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

Optimal. Leaf size=57 \[ \frac {a x^2}{2}+\frac {b \sqrt {1-\left (c+d x^2\right )^2}}{2 d}+\frac {b \left (c+d x^2\right ) \sin ^{-1}\left (c+d x^2\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6715, 4803, 4619, 261} \[ \frac {a x^2}{2}+\frac {b \sqrt {1-\left (c+d x^2\right )^2}}{2 d}+\frac {b \left (c+d x^2\right ) \sin ^{-1}\left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcSin[c + d*x^2]),x]

[Out]

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 4619

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

Rule 4803

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSin[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 70, normalized size = 1.23 \[ \frac {a x^2}{2}+\frac {b \left (\sqrt {-c^2-2 c d x^2-d^2 x^4+1}+c \sin ^{-1}\left (c+d x^2\right )\right )}{2 d}+\frac {1}{2} b x^2 \sin ^{-1}\left (c+d x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcSin[c + d*x^2]),x]

[Out]

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

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fricas [A]  time = 0.44, size = 57, normalized size = 1.00 \[ \frac {a d x^{2} + {\left (b d x^{2} + b c\right )} \arcsin \left (d x^{2} + c\right ) + \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} b}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 2.46, size = 49, normalized size = 0.86 \[ \frac {{\left (d x^{2} + c\right )} a + {\left ({\left (d x^{2} + c\right )} \arcsin \left (d x^{2} + c\right ) + \sqrt {-{\left (d x^{2} + c\right )}^{2} + 1}\right )} b}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 50, normalized size = 0.88 \[ \frac {a \left (d \,x^{2}+c \right )+b \left (\left (d \,x^{2}+c \right ) \arcsin \left (d \,x^{2}+c \right )+\sqrt {1-\left (d \,x^{2}+c \right )^{2}}\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 45, normalized size = 0.79 \[ \frac {1}{2} \, a x^{2} + \frac {{\left ({\left (d x^{2} + c\right )} \arcsin \left (d x^{2} + c\right ) + \sqrt {-{\left (d x^{2} + c\right )}^{2} + 1}\right )} b}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.73, size = 108, normalized size = 1.89 \[ \frac {a\,x^2}{2}+\frac {b\,x^2\,\mathrm {asin}\left (d\,x^2+c\right )}{2}+\frac {b\,\sqrt {-c^2-2\,c\,d\,x^2-d^2\,x^4+1}}{2\,d}+\frac {b\,c\,\ln \left (\sqrt {-c^2-2\,c\,d\,x^2-d^2\,x^4+1}-\frac {d^2\,x^2+c\,d}{\sqrt {-d^2}}\right )}{2\,\sqrt {-d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*asin(c + d*x^2)),x)

[Out]

(a*x^2)/2 + (b*x^2*asin(c + d*x^2))/2 + (b*(1 - d^2*x^4 - 2*c*d*x^2 - c^2)^(1/2))/(2*d) + (b*c*log((1 - d^2*x^
4 - 2*c*d*x^2 - c^2)^(1/2) - (c*d + d^2*x^2)/(-d^2)^(1/2)))/(2*(-d^2)^(1/2))

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sympy [A]  time = 0.25, size = 76, normalized size = 1.33 \[ \begin {cases} \frac {a x^{2}}{2} + \frac {b c \operatorname {asin}{\left (c + d x^{2} \right )}}{2 d} + \frac {b x^{2} \operatorname {asin}{\left (c + d x^{2} \right )}}{2} + \frac {b \sqrt {- c^{2} - 2 c d x^{2} - d^{2} x^{4} + 1}}{2 d} & \text {for}\: d \neq 0 \\\frac {x^{2} \left (a + b \operatorname {asin}{\relax (c )}\right )}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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