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

Optimal. Leaf size=132 \[ \frac{2 i e^2}{45 d \left (a^3+i a^3 \tan (c+d x)\right ) \sqrt{e \sec (c+d x)}}+\frac{2 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2}{9 a d (a+i a \tan (c+d x))^2 \sqrt{e \sec (c+d x)}} \]

[Out]

(2*e^2*EllipticE[(c + d*x)/2, 2])/(15*a^3*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((4*I)/9)*e^2)/(a*d*Sq
rt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^2) + (((2*I)/45)*e^2)/(d*Sqrt[e*Sec[c + d*x]]*(a^3 + I*a^3*Tan[c + d
*x]))

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Rubi [A]  time = 0.128976, antiderivative size = 132, 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, 3502, 3771, 2639} \[ \frac{2 i e^2}{45 d \left (a^3+i a^3 \tan (c+d x)\right ) \sqrt{e \sec (c+d x)}}+\frac{2 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2}{9 a d (a+i a \tan (c+d x))^2 \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^2*EllipticE[(c + d*x)/2, 2])/(15*a^3*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((4*I)/9)*e^2)/(a*d*Sq
rt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^2) + (((2*I)/45)*e^2)/(d*Sqrt[e*Sec[c + d*x]]*(a^3 + I*a^3*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 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{(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^3} \, dx &=\frac{4 i e^2}{9 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^2}+\frac{e^2 \int \frac{1}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))} \, dx}{9 a^2}\\ &=\frac{4 i e^2}{9 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^2}+\frac{2 i e^2}{45 d \sqrt{e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{e^2 \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{15 a^3}\\ &=\frac{4 i e^2}{9 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^2}+\frac{2 i e^2}{45 d \sqrt{e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{e^2 \int \sqrt{\cos (c+d x)} \, dx}{15 a^3 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{2 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2}{9 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^2}+\frac{2 i e^2}{45 d \sqrt{e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.743169, size = 140, normalized size = 1.06 \[ -\frac{e^{-i d x} \sec ^2(c+d x) (\cos (d x)+i \sin (d x)) (e \sec (c+d x))^{3/2} \left (6 e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+3 i \sin (2 (c+d x))+8 \cos (2 (c+d x))+8\right )}{45 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^2*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I*Sin[d*x])*(8 + 8*Cos[2*(c + d*x)] + 6*E^((2*I)*(c + d*x)
)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + (3*I)*Sin[2*(c + d*x
)]))/(45*a^3*d*E^(I*d*x)*(-I + Tan[c + d*x])^3)

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Maple [B]  time = 0.299, size = 368, normalized size = 2.8 \begin{align*} -{\frac{2\,\cos \left ( dx+c \right ) }{45\,d{a}^{3}\sin \left ( dx+c \right ) } \left ( -20\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +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 ) -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}}}+20\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}+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}}}-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}}}+9\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -19\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-3\,\cos \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45/a^3/d*(-20*I*cos(d*x+c)^5*sin(d*x+c)+3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elli
pticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)-3*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)+20*cos(d*x+c)^6+3*I*sin(d*x+c)*Elli
pticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3*I*sin(d*x+c)
*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+9*I*cos(d
*x+c)^3*sin(d*x+c)-19*cos(d*x+c)^4+2*cos(d*x+c)^2-3*cos(d*x+c))*(e/cos(d*x+c))^(3/2)*cos(d*x+c)/sin(d*x+c)

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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((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 (90 \, a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}{\rm integral}\left (-\frac{i \, \sqrt{2} e \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 )}}{15 \, a^{3} d}, x\right ) + \sqrt{2}{\left (12 i \, e e^{\left (6 i \, d x + 6 i \, c\right )} + 23 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e\right )} \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 )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{90 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/90*(90*a^3*d*e^(5*I*d*x + 5*I*c)*integral(-1/15*I*sqrt(2)*e*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x +
1/2*I*c)/(a^3*d), x) + sqrt(2)*(12*I*e*e^(6*I*d*x + 6*I*c) + 23*I*e*e^(4*I*d*x + 4*I*c) + 16*I*e*e^(2*I*d*x +
2*I*c) + 5*I*e)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-5*I*d*x - 5*I*c)/(a^3*d)

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

[Out]

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