3.187 \(\int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0605921, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3486, 3768, 3771, 2639} \[ -\frac{2 a e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i a (e \sec (c+d x))^{3/2}}{3 d}+\frac{2 a e \sin (c+d x) \sqrt{e \sec (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x]),x]

[Out]

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

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

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

Mathematica [C]  time = 0.801333, size = 102, normalized size = 1.13 \[ \frac{2 a e e^{-2 i d x} \sqrt{e \sec (c+d x)} (\cos (c+3 d x)+i \sin (c+3 d x)) \left (i \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+\tan (c+d x)-2 i\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x]),x]

[Out]

(2*a*e*Sqrt[e*Sec[c + d*x]]*(Cos[c + 3*d*x] + I*Sin[c + 3*d*x])*(-2*I + I*Sqrt[1 + E^((2*I)*(c + d*x))]*Hyperg
eometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + Tan[c + d*x]))/(3*d*E^((2*I)*d*x))

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Maple [B]  time = 0.21, size = 351, normalized size = 3.9 \begin{align*}{\frac{2\,a \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}} \left ( 3\,i\sin \left ( dx+c \right ) \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 EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -3\,i\sin \left ( dx+c \right ) \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 ) +3\,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 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 ) -3\,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 ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +i\sin \left ( dx+c \right ) -3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,\cos \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \sec \left (d x + c\right )\right )^{\frac{3}{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*(sqrt(2)*(-6*I*a*e*e^(3*I*d*x + 3*I*c) - 2*I*a*e*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2
*I*d*x + 1/2*I*c) + 3*(d*e^(2*I*d*x + 2*I*c) + d)*integral(I*sqrt(2)*a*e*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(
1/2*I*d*x + 1/2*I*c)/d, x))/(d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \left (e \sec{\left (c + d x \right )}\right )^{\frac{3}{2}}\, dx + \int i \left (e \sec{\left (c + d x \right )}\right )^{\frac{3}{2}} \tan{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(3/2)*(a+I*a*tan(d*x+c)),x)

[Out]

a*(Integral((e*sec(c + d*x))**(3/2), x) + Integral(I*(e*sec(c + d*x))**(3/2)*tan(c + d*x), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \sec \left (d x + c\right )\right )^{\frac{3}{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a), x)