3.505 \(\int \frac{(c+d \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=186 \[ \frac{\left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f \sqrt{c+d \sin (e+f x)}}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{f (a \sin (e+f x)+a)}-\frac{(c-3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

[Out]

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

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Rubi [A]  time = 0.257038, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2767, 2752, 2663, 2661, 2655, 2653} \[ \frac{\left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f \sqrt{c+d \sin (e+f x)}}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{f (a \sin (e+f x)+a)}-\frac{(c-3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x]),x]

[Out]

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

Rule 2767

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

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

Mathematica [A]  time = 1.56371, size = 223, normalized size = 1.2 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (2 (c-d) \sin \left (\frac{1}{2} (e+f x)\right ) (c+d \sin (e+f x))-\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left ((c-d) \left ((c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )+c+d \sin (e+f x)\right )-\left (c^2-2 c d-3 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )\right )}{a f (\sin (e+f x)+1) \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x]),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)*Sin[(e + f*x)/2]*(c + d*Sin[e + f*x]) - (Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])*(-((c^2 - 2*c*d - 3*d^2)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e +
f*x])/(c + d)]) + (c - d)*(c + d*Sin[e + f*x] + (c + d)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[
(c + d*Sin[e + f*x])/(c + d)]))))/(a*f*(1 + Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]])

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Maple [B]  time = 1.28, size = 925, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e)),x)

[Out]

(cos(f*x+e)^2*sin(f*x+e)*d+c*cos(f*x+e)^2)^(1/2)*((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/
(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^
(1/2))*c^3-3*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(
c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d-(d/(c-d)*sin(f*x+e)+1/(c
-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*
x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c*d^2+3*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d
/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))
^(1/2))*d^3+2*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/
(c-d))^(1/2)*EllipticF((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d-2*(d/(c-d)*sin(f*x+e)+1
/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticF((d/(c-d)*sin
(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*d^3-c*cos(f*x+e)^2*d^2+cos(f*x+e)^2*d^3+c^2*d*sin(f*x+e)-2*sin(f
*x+e)*d^2*c+d^3*sin(f*x+e)-c^2*d+2*c*d^2-d^3)/d/(-(c+d*sin(f*x+e))*(-1+sin(f*x+e))*(1+sin(f*x+e)))^(1/2)/a/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{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{a \sin \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + e) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((d*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + e) + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c \sqrt{c + d \sin{\left (e + f x \right )}}}{\sin{\left (e + f x \right )} + 1}\, dx + \int \frac{d \sqrt{c + d \sin{\left (e + f x \right )}} \sin{\left (e + f x \right )}}{\sin{\left (e + f x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e)),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{a \sin \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + e) + a), x)