3.256 \(\int \frac{(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=157 \[ -\frac{10 e^5 \sin (c+d x) (e \sec (c+d x))^{3/2}}{a^4 d}+\frac{12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{10 e^6 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{a^4 d}+\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3} \]

[Out]

(-10*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(a^4*d) - (10*e^5*(e*Sec[c + d*x])
^(3/2)*Sin[c + d*x])/(a^4*d) + (((4*I)/3)*e^2*(e*Sec[c + d*x])^(9/2))/(a*d*(a + I*a*Tan[c + d*x])^3) + ((12*I)
*e^4*(e*Sec[c + d*x])^(5/2))/(d*(a^4 + I*a^4*Tan[c + d*x]))

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Rubi [A]  time = 0.154751, antiderivative size = 157, 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} \[ -\frac{10 e^5 \sin (c+d x) (e \sec (c+d x))^{3/2}}{a^4 d}+\frac{12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{10 e^6 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{a^4 d}+\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(13/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-10*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(a^4*d) - (10*e^5*(e*Sec[c + d*x])
^(3/2)*Sin[c + d*x])/(a^4*d) + (((4*I)/3)*e^2*(e*Sec[c + d*x])^(9/2))/(a*d*(a + I*a*Tan[c + d*x])^3) + ((12*I)
*e^4*(e*Sec[c + d*x])^(5/2))/(d*(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 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))^4} \, dx &=\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}-\frac{\left (3 e^2\right ) \int \frac{(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^2} \, dx}{a^2}\\ &=\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac{12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{\left (15 e^4\right ) \int (e \sec (c+d x))^{5/2} \, dx}{a^4}\\ &=-\frac{10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac{12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{\left (5 e^6\right ) \int \sqrt{e \sec (c+d x)} \, dx}{a^4}\\ &=-\frac{10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac{12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{\left (5 e^6 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{a^4}\\ &=-\frac{10 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)}}{a^4 d}-\frac{10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac{4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac{12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.500015, size = 134, normalized size = 0.85 \[ \frac{i e^6 \sec ^5(c+d x) \sqrt{e \sec (c+d x)} (\cos (3 (c+d x))+i \sin (3 (c+d x))) \left (11 i \sin (2 (c+d x))+19 \cos (2 (c+d x))+30 i \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (c+d x)+i \sin (c+d x))+21\right )}{3 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(13/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I/3)*e^6*Sec[c + d*x]^5*Sqrt[e*Sec[c + d*x]]*(21 + 19*Cos[2*(c + d*x)] + (30*I)*Cos[c + d*x]^(3/2)*EllipticF
[(c + d*x)/2, 2]*(Cos[c + d*x] + I*Sin[c + d*x]) + (11*I)*Sin[2*(c + d*x)])*(Cos[3*(c + d*x)] + I*Sin[3*(c + d
*x)]))/(a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]  time = 0.3, size = 198, normalized size = 1.3 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{3\,{a}^{4}d} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{13}{2}}} \left ( -15\,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 ) -15\,i\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}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +8\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +12\,i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) \right ) } \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))^4,x)

[Out]

2/3/a^4/d*(e/cos(d*x+c))^(13/2)*(-15*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)-15*I*cos(d*x+c)*(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)+8*I*cos(d*x+c)^3+8*cos(d*x+c)^2*sin(d*x+c)+12*I*cos(d*x+c)+sin
(d*x+c))*cos(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))^(13/2)/(a+I*a*tan(d*x+c))^4,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 (4 i \, d x + 4 i \, c\right )} + 42 i \, e^{6} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 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 )} + 3 \,{\left (a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}{\rm integral}\left (\frac{5 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 )}}{a^{4} d}, x\right )}{3 \,{\left (a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )}\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))^4,x, algorithm="fricas")

[Out]

1/3*(sqrt(2)*(30*I*e^6*e^(4*I*d*x + 4*I*c) + 42*I*e^6*e^(2*I*d*x + 2*I*c) + 8*I*e^6)*sqrt(e/(e^(2*I*d*x + 2*I*
c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 3*(a^4*d*e^(4*I*d*x + 4*I*c) + a^4*d*e^(2*I*d*x + 2*I*c))*integral(5*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^4*d), x))/(a^4*d*e^(4*I*d*x + 4*I*c) + a
^4*d*e^(2*I*d*x + 2*I*c))

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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))**4,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 )}^{4}}\,{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))^4,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(13/2)/(I*a*tan(d*x + c) + a)^4, x)