Integrand size = 17, antiderivative size = 122 \[ \int x \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {3 b \left (2 c^2 d+e\right ) x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b x \sqrt {1-c^2 x^2} \left (d+e x^2\right )}{16 c}-\frac {b \left (8 c^4 d^2+8 c^2 d e+3 e^2\right ) \arcsin (c x)}{32 c^4 e}+\frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{4 e} \]
-1/32*b*(8*c^4*d^2+8*c^2*d*e+3*e^2)*arcsin(c*x)/c^4/e+1/4*(e*x^2+d)^2*(a+b *arcsin(c*x))/e+3/32*b*(2*c^2*d+e)*x*(-c^2*x^2+1)^(1/2)/c^3+1/16*b*x*(e*x^ 2+d)*(-c^2*x^2+1)^(1/2)/c
Time = 0.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.78 \[ \int x \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {c x \left (8 a c^3 x \left (2 d+e x^2\right )+b \sqrt {1-c^2 x^2} \left (3 e+2 c^2 \left (4 d+e x^2\right )\right )\right )+b \left (-8 c^2 d-3 e+8 c^4 \left (2 d x^2+e x^4\right )\right ) \arcsin (c x)}{32 c^4} \]
(c*x*(8*a*c^3*x*(2*d + e*x^2) + b*Sqrt[1 - c^2*x^2]*(3*e + 2*c^2*(4*d + e* x^2))) + b*(-8*c^2*d - 3*e + 8*c^4*(2*d*x^2 + e*x^4))*ArcSin[c*x])/(32*c^4 )
Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5228, 318, 25, 299, 223}
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.
\(\displaystyle \int x \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx\) |
\(\Big \downarrow \) 5228 |
\(\displaystyle \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{4 e}-\frac {b c \int \frac {\left (e x^2+d\right )^2}{\sqrt {1-c^2 x^2}}dx}{4 e}\) |
\(\Big \downarrow \) 318 |
\(\displaystyle \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{4 e}-\frac {b c \left (-\frac {\int -\frac {3 e \left (2 d c^2+e\right ) x^2+d \left (4 d c^2+e\right )}{\sqrt {1-c^2 x^2}}dx}{4 c^2}-\frac {e x \sqrt {1-c^2 x^2} \left (d+e x^2\right )}{4 c^2}\right )}{4 e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{4 e}-\frac {b c \left (\frac {\int \frac {3 e \left (2 d c^2+e\right ) x^2+d \left (4 d c^2+e\right )}{\sqrt {1-c^2 x^2}}dx}{4 c^2}-\frac {e x \sqrt {1-c^2 x^2} \left (d+e x^2\right )}{4 c^2}\right )}{4 e}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{4 e}-\frac {b c \left (\frac {\frac {\left (8 c^4 d^2+8 c^2 d e+3 e^2\right ) \int \frac {1}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {3 e x \sqrt {1-c^2 x^2} \left (2 c^2 d+e\right )}{2 c^2}}{4 c^2}-\frac {e x \sqrt {1-c^2 x^2} \left (d+e x^2\right )}{4 c^2}\right )}{4 e}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{4 e}-\frac {b c \left (\frac {\frac {\arcsin (c x) \left (8 c^4 d^2+8 c^2 d e+3 e^2\right )}{2 c^3}-\frac {3 e x \sqrt {1-c^2 x^2} \left (2 c^2 d+e\right )}{2 c^2}}{4 c^2}-\frac {e x \sqrt {1-c^2 x^2} \left (d+e x^2\right )}{4 c^2}\right )}{4 e}\) |
((d + e*x^2)^2*(a + b*ArcSin[c*x]))/(4*e) - (b*c*(-1/4*(e*x*Sqrt[1 - c^2*x ^2]*(d + e*x^2))/c^2 + ((-3*e*(2*c^2*d + e)*x*Sqrt[1 - c^2*x^2])/(2*c^2) + ((8*c^4*d^2 + 8*c^2*d*e + 3*e^2)*ArcSin[c*x])/(2*c^3))/(4*c^2)))/(4*e)
3.6.99.3.1 Defintions of rubi rules used
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]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])/(2*e*(p + 1))), x] - Simp[b*(c/(2*e*(p + 1))) Int[(d + e*x^2)^(p + 1)/Sqrt[1 - c^2*x^2], x] , x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[c^2*d + e, 0] && NeQ[p, -1]
Time = 0.20 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.32
method | result | size |
parts | \(\frac {a \left (e \,x^{2}+d \right )^{2}}{4 e}+\frac {b \left (\frac {c^{2} e \arcsin \left (c x \right ) x^{4}}{4}+\frac {\arcsin \left (c x \right ) c^{2} x^{2} d}{2}+\frac {c^{2} \arcsin \left (c x \right ) d^{2}}{4 e}-\frac {c^{4} d^{2} \arcsin \left (c x \right )+e^{2} \left (-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4}-\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{8}+\frac {3 \arcsin \left (c x \right )}{8}\right )+2 d \,c^{2} e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{4 c^{2} e}\right )}{c^{2}}\) | \(161\) |
derivativedivides | \(\frac {\frac {a \left (c^{2} e \,x^{2}+c^{2} d \right )^{2}}{4 c^{2} e}+\frac {b \left (\frac {\arcsin \left (c x \right ) c^{4} d^{2}}{4 e}+\frac {\arcsin \left (c x \right ) c^{4} d \,x^{2}}{2}+\frac {e \arcsin \left (c x \right ) c^{4} x^{4}}{4}-\frac {c^{4} d^{2} \arcsin \left (c x \right )+e^{2} \left (-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4}-\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{8}+\frac {3 \arcsin \left (c x \right )}{8}\right )+2 d \,c^{2} e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{4 e}\right )}{c^{2}}}{c^{2}}\) | \(172\) |
default | \(\frac {\frac {a \left (c^{2} e \,x^{2}+c^{2} d \right )^{2}}{4 c^{2} e}+\frac {b \left (\frac {\arcsin \left (c x \right ) c^{4} d^{2}}{4 e}+\frac {\arcsin \left (c x \right ) c^{4} d \,x^{2}}{2}+\frac {e \arcsin \left (c x \right ) c^{4} x^{4}}{4}-\frac {c^{4} d^{2} \arcsin \left (c x \right )+e^{2} \left (-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4}-\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{8}+\frac {3 \arcsin \left (c x \right )}{8}\right )+2 d \,c^{2} e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{4 e}\right )}{c^{2}}}{c^{2}}\) | \(172\) |
1/4*a*(e*x^2+d)^2/e+b/c^2*(1/4*c^2*e*arcsin(c*x)*x^4+1/2*arcsin(c*x)*c^2*x ^2*d+1/4*c^2/e*arcsin(c*x)*d^2-1/4/c^2/e*(c^4*d^2*arcsin(c*x)+e^2*(-1/4*c^ 3*x^3*(-c^2*x^2+1)^(1/2)-3/8*c*x*(-c^2*x^2+1)^(1/2)+3/8*arcsin(c*x))+2*d*c ^2*e*(-1/2*c*x*(-c^2*x^2+1)^(1/2)+1/2*arcsin(c*x))))
Time = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.84 \[ \int x \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {8 \, a c^{4} e x^{4} + 16 \, a c^{4} d x^{2} + {\left (8 \, b c^{4} e x^{4} + 16 \, b c^{4} d x^{2} - 8 \, b c^{2} d - 3 \, b e\right )} \arcsin \left (c x\right ) + {\left (2 \, b c^{3} e x^{3} + {\left (8 \, b c^{3} d + 3 \, b c e\right )} x\right )} \sqrt {-c^{2} x^{2} + 1}}{32 \, c^{4}} \]
1/32*(8*a*c^4*e*x^4 + 16*a*c^4*d*x^2 + (8*b*c^4*e*x^4 + 16*b*c^4*d*x^2 - 8 *b*c^2*d - 3*b*e)*arcsin(c*x) + (2*b*c^3*e*x^3 + (8*b*c^3*d + 3*b*c*e)*x)* sqrt(-c^2*x^2 + 1))/c^4
Time = 0.33 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.25 \[ \int x \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\begin {cases} \frac {a d x^{2}}{2} + \frac {a e x^{4}}{4} + \frac {b d x^{2} \operatorname {asin}{\left (c x \right )}}{2} + \frac {b e x^{4} \operatorname {asin}{\left (c x \right )}}{4} + \frac {b d x \sqrt {- c^{2} x^{2} + 1}}{4 c} + \frac {b e x^{3} \sqrt {- c^{2} x^{2} + 1}}{16 c} - \frac {b d \operatorname {asin}{\left (c x \right )}}{4 c^{2}} + \frac {3 b e x \sqrt {- c^{2} x^{2} + 1}}{32 c^{3}} - \frac {3 b e \operatorname {asin}{\left (c x \right )}}{32 c^{4}} & \text {for}\: c \neq 0 \\a \left (\frac {d x^{2}}{2} + \frac {e x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]
Piecewise((a*d*x**2/2 + a*e*x**4/4 + b*d*x**2*asin(c*x)/2 + b*e*x**4*asin( c*x)/4 + b*d*x*sqrt(-c**2*x**2 + 1)/(4*c) + b*e*x**3*sqrt(-c**2*x**2 + 1)/ (16*c) - b*d*asin(c*x)/(4*c**2) + 3*b*e*x*sqrt(-c**2*x**2 + 1)/(32*c**3) - 3*b*e*asin(c*x)/(32*c**4), Ne(c, 0)), (a*(d*x**2/2 + e*x**4/4), True))
Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00 \[ \int x \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {1}{4} \, a e x^{4} + \frac {1}{2} \, a d x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b d + \frac {1}{32} \, {\left (8 \, x^{4} \arcsin \left (c x\right ) + {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{4}} - \frac {3 \, \arcsin \left (c x\right )}{c^{5}}\right )} c\right )} b e \]
1/4*a*e*x^4 + 1/2*a*d*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*d + 1/32*(8*x^4*arcsin(c*x) + (2*sqrt(-c^2*x^ 2 + 1)*x^3/c^2 + 3*sqrt(-c^2*x^2 + 1)*x/c^4 - 3*arcsin(c*x)/c^5)*c)*b*e
Time = 0.29 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.38 \[ \int x \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {1}{4} \, a e x^{4} + \frac {\sqrt {-c^{2} x^{2} + 1} b d x}{4 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} b d \arcsin \left (c x\right )}{2 \, c^{2}} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b e x}{16 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )} a d}{2 \, c^{2}} + \frac {b d \arcsin \left (c x\right )}{4 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b e \arcsin \left (c x\right )}{4 \, c^{4}} + \frac {5 \, \sqrt {-c^{2} x^{2} + 1} b e x}{32 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )} b e \arcsin \left (c x\right )}{2 \, c^{4}} + \frac {5 \, b e \arcsin \left (c x\right )}{32 \, c^{4}} \]
1/4*a*e*x^4 + 1/4*sqrt(-c^2*x^2 + 1)*b*d*x/c + 1/2*(c^2*x^2 - 1)*b*d*arcsi n(c*x)/c^2 - 1/16*(-c^2*x^2 + 1)^(3/2)*b*e*x/c^3 + 1/2*(c^2*x^2 - 1)*a*d/c ^2 + 1/4*b*d*arcsin(c*x)/c^2 + 1/4*(c^2*x^2 - 1)^2*b*e*arcsin(c*x)/c^4 + 5 /32*sqrt(-c^2*x^2 + 1)*b*e*x/c^3 + 1/2*(c^2*x^2 - 1)*b*e*arcsin(c*x)/c^4 + 5/32*b*e*arcsin(c*x)/c^4
Timed out. \[ \int x \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int x\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (e\,x^2+d\right ) \,d x \]