3.149 \(\int x (a+b \sin ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=76 \[ \frac {b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{2} x^2 \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{4} b^2 x^2 \]

[Out]

-1/4*b^2*x^2-1/4*(a+b*arcsin(c*x))^2/c^2+1/2*x^2*(a+b*arcsin(c*x))^2+1/2*b*x*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1
/2)/c

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Rubi [A]  time = 0.12, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4627, 4707, 4641, 30} \[ \frac {b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{2} x^2 \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{4} b^2 x^2 \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcSin[c*x])^2,x]

[Out]

-(b^2*x^2)/4 + (b*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(2*c) - (a + b*ArcSin[c*x])^2/(4*c^2) + (x^2*(a + b
*ArcSin[c*x])^2)/2

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4627

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

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]

Rule 4707

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 73, normalized size = 0.96 \[ -\frac {-2 c^2 x^2 \left (a+b \sin ^{-1}(c x)\right )^2-2 b c x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+\left (a+b \sin ^{-1}(c x)\right )^2+b^2 c^2 x^2}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcSin[c*x])^2,x]

[Out]

-1/4*(b^2*c^2*x^2 - 2*b*c*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]) + (a + b*ArcSin[c*x])^2 - 2*c^2*x^2*(a + b*A
rcSin[c*x])^2)/c^2

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fricas [A]  time = 1.09, size = 99, normalized size = 1.30 \[ \frac {{\left (2 \, a^{2} - b^{2}\right )} c^{2} x^{2} + {\left (2 \, b^{2} c^{2} x^{2} - b^{2}\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (2 \, a b c^{2} x^{2} - a b\right )} \arcsin \left (c x\right ) + 2 \, {\left (b^{2} c x \arcsin \left (c x\right ) + a b c x\right )} \sqrt {-c^{2} x^{2} + 1}}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsin(c*x))^2,x, algorithm="fricas")

[Out]

1/4*((2*a^2 - b^2)*c^2*x^2 + (2*b^2*c^2*x^2 - b^2)*arcsin(c*x)^2 + 2*(2*a*b*c^2*x^2 - a*b)*arcsin(c*x) + 2*(b^
2*c*x*arcsin(c*x) + a*b*c*x)*sqrt(-c^2*x^2 + 1))/c^2

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giac [B]  time = 0.26, size = 155, normalized size = 2.04 \[ \frac {\sqrt {-c^{2} x^{2} + 1} b^{2} x \arcsin \left (c x\right )}{2 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} b^{2} \arcsin \left (c x\right )^{2}}{2 \, c^{2}} + \frac {\sqrt {-c^{2} x^{2} + 1} a b x}{2 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} a b \arcsin \left (c x\right )}{c^{2}} + \frac {b^{2} \arcsin \left (c x\right )^{2}}{4 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )} a^{2}}{2 \, c^{2}} - \frac {{\left (c^{2} x^{2} - 1\right )} b^{2}}{4 \, c^{2}} + \frac {a b \arcsin \left (c x\right )}{2 \, c^{2}} - \frac {b^{2}}{8 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsin(c*x))^2,x, algorithm="giac")

[Out]

1/2*sqrt(-c^2*x^2 + 1)*b^2*x*arcsin(c*x)/c + 1/2*(c^2*x^2 - 1)*b^2*arcsin(c*x)^2/c^2 + 1/2*sqrt(-c^2*x^2 + 1)*
a*b*x/c + (c^2*x^2 - 1)*a*b*arcsin(c*x)/c^2 + 1/4*b^2*arcsin(c*x)^2/c^2 + 1/2*(c^2*x^2 - 1)*a^2/c^2 - 1/4*(c^2
*x^2 - 1)*b^2/c^2 + 1/2*a*b*arcsin(c*x)/c^2 - 1/8*b^2/c^2

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maple [A]  time = 0.04, size = 120, normalized size = 1.58 \[ \frac {\frac {a^{2} c^{2} x^{2}}{2}+b^{2} \left (\frac {\left (c^{2} x^{2}-1\right ) \arcsin \left (c x \right )^{2}}{2}+\frac {\arcsin \left (c x \right ) \left (c x \sqrt {-c^{2} x^{2}+1}+\arcsin \left (c x \right )\right )}{2}-\frac {\arcsin \left (c x \right )^{2}}{4}-\frac {c^{2} x^{2}}{4}\right )+2 a b \left (\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}-\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arcsin(c*x))^2,x)

[Out]

1/c^2*(1/2*a^2*c^2*x^2+b^2*(1/2*(c^2*x^2-1)*arcsin(c*x)^2+1/2*arcsin(c*x)*(c*x*(-c^2*x^2+1)^(1/2)+arcsin(c*x))
-1/4*arcsin(c*x)^2-1/4*c^2*x^2)+2*a*b*(1/2*c^2*x^2*arcsin(c*x)+1/4*c*x*(-c^2*x^2+1)^(1/2)-1/4*arcsin(c*x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a^{2} x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} a b + \frac {1}{2} \, {\left (x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, c \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{2} x^{2} - 1}\,{d x}\right )} b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsin(c*x))^2,x, algorithm="maxima")

[Out]

1/2*a^2*x^2 + 1/2*(2*x^2*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x/c^2 - arcsin(c*x)/c^3))*a*b + 1/2*(x^2*arctan2(
c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*c*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*x^2*arctan2(c*x, sqrt(c*x +
1)*sqrt(-c*x + 1))/(c^2*x^2 - 1), x))*b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*asin(c*x))^2,x)

[Out]

int(x*(a + b*asin(c*x))^2, x)

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sympy [A]  time = 0.55, size = 126, normalized size = 1.66 \[ \begin {cases} \frac {a^{2} x^{2}}{2} + a b x^{2} \operatorname {asin}{\left (c x \right )} + \frac {a b x \sqrt {- c^{2} x^{2} + 1}}{2 c} - \frac {a b \operatorname {asin}{\left (c x \right )}}{2 c^{2}} + \frac {b^{2} x^{2} \operatorname {asin}^{2}{\left (c x \right )}}{2} - \frac {b^{2} x^{2}}{4} + \frac {b^{2} x \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{2 c} - \frac {b^{2} \operatorname {asin}^{2}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asin(c*x))**2,x)

[Out]

Piecewise((a**2*x**2/2 + a*b*x**2*asin(c*x) + a*b*x*sqrt(-c**2*x**2 + 1)/(2*c) - a*b*asin(c*x)/(2*c**2) + b**2
*x**2*asin(c*x)**2/2 - b**2*x**2/4 + b**2*x*sqrt(-c**2*x**2 + 1)*asin(c*x)/(2*c) - b**2*asin(c*x)**2/(4*c**2),
Ne(c, 0)), (a**2*x**2/2, True))

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