Integrand size = 12, antiderivative size = 76 \[ \int x (a+b \arcsin (c x))^2 \, dx=-\frac {1}{4} b^2 x^2+\frac {b x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c}-\frac {(a+b \arcsin (c x))^2}{4 c^2}+\frac {1}{2} x^2 (a+b \arcsin (c x))^2 \] Output:
-1/4*b^2*x^2+1/2*b*x*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))/c-1/4*(a+b*arcsi n(c*x))^2/c^2+1/2*x^2*(a+b*arcsin(c*x))^2
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int x (a+b \arcsin (c x))^2 \, dx=-\frac {b^2 c^2 x^2-2 b c x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+(a+b \arcsin (c x))^2-2 c^2 x^2 (a+b \arcsin (c x))^2}{4 c^2} \] Input:
Integrate[x*(a + b*ArcSin[c*x])^2,x]
Output:
-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*ArcSin[c*x])^2)/c^2
Time = 0.39 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5138, 5210, 15, 5152}
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.
\(\displaystyle \int x (a+b \arcsin (c x))^2 \, dx\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {1}{2} x^2 (a+b \arcsin (c x))^2-b c \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx\) |
\(\Big \downarrow \) 5210 |
\(\displaystyle \frac {1}{2} x^2 (a+b \arcsin (c x))^2-b c \left (\frac {\int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}dx}{2 c^2}+\frac {b \int xdx}{2 c}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c^2}\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{2} x^2 (a+b \arcsin (c x))^2-b c \left (\frac {\int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c^2}+\frac {b x^2}{4 c}\right )\) |
\(\Big \downarrow \) 5152 |
\(\displaystyle \frac {1}{2} x^2 (a+b \arcsin (c x))^2-b c \left (\frac {(a+b \arcsin (c x))^2}{4 b c^3}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c^2}+\frac {b x^2}{4 c}\right )\) |
Input:
Int[x*(a + b*ArcSin[c*x])^2,x]
Output:
(x^2*(a + b*ArcSin[c*x])^2)/2 - b*c*((b*x^2)/(4*c) - (x*Sqrt[1 - c^2*x^2]* (a + b*ArcSin[c*x]))/(2*c^2) + (a + b*ArcSin[c*x])^2/(4*b*c^3))
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[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]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && NeQ[n, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(e*(m + 2*p + 1))), x] + (Simp[f^2*((m - 1)/(c^2*(m + 2*p + 1))) Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + S imp[b*f*(n/(c*(m + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f* x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m , 1] && NeQ[m + 2*p + 1, 0]
Time = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.28
method | result | size |
parts | \(\frac {a^{2} x^{2}}{2}+\frac {b^{2} \left (-\frac {\cos \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )^{2}}{4}+\frac {\cos \left (2 \arcsin \left (c x \right )\right )}{8}+\frac {\sin \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )}{4}\right )}{c^{2}}+\frac {2 a b \left (\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}-\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}}\) | \(97\) |
derivativedivides | \(\frac {\frac {c^{2} x^{2} a^{2}}{2}+b^{2} \left (-\frac {\cos \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )^{2}}{4}+\frac {\cos \left (2 \arcsin \left (c x \right )\right )}{8}+\frac {\sin \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )}{4}\right )+2 a b \left (\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}-\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}}\) | \(98\) |
default | \(\frac {\frac {c^{2} x^{2} a^{2}}{2}+b^{2} \left (-\frac {\cos \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )^{2}}{4}+\frac {\cos \left (2 \arcsin \left (c x \right )\right )}{8}+\frac {\sin \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )}{4}\right )+2 a b \left (\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}-\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}}\) | \(98\) |
orering | \(\frac {\left (7 c^{2} x^{2}-6\right ) \left (a +b \arcsin \left (c x \right )\right )^{2}}{8 c^{2}}-\frac {\left (3 c^{2} x^{2}-4\right ) \left (\left (a +b \arcsin \left (c x \right )\right )^{2}+\frac {2 x \left (a +b \arcsin \left (c x \right )\right ) b c}{\sqrt {-c^{2} x^{2}+1}}\right )}{8 c^{2}}+\frac {x \left (c x -1\right ) \left (c x +1\right ) \left (\frac {4 \left (a +b \arcsin \left (c x \right )\right ) b c}{\sqrt {-c^{2} x^{2}+1}}+\frac {2 x \,b^{2} c^{2}}{-c^{2} x^{2}+1}+\frac {2 x^{2} \left (a +b \arcsin \left (c x \right )\right ) b \,c^{3}}{\left (-c^{2} x^{2}+1\right )^{\frac {3}{2}}}\right )}{8 c^{2}}\) | \(169\) |
Input:
int(x*(a+b*arcsin(c*x))^2,x,method=_RETURNVERBOSE)
Output:
1/2*a^2*x^2+b^2/c^2*(-1/4*cos(2*arcsin(c*x))*arcsin(c*x)^2+1/8*cos(2*arcsi n(c*x))+1/4*sin(2*arcsin(c*x))*arcsin(c*x))+2*a*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))
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.30 \[ \int x (a+b \arcsin (c x))^2 \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} c^{2} x^{2} + {\left (2 \, b^{2} c^{2} x^{2} - b^{2}\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (2 \, a b c^{2} x^{2} - a b\right )} \arcsin \left (c x\right ) + 2 \, {\left (b^{2} c x \arcsin \left (c x\right ) + a b c x\right )} \sqrt {-c^{2} x^{2} + 1}}{4 \, c^{2}} \] Input:
integrate(x*(a+b*arcsin(c*x))^2,x, algorithm="fricas")
Output:
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
Time = 0.23 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.66 \[ \int x (a+b \arcsin (c x))^2 \, dx=\begin {cases} \frac {a^{2} x^{2}}{2} + a b x^{2} \operatorname {asin}{\left (c x \right )} + \frac {a b x \sqrt {- c^{2} x^{2} + 1}}{2 c} - \frac {a b \operatorname {asin}{\left (c x \right )}}{2 c^{2}} + \frac {b^{2} x^{2} \operatorname {asin}^{2}{\left (c x \right )}}{2} - \frac {b^{2} x^{2}}{4} + \frac {b^{2} x \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{2 c} - \frac {b^{2} \operatorname {asin}^{2}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{2}}{2} & \text {otherwise} \end {cases} \] Input:
integrate(x*(a+b*asin(c*x))**2,x)
Output:
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), N e(c, 0)), (a**2*x**2/2, True))
\[ \int x (a+b \arcsin (c x))^2 \, dx=\int { {\left (b \arcsin \left (c x\right ) + a\right )}^{2} x \,d x } \] Input:
integrate(x*(a+b*arcsin(c*x))^2,x, algorithm="maxima")
Output:
1/2*a^2*x^2 + 1/2*(2*x^2*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x/c^2 - arcsi n(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
Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (66) = 132\).
Time = 0.13 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.04 \[ \int x (a+b \arcsin (c x))^2 \, dx=\frac {\sqrt {-c^{2} x^{2} + 1} b^{2} x \arcsin \left (c x\right )}{2 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} b^{2} \arcsin \left (c x\right )^{2}}{2 \, c^{2}} + \frac {\sqrt {-c^{2} x^{2} + 1} a b x}{2 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} a b \arcsin \left (c x\right )}{c^{2}} + \frac {b^{2} \arcsin \left (c x\right )^{2}}{4 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )} a^{2}}{2 \, c^{2}} - \frac {{\left (c^{2} x^{2} - 1\right )} b^{2}}{4 \, c^{2}} + \frac {a b \arcsin \left (c x\right )}{2 \, c^{2}} - \frac {b^{2}}{8 \, c^{2}} \] Input:
integrate(x*(a+b*arcsin(c*x))^2,x, algorithm="giac")
Output:
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
Timed out. \[ \int x (a+b \arcsin (c x))^2 \, dx=\int x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2 \,d x \] Input:
int(x*(a + b*asin(c*x))^2,x)
Output:
int(x*(a + b*asin(c*x))^2, x)
Time = 0.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.54 \[ \int x (a+b \arcsin (c x))^2 \, dx=\frac {2 \mathit {asin} \left (c x \right )^{2} b^{2} c^{2} x^{2}-\mathit {asin} \left (c x \right )^{2} b^{2}+2 \sqrt {-c^{2} x^{2}+1}\, \mathit {asin} \left (c x \right ) b^{2} c x +4 \mathit {asin} \left (c x \right ) a b \,c^{2} x^{2}-2 \mathit {asin} \left (c x \right ) a b +2 \sqrt {-c^{2} x^{2}+1}\, a b c x +2 a^{2} c^{2} x^{2}-b^{2} c^{2} x^{2}}{4 c^{2}} \] Input:
int(x*(a+b*asin(c*x))^2,x)
Output:
(2*asin(c*x)**2*b**2*c**2*x**2 - asin(c*x)**2*b**2 + 2*sqrt( - c**2*x**2 + 1)*asin(c*x)*b**2*c*x + 4*asin(c*x)*a*b*c**2*x**2 - 2*asin(c*x)*a*b + 2*s qrt( - c**2*x**2 + 1)*a*b*c*x + 2*a**2*c**2*x**2 - b**2*c**2*x**2)/(4*c**2 )