3.248 \(\int \frac{\sqrt{e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=116 \[ -\frac{2 (e \cos (c+d x))^{3/2}}{5 d e \left (a^2 \sin (c+d x)+a^2\right )}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 a^2 d \sqrt{\cos (c+d x)}}-\frac{2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.119396, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2681, 2683, 2640, 2639} \[ -\frac{2 (e \cos (c+d x))^{3/2}}{5 d e \left (a^2 \sin (c+d x)+a^2\right )}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 a^2 d \sqrt{\cos (c+d x)}}-\frac{2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Cos[c + d*x]]/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 2681

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m + p + 1)), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rule 2683

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(g*Cos[e
+ f*x])^(p + 1))/(a*f*g*(p - 1)*(a + b*Sin[e + f*x])), x] + Dist[p/(a*(p - 1)), Int[(g*Cos[e + f*x])^p, x], x
] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] &&  !GeQ[p, 1] && IntegerQ[2*p]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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{\sqrt{e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx &=-\frac{2 (e \cos (c+d x))^{3/2}}{5 d e (a+a \sin (c+d x))^2}+\frac{\int \frac{\sqrt{e \cos (c+d x)}}{a+a \sin (c+d x)} \, dx}{5 a}\\ &=-\frac{2 (e \cos (c+d x))^{3/2}}{5 d e (a+a \sin (c+d x))^2}-\frac{2 (e \cos (c+d x))^{3/2}}{5 d e \left (a^2+a^2 \sin (c+d x)\right )}-\frac{\int \sqrt{e \cos (c+d x)} \, dx}{5 a^2}\\ &=-\frac{2 (e \cos (c+d x))^{3/2}}{5 d e (a+a \sin (c+d x))^2}-\frac{2 (e \cos (c+d x))^{3/2}}{5 d e \left (a^2+a^2 \sin (c+d x)\right )}-\frac{\sqrt{e \cos (c+d x)} \int \sqrt{\cos (c+d x)} \, dx}{5 a^2 \sqrt{\cos (c+d x)}}\\ &=-\frac{2 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d \sqrt{\cos (c+d x)}}-\frac{2 (e \cos (c+d x))^{3/2}}{5 d e (a+a \sin (c+d x))^2}-\frac{2 (e \cos (c+d x))^{3/2}}{5 d e \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.0416356, size = 66, normalized size = 0.57 \[ -\frac{(e \cos (c+d x))^{3/2} \, _2F_1\left (\frac{3}{4},\frac{9}{4};\frac{7}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{3 \sqrt [4]{2} a^2 d e (\sin (c+d x)+1)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[e*Cos[c + d*x]]/(a + a*Sin[c + d*x])^2,x]

[Out]

-((e*Cos[c + d*x])^(3/2)*Hypergeometric2F1[3/4, 9/4, 7/4, (1 - Sin[c + d*x])/2])/(3*2^(1/4)*a^2*d*e*(1 + Sin[c
+ d*x])^(3/4))

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Maple [B]  time = 1.456, size = 303, normalized size = 2.6 \begin{align*} -{\frac{2\,e}{5\,{a}^{2}d} \left ( 4\,{\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}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) -4\,{\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}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -6\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +2\,\sin \left ( 1/2\,dx+c/2 \right ) \right ) \left ( 4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-1} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1
/2)*(4*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)*sin
(1/2*d*x+1/2*c)^4-8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-4*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)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/
2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-6*sin
(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2*sin(1/2*d*x+1/2*c))*e/d

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

[Out]

integrate(sqrt(e*cos(d*x + c))/(a*sin(d*x + c) + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(e*cos(d*x + c))/(a^2*cos(d*x + c)^2 - 2*a^2*sin(d*x + c) - 2*a^2), x)

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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+a*sin(d*x+c))**2,x)

[Out]

Timed out

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

[Out]

integrate(sqrt(e*cos(d*x + c))/(a*sin(d*x + c) + a)^2, x)