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

Optimal. Leaf size=74 \[ -\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{a d \sqrt {\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2683, 2640, 2639} \[ -\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{a d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(a*d*Sqrt[Cos[c + d*x]]) - (2*(e*Cos[c + d*x])^(3/2))/(d*e
*(a + a*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 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)} \, dx &=-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a+a \sin (c+d x))}-\frac {\int \sqrt {e \cos (c+d x)} \, dx}{a}\\ &=-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a+a \sin (c+d x))}-\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)} \, dx}{a \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 )}{a d \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 66, normalized size = 0.89 \[ -\frac {2^{3/4} (e \cos (c+d x))^{3/2} \, _2F_1\left (\frac {3}{4},\frac {5}{4};\frac {7}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{3 a d e (\sin (c+d x)+1)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(2^(3/4)*(e*Cos[c + d*x])^(3/2)*Hypergeometric2F1[3/4, 5/4, 7/4, (1 - Sin[c + d*x])/2])/(a*d*e*(1 + Sin[c
+ d*x])^(3/4))

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \cos \left (d x + c\right )}}{a \sin \left (d x + c\right ) + a}, x\right ) \]

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)),x, algorithm="fricas")

[Out]

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

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

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)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 1.14, size = 115, normalized size = 1.55 \[ -\frac {2 \left (\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}{\sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a d} \]

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)),x)

[Out]

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

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)),x, algorithm="maxima")

[Out]

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

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

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)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {e \cos {\left (c + d x \right )}}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

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)),x)

[Out]

Integral(sqrt(e*cos(c + d*x))/(sin(c + d*x) + 1), x)/a

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