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

Optimal. Leaf size=228 \[ \frac{2 \left (d^2 \left (a^2-b^2\right )-2 a b c d+2 b^2 c^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{d^2 f \left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{2 (b c-a d)^2 \cos (e+f x)}{d f \left (c^2-d^2\right ) \sqrt{c+d \sin (e+f x)}}-\frac{4 b (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{d^2 f \sqrt{c+d \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.307907, antiderivative size = 228, 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 = {2790, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (d^2 \left (a^2-b^2\right )-2 a b c d+2 b^2 c^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{d^2 f \left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{2 (b c-a d)^2 \cos (e+f x)}{d f \left (c^2-d^2\right ) \sqrt{c+d \sin (e+f x)}}-\frac{4 b (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{d^2 f \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2790

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

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))^2}{(c+d \sin (e+f x))^{3/2}} \, dx &=\frac{2 (b c-a d)^2 \cos (e+f x)}{d \left (c^2-d^2\right ) f \sqrt{c+d \sin (e+f x)}}+\frac{2 \int \frac{\frac{1}{2} d \left (a^2 c+b^2 c-2 a b d\right )+\frac{1}{2} \left (2 b^2 c^2-2 a b c d+\left (a^2-b^2\right ) d^2\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{d \left (c^2-d^2\right )}\\ &=\frac{2 (b c-a d)^2 \cos (e+f x)}{d \left (c^2-d^2\right ) f \sqrt{c+d \sin (e+f x)}}-\frac{(2 b (b c-a d)) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{d^2}-\frac{\left (2 a b c d-a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{d^2 \left (c^2-d^2\right )}\\ &=\frac{2 (b c-a d)^2 \cos (e+f x)}{d \left (c^2-d^2\right ) f \sqrt{c+d \sin (e+f x)}}-\frac{\left (\left (2 a b c d-a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{d^2 \left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{\left (2 b (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^2 \sqrt{c+d \sin (e+f x)}}\\ &=\frac{2 (b c-a d)^2 \cos (e+f x)}{d \left (c^2-d^2\right ) f \sqrt{c+d \sin (e+f x)}}-\frac{2 \left (2 a b c d-a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) 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^2 \left (c^2-d^2\right ) f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{4 b (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^2 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.865774, size = 172, normalized size = 0.75 \[ \frac{2 \left (\frac{\sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left (\left (-a^2 d^2+2 a b c d+b^2 \left (d^2-2 c^2\right )\right ) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )+2 b (c-d) (b c-a d) F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )}{d (c-d)}+\frac{(b c-a d)^2 \cos (e+f x)}{c^2-d^2}\right )}{d f \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 3.56, size = 888, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(b/d^2*(2*b*d*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))
/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+
d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))
+4*d*a*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(
-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-2*c*b*(c/
d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f
*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))+(a^2*d^2-2*a*b*c*d
+b^2*c^2)/d^2*(2*d*cos(f*x+e)^2/(c^2-d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*c/(c^2-d^2)*(c/d-1)*((c+d*
sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*co
s(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+2/(c^2-d^2)*d*(c/d-1)*((c+d*si
n(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(
f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*
x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))))/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{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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