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

Optimal. Leaf size=81 \[ d x \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {b \sqrt {1-c^2 x^2} \left (3 c^2 d+e\right )}{3 c^3}-\frac {b e \left (1-c^2 x^2\right )^{3/2}}{9 c^3} \]

[Out]

-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))+1/3*b*(3*c^2*d+e)*(-c^2*x^2+
1)^(1/2)/c^3

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Rubi [A]  time = 0.07, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {4665, 444, 43} \[ d x \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {b \sqrt {1-c^2 x^2} \left (3 c^2 d+e\right )}{3 c^3}-\frac {b e \left (1-c^2 x^2\right )^{3/2}}{9 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(3*c^2*d + e)*Sqrt[1 - c^2*x^2])/(3*c^3) - (b*e*(1 - c^2*x^2)^(3/2))/(9*c^3) + d*x*(a + b*ArcSin[c*x]) + (e
*x^3*(a + b*ArcSin[c*x]))/3

Rule 43

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] && NeQ[b*c - a*d, 0] && 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])

Rule 444

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

Rule 4665

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2)
^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; F
reeQ[{a, b, c, d, e}, x] && NeQ[c^2*d + e, 0] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 71, normalized size = 0.88 \[ \frac {1}{9} \left (3 a x \left (3 d+e x^2\right )+\frac {b \sqrt {1-c^2 x^2} \left (c^2 \left (9 d+e x^2\right )+2 e\right )}{c^3}+3 b x \sin ^{-1}(c x) \left (3 d+e x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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

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fricas [A]  time = 0.94, size = 82, normalized size = 1.01 \[ \frac {3 \, a c^{3} e x^{3} + 9 \, a c^{3} d x + 3 \, {\left (b c^{3} e x^{3} + 3 \, b c^{3} d x\right )} \arcsin \left (c x\right ) + {\left (b c^{2} e x^{2} + 9 \, b c^{2} d + 2 \, b e\right )} \sqrt {-c^{2} x^{2} + 1}}{9 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.28, size = 114, normalized size = 1.41 \[ \frac {1}{3} \, a x^{3} e + b d x \arcsin \left (c x\right ) + a d x + \frac {{\left (c^{2} x^{2} - 1\right )} b x \arcsin \left (c x\right ) e}{3 \, c^{2}} + \frac {b x \arcsin \left (c x\right ) e}{3 \, c^{2}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d}{c} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b e}{9 \, c^{3}} + \frac {\sqrt {-c^{2} x^{2} + 1} b e}{3 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3*e + b*d*x*arcsin(c*x) + a*d*x + 1/3*(c^2*x^2 - 1)*b*x*arcsin(c*x)*e/c^2 + 1/3*b*x*arcsin(c*x)*e/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

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maple [A]  time = 0.00, size = 111, normalized size = 1.37 \[ \frac {\frac {a \left (\frac {1}{3} c^{3} x^{3} e +c^{3} d x \right )}{c^{2}}+\frac {b \left (\frac {\arcsin \left (c x \right ) c^{3} x^{3} e}{3}+\arcsin \left (c x \right ) c^{3} d x -\frac {e \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}+c^{2} d \sqrt {-c^{2} x^{2}+1}\right )}{c^{2}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(a/c^2*(1/3*c^3*x^3*e+c^3*d*x)+b/c^2*(1/3*arcsin(c*x)*c^3*x^3*e+arcsin(c*x)*c^3*d*x-1/3*e*(-1/3*c^2*x^2*(-
c^2*x^2+1)^(1/2)-2/3*(-c^2*x^2+1)^(1/2))+c^2*d*(-c^2*x^2+1)^(1/2)))

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maxima [A]  time = 0.49, size = 91, normalized size = 1.12 \[ \frac {1}{3} \, a e x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b e + a d x + \frac {{\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*sqrt(-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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \left \{\begin {array}{cl} b\,e\,\left (\frac {\sqrt {\frac {1}{c^2}-x^2}\,\left (\frac {2}{c^2}+x^2\right )}{9}+\frac {x^3\,\mathrm {asin}\left (c\,x\right )}{3}\right )+\frac {a\,x\,\left (e\,x^2+3\,d\right )}{3}+\frac {b\,d\,\left (\sqrt {1-c^2\,x^2}+c\,x\,\mathrm {asin}\left (c\,x\right )\right )}{c} & \text {\ if\ \ }0<c\\ \int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (e\,x^2+d\right ) \,d x & \text {\ if\ \ }\neg 0<c \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))*(d + e*x^2),x)

[Out]

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))

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sympy [A]  time = 0.55, size = 109, normalized size = 1.35 \[ \begin {cases} a d x + \frac {a e x^{3}}{3} + b d x \operatorname {asin}{\left (c x \right )} + \frac {b e x^{3} \operatorname {asin}{\left (c x \right )}}{3} + \frac {b d \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {b e x^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c} + \frac {2 b e \sqrt {- c^{2} x^{2} + 1}}{9 c^{3}} & \text {for}\: c \neq 0 \\a \left (d x + \frac {e x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*sqrt(-c**2*x**2 + 1)/(9*c**3), Ne(c, 0)), (a*(d*x + e*x**3/3), True))

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