3.233 \(\int \frac{(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=183 \[ \frac{22 e^7 \sin (c+d x) \sqrt{e \sec (c+d x)}}{15 a^2 d}+\frac{22 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{45 a^2 d}+\frac{22 e^3 \sin (c+d x) (e \sec (c+d x))^{9/2}}{63 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{22 e^8 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]

[Out]

(-22*e^8*EllipticE[(c + d*x)/2, 2])/(15*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (22*e^7*Sqrt[e*Sec[c
+ d*x]]*Sin[c + d*x])/(15*a^2*d) + (22*e^5*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(45*a^2*d) + (22*e^3*(e*Sec[c
+ d*x])^(9/2)*Sin[c + d*x])/(63*a^2*d) - (((4*I)/7)*e^2*(e*Sec[c + d*x])^(11/2))/(d*(a^2 + I*a^2*Tan[c + d*x])
)

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Rubi [A]  time = 0.127274, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3500, 3768, 3771, 2639} \[ \frac{22 e^7 \sin (c+d x) \sqrt{e \sec (c+d x)}}{15 a^2 d}+\frac{22 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{45 a^2 d}+\frac{22 e^3 \sin (c+d x) (e \sec (c+d x))^{9/2}}{63 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{22 e^8 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(15/2)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(-22*e^8*EllipticE[(c + d*x)/2, 2])/(15*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (22*e^7*Sqrt[e*Sec[c
+ d*x]]*Sin[c + d*x])/(15*a^2*d) + (22*e^5*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(45*a^2*d) + (22*e^3*(e*Sec[c
+ d*x])^(9/2)*Sin[c + d*x])/(63*a^2*d) - (((4*I)/7)*e^2*(e*Sec[c + d*x])^(11/2))/(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))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (11 e^2\right ) \int (e \sec (c+d x))^{11/2} \, dx}{7 a^2}\\ &=\frac{22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (11 e^4\right ) \int (e \sec (c+d x))^{7/2} \, dx}{9 a^2}\\ &=\frac{22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac{22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (11 e^6\right ) \int (e \sec (c+d x))^{3/2} \, dx}{15 a^2}\\ &=\frac{22 e^7 \sqrt{e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac{22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac{22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{\left (11 e^8\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{15 a^2}\\ &=\frac{22 e^7 \sqrt{e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac{22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac{22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{\left (11 e^8\right ) \int \sqrt{\cos (c+d x)} \, dx}{15 a^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=-\frac{22 e^8 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{22 e^7 \sqrt{e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac{22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac{22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 2.28387, size = 302, normalized size = 1.65 \[ \frac{(\cos (d x)+i \sin (d x))^2 (e \sec (c+d x))^{15/2} \left (\frac{22 i \sqrt{2} e^{3 i c-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )}{-1+e^{2 i c}}+\frac{1}{56} \csc (c) (\cos (2 c)+i \sin (2 c)) \sec ^{\frac{9}{2}}(c+d x) (-720 i \sin (2 c+d x)+1050 \cos (2 c+d x)+1078 \cos (2 c+3 d x)+77 \cos (4 c+3 d x)+231 \cos (4 c+5 d x)+720 i \sin (d x)+1260 \cos (d x))\right )}{45 d \sec ^{\frac{11}{2}}(c+d x) (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(15/2)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

((e*Sec[c + d*x])^(15/2)*(Cos[d*x] + I*Sin[d*x])^2*(((22*I)*Sqrt[2]*E^((3*I)*c - I*d*x)*Sqrt[E^(I*(c + d*x))/(
1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1
+ E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(-1 + E^((2*I)*c)) + (Csc[c]*Sec[c + d
*x]^(9/2)*(Cos[2*c] + I*Sin[2*c])*(1260*Cos[d*x] + 1050*Cos[2*c + d*x] + 1078*Cos[2*c + 3*d*x] + 77*Cos[4*c +
3*d*x] + 231*Cos[4*c + 5*d*x] + (720*I)*Sin[d*x] - (720*I)*Sin[2*c + d*x]))/56))/(45*d*Sec[c + d*x]^(11/2)*(a
+ I*a*Tan[c + d*x])^2)

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Maple [B]  time = 0.312, size = 384, normalized size = 2.1 \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}}{315\,{a}^{2}d \left ( \sin \left ( dx+c \right ) \right ) ^{5}} \left ( 231\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}\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 ) -231\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}\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 ) +231\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}\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 ) -231\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}\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 ) -231\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}+154\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-90\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +112\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-35 \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{15}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^2,x)

[Out]

2/315/a^2/d*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(231*I*cos(d*x+c)^5*(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)-231*I*cos(d*x+c)^5*(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)+231*I*sin(d*x+c)*cos(d*
x+c)^4*(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)-231
*I*sin(d*x+c)*cos(d*x+c)^4*(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)-231*cos(d*x+c)^5+154*cos(d*x+c)^4-90*I*cos(d*x+c)*sin(d*x+c)+112*cos(d*x+c)^2-35)*(e/cos(d*x+
c))^(15/2)*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))^(15/2)/(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 (-462 i \, e^{7} e^{\left (9 i \, d x + 9 i \, c\right )} - 2156 i \, e^{7} e^{\left (7 i \, d x + 7 i \, c\right )} - 3960 i \, e^{7} e^{\left (5 i \, d x + 5 i \, c\right )} - 3540 i \, e^{7} e^{\left (3 i \, d x + 3 i \, c\right )} - 154 i \, e^{7} 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 )} + 315 \,{\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}{\rm integral}\left (\frac{11 i \, \sqrt{2} e^{7} \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^{2} d}, x\right )}{315 \,{\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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))^(15/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/315*(sqrt(2)*(-462*I*e^7*e^(9*I*d*x + 9*I*c) - 2156*I*e^7*e^(7*I*d*x + 7*I*c) - 3960*I*e^7*e^(5*I*d*x + 5*I*
c) - 3540*I*e^7*e^(3*I*d*x + 3*I*c) - 154*I*e^7*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) + 315*(a^2*d*e^(8*I*d*x + 8*I*c) + 4*a^2*d*e^(6*I*d*x + 6*I*c) + 6*a^2*d*e^(4*I*d*x + 4*I*c) + 4*
a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*integral(11/15*I*sqrt(2)*e^7*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^(8*I*d*x + 8*I*c) + 4*a^2*d*e^(6*I*d*x + 6*I*c) + 6*a^2*d*e^(4*I*d*x + 4*I*c
) + 4*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))**(15/2)/(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{15}{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))^(15/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(15/2)/(I*a*tan(d*x + c) + a)^2, x)