Integrand size = 10, antiderivative size = 130 \[ \int x \arcsin (a+b x)^2 \, dx=\frac {2 a x}{b}-\frac {(a+b x)^2}{4 b^2}-\frac {2 a \sqrt {1-(a+b x)^2} \arcsin (a+b x)}{b^2}+\frac {(a+b x) \sqrt {1-(a+b x)^2} \arcsin (a+b x)}{2 b^2}-\frac {\arcsin (a+b x)^2}{4 b^2}-\frac {a^2 \arcsin (a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \arcsin (a+b x)^2 \]
2*a*x/b-1/4*(b*x+a)^2/b^2-1/4*arcsin(b*x+a)^2/b^2-1/2*a^2*arcsin(b*x+a)^2/ b^2+1/2*x^2*arcsin(b*x+a)^2-2*a*arcsin(b*x+a)*(1-(b*x+a)^2)^(1/2)/b^2+1/2* (b*x+a)*arcsin(b*x+a)*(1-(b*x+a)^2)^(1/2)/b^2
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.64 \[ \int x \arcsin (a+b x)^2 \, dx=\frac {b x (6 a-b x)-2 (3 a-b x) \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)+\left (-1-2 a^2+2 b^2 x^2\right ) \arcsin (a+b x)^2}{4 b^2} \]
(b*x*(6*a - b*x) - 2*(3*a - b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[ a + b*x] + (-1 - 2*a^2 + 2*b^2*x^2)*ArcSin[a + b*x]^2)/(4*b^2)
Time = 0.45 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5304, 25, 27, 5242, 5262, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x \arcsin (a+b x)^2 \, dx\) |
\(\Big \downarrow \) 5304 |
\(\displaystyle \frac {\int x \arcsin (a+b x)^2d(a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int -x \arcsin (a+b x)^2d(a+b x)}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int -b x \arcsin (a+b x)^2d(a+b x)}{b^2}\) |
\(\Big \downarrow \) 5242 |
\(\displaystyle -\frac {\int \frac {b^2 x^2 \arcsin (a+b x)}{\sqrt {1-(a+b x)^2}}d(a+b x)-\frac {1}{2} b^2 x^2 \arcsin (a+b x)^2}{b^2}\) |
\(\Big \downarrow \) 5262 |
\(\displaystyle -\frac {\int \left (\frac {\arcsin (a+b x) a^2}{\sqrt {1-(a+b x)^2}}-\frac {2 (a+b x) \arcsin (a+b x) a}{\sqrt {1-(a+b x)^2}}+\frac {(a+b x)^2 \arcsin (a+b x)}{\sqrt {1-(a+b x)^2}}\right )d(a+b x)-\frac {1}{2} b^2 x^2 \arcsin (a+b x)^2}{b^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{2} a^2 \arcsin (a+b x)^2-\frac {1}{2} b^2 x^2 \arcsin (a+b x)^2-\frac {1}{2} \sqrt {1-(a+b x)^2} (a+b x) \arcsin (a+b x)+\frac {1}{4} \arcsin (a+b x)^2+2 a \sqrt {1-(a+b x)^2} \arcsin (a+b x)+\frac {1}{4} (a+b x)^2-2 a (a+b x)}{b^2}\) |
-((-2*a*(a + b*x) + (a + b*x)^2/4 + 2*a*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b *x] - ((a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/2 + ArcSin[a + b*x ]^2/4 + (a^2*ArcSin[a + b*x]^2)/2 - (b^2*x^2*ArcSin[a + b*x]^2)/2)/b^2)
3.2.33.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(m_.), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(e*(m + 1))), x] - Simp[b*c*(n/(e*(m + 1))) Int[(d + e*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ ) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] & & EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ [n, 0] && (m == 1 || p > 0 || (n == 1 && p > -1) || (m == 2 && p < -2))
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A rcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.39 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {\frac {\left (-1+\left (b x +a \right )^{2}\right ) \arcsin \left (b x +a \right )^{2}}{2}+\frac {\arcsin \left (b x +a \right ) \left (\left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}+\arcsin \left (b x +a \right )\right )}{2}-\frac {\arcsin \left (b x +a \right )^{2}}{4}-\frac {\left (b x +a \right )^{2}}{4}-a \left (\arcsin \left (b x +a \right )^{2} \left (b x +a \right )-2 b x -2 a +2 \arcsin \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}\right )}{b^{2}}\) | \(124\) |
default | \(\frac {\frac {\left (-1+\left (b x +a \right )^{2}\right ) \arcsin \left (b x +a \right )^{2}}{2}+\frac {\arcsin \left (b x +a \right ) \left (\left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}+\arcsin \left (b x +a \right )\right )}{2}-\frac {\arcsin \left (b x +a \right )^{2}}{4}-\frac {\left (b x +a \right )^{2}}{4}-a \left (\arcsin \left (b x +a \right )^{2} \left (b x +a \right )-2 b x -2 a +2 \arcsin \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}\right )}{b^{2}}\) | \(124\) |
1/b^2*(1/2*(-1+(b*x+a)^2)*arcsin(b*x+a)^2+1/2*arcsin(b*x+a)*((b*x+a)*(1-(b *x+a)^2)^(1/2)+arcsin(b*x+a))-1/4*arcsin(b*x+a)^2-1/4*(b*x+a)^2-a*(arcsin( b*x+a)^2*(b*x+a)-2*b*x-2*a+2*arcsin(b*x+a)*(1-(b*x+a)^2)^(1/2)))
Time = 0.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62 \[ \int x \arcsin (a+b x)^2 \, dx=-\frac {b^{2} x^{2} - 6 \, a b x - {\left (2 \, b^{2} x^{2} - 2 \, a^{2} - 1\right )} \arcsin \left (b x + a\right )^{2} - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x - 3 \, a\right )} \arcsin \left (b x + a\right )}{4 \, b^{2}} \]
-1/4*(b^2*x^2 - 6*a*b*x - (2*b^2*x^2 - 2*a^2 - 1)*arcsin(b*x + a)^2 - 2*sq rt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x - 3*a)*arcsin(b*x + a))/b^2
Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.06 \[ \int x \arcsin (a+b x)^2 \, dx=\begin {cases} - \frac {a^{2} \operatorname {asin}^{2}{\left (a + b x \right )}}{2 b^{2}} + \frac {3 a x}{2 b} - \frac {3 a \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}{\left (a + b x \right )}}{2 b^{2}} + \frac {x^{2} \operatorname {asin}^{2}{\left (a + b x \right )}}{2} - \frac {x^{2}}{4} + \frac {x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}{\left (a + b x \right )}}{2 b} - \frac {\operatorname {asin}^{2}{\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {asin}^{2}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \]
Piecewise((-a**2*asin(a + b*x)**2/(2*b**2) + 3*a*x/(2*b) - 3*a*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/(2*b**2) + x**2*asin(a + b*x)**2/ 2 - x**2/4 + x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/(2*b) - asin(a + b*x)**2/(4*b**2), Ne(b, 0)), (x**2*asin(a)**2/2, True))
\[ \int x \arcsin (a+b x)^2 \, dx=\int { x \arcsin \left (b x + a\right )^{2} \,d x } \]
1/2*x^2*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2 + b*integ rate(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*x^2*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))/(b^2*x^2 + 2*a*b*x + a^2 - 1), x)
Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.07 \[ \int x \arcsin (a+b x)^2 \, dx=-\frac {{\left (b x + a\right )} a \arcsin \left (b x + a\right )^{2}}{b^{2}} + \frac {{\left ({\left (b x + a\right )}^{2} - 1\right )} \arcsin \left (b x + a\right )^{2}}{2 \, b^{2}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right )}{2 \, b^{2}} - \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} a \arcsin \left (b x + a\right )}{b^{2}} + \frac {2 \, {\left (b x + a\right )} a}{b^{2}} + \frac {\arcsin \left (b x + a\right )^{2}}{4 \, b^{2}} - \frac {{\left (b x + a\right )}^{2} - 1}{4 \, b^{2}} - \frac {1}{8 \, b^{2}} \]
-(b*x + a)*a*arcsin(b*x + a)^2/b^2 + 1/2*((b*x + a)^2 - 1)*arcsin(b*x + a) ^2/b^2 + 1/2*sqrt(-(b*x + a)^2 + 1)*(b*x + a)*arcsin(b*x + a)/b^2 - 2*sqrt (-(b*x + a)^2 + 1)*a*arcsin(b*x + a)/b^2 + 2*(b*x + a)*a/b^2 + 1/4*arcsin( b*x + a)^2/b^2 - 1/4*((b*x + a)^2 - 1)/b^2 - 1/8/b^2
Timed out. \[ \int x \arcsin (a+b x)^2 \, dx=\int x\,{\mathrm {asin}\left (a+b\,x\right )}^2 \,d x \]