3.54 \(\int \frac{(f+g x)^3 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=410 \[ \frac{2 (c f+g) (c f-g) \left (c^2 f x+g\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{(f+g x)^2 \left (c^2 f x+g\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}-\frac{b (f+g x) \left (c^2 f^2+2 c^2 f g x+g^2\right )}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{b g \sqrt{1-c^2 x^2} (c f-g)^2 \log (c x+1)}{12 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} (c f+g) (c f-g)^2 \log (c x+1)}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} (c f+g)^2 (c f-g) \log (1-c x)}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{b g \sqrt{1-c^2 x^2} (c f+g)^2 \log (1-c x)}{12 c^4 d^2 \sqrt{d-c^2 d x^2}} \]

[Out]

-(b*(f + g*x)*(c^2*f^2 + g^2 + 2*c^2*f*g*x))/(6*c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (2*(c*f - g)*
(c*f + g)*(g + c^2*f*x)*(a + b*ArcSin[c*x]))/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) + ((g + c^2*f*x)*(f + g*x)^2*(a +
b*ArcSin[c*x]))/(3*c^2*d^2*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]) + (b*(c*f - g)*(c*f + g)^2*Sqrt[1 - c^2*x^2]*Lo
g[1 - c*x])/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) - (b*g*(c*f + g)^2*Sqrt[1 - c^2*x^2]*Log[1 - c*x])/(12*c^4*d^2*Sqr
t[d - c^2*d*x^2]) + (b*(c*f - g)^2*g*Sqrt[1 - c^2*x^2]*Log[1 + c*x])/(12*c^4*d^2*Sqrt[d - c^2*d*x^2]) + (b*(c*
f - g)^2*(c*f + g)*Sqrt[1 - c^2*x^2]*Log[1 + c*x])/(3*c^4*d^2*Sqrt[d - c^2*d*x^2])

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Rubi [A]  time = 0.43268, antiderivative size = 410, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4777, 723, 637, 4761, 819, 633, 31} \[ \frac{2 (c f+g) (c f-g) \left (c^2 f x+g\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{(f+g x)^2 \left (c^2 f x+g\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}-\frac{b (f+g x) \left (c^2 f^2+2 c^2 f g x+g^2\right )}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{b g \sqrt{1-c^2 x^2} (c f-g)^2 \log (c x+1)}{12 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} (c f+g) (c f-g)^2 \log (c x+1)}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} (c f+g)^2 (c f-g) \log (1-c x)}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{b g \sqrt{1-c^2 x^2} (c f+g)^2 \log (1-c x)}{12 c^4 d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x)^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

-(b*(f + g*x)*(c^2*f^2 + g^2 + 2*c^2*f*g*x))/(6*c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (2*(c*f - g)*
(c*f + g)*(g + c^2*f*x)*(a + b*ArcSin[c*x]))/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) + ((g + c^2*f*x)*(f + g*x)^2*(a +
b*ArcSin[c*x]))/(3*c^2*d^2*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]) + (b*(c*f - g)*(c*f + g)^2*Sqrt[1 - c^2*x^2]*Lo
g[1 - c*x])/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) - (b*g*(c*f + g)^2*Sqrt[1 - c^2*x^2]*Log[1 - c*x])/(12*c^4*d^2*Sqr
t[d - c^2*d*x^2]) + (b*(c*f - g)^2*g*Sqrt[1 - c^2*x^2]*Log[1 + c*x])/(12*c^4*d^2*Sqrt[d - c^2*d*x^2]) + (b*(c*
f - g)^2*(c*f + g)*Sqrt[1 - c^2*x^2]*Log[1 + c*x])/(3*c^4*d^2*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 723

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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]

Rubi steps

\begin{align*} \int \frac{(f+g x)^3 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac{\sqrt{1-c^2 x^2} \int \frac{(f+g x)^3 \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{2 (c f-g) (c f+g) \left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g+c^2 f x\right ) (f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}-\frac{\left (b c \sqrt{1-c^2 x^2}\right ) \int \left (\frac{\left (g+c^2 f x\right ) (f+g x)^2}{3 c^2 \left (1-c^2 x^2\right )^2}+\frac{2 (c f-g) (c f+g) \left (g+c^2 f x\right )}{3 c^4 \left (1-c^2 x^2\right )}\right ) \, dx}{d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{2 (c f-g) (c f+g) \left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g+c^2 f x\right ) (f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}-\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int \frac{\left (g+c^2 f x\right ) (f+g x)^2}{\left (1-c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (2 b (c f-g) (c f+g) \sqrt{1-c^2 x^2}\right ) \int \frac{g+c^2 f x}{1-c^2 x^2} \, dx}{3 c^3 d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b (f+g x) \left (c^2 f^2+g^2+2 c^2 f g x\right )}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{2 (c f-g) (c f+g) \left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g+c^2 f x\right ) (f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int \frac{g \left (c^2 f^2+g^2\right )+2 c^2 f g^2 x}{1-c^2 x^2} \, dx}{6 c^3 d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (b (c f-g)^2 (c f+g) \sqrt{1-c^2 x^2}\right ) \int \frac{1}{-c-c^2 x} \, dx}{3 c^2 d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (b (c f-g) (c f+g)^2 \sqrt{1-c^2 x^2}\right ) \int \frac{1}{c-c^2 x} \, dx}{3 c^2 d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b (f+g x) \left (c^2 f^2+g^2+2 c^2 f g x\right )}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{2 (c f-g) (c f+g) \left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g+c^2 f x\right ) (f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{b (c f-g) (c f+g)^2 \sqrt{1-c^2 x^2} \log (1-c x)}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{b (c f-g)^2 (c f+g) \sqrt{1-c^2 x^2} \log (1+c x)}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (b (c f-g)^2 g \sqrt{1-c^2 x^2}\right ) \int \frac{1}{-c-c^2 x} \, dx}{12 c^2 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (b g (c f+g)^2 \sqrt{1-c^2 x^2}\right ) \int \frac{1}{c-c^2 x} \, dx}{12 c^2 d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b (f+g x) \left (c^2 f^2+g^2+2 c^2 f g x\right )}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{2 (c f-g) (c f+g) \left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g+c^2 f x\right ) (f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{b (c f-g) (c f+g)^2 \sqrt{1-c^2 x^2} \log (1-c x)}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{b g (c f+g)^2 \sqrt{1-c^2 x^2} \log (1-c x)}{12 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{b (c f-g)^2 g \sqrt{1-c^2 x^2} \log (1+c x)}{12 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{b (c f-g)^2 (c f+g) \sqrt{1-c^2 x^2} \log (1+c x)}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [C]  time = 1.27782, size = 366, normalized size = 0.89 \[ \frac{\sqrt{d-c^2 d x^2} \left (-\sqrt{-c^2} \left (-6 a c^2 f^2 g+4 a c^6 f^3 x^3-6 a c^4 f^3 x-6 a c^4 f g^2 x^3-6 a c^2 g^3 x^2+4 a g^3-b c f \left (1-c^2 x^2\right )^{3/2} \left (2 c^2 f^2-3 g^2\right ) \log \left (c^2 x^2-1\right )+2 b \sin ^{-1}(c x) \left (-3 c^4 f x \left (f^2+g^2 x^2\right )-3 c^2 g \left (f^2+g^2 x^2\right )+2 c^6 f^3 x^3+2 g^3\right )+3 b c^3 f^2 g x \sqrt{1-c^2 x^2}+b c^3 f^3 \sqrt{1-c^2 x^2}+3 b c f g^2 \sqrt{1-c^2 x^2}+b c g^3 x \sqrt{1-c^2 x^2}\right )+i b c g \left (1-c^2 x^2\right )^{3/2} \left (3 c^2 f^2-5 g^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-c^2} x\right ),1\right )\right )}{6 c^4 \sqrt{-c^2} d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(Sqrt[d - c^2*d*x^2]*(I*b*c*g*(3*c^2*f^2 - 5*g^2)*(1 - c^2*x^2)^(3/2)*EllipticF[I*ArcSinh[Sqrt[-c^2]*x], 1] -
Sqrt[-c^2]*(-6*a*c^2*f^2*g + 4*a*g^3 - 6*a*c^4*f^3*x - 6*a*c^2*g^3*x^2 + 4*a*c^6*f^3*x^3 - 6*a*c^4*f*g^2*x^3 +
b*c^3*f^3*Sqrt[1 - c^2*x^2] + 3*b*c*f*g^2*Sqrt[1 - c^2*x^2] + 3*b*c^3*f^2*g*x*Sqrt[1 - c^2*x^2] + b*c*g^3*x*S
qrt[1 - c^2*x^2] + 2*b*(2*g^3 + 2*c^6*f^3*x^3 - 3*c^2*g*(f^2 + g^2*x^2) - 3*c^4*f*x*(f^2 + g^2*x^2))*ArcSin[c*
x] - b*c*f*(2*c^2*f^2 - 3*g^2)*(1 - c^2*x^2)^(3/2)*Log[-1 + c^2*x^2])))/(6*c^4*Sqrt[-c^2]*d^3*(-1 + c^2*x^2)^2
)

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Maple [C]  time = 0.575, size = 5098, normalized size = 12.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x)

[Out]

result too large to display

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Maxima [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((g*x+f)^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a g^{3} x^{3} + 3 \, a f g^{2} x^{2} + 3 \, a f^{2} g x + a f^{3} +{\left (b g^{3} x^{3} + 3 \, b f g^{2} x^{2} + 3 \, b f^{2} g x + b f^{3}\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(-(a*g^3*x^3 + 3*a*f*g^2*x^2 + 3*a*f^2*g*x + a*f^3 + (b*g^3*x^3 + 3*b*f*g^2*x^2 + 3*b*f^2*g*x + b*f^3)
*arcsin(c*x))*sqrt(-c^2*d*x^2 + d)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**3*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x + f\right )}^{3}{\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((g*x + f)^3*(b*arcsin(c*x) + a)/(-c^2*d*x^2 + d)^(5/2), x)