3.238 \(\int \frac{(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=90 \[ -\frac{2 e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 a^2 d}+\frac{4 i e^2 \sqrt{e \sec (c+d x)}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

(-2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*a^2*d) + (((4*I)/3)*e^2*Sqrt[e*S
ec[c + d*x]])/(d*(a^2 + I*a^2*Tan[c + d*x]))

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Rubi [A]  time = 0.073417, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3500, 3771, 2641} \[ -\frac{2 e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 a^2 d}+\frac{4 i e^2 \sqrt{e \sec (c+d x)}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(5/2)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Mathematica [A]  time = 0.36464, size = 101, normalized size = 1.12 \[ \frac{2 (e \sec (c+d x))^{5/2} (\cos (c+d x)+i \sin (c+d x)) \left (\sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (c+d x)+i \sin (c+d x))-2 i \cos (c+d x)\right )}{3 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(5/2)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.24, size = 173, normalized size = 1.9 \begin{align*} -{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,{a}^{2}d} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}} \left ( 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 ) +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 ) -2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,\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))^(5/2)/(a+I*a*tan(d*x+c))^2,x)

[Out]

-2/3/a^2/d*(e/cos(d*x+c))^(5/2)*cos(d*x+c)^2*(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)+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)-2*I*cos(d*x+c)^2-2*cos(d*x+c)*sin(d*x+c))

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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))^(5/2)/(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{{\left (3 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )}{\rm integral}\left (\frac{i \, \sqrt{2} e^{2} \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 \, a^{2} d}, x\right ) + \sqrt{2}{\left (2 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, e^{2}\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 (-2 i \, d x - 2 i \, c\right )}}{3 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(3*a^2*d*e^(2*I*d*x + 2*I*c)*integral(1/3*I*sqrt(2)*e^2*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) + sqrt(2)*(2*I*e^2*e^(2*I*d*x + 2*I*c) + 2*I*e^2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/
2*I*d*x + 1/2*I*c))*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))**(5/2)/(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{5}{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))^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(5/2)/(I*a*tan(d*x + c) + a)^2, x)