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]

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

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Rubi [A]  time = 0.12, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.04, 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]

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

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e \cos \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \cos \left (d x + c\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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)

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maple [B]  time = 2.17, size = 303, normalized size = 2.61 \[ -\frac {2 \left (4 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-4 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}{5 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*EllipticE(cos(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)+EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \cos \left (d x + c\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {e\,\cos \left (c+d\,x\right )}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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