3.254 \(\int \frac{1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=186 \[ \frac{12 i e^2}{55 d \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}+\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{11 a^3 d e^2}+\frac{6 e \sin (c+d x)}{55 a^3 d (e \sec (c+d x))^{5/2}}+\frac{2 \sin (c+d x)}{11 a^3 d e \sqrt{e \sec (c+d x)}}+\frac{2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}} \]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(11*a^3*d*e^2) + (6*e*Sin[c + d*x])/(55*
a^3*d*(e*Sec[c + d*x])^(5/2)) + (2*Sin[c + d*x])/(11*a^3*d*e*Sqrt[e*Sec[c + d*x]]) + ((2*I)/15)/(d*(e*Sec[c +
d*x])^(3/2)*(a + I*a*Tan[c + d*x])^3) + (((12*I)/55)*e^2)/(d*(e*Sec[c + d*x])^(7/2)*(a^3 + I*a^3*Tan[c + d*x])
)

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Rubi [A]  time = 0.18169, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3502, 3500, 3769, 3771, 2641} \[ \frac{12 i e^2}{55 d \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}+\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{11 a^3 d e^2}+\frac{6 e \sin (c+d x)}{55 a^3 d (e \sec (c+d x))^{5/2}}+\frac{2 \sin (c+d x)}{11 a^3 d e \sqrt{e \sec (c+d x)}}+\frac{2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(11*a^3*d*e^2) + (6*e*Sin[c + d*x])/(55*
a^3*d*(e*Sec[c + d*x])^(5/2)) + (2*Sin[c + d*x])/(11*a^3*d*e*Sqrt[e*Sec[c + d*x]]) + ((2*I)/15)/(d*(e*Sec[c +
d*x])^(3/2)*(a + I*a*Tan[c + d*x])^3) + (((12*I)/55)*e^2)/(d*(e*Sec[c + d*x])^(7/2)*(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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3} \, dx &=\frac{2 i}{15 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}+\frac{3 \int \frac{1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx}{5 a}\\ &=\frac{2 i}{15 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}+\frac{12 i e^2}{55 d (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{\left (21 e^2\right ) \int \frac{1}{(e \sec (c+d x))^{7/2}} \, dx}{55 a^3}\\ &=\frac{6 e \sin (c+d x)}{55 a^3 d (e \sec (c+d x))^{5/2}}+\frac{2 i}{15 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}+\frac{12 i e^2}{55 d (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{3 \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{11 a^3}\\ &=\frac{6 e \sin (c+d x)}{55 a^3 d (e \sec (c+d x))^{5/2}}+\frac{2 \sin (c+d x)}{11 a^3 d e \sqrt{e \sec (c+d x)}}+\frac{2 i}{15 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}+\frac{12 i e^2}{55 d (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{\int \sqrt{e \sec (c+d x)} \, dx}{11 a^3 e^2}\\ &=\frac{6 e \sin (c+d x)}{55 a^3 d (e \sec (c+d x))^{5/2}}+\frac{2 \sin (c+d x)}{11 a^3 d e \sqrt{e \sec (c+d x)}}+\frac{2 i}{15 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}+\frac{12 i e^2}{55 d (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{11 a^3 e^2}\\ &=\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{11 a^3 d e^2}+\frac{6 e \sin (c+d x)}{55 a^3 d (e \sec (c+d x))^{5/2}}+\frac{2 \sin (c+d x)}{11 a^3 d e \sqrt{e \sec (c+d x)}}+\frac{2 i}{15 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}+\frac{12 i e^2}{55 d (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.582232, size = 151, normalized size = 0.81 \[ \frac{\sec ^5(c+d x) \left (-114 i \sin (c+d x)-81 i \sin (3 (c+d x))+33 i \sin (5 (c+d x))-332 \cos (c+d x)-154 \cos (3 (c+d x))+22 \cos (5 (c+d x))+240 i \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))\right )}{1320 a^3 d (\tan (c+d x)-i)^3 (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

(Sec[c + d*x]^5*(-332*Cos[c + d*x] - 154*Cos[3*(c + d*x)] + 22*Cos[5*(c + d*x)] - (114*I)*Sin[c + d*x] + (240*
I)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)]) - (81*I)*Sin[3*(c + d*
x)] + (33*I)*Sin[5*(c + d*x)]))/(1320*a^3*d*(e*Sec[c + d*x])^(3/2)*(-I + Tan[c + d*x])^3)

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Maple [A]  time = 0.411, size = 261, normalized size = 1.4 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}\cos \left ( dx+c \right ) }{165\,d{a}^{3}{e}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}} \left ( 44\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}+44\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}-15\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+7\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +15\,i\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}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +9\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +15\,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 ) +15\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/165/a^3/d*(e/cos(d*x+c))^(3/2)*(cos(d*x+c)-1)^2*(cos(d*x+c)+1)^2*cos(d*x+c)*(44*I*cos(d*x+c)^8+44*sin(d*x+c)
*cos(d*x+c)^7-15*I*cos(d*x+c)^6+7*cos(d*x+c)^5*sin(d*x+c)+15*I*cos(d*x+c)*(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)+9*cos(d*x+c)^3*sin(d*x+c)+15*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)+15*cos(d*x+c)*sin(d*x+c))
/e^3/sin(d*x+c)^4

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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))^(3/2)/(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{{\left (2640 \, a^{3} d e^{2} e^{\left (8 i \, d x + 8 i \, c\right )}{\rm integral}\left (-\frac{i \, \sqrt{2} \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 )}}{11 \, a^{3} d e^{2}}, x\right ) + \sqrt{2} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-55 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 235 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 446 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 218 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 73 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 11 i\right )} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{2640 \, a^{3} d e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2640*(2640*a^3*d*e^2*e^(8*I*d*x + 8*I*c)*integral(-1/11*I*sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*
I*d*x - 1/2*I*c)/(a^3*d*e^2), x) + sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-55*I*e^(10*I*d*x + 10*I*c) + 23
5*I*e^(8*I*d*x + 8*I*c) + 446*I*e^(6*I*d*x + 6*I*c) + 218*I*e^(4*I*d*x + 4*I*c) + 73*I*e^(2*I*d*x + 2*I*c) + 1
1*I)*e^(1/2*I*d*x + 1/2*I*c))*e^(-8*I*d*x - 8*I*c)/(a^3*d*e^2)

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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))**(3/2)/(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}{\left (e \sec \left (d x + c\right )\right )^{\frac{3}{2}}{\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))^(3/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a)^3), x)