4 (c+d x))+62\right )}{468 a^3 d e} \]

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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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-117 i \, e^{\left (9 i \, d x + 9 i \, c\right )} - 219 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 34 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 302 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 124 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 124 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 50 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 50 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 9 i \, e^{\left (i \, d x + i \, c\right )} - 9 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 936 \, {\left (a^{3} d e e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e e^{\left (7 i \, d x + 7 i \, c\right )}\right )} {\rm integral}\left (\frac {\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 14 i \, e^{\left (i \, d x + i \, c\right )} - 7 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{39 \, {\left (a^{3} d e e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, a^{3} d e e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d e e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{936 \, {\left (a^{3} d e e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e e^{\left (7 i \, d x + 7 i \, c\right )}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

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)

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maple [B]  time = 1.72, size = 395, normalized size = 2.60 \[ \frac {2 \left (36 i \left (\cos ^{7}\left (d x +c \right )\right ) \sin \left (d x +c \right )-36 \left (\cos ^{8}\left (d x +c \right )\right )-13 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+21 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-21 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+31 \left (\cos ^{6}\left (d x +c \right )\right )+21 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-21 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-2 \left (\cos ^{4}\left (d x +c \right )\right )-14 \left (\cos ^{2}\left (d x +c \right )\right )+21 \cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {e}{\cos \left (d x +c \right )}}}{117 a^{3} d \sin \left (d x +c \right )^{5} e} \]

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/(1+cos(d*x+c)))
^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-21*I*
EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c
)*sin(d*x+c)+31*cos(d*x+c)^6+21*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*Ellipt
icF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-21*I*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-2*cos(d*x+c)^4-14*cos(d*x+c)^2+21*cos(d*x+c))*(1+cos(d*x+c))^2*(-1+
cos(d*x+c))^2*(e/cos(d*x+c))^(1/2)/sin(d*x+c)^5/e

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e/cos(c + d*x))^(1/2)*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

int(1/((e/cos(c + d*x))^(1/2)*(a + a*tan(c + d*x)*1i)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \int \frac {1}{\sqrt {e \sec {\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )} - 3 i \sqrt {e \sec {\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )} - 3 \sqrt {e \sec {\left (c + d x \right )}} \tan {\left (c + d x \right )} + i \sqrt {e \sec {\left (c + d x \right )}}}\, dx}{a^{3}} \]

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]

I*Integral(1/(sqrt(e*sec(c + d*x))*tan(c + d*x)**3 - 3*I*sqrt(e*sec(c + d*x))*tan(c + d*x)**2 - 3*sqrt(e*sec(c
+ d*x))*tan(c + d*x) + I*sqrt(e*sec(c + d*x))), x)/a**3

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