3.50 \(\int \frac{(f+g x)^2 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=213 \[ \frac{\left (x \left (c^2 f^2+g^2\right )+2 f g\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}}-\frac{g^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} (c f-g)^2 \log (c x+1)}{2 c^3 d \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} (c f+g)^2 \log (1-c x)}{2 c^3 d \sqrt{d-c^2 d x^2}} \]

[Out]

((2*f*g + (c^2*f^2 + g^2)*x)*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2]) - (g^2*Sqrt[1 - c^2*x^2]*(a + b*
ArcSin[c*x])^2)/(2*b*c^3*d*Sqrt[d - c^2*d*x^2]) + (b*(c*f + g)^2*Sqrt[1 - c^2*x^2]*Log[1 - c*x])/(2*c^3*d*Sqrt
[d - c^2*d*x^2]) + (b*(c*f - g)^2*Sqrt[1 - c^2*x^2]*Log[1 + c*x])/(2*c^3*d*Sqrt[d - c^2*d*x^2])

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Rubi [A]  time = 0.428221, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4777, 4775, 637, 4761, 633, 31, 4641} \[ \frac{\left (x \left (c^2 f^2+g^2\right )+2 f g\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}}-\frac{g^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} (c f-g)^2 \log (c x+1)}{2 c^3 d \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} (c f+g)^2 \log (1-c x)}{2 c^3 d \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x)^2*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(3/2),x]

[Out]

((2*f*g + (c^2*f^2 + g^2)*x)*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2]) - (g^2*Sqrt[1 - c^2*x^2]*(a + b*
ArcSin[c*x])^2)/(2*b*c^3*d*Sqrt[d - c^2*d*x^2]) + (b*(c*f + g)^2*Sqrt[1 - c^2*x^2]*Log[1 - c*x])/(2*c^3*d*Sqrt
[d - c^2*d*x^2]) + (b*(c*f - g)^2*Sqrt[1 - c^2*x^2]*Log[1 + c*x])/(2*c^3*d*Sqrt[d - c^2*d*x^2])

Rule 4777

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Dist[(d^IntPart[p]*(d + e*x^2)^FracPart[p])/(1 - c^2*x^2)^FracPart[p], Int[(f + g*x)^m*(1 - c^2*x^2)^p*(a +
b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && IntegerQ
[p - 1/2] &&  !GtQ[d, 0]

Rule 4775

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(a + b*ArcSin[c*x])^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; Free
Q[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && IGtQ[n, 0]

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 4761

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With
[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[Dist[1/Sqrt[1 - c^
2*x^2], u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && ILtQ[p + 1/2,
0] && GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3])

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 4641

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

Rubi steps

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

Mathematica [A]  time = 0.702019, size = 156, normalized size = 0.73 \[ \frac{\sqrt{1-c^2 x^2} \left ((g-c f)^2 \left (2 b \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-\cot \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right ) \left (a+b \sin ^{-1}(c x)\right )\right )+(c f+g)^2 \left (\tan \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right ) \left (a+b \sin ^{-1}(c x)\right )+2 b \log \left (\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )\right )-\frac{g^2 \left (a+b \sin ^{-1}(c x)\right )^2}{b}\right )}{2 c^3 d \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)^2*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(3/2),x]

[Out]

(Sqrt[1 - c^2*x^2]*(-((g^2*(a + b*ArcSin[c*x])^2)/b) + (-(c*f) + g)^2*(-((a + b*ArcSin[c*x])*Cot[(Pi + 2*ArcSi
n[c*x])/4]) + 2*b*Log[Sin[(Pi + 2*ArcSin[c*x])/4]]) + (c*f + g)^2*(2*b*Log[Cos[(Pi + 2*ArcSin[c*x])/4]] + (a +
b*ArcSin[c*x])*Tan[(Pi + 2*ArcSin[c*x])/4])))/(2*c^3*d*Sqrt[d - c^2*d*x^2])

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Maple [C]  time = 0.364, size = 867, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x)

[Out]

a*g^2*x/c^2/d/(-c^2*d*x^2+d)^(1/2)-a*g^2/c^2/d/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+2*a*
f*g/c^2/d/(-c^2*d*x^2+d)^(1/2)+a*f^2/d*x/(-c^2*d*x^2+d)^(1/2)+1/2*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/
c^3/d^2/(c^2*x^2-1)*g^2*arcsin(c*x)^2+I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^3/d^2/(c^2*x^2-1)*arcsin
(c*x)*g^2+I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c/d^2/(c^2*x^2-1)*arcsin(c*x)*f^2-2*b*(-d*(c^2*x^2-1))
^(1/2)*arcsin(c*x)/c^2/d^2/(c^2*x^2-1)*(-c^2*x^2+1)*f*g-2*b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)/d^2/(c^2*x^2-1)
*x^2*f*g-b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)/d^2/(c^2*x^2-1)*x*f^2-b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)/c^2/d
^2/(c^2*x^2-1)*x*g^2-b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/d^2/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)
-I)*f^2-2*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^2/d^2/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)-I)*f*g-b
*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^3/d^2/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)-I)*g^2-b*(-d*(c^2*x
^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/d^2/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*f^2+2*b*(-d*(c^2*x^2-1))^(1/2
)*(-c^2*x^2+1)^(1/2)/c^2/d^2/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*f*g-b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2
+1)^(1/2)/c^3/d^2/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*g^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-c^{2} d x^{2} + d}{\left (a g^{2} x^{2} + 2 \, a f g x + a f^{2} +{\left (b g^{2} x^{2} + 2 \, b f g x + b f^{2}\right )} \arcsin \left (c x\right )\right )}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(a*g^2*x^2 + 2*a*f*g*x + a*f^2 + (b*g^2*x^2 + 2*b*f*g*x + b*f^2)*arcsin(c*x))/(c
^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asin}{\left (c x \right )}\right ) \left (f + g x\right )^{2}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**2*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(3/2),x)

[Out]

Integral((a + b*asin(c*x))*(f + g*x)**2/(-d*(c*x - 1)*(c*x + 1))**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*x + f)^2*(b*arcsin(c*x) + a)/(-c^2*d*x^2 + d)^(3/2), x)