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

Optimal. Leaf size=85 \[ \frac{2 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{5 d (e \sec (c+d x))^{5/2}} \]

[Out]

(2*a^2*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (((4*I)/5)*(a^2 + I*a^2*
Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(5/2))

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Rubi [A]  time = 0.0715836, antiderivative size = 85, 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 = {3496, 3771, 2639} \[ \frac{2 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{5 d (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (((4*I)/5)*(a^2 + I*a^2*
Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(5/2))

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && 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 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{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{5/2}} \, dx &=-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{5 d (e \sec (c+d x))^{5/2}}+\frac{a^2 \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{5 e^2}\\ &=-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{5 d (e \sec (c+d x))^{5/2}}+\frac{a^2 \int \sqrt{\cos (c+d x)} \, dx}{5 e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{2 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{5 d (e \sec (c+d x))^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.982941, size = 114, normalized size = 1.34 \[ -\frac{i \sqrt{2} a^2 \left (1+e^{2 i (c+d x)}\right )^{3/2} \left (\frac{e e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (2 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+3 \sqrt{1+e^{2 i (c+d x)}}\right )}{15 d e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.203, size = 343, normalized size = 4. \begin{align*} -{\frac{2\,{a}^{2}}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) } \left ( i{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \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}}}-i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \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}}}+i{\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 ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-i{\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 ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+2\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+2\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}- \left ( \cos \left ( dx+c \right ) \right ) ^{2}-\cos \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*a^2/d*(I*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)-I*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)*(1/(cos(d*x+c)+1))^
(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1
))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-I*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c
)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*I*sin(d*x+c)*cos(d*x+c)^3+2*cos(d*x+c)^4-cos(d*x+c)^2-cos(d*x+
c))/cos(d*x+c)^3/sin(d*x+c)/(e/cos(d*x+c))^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (-i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 3 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} e^{\left (i \, d x + i \, c\right )} - 2 i \, a^{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 )} + 5 \,{\left (d e^{3} e^{\left (i \, d x + i \, c\right )} - d e^{3}\right )}{\rm integral}\left (\frac{\sqrt{2}{\left (-i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a^{2} e^{\left (i \, d x + i \, c\right )} - i \, a^{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 )}}{5 \,{\left (d e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{5 \,{\left (d e^{3} e^{\left (i \, d x + i \, c\right )} - d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{1}{\left (e \sec{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx + \int - \frac{\tan ^{2}{\left (c + d x \right )}}{\left (e \sec{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx + \int \frac{2 i \tan{\left (c + d x \right )}}{\left (e \sec{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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