3.235 \(\int \frac{(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=152 \[ \frac{14 e^5 \sin (c+d x) \sqrt{e \sec (c+d x)}}{5 a^2 d}+\frac{14 e^3 \sin (c+d x) (e \sec (c+d x))^{5/2}}{15 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{14 e^6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]

[Out]

(-14*e^6*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (14*e^5*Sqrt[e*Sec[c +
d*x]]*Sin[c + d*x])/(5*a^2*d) + (14*e^3*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(15*a^2*d) - (((4*I)/3)*e^2*(e*S
ec[c + d*x])^(7/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))

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Rubi [A]  time = 0.107401, antiderivative size = 152, 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, 3768, 3771, 2639} \[ \frac{14 e^5 \sin (c+d x) \sqrt{e \sec (c+d x)}}{5 a^2 d}+\frac{14 e^3 \sin (c+d x) (e \sec (c+d x))^{5/2}}{15 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{14 e^6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(11/2)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(-14*e^6*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (14*e^5*Sqrt[e*Sec[c +
d*x]]*Sin[c + d*x])/(5*a^2*d) + (14*e^3*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(15*a^2*d) - (((4*I)/3)*e^2*(e*S
ec[c + d*x])^(7/2))/(d*(a^2 + I*a^2*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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))^{11/2}}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (7 e^2\right ) \int (e \sec (c+d x))^{7/2} \, dx}{3 a^2}\\ &=\frac{14 e^3 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (7 e^4\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 a^2}\\ &=\frac{14 e^5 \sqrt{e \sec (c+d x)} \sin (c+d x)}{5 a^2 d}+\frac{14 e^3 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{\left (7 e^6\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{5 a^2}\\ &=\frac{14 e^5 \sqrt{e \sec (c+d x)} \sin (c+d x)}{5 a^2 d}+\frac{14 e^3 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{\left (7 e^6\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=-\frac{14 e^6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{14 e^5 \sqrt{e \sec (c+d x)} \sin (c+d x)}{5 a^2 d}+\frac{14 e^3 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.870055, size = 123, normalized size = 0.81 \[ \frac{2 i e^5 e^{i (c+d x)} \left (7 \left (1+e^{2 i (c+d x)}\right )^{5/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-56 e^{2 i (c+d x)}-21 e^{4 i (c+d x)}-47\right ) \sqrt{e \sec (c+d x)}}{15 a^2 d \left (1+e^{2 i (c+d x)}\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(11/2)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((2*I)/15)*e^5*E^(I*(c + d*x))*(-47 - 56*E^((2*I)*(c + d*x)) - 21*E^((4*I)*(c + d*x)) + 7*(1 + E^((2*I)*(c +
d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sqrt[e*Sec[c + d*x]])/(a^2*d*(1 + E^((2*I
)*(c + d*x)))^2)

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Maple [B]  time = 0.284, size = 374, normalized size = 2.5 \begin{align*} -{\frac{2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15\,{a}^{2}d \left ( \sin \left ( dx+c \right ) \right ) ^{5}} \left ( 21\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}\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 \left ( \cos \left ( dx+c \right ) \right ) ^{3}\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}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +21\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}\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 \left ( \cos \left ( dx+c \right ) \right ) ^{2}\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 ) \sin \left ( dx+c \right ) +21\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+10\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -24\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3 \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x)

[Out]

-2/15/a^2/d*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(21*I*cos(d*x+c)^3*(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*cos(d*x+c)^3*(1/(cos(d*x+c)+1))^(1/2)*(
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)+21*I*cos(d*x+c)^2*(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*c
os(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I
)*sin(d*x+c)+21*cos(d*x+c)^3+10*I*cos(d*x+c)*sin(d*x+c)-24*cos(d*x+c)^2+3)*(e/cos(d*x+c))^(11/2)*cos(d*x+c)^3/
sin(d*x+c)^5

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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))^(11/2)/(a+I*a*tan(d*x+c))^2,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}{\left (-42 i \, e^{5} e^{\left (5 i \, d x + 5 i \, c\right )} - 112 i \, e^{5} e^{\left (3 i \, d x + 3 i \, c\right )} - 94 i \, e^{5} e^{\left (i \, d x + i \, c\right )}\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 )} + 15 \,{\left (a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}{\rm integral}\left (\frac{7 i \, \sqrt{2} e^{5} \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 )}}{5 \, a^{2} d}, x\right )}{15 \,{\left (a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/15*(sqrt(2)*(-42*I*e^5*e^(5*I*d*x + 5*I*c) - 112*I*e^5*e^(3*I*d*x + 3*I*c) - 94*I*e^5*e^(I*d*x + I*c))*sqrt(
e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 15*(a^2*d*e^(4*I*d*x + 4*I*c) + 2*a^2*d*e^(2*I*d*x + 2*
I*c) + a^2*d)*integral(7/5*I*sqrt(2)*e^5*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(a^2*d), x)
)/(a^2*d*e^(4*I*d*x + 4*I*c) + 2*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*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))**(11/2)/(a+I*a*tan(d*x+c))**2,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{11}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(11/2)/(I*a*tan(d*x + c) + a)^2, x)