3.599 \(\int x (d+e x^2) (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=122 \[ \frac {\left (d+e x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 e}+\frac {b x \sqrt {1-c^2 x^2} \left (d+e x^2\right )}{16 c}-\frac {b \left (8 c^4 d^2+8 c^2 d e+3 e^2\right ) \sin ^{-1}(c x)}{32 c^4 e}+\frac {3 b x \sqrt {1-c^2 x^2} \left (2 c^2 d+e\right )}{32 c^3} \]

[Out]

-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

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Rubi [A]  time = 0.09, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {4729, 416, 388, 216} \[ \frac {\left (d+e x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 e}-\frac {b \left (8 c^4 d^2+8 c^2 d e+3 e^2\right ) \sin ^{-1}(c x)}{32 c^4 e}+\frac {b x \sqrt {1-c^2 x^2} \left (d+e x^2\right )}{16 c}+\frac {3 b x \sqrt {1-c^2 x^2} \left (2 c^2 d+e\right )}{32 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x*(d + e*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

(3*b*(2*c^2*d + e)*x*Sqrt[1 - c^2*x^2])/(32*c^3) + (b*x*Sqrt[1 - c^2*x^2]*(d + e*x^2))/(16*c) - (b*(8*c^4*d^2
+ 8*c^2*d*e + 3*e^2)*ArcSin[c*x])/(32*c^4*e) + ((d + e*x^2)^2*(a + b*ArcSin[c*x]))/(4*e)

Rule 216

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 388

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

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 4729

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] - Dist[(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]

Rubi steps

\begin {align*} \int x \left (d+e x^2\right ) \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {\left (d+e x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 e}-\frac {(b c) \int \frac {\left (d+e x^2\right )^2}{\sqrt {1-c^2 x^2}} \, dx}{4 e}\\ &=\frac {b x \sqrt {1-c^2 x^2} \left (d+e x^2\right )}{16 c}+\frac {\left (d+e x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 e}+\frac {b \int \frac {-d \left (4 c^2 d+e\right )-3 e \left (2 c^2 d+e\right ) x^2}{\sqrt {1-c^2 x^2}} \, dx}{16 c e}\\ &=\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 {\left (d+e x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 e}-\frac {\left (b \left (8 c^4 d^2+8 c^2 d e+3 e^2\right )\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{32 c^3 e}\\ &=\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 ) \sin ^{-1}(c x)}{32 c^4 e}+\frac {\left (d+e x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 e}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 95, normalized size = 0.78 \[ \frac {c x \left (8 a c^3 x \left (2 d+e x^2\right )+b \sqrt {1-c^2 x^2} \left (2 c^2 \left (4 d+e x^2\right )+3 e\right )\right )+b \sin ^{-1}(c x) \left (8 c^4 \left (2 d x^2+e x^4\right )-8 c^2 d-3 e\right )}{32 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d + e*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

(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)

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fricas [A]  time = 0.64, size = 102, normalized size = 0.84 \[ \frac {8 \, a c^{4} e x^{4} + 16 \, a c^{4} d x^{2} + {\left (8 \, b c^{4} e x^{4} + 16 \, b c^{4} d x^{2} - 8 \, b c^{2} d - 3 \, b e\right )} \arcsin \left (c x\right ) + {\left (2 \, b c^{3} e x^{3} + {\left (8 \, b c^{3} d + 3 \, b c e\right )} x\right )} \sqrt {-c^{2} x^{2} + 1}}{32 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.32, size = 174, normalized size = 1.43 \[ \frac {1}{4} \, a x^{4} e + \frac {\sqrt {-c^{2} x^{2} + 1} b d x}{4 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} b d \arcsin \left (c x\right )}{2 \, c^{2}} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b x e}{16 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )} a d}{2 \, c^{2}} + \frac {b d \arcsin \left (c x\right )}{4 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b \arcsin \left (c x\right ) e}{4 \, c^{4}} + \frac {5 \, \sqrt {-c^{2} x^{2} + 1} b x e}{32 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )} b \arcsin \left (c x\right ) e}{2 \, c^{4}} + \frac {5 \, b \arcsin \left (c x\right ) e}{32 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

1/4*a*x^4*e + 1/4*sqrt(-c^2*x^2 + 1)*b*d*x/c + 1/2*(c^2*x^2 - 1)*b*d*arcsin(c*x)/c^2 - 1/16*(-c^2*x^2 + 1)^(3/
2)*b*x*e/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*arcsin(c*x)*e/c^4 +
5/32*sqrt(-c^2*x^2 + 1)*b*x*e/c^3 + 1/2*(c^2*x^2 - 1)*b*arcsin(c*x)*e/c^4 + 5/32*b*arcsin(c*x)*e/c^4

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maple [A]  time = 0.00, size = 137, normalized size = 1.12 \[ \frac {\frac {a \left (\frac {1}{4} e \,c^{4} x^{4}+\frac {1}{2} x^{2} c^{4} d \right )}{c^{2}}+\frac {b \left (\frac {\arcsin \left (c x \right ) e \,c^{4} x^{4}}{4}+\frac {\arcsin \left (c x \right ) c^{4} d \,x^{2}}{2}-\frac {e \left (-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4}-\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{8}+\frac {3 \arcsin \left (c x \right )}{8}\right )}{4}-\frac {c^{2} d \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{2}\right )}{c^{2}}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x^2+d)*(a+b*arcsin(c*x)),x)

[Out]

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

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maxima [A]  time = 1.24, size = 122, normalized size = 1.00 \[ \frac {1}{4} \, a e x^{4} + \frac {1}{2} \, a d x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b d + \frac {1}{32} \, {\left (8 \, x^{4} \arcsin \left (c x\right ) + {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{4}} - \frac {3 \, \arcsin \left (c x\right )}{c^{5}}\right )} c\right )} b e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

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 + 1/3
2*(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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (e\,x^2+d\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*asin(c*x))*(d + e*x^2),x)

[Out]

int(x*(a + b*asin(c*x))*(d + e*x^2), x)

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sympy [A]  time = 1.14, size = 153, normalized size = 1.25 \[ \begin {cases} \frac {a d x^{2}}{2} + \frac {a e x^{4}}{4} + \frac {b d x^{2} \operatorname {asin}{\left (c x \right )}}{2} + \frac {b e x^{4} \operatorname {asin}{\left (c x \right )}}{4} + \frac {b d x \sqrt {- c^{2} x^{2} + 1}}{4 c} + \frac {b e x^{3} \sqrt {- c^{2} x^{2} + 1}}{16 c} - \frac {b d \operatorname {asin}{\left (c x \right )}}{4 c^{2}} + \frac {3 b e x \sqrt {- c^{2} x^{2} + 1}}{32 c^{3}} - \frac {3 b e \operatorname {asin}{\left (c x \right )}}{32 c^{4}} & \text {for}\: c \neq 0 \\a \left (\frac {d x^{2}}{2} + \frac {e x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x**2+d)*(a+b*asin(c*x)),x)

[Out]

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))

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