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

Optimal. Leaf size=189 \[ \frac{4 a^2 (c-2 d) (c-d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f \sqrt{c+d \sin (e+f x)}}-\frac{4 a^2 (c-3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 a^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 d f} \]

[Out]

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

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Rubi [A]  time = 0.246494, antiderivative size = 189, 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 = {2763, 2752, 2663, 2661, 2655, 2653} \[ \frac{4 a^2 (c-2 d) (c-d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f \sqrt{c+d \sin (e+f x)}}-\frac{4 a^2 (c-3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 a^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && 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{(a+a \sin (e+f x))^2}{\sqrt{c+d \sin (e+f x)}} \, dx &=-\frac{2 a^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 d f}+\frac{2 \int \frac{2 a^2 d-a^2 (c-3 d) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 d}\\ &=-\frac{2 a^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 d f}-\frac{\left (2 a^2 (c-3 d)\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{3 d^2}+\frac{\left (2 a^2 (c-2 d) (c-d)\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 d^2}\\ &=-\frac{2 a^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 d f}-\frac{\left (2 a^2 (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}{3 d^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (2 a^2 (c-2 d) (c-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}{3 d^2 \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{2 a^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{3 d f}-\frac{4 a^2 (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)}}{3 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 a^2 (c-2 d) (c-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}}}{3 d^2 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.08974, size = 193, normalized size = 1.02 \[ -\frac{2 a^2 (\sin (e+f x)+1)^2 \left (2 \left (c^2-3 c d+2 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-2 \left (c^2-2 c d-3 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )+d \cos (e+f x) (c+d \sin (e+f x))\right )}{3 d^2 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.998, size = 758, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a^2*(2*((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)*El
lipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^2*d-12*((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)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^
(1/2))*d^2*c+10*((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
)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^3-2*((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)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))
^(1/2))*c^3+6*((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)*
EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^2*d+2*((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)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))
^(1/2))*c*d^2-6*((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
)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^3-d^3*sin(f*x+e)^3-c*d^2*sin(f*x+e)^2+d^3*si
n(f*x+e)+c*d^2)/d^3/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 (a \sin \left (f x + e\right ) + a\right )}^{2}}{\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+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^2/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{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}{\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+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{2 \sin{\left (e + f x \right )}}{\sqrt{c + d \sin{\left (e + f x \right )}}}\, dx + \int \frac{\sin ^{2}{\left (e + f x \right )}}{\sqrt{c + d \sin{\left (e + f x \right )}}}\, dx + \int \frac{1}{\sqrt{c + d \sin{\left (e + f x \right )}}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(2*sin(e + f*x)/sqrt(c + d*sin(e + f*x)), x) + Integral(sin(e + f*x)**2/sqrt(c + d*sin(e + f*x))
, x) + Integral(1/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{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{\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+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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