3.726 \(\int \frac{a+b \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx\)

Optimal. Leaf size=140 \[ \frac{2 b \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{d f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{d f \sqrt{c+d \sin (e+f x)}} \]

[Out]

(2*b*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(d*f*Sqrt[(c + d*Sin[e + f*x])/(c
+ d)]) - (2*(b*c - a*d)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(d*f*
Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.124015, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2752, 2663, 2661, 2655, 2653} \[ \frac{2 b \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{d f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{d f \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(d*f*Sqrt[(c + d*Sin[e + f*x])/(c
+ d)]) - (2*(b*c - a*d)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(d*f*
Sqrt[c + d*Sin[e + f*x]])

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{a+b \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx &=\frac{b \int \sqrt{c+d \sin (e+f x)} \, dx}{d}+\frac{(-b c+a d) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{d}\\ &=\frac{\left (b \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{d \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left ((-b c+a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}\right ) \int \frac{1}{\sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}}} \, dx}{d \sqrt{c+d \sin (e+f x)}}\\ &=\frac{2 b E\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{c+d \sin (e+f x)}}{d f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 (b c-a d) F\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}{d f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.55958, size = 101, normalized size = 0.72 \[ -\frac{2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left ((a d-b c) F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )+b (c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )}{d f \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*(c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] + (-(b*c) + a*d)*EllipticF[(-2*e + Pi - 2*f*x)/
4, (2*d)/(c + d)])*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(d*f*Sqrt[c + d*Sin[e + f*x]])

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Maple [A]  time = 1.069, size = 243, normalized size = 1.7 \begin{align*} -2\,{\frac{c-d}{{d}^{2}\cos \left ( fx+e \right ) \sqrt{c+d\sin \left ( fx+e \right ) }f}\sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}}\sqrt{-{\frac{ \left ( -1+\sin \left ( fx+e \right ) \right ) d}{c+d}}}\sqrt{-{\frac{d \left ( 1+\sin \left ( fx+e \right ) \right ) }{c-d}}} \left ({\it EllipticE} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},\sqrt{{\frac{c-d}{c+d}}} \right ) bc+{\it EllipticE} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},\sqrt{{\frac{c-d}{c+d}}} \right ) bd-a{\it EllipticF} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},\sqrt{{\frac{c-d}{c+d}}} \right ) d-{\it EllipticF} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},\sqrt{{\frac{c-d}{c+d}}} \right ) bd \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(c-d)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*(Elli
pticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*c+EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/
(c+d))^(1/2))*b*d-a*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d-EllipticF(((c+d*sin(f*x+e)
)/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*d)/d^2/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \sin \left (f x + e\right ) + a}{\sqrt{d \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \sin{\left (e + f x \right )}}{\sqrt{c + d \sin{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))/(c+d*sin(f*x+e))**(1/2),x)

[Out]

Integral((a + b*sin(e + f*x))/sqrt(c + d*sin(e + f*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \sin \left (f x + e\right ) + a}{\sqrt{d \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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