3.234 \(\int \frac{(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=152 \[ \frac{6 e^5 \sin (c+d x) (e \sec (c+d x))^{3/2}}{7 a^2 d}+\frac{18 e^3 \sin (c+d x) (e \sec (c+d x))^{7/2}}{35 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{6 e^6 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 a^2 d} \]

[Out]

(6*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(7*a^2*d) + (6*e^5*(e*Sec[c + d*x])^
(3/2)*Sin[c + d*x])/(7*a^2*d) + (18*e^3*(e*Sec[c + d*x])^(7/2)*Sin[c + d*x])/(35*a^2*d) - (((4*I)/5)*e^2*(e*Se
c[c + d*x])^(9/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))

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Rubi [A]  time = 0.108869, 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, 2641} \[ \frac{6 e^5 \sin (c+d x) (e \sec (c+d x))^{3/2}}{7 a^2 d}+\frac{18 e^3 \sin (c+d x) (e \sec (c+d x))^{7/2}}{35 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{6 e^6 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(13/2)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(6*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(7*a^2*d) + (6*e^5*(e*Sec[c + d*x])^
(3/2)*Sin[c + d*x])/(7*a^2*d) + (18*e^3*(e*Sec[c + d*x])^(7/2)*Sin[c + d*x])/(35*a^2*d) - (((4*I)/5)*e^2*(e*Se
c[c + d*x])^(9/2))/(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 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{(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (9 e^2\right ) \int (e \sec (c+d x))^{9/2} \, dx}{5 a^2}\\ &=\frac{18 e^3 (e \sec (c+d x))^{7/2} \sin (c+d x)}{35 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (9 e^4\right ) \int (e \sec (c+d x))^{5/2} \, dx}{7 a^2}\\ &=\frac{6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{7 a^2 d}+\frac{18 e^3 (e \sec (c+d x))^{7/2} \sin (c+d x)}{35 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (3 e^6\right ) \int \sqrt{e \sec (c+d x)} \, dx}{7 a^2}\\ &=\frac{6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{7 a^2 d}+\frac{18 e^3 (e \sec (c+d x))^{7/2} \sin (c+d x)}{35 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (3 e^6 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{7 a^2}\\ &=\frac{6 e^6 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 a^2 d}+\frac{6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{7 a^2 d}+\frac{18 e^3 (e \sec (c+d x))^{7/2} \sin (c+d x)}{35 a^2 d}-\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.486882, size = 85, normalized size = 0.56 \[ \frac{e^6 \sec ^3(c+d x) \sqrt{e \sec (c+d x)} \left (-5 \sin (c+d x)+15 \sin (3 (c+d x))-56 i \cos (c+d x)+60 \cos ^{\frac{7}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{70 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(13/2)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(e^6*Sec[c + d*x]^3*Sqrt[e*Sec[c + d*x]]*((-56*I)*Cos[c + d*x] + 60*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2,
2] - 5*Sin[c + d*x] + 15*Sin[3*(c + d*x)]))/(70*a^2*d)

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Maple [A]  time = 0.279, size = 219, 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} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{35\,{a}^{2}d \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ( 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 ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+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}}} \left ( \cos \left ( dx+c \right ) \right ) ^{3}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +15\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -14\,i\cos \left ( dx+c \right ) -5\,\sin \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^2,x)

[Out]

2/35/a^2/d*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(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)*cos(d*x+c)^4+15*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2)*cos(d*x+c)^3*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+15*cos(d*x+c)^2*sin(d*x+c)-14*I*cos(d*x+c)-5*sin
(d*x+c))*(e/cos(d*x+c))^(13/2)*cos(d*x+c)^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((e*sec(d*x+c))^(13/2)/(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 (-30 i \, e^{6} e^{\left (6 i \, d x + 6 i \, c\right )} - 102 i \, e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} - 122 i \, e^{6} e^{\left (2 i \, d x + 2 i \, c\right )} + 30 i \, e^{6}\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 )} + 35 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}{\rm integral}\left (-\frac{3 i \, \sqrt{2} e^{6} \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 )}}{7 \, a^{2} d}, x\right )}{35 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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))^(13/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/35*(sqrt(2)*(-30*I*e^6*e^(6*I*d*x + 6*I*c) - 102*I*e^6*e^(4*I*d*x + 4*I*c) - 122*I*e^6*e^(2*I*d*x + 2*I*c) +
30*I*e^6)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 35*(a^2*d*e^(6*I*d*x + 6*I*c) + 3*a^2*d
*e^(4*I*d*x + 4*I*c) + 3*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*integral(-3/7*I*sqrt(2)*e^6*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^(6*I*d*x + 6*I*c) + 3*a^2*d*e^(4*I*d*x + 4*I*c) + 3
*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))**(13/2)/(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{13}{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))^(13/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(13/2)/(I*a*tan(d*x + c) + a)^2, x)