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

Optimal. Leaf size=107 \[ -\frac{6 a^2 \sin (c+d x) \sqrt{e \sec (c+d x)}}{d e}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt{e \sec (c+d x)}}+\frac{6 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]

[Out]

(6*a^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (6*a^2*Sqrt[e*Sec[c + d*x]]*Si
n[c + d*x])/(d*e) - ((4*I)*(a^2 + I*a^2*Tan[c + d*x]))/(d*Sqrt[e*Sec[c + d*x]])

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Rubi [A]  time = 0.0806821, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3496, 3768, 3771, 2639} \[ -\frac{6 a^2 \sin (c+d x) \sqrt{e \sec (c+d x)}}{d e}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt{e \sec (c+d x)}}+\frac{6 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (6*a^2*Sqrt[e*Sec[c + d*x]]*Si
n[c + d*x])/(d*e) - ((4*I)*(a^2 + I*a^2*Tan[c + d*x]))/(d*Sqrt[e*Sec[c + d*x]])

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 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{(a+i a \tan (c+d x))^2}{\sqrt{e \sec (c+d x)}} \, dx &=-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt{e \sec (c+d x)}}-\frac{\left (3 a^2\right ) \int (e \sec (c+d x))^{3/2} \, dx}{e^2}\\ &=-\frac{6 a^2 \sqrt{e \sec (c+d x)} \sin (c+d x)}{d e}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt{e \sec (c+d x)}}+\left (3 a^2\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx\\ &=-\frac{6 a^2 \sqrt{e \sec (c+d x)} \sin (c+d x)}{d e}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt{e \sec (c+d x)}}+\frac{\left (3 a^2\right ) \int \sqrt{\cos (c+d x)} \, dx}{\sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{6 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{6 a^2 \sqrt{e \sec (c+d x)} \sin (c+d x)}{d e}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt{e \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.978643, size = 132, normalized size = 1.23 \[ -\frac{2 i \sqrt{2} a^2 e^{2 i (c+d x)} \left (\left (1+e^{2 i (c+d x)}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-\sqrt{1+e^{2 i (c+d x)}}\right )}{d \left (1+e^{2 i (c+d x)}\right )^{3/2} \sqrt{\frac{e e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[2]*a^2*E^((2*I)*(c + d*x))*(-Sqrt[1 + E^((2*I)*(c + d*x))] + (1 + E^((2*I)*(c + d*x)))*Hypergeome
tric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(d*Sqrt[(e*E^(I*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^(
(2*I)*(c + d*x)))^(3/2))

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Maple [B]  time = 0.266, size = 1099, normalized size = 10.3 \begin{align*} \text{result too large to display} \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))^(1/2),x)

[Out]

a^2/d*(cos(d*x+c)-1)*(I*ln(-2*(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)
/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*cos(d*x+c)^2*sin(d*x+c)+12*I*(-cos(d*x+c)/(cos(d*x+c)+1
)^2)^(3/2)*cos(d*x+c)^2*sin(d*x+c)+12*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^3*sin(d*x+c)-6*I*(cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*
x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^3*sin(d*x+c)-I*cos(d*x+c)^2*ln(-(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d
*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)+6*I*(1/(c
os(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)/(c
os(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*sin(d*x+c)+4*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^5+4*I*(-cos(d*x+
c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)*sin(d*x+c)+12*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+
c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^2*sin(d*x+c)-6*I*(
1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*Elli
pticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)+6*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^4+6*I*(cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x
+c)-1)/sin(d*x+c),I)*cos(d*x+c)^3*sin(d*x+c)-4*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^3+4*I*(-cos(d*x
+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^4*sin(d*x+c)-12*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d
*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^2*sin(d*x+c)-8*
(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^2+2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2))/(cos(d*x+c)+1)^2/(-c
os(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)/cos(d*x+c)/sin(d*x+c)^3/(e/cos(d*x+c))^(1/2)

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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}}{\sqrt{e \sec \left (d x + c\right )}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (-4 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 )} - 6 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 )} +{\left (d e e^{\left (i \, d x + i \, c\right )} - d e\right )}{\rm integral}\left (\frac{\sqrt{2}{\left (-3 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, a^{2} e^{\left (i \, d x + i \, c\right )} - 3 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 )}}{d e e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e e^{\left (2 i \, d x + 2 i \, c\right )} + d e e^{\left (i \, d x + i \, c\right )}}, x\right )}{d e e^{\left (i \, d x + i \, c\right )} - d e} \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))^(1/2),x, algorithm="fricas")

[Out]

(sqrt(2)*(-4*I*a^2*e^(2*I*d*x + 2*I*c) - 2*I*a^2*e^(I*d*x + I*c) - 6*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*e^(I*d*x + I*c) - d*e)*integral(sqrt(2)*(-3*I*a^2*e^(2*I*d*x + 2*I*c) - 6*I*a^2
*e^(I*d*x + I*c) - 3*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*e^(3*I*d*x + 3*I*c)
- 2*d*e*e^(2*I*d*x + 2*I*c) + d*e*e^(I*d*x + I*c)), x))/(d*e*e^(I*d*x + I*c) - d*e)

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

[Out]

a**2*(Integral(1/sqrt(e*sec(c + d*x)), x) + Integral(-tan(c + d*x)**2/sqrt(e*sec(c + d*x)), x) + Integral(2*I*
tan(c + d*x)/sqrt(e*sec(c + d*x)), 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}}{\sqrt{e \sec \left (d x + c\right )}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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