3.261 \(\int \frac{(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=163 \[ \frac{2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac{4 i e^4}{117 d \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}+\frac{2 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{39 a^4 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt{e \sec (c+d x)}} \]

[Out]

(2*e^2*EllipticE[(c + d*x)/2, 2])/(39*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*e^3*Sin[c + d*x])/(1
17*a^4*d*(e*Sec[c + d*x])^(3/2)) + (((4*I)/13)*e^2)/(a*d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3) + (((4
*I)/117)*e^4)/(d*(e*Sec[c + d*x])^(5/2)*(a^4 + I*a^4*Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.146264, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3500, 3769, 3771, 2639} \[ \frac{2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac{4 i e^4}{117 d \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}+\frac{2 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{39 a^4 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2}{13 a d (a+i a \tan (c+d x))^3 \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(2*e^2*EllipticE[(c + d*x)/2, 2])/(39*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*e^3*Sin[c + d*x])/(1
17*a^4*d*(e*Sec[c + d*x])^(3/2)) + (((4*I)/13)*e^2)/(a*d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3) + (((4
*I)/117)*e^4)/(d*(e*Sec[c + d*x])^(5/2)*(a^4 + I*a^4*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{(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^4} \, dx &=\frac{4 i e^2}{13 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac{e^2 \int \frac{1}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^2} \, dx}{13 a^2}\\ &=\frac{4 i e^2}{13 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac{4 i e^4}{117 d (e \sec (c+d x))^{5/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\left (5 e^4\right ) \int \frac{1}{(e \sec (c+d x))^{5/2}} \, dx}{117 a^4}\\ &=\frac{2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac{4 i e^2}{13 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac{4 i e^4}{117 d (e \sec (c+d x))^{5/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{e^2 \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{39 a^4}\\ &=\frac{2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac{4 i e^2}{13 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac{4 i e^4}{117 d (e \sec (c+d x))^{5/2} \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{e^2 \int \sqrt{\cos (c+d x)} \, dx}{39 a^4 \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 )}{39 a^4 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 e^3 \sin (c+d x)}{117 a^4 d (e \sec (c+d x))^{3/2}}+\frac{4 i e^2}{13 a d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}+\frac{4 i e^4}{117 d (e \sec (c+d x))^{5/2} \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 1.4279, size = 142, normalized size = 0.87 \[ \frac{i e^{-i d x} \sec ^2(c+d x) (\cos (d x)+i \sin (d x)) (e \sec (c+d x))^{3/2} \left (\frac{24 e^{4 i (c+d x)} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}+22 i \sin (2 (c+d x))+40 \cos (2 (c+d x))+28\right )}{234 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I/234)*Sec[c + d*x]^2*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I*Sin[d*x])*(28 + 40*Cos[2*(c + d*x)] + (24*E^((4*I
)*(c + d*x))*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + (22*I)*S
in[2*(c + d*x)]))/(a^4*d*E^(I*d*x)*(-I + Tan[c + d*x])^4)

________________________________________________________________________________________

Maple [B]  time = 0.381, size = 378, normalized size = 2.3 \begin{align*}{\frac{2\,\cos \left ( dx+c \right ) }{117\,{a}^{4}d\sin \left ( dx+c \right ) } \left ( 72\,i \left ( \cos \left ( dx+c \right ) \right ) ^{7}\sin \left ( dx+c \right ) -72\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}-52\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}-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}}}+88\, \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}}}-17\, \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))^4,x)

[Out]

2/117/a^4/d*(72*I*cos(d*x+c)^7*sin(d*x+c)-72*cos(d*x+c)^8-52*I*cos(d*x+c)^5*sin(d*x+c)-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)+3*I*Ell
ipticF(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)+88*cos(d*x+c)^6-3*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)+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)-17*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)

________________________________________________________________________________________

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))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (468 \, a^{4} d e^{\left (7 i \, d x + 7 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 )}}{39 \, a^{4} d}, x\right ) + \sqrt{2}{\left (24 i \, e e^{\left (8 i \, d x + 8 i \, c\right )} + 55 i \, e e^{\left (6 i \, d x + 6 i \, c\right )} + 59 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 37 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 9 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 (-7 i \, d x - 7 i \, c\right )}}{468 \, a^{4} 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))^4,x, algorithm="fricas")

[Out]

1/468*(468*a^4*d*e^(7*I*d*x + 7*I*c)*integral(-1/39*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^4*d), x) + sqrt(2)*(24*I*e*e^(8*I*d*x + 8*I*c) + 55*I*e*e^(6*I*d*x + 6*I*c) + 59*I*e*e^(4*I*d*x
+ 4*I*c) + 37*I*e*e^(2*I*d*x + 2*I*c) + 9*I*e)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(
-7*I*d*x - 7*I*c)/(a^4*d)

________________________________________________________________________________________

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))**4,x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

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 )}^{4}}\,{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))^4,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(3/2)/(I*a*tan(d*x + c) + a)^4, x)