3.658 \(\int \sqrt{e \cos (c+d x)} (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=60 \[ \frac{2 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}-\frac{2 i a \sqrt{e \cos (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.0782001, antiderivative size = 60, 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 = {3515, 3486, 3771, 2639} \[ \frac{2 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}-\frac{2 i a \sqrt{e \cos (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Cos[c + d*x]]*(a + I*a*Tan[c + d*x]),x]

[Out]

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

Rule 3515

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

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

Mathematica [C]  time = 0.690825, size = 244, normalized size = 4.07 \[ \frac{a e (\cot (c)+i) e^{-i (c+d x)} \left (e^{2 i d x} \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1-i e^{i (c+d x)}} \sqrt{e^{i (c+d x)} \left (e^{i (c+d x)}-i\right )} F\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )+3 \sqrt{1-i e^{i (c+d x)}} \sqrt{e^{i (c+d x)} \left (e^{i (c+d x)}-i\right )} E\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )\right )}{3 d \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[e*Cos[c + d*x]]*(a + I*a*Tan[c + d*x]),x]

[Out]

(a*e*(I + Cot[c])*(3*Sqrt[1 - I*E^(I*(c + d*x))]*Sqrt[E^(I*(c + d*x))*(-I + E^(I*(c + d*x)))]*EllipticE[ArcSin
[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1] - 3*Sqrt[1 - I*E^(I*(c + d*x))]*Sqrt[E^(I*(c + d*x))*(-I + E^(I*
(c + d*x)))]*EllipticF[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1] + E^((2*I)*d*x)*Sqrt[1 + E^((2*I)*(
c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(3*d*E^(I*(c + d*x))*Sqrt[e*Cos[c + d*x]])

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Maple [A]  time = 1.352, size = 108, normalized size = 1.8 \begin{align*} 2\,{\frac{ae \left ( 2\,i \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-i\sin \left ( 1/2\,dx+c/2 \right ) \right ) }{\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(1/2)*(a+I*a*tan(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \cos \left (d x + c\right )}{\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*cos(d*x+c))^(1/2)*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate(sqrt(e*cos(d*x + c))*(I*a*tan(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(1/2)*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(1/2)*(a+I*a*tan(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \cos \left (d x + c\right )}{\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*cos(d*x+c))^(1/2)*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate(sqrt(e*cos(d*x + c))*(I*a*tan(d*x + c) + a), x)