Integrand size = 16, antiderivative size = 81 \[ \int \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {b \left (3 c^2 d+e\right ) \sqrt {1-c^2 x^2}}{3 c^3}-\frac {b e \left (1-c^2 x^2\right )^{3/2}}{9 c^3}+d x (a+b \arcsin (c x))+\frac {1}{3} e x^3 (a+b \arcsin (c x)) \] Output:
1/3*b*(3*c^2*d+e)*(-c^2*x^2+1)^(1/2)/c^3-1/9*b*e*(-c^2*x^2+1)^(3/2)/c^3+d* x*(a+b*arcsin(c*x))+1/3*e*x^3*(a+b*arcsin(c*x))
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {1}{9} \left (3 a x \left (3 d+e x^2\right )+\frac {b \sqrt {1-c^2 x^2} \left (2 e+c^2 \left (9 d+e x^2\right )\right )}{c^3}+3 b x \left (3 d+e x^2\right ) \arcsin (c x)\right ) \] Input:
Integrate[(d + e*x^2)*(a + b*ArcSin[c*x]),x]
Output:
(3*a*x*(3*d + e*x^2) + (b*Sqrt[1 - c^2*x^2]*(2*e + c^2*(9*d + e*x^2)))/c^3 + 3*b*x*(3*d + e*x^2)*ArcSin[c*x])/9
Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5170, 27, 353, 53, 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.
\(\displaystyle \int \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx\) |
\(\Big \downarrow \) 5170 |
\(\displaystyle -b c \int \frac {x \left (e x^2+3 d\right )}{3 \sqrt {1-c^2 x^2}}dx+d x (a+b \arcsin (c x))+\frac {1}{3} e x^3 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{3} b c \int \frac {x \left (e x^2+3 d\right )}{\sqrt {1-c^2 x^2}}dx+d x (a+b \arcsin (c x))+\frac {1}{3} e x^3 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 353 |
\(\displaystyle -\frac {1}{6} b c \int \frac {e x^2+3 d}{\sqrt {1-c^2 x^2}}dx^2+d x (a+b \arcsin (c x))+\frac {1}{3} e x^3 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {1}{6} b c \int \left (\frac {3 d c^2+e}{c^2 \sqrt {1-c^2 x^2}}-\frac {e \sqrt {1-c^2 x^2}}{c^2}\right )dx^2+d x (a+b \arcsin (c x))+\frac {1}{3} e x^3 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle d x (a+b \arcsin (c x))+\frac {1}{3} e x^3 (a+b \arcsin (c x))-\frac {1}{6} b c \left (\frac {2 e \left (1-c^2 x^2\right )^{3/2}}{3 c^4}-\frac {2 \sqrt {1-c^2 x^2} \left (3 c^2 d+e\right )}{c^4}\right )\) |
Input:
Int[(d + e*x^2)*(a + b*ArcSin[c*x]),x]
Output:
-1/6*(b*c*((-2*(3*c^2*d + e)*Sqrt[1 - c^2*x^2])/c^4 + (2*e*(1 - c^2*x^2)^( 3/2))/(3*c^4))) + d*x*(a + b*ArcSin[c*x]) + (e*x^3*(a + b*ArcSin[c*x]))/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ {a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbo l] :> With[{u = IntHide[(d + e*x^2)^p, x]}, Simp[(a + b*ArcSin[c*x]) u, x ] - Simp[b*c Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; Fr eeQ[{a, b, c, d, e}, x] && NeQ[c^2*d + e, 0] && (IGtQ[p, 0] || ILtQ[p + 1/2 , 0])
Time = 0.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.23
method | result | size |
parts | \(a \left (\frac {1}{3} e \,x^{3}+d x \right )+\frac {b \left (\frac {c \arcsin \left (c x \right ) e \,x^{3}}{3}+\arcsin \left (c x \right ) c x d -\frac {e \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )-3 d \,c^{2} \sqrt {-c^{2} x^{2}+1}}{3 c^{2}}\right )}{c}\) | \(100\) |
derivativedivides | \(\frac {\frac {a \left (c^{3} d x +\frac {1}{3} e \,c^{3} x^{3}\right )}{c^{2}}+\frac {b \left (\arcsin \left (c x \right ) d \,c^{3} x +\frac {\arcsin \left (c x \right ) e \,c^{3} x^{3}}{3}-\frac {e \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}+d \,c^{2} \sqrt {-c^{2} x^{2}+1}\right )}{c^{2}}}{c}\) | \(111\) |
default | \(\frac {\frac {a \left (c^{3} d x +\frac {1}{3} e \,c^{3} x^{3}\right )}{c^{2}}+\frac {b \left (\arcsin \left (c x \right ) d \,c^{3} x +\frac {\arcsin \left (c x \right ) e \,c^{3} x^{3}}{3}-\frac {e \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}+d \,c^{2} \sqrt {-c^{2} x^{2}+1}\right )}{c^{2}}}{c}\) | \(111\) |
orering | \(\frac {x \left (5 e^{2} x^{4} c^{4}+30 c^{4} d e \,x^{2}+9 c^{4} d^{2}+2 c^{2} e^{2} x^{2}-18 c^{2} d e -4 e^{2}\right ) \left (a +b \arcsin \left (c x \right )\right )}{9 \left (e \,x^{2}+d \right ) c^{4}}-\frac {\left (c^{2} e \,x^{2}+9 c^{2} d +2 e \right ) \left (c x -1\right ) \left (c x +1\right ) \left (2 e x \left (a +b \arcsin \left (c x \right )\right )+\frac {\left (e \,x^{2}+d \right ) b c}{\sqrt {-c^{2} x^{2}+1}}\right )}{9 c^{4} \left (e \,x^{2}+d \right )}\) | \(155\) |
Input:
int((e*x^2+d)*(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)
Output:
a*(1/3*e*x^3+d*x)+b/c*(1/3*c*arcsin(c*x)*e*x^3+arcsin(c*x)*c*x*d-1/3/c^2*( e*(-1/3*c^2*x^2*(-c^2*x^2+1)^(1/2)-2/3*(-c^2*x^2+1)^(1/2))-3*d*c^2*(-c^2*x ^2+1)^(1/2)))
Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01 \[ \int \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {3 \, a c^{3} e x^{3} + 9 \, a c^{3} d x + 3 \, {\left (b c^{3} e x^{3} + 3 \, b c^{3} d x\right )} \arcsin \left (c x\right ) + {\left (b c^{2} e x^{2} + 9 \, b c^{2} d + 2 \, b e\right )} \sqrt {-c^{2} x^{2} + 1}}{9 \, c^{3}} \] Input:
integrate((e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="fricas")
Output:
1/9*(3*a*c^3*e*x^3 + 9*a*c^3*d*x + 3*(b*c^3*e*x^3 + 3*b*c^3*d*x)*arcsin(c* x) + (b*c^2*e*x^2 + 9*b*c^2*d + 2*b*e)*sqrt(-c^2*x^2 + 1))/c^3
Time = 0.26 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.35 \[ \int \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\begin {cases} a d x + \frac {a e x^{3}}{3} + b d x \operatorname {asin}{\left (c x \right )} + \frac {b e x^{3} \operatorname {asin}{\left (c x \right )}}{3} + \frac {b d \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {b e x^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c} + \frac {2 b e \sqrt {- c^{2} x^{2} + 1}}{9 c^{3}} & \text {for}\: c \neq 0 \\a \left (d x + \frac {e x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((e*x**2+d)*(a+b*asin(c*x)),x)
Output:
Piecewise((a*d*x + a*e*x**3/3 + b*d*x*asin(c*x) + b*e*x**3*asin(c*x)/3 + b *d*sqrt(-c**2*x**2 + 1)/c + b*e*x**2*sqrt(-c**2*x**2 + 1)/(9*c) + 2*b*e*sq rt(-c**2*x**2 + 1)/(9*c**3), Ne(c, 0)), (a*(d*x + e*x**3/3), True))
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12 \[ \int \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {1}{3} \, a e x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b e + a d x + \frac {{\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d}{c} \] Input:
integrate((e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="maxima")
Output:
1/3*a*e*x^3 + 1/9*(3*x^3*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*s qrt(-c^2*x^2 + 1)/c^4))*b*e + a*d*x + (c*x*arcsin(c*x) + sqrt(-c^2*x^2 + 1 ))*b*d/c
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.35 \[ \int \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {1}{3} \, a e x^{3} + b d x \arcsin \left (c x\right ) + a d x + \frac {{\left (c^{2} x^{2} - 1\right )} b e x \arcsin \left (c x\right )}{3 \, c^{2}} + \frac {b e x \arcsin \left (c x\right )}{3 \, c^{2}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d}{c} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b e}{9 \, c^{3}} + \frac {\sqrt {-c^{2} x^{2} + 1} b e}{3 \, c^{3}} \] Input:
integrate((e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="giac")
Output:
1/3*a*e*x^3 + b*d*x*arcsin(c*x) + a*d*x + 1/3*(c^2*x^2 - 1)*b*e*x*arcsin(c *x)/c^2 + 1/3*b*e*x*arcsin(c*x)/c^2 + sqrt(-c^2*x^2 + 1)*b*d/c - 1/9*(-c^2 *x^2 + 1)^(3/2)*b*e/c^3 + 1/3*sqrt(-c^2*x^2 + 1)*b*e/c^3
Timed out. \[ \int \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\left \{\begin {array}{cl} b\,e\,\left (\frac {\sqrt {\frac {1}{c^2}-x^2}\,\left (\frac {2}{c^2}+x^2\right )}{9}+\frac {x^3\,\mathrm {asin}\left (c\,x\right )}{3}\right )+\frac {a\,x\,\left (e\,x^2+3\,d\right )}{3}+\frac {b\,d\,\left (\sqrt {1-c^2\,x^2}+c\,x\,\mathrm {asin}\left (c\,x\right )\right )}{c} & \text {\ if\ \ }0<c\\ \int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (e\,x^2+d\right ) \,d x & \text {\ if\ \ }\neg 0<c \end {array}\right . \] Input:
int((a + b*asin(c*x))*(d + e*x^2),x)
Output:
piecewise(0 < c, b*e*(((1/c^2 - x^2)^(1/2)*(2/c^2 + x^2))/9 + (x^3*asin(c* x))/3) + (a*x*(3*d + e*x^2))/3 + (b*d*((- c^2*x^2 + 1)^(1/2) + c*x*asin(c* x)))/c, ~0 < c, int((a + b*asin(c*x))*(d + e*x^2), x))
Time = 0.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.27 \[ \int \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {9 \mathit {asin} \left (c x \right ) b \,c^{3} d x +3 \mathit {asin} \left (c x \right ) b \,c^{3} e \,x^{3}+9 \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} d +\sqrt {-c^{2} x^{2}+1}\, b \,c^{2} e \,x^{2}+2 \sqrt {-c^{2} x^{2}+1}\, b e +9 a \,c^{3} d x +3 a \,c^{3} e \,x^{3}}{9 c^{3}} \] Input:
int((e*x^2+d)*(a+b*asin(c*x)),x)
Output:
(9*asin(c*x)*b*c**3*d*x + 3*asin(c*x)*b*c**3*e*x**3 + 9*sqrt( - c**2*x**2 + 1)*b*c**2*d + sqrt( - c**2*x**2 + 1)*b*c**2*e*x**2 + 2*sqrt( - c**2*x**2 + 1)*b*e + 9*a*c**3*d*x + 3*a*c**3*e*x**3)/(9*c**3)