\(\int x (a+b \arcsin (c x)) \, dx\) [142]

Optimal result

Integrand size = 10, antiderivative size = 51 \[ \int x (a+b \arcsin (c x)) \, dx=\frac {b x \sqrt {1-c^2 x^2}}{4 c}-\frac {b \arcsin (c x)}{4 c^2}+\frac {1}{2} x^2 (a+b \arcsin (c x)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4723, 327, 222} \[ \int x (a+b \arcsin (c x)) \, dx=\frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {b \arcsin (c x)}{4 c^2}+\frac {b x \sqrt {1-c^2 x^2}}{4 c} \]

[In]

Int[x*(a + b*ArcSin[c*x]),x]

[Out]

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

Rule 222

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

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4723

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {1}{2} (b c) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx \\ & = \frac {b x \sqrt {1-c^2 x^2}}{4 c}+\frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {b \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 c} \\ & = \frac {b x \sqrt {1-c^2 x^2}}{4 c}-\frac {b \arcsin (c x)}{4 c^2}+\frac {1}{2} x^2 (a+b \arcsin (c x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10 \[ \int x (a+b \arcsin (c x)) \, dx=\frac {a x^2}{2}+\frac {b x \sqrt {1-c^2 x^2}}{4 c}-\frac {b \arcsin (c x)}{4 c^2}+\frac {1}{2} b x^2 \arcsin (c x) \]

[In]

Integrate[x*(a + b*ArcSin[c*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94

method	result	size
parts	\(\frac {a \,x^{2}}{2}+\frac {b \left (\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}-\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}}\)	\(48\)
derivativedivides	\(\frac {\frac {c^{2} x^{2} a}{2}+b \left (\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}-\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}}\)	\(52\)
default	\(\frac {\frac {c^{2} x^{2} a}{2}+b \left (\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}-\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}}\)	\(52\)

[In]

int(x*(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int x (a+b \arcsin (c x)) \, dx=\frac {2 \, a c^{2} x^{2} + \sqrt {-c^{2} x^{2} + 1} b c x + {\left (2 \, b c^{2} x^{2} - b\right )} \arcsin \left (c x\right )}{4 \, c^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06 \[ \int x (a+b \arcsin (c x)) \, dx=\begin {cases} \frac {a x^{2}}{2} + \frac {b x^{2} \operatorname {asin}{\left (c x \right )}}{2} + \frac {b x \sqrt {- c^{2} x^{2} + 1}}{4 c} - \frac {b \operatorname {asin}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {a x^{2}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int x (a+b \arcsin (c x)) \, dx=\frac {1}{2} \, a x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b \]

[In]

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

[Out]

1/2*a*x^2 + 1/4*(2*x^2*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x/c^2 - arcsin(c*x)/c^3))*b

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.25 \[ \int x (a+b \arcsin (c x)) \, dx=\frac {\sqrt {-c^{2} x^{2} + 1} b x}{4 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} b \arcsin \left (c x\right )}{2 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )} a}{2 \, c^{2}} + \frac {b \arcsin \left (c x\right )}{4 \, c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88 \[ \int x (a+b \arcsin (c x)) \, dx=\frac {a\,x^2}{2}+\frac {b\,\left (\frac {\mathrm {asin}\left (c\,x\right )\,\left (2\,c^2\,x^2-1\right )}{4}+\frac {c\,x\,\sqrt {1-c^2\,x^2}}{4}\right )}{c^2} \]

[In]

int(x*(a + b*asin(c*x)),x)

[Out]

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