3.2.59 \(\int x (d-c^2 d x^2) (a+b \text {ArcSin}(c x))^2 \, dx\) [159]

Optimal. Leaf size=138 \[ -\frac {5}{32} b^2 d x^2+\frac {1}{32} b^2 c^2 d x^4+\frac {3 b d x \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{16 c}+\frac {b d x \left (1-c^2 x^2\right )^{3/2} (a+b \text {ArcSin}(c x))}{8 c}+\frac {3 d (a+b \text {ArcSin}(c x))^2}{32 c^2}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \text {ArcSin}(c x))^2}{4 c^2} \]

[Out]

-5/32*b^2*d*x^2+1/32*b^2*c^2*d*x^4+1/8*b*d*x*(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))/c+3/32*d*(a+b*arcsin(c*x))^2
/c^2-1/4*d*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))^2/c^2+3/16*b*d*x*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c

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Rubi [A]
time = 0.10, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4767, 4743, 4741, 4737, 30, 14} \begin {gather*} \frac {b d x \left (1-c^2 x^2\right )^{3/2} (a+b \text {ArcSin}(c x))}{8 c}+\frac {3 b d x \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{16 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \text {ArcSin}(c x))^2}{4 c^2}+\frac {3 d (a+b \text {ArcSin}(c x))^2}{32 c^2}+\frac {1}{32} b^2 c^2 d x^4-\frac {5}{32} b^2 d x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b^2*d*x^2)/32 + (b^2*c^2*d*x^4)/32 + (3*b*d*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(16*c) + (b*d*x*(1 -
c^2*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(8*c) + (3*d*(a + b*ArcSin[c*x])^2)/(32*c^2) - (d*(1 - c^2*x^2)^2*(a + b*A
rcSin[c*x])^2)/(4*c^2)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4737

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*Si
mp[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]

Rule 4741

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[x*Sqrt[d + e*x^2]*((
a + b*ArcSin[c*x])^n/2), x] + (Dist[(1/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(a + b*ArcSin[c*x])^n/S
qrt[1 - c^2*x^2], x], x] - Dist[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[x*(a + b*ArcSin[c*x])^(
n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4743

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[x*(d + e*x^2)^p*((
a + b*ArcSin[c*x])^n/(2*p + 1)), x] + (Dist[2*d*(p/(2*p + 1)), Int[(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n,
x], x] - Dist[b*c*(n/(2*p + 1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcS
in[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0]

Rule 4767

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^(
p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p
], Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*
d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int x \left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=-\frac {d \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2}+\frac {(b d) \int \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{2 c}\\ &=\frac {b d x \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}-\frac {d \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2}-\frac {1}{8} \left (b^2 d\right ) \int x \left (1-c^2 x^2\right ) \, dx+\frac {(3 b d) \int \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{8 c}\\ &=\frac {3 b d x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{16 c}+\frac {b d x \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}-\frac {d \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2}-\frac {1}{8} \left (b^2 d\right ) \int \left (x-c^2 x^3\right ) \, dx-\frac {1}{16} \left (3 b^2 d\right ) \int x \, dx+\frac {(3 b d) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{16 c}\\ &=-\frac {5}{32} b^2 d x^2+\frac {1}{32} b^2 c^2 d x^4+\frac {3 b d x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{16 c}+\frac {b d x \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}+\frac {3 d \left (a+b \sin ^{-1}(c x)\right )^2}{32 c^2}-\frac {d \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 157, normalized size = 1.14 \begin {gather*} -\frac {d \left (c x \left (b^2 c x \left (5-c^2 x^2\right )+8 a^2 c x \left (-2+c^2 x^2\right )+2 a b \sqrt {1-c^2 x^2} \left (-5+2 c^2 x^2\right )\right )+2 b \left (b c x \sqrt {1-c^2 x^2} \left (-5+2 c^2 x^2\right )+a \left (5-16 c^2 x^2+8 c^4 x^4\right )\right ) \text {ArcSin}(c x)+b^2 \left (5-16 c^2 x^2+8 c^4 x^4\right ) \text {ArcSin}(c x)^2\right )}{32 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(d*(c*x*(b^2*c*x*(5 - c^2*x^2) + 8*a^2*c*x*(-2 + c^2*x^2) + 2*a*b*Sqrt[1 - c^2*x^2]*(-5 + 2*c^2*x^2)) +
2*b*(b*c*x*Sqrt[1 - c^2*x^2]*(-5 + 2*c^2*x^2) + a*(5 - 16*c^2*x^2 + 8*c^4*x^4))*ArcSin[c*x] + b^2*(5 - 16*c^2*
x^2 + 8*c^4*x^4)*ArcSin[c*x]^2))/c^2

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Maple [A]
time = 0.12, size = 192, normalized size = 1.39

method	result	size
derivativedivides	\(\frac {-\frac {d \left (c^{2} x^{2}-1\right )^{2} a^{2}}{4}-d \,b^{2} \left (\frac {\arcsin \left (c x \right )^{2} \left (c^{2} x^{2}-1\right )^{2}}{4}-\frac {\arcsin \left (c x \right ) \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}+5 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{16}+\frac {3 \arcsin \left (c x \right )^{2}}{32}-\frac {\left (2 c^{2} x^{2}-5\right )^{2}}{128}\right )-2 d a b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}-\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {5 \arcsin \left (c x \right )}{32}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}-\frac {5 c x \sqrt {-c^{2} x^{2}+1}}{32}\right )}{c^{2}}\)	\(192\)
default	\(\frac {-\frac {d \left (c^{2} x^{2}-1\right )^{2} a^{2}}{4}-d \,b^{2} \left (\frac {\arcsin \left (c x \right )^{2} \left (c^{2} x^{2}-1\right )^{2}}{4}-\frac {\arcsin \left (c x \right ) \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}+5 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{16}+\frac {3 \arcsin \left (c x \right )^{2}}{32}-\frac {\left (2 c^{2} x^{2}-5\right )^{2}}{128}\right )-2 d a b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}-\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {5 \arcsin \left (c x \right )}{32}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}-\frac {5 c x \sqrt {-c^{2} x^{2}+1}}{32}\right )}{c^{2}}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(-c^2*d*x^2+d)*(a+b*arcsin(c*x))^2,x,method=_RETURNVERBOSE)

[Out]

1/c^2*(-1/4*d*(c^2*x^2-1)^2*a^2-d*b^2*(1/4*arcsin(c*x)^2*(c^2*x^2-1)^2-1/16*arcsin(c*x)*(-2*c^3*x^3*(-c^2*x^2+
1)^(1/2)+5*c*x*(-c^2*x^2+1)^(1/2)+3*arcsin(c*x))+3/32*arcsin(c*x)^2-1/128*(2*c^2*x^2-5)^2)-2*d*a*b*(1/4*c^4*x^
4*arcsin(c*x)-1/2*c^2*x^2*arcsin(c*x)+5/32*arcsin(c*x)+1/16*c^3*x^3*(-c^2*x^2+1)^(1/2)-5/32*c*x*(-c^2*x^2+1)^(
1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*c^2*d*x^4 - 1/16*(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)*a*b*c^2*d + 1/2*a^2*d*x^2 + 1/2*(2*x^2*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x/c^2 - arcsin(
c*x)/c^3))*a*b*d - 1/4*(b^2*c^2*d*x^4 - 2*b^2*d*x^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 - integrate(
1/2*(b^2*c^3*d*x^4 - 2*b^2*c*d*x^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c
^2*x^2 - 1), x)

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Fricas [A]
time = 2.76, size = 176, normalized size = 1.28 \begin {gather*} -\frac {{\left (8 \, a^{2} - b^{2}\right )} c^{4} d x^{4} - {\left (16 \, a^{2} - 5 \, b^{2}\right )} c^{2} d x^{2} + {\left (8 \, b^{2} c^{4} d x^{4} - 16 \, b^{2} c^{2} d x^{2} + 5 \, b^{2} d\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (8 \, a b c^{4} d x^{4} - 16 \, a b c^{2} d x^{2} + 5 \, a b d\right )} \arcsin \left (c x\right ) + 2 \, {\left (2 \, a b c^{3} d x^{3} - 5 \, a b c d x + {\left (2 \, b^{2} c^{3} d x^{3} - 5 \, b^{2} c d x\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{32 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*((8*a^2 - b^2)*c^4*d*x^4 - (16*a^2 - 5*b^2)*c^2*d*x^2 + (8*b^2*c^4*d*x^4 - 16*b^2*c^2*d*x^2 + 5*b^2*d)*a
rcsin(c*x)^2 + 2*(8*a*b*c^4*d*x^4 - 16*a*b*c^2*d*x^2 + 5*a*b*d)*arcsin(c*x) + 2*(2*a*b*c^3*d*x^3 - 5*a*b*c*d*x
+ (2*b^2*c^3*d*x^3 - 5*b^2*c*d*x)*arcsin(c*x))*sqrt(-c^2*x^2 + 1))/c^2

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (129) = 258\).
time = 0.41, size = 269, normalized size = 1.95 \begin {gather*} \begin {cases} - \frac {a^{2} c^{2} d x^{4}}{4} + \frac {a^{2} d x^{2}}{2} - \frac {a b c^{2} d x^{4} \operatorname {asin}{\left (c x \right )}}{2} - \frac {a b c d x^{3} \sqrt {- c^{2} x^{2} + 1}}{8} + a b d x^{2} \operatorname {asin}{\left (c x \right )} + \frac {5 a b d x \sqrt {- c^{2} x^{2} + 1}}{16 c} - \frac {5 a b d \operatorname {asin}{\left (c x \right )}}{16 c^{2}} - \frac {b^{2} c^{2} d x^{4} \operatorname {asin}^{2}{\left (c x \right )}}{4} + \frac {b^{2} c^{2} d x^{4}}{32} - \frac {b^{2} c d x^{3} \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{8} + \frac {b^{2} d x^{2} \operatorname {asin}^{2}{\left (c x \right )}}{2} - \frac {5 b^{2} d x^{2}}{32} + \frac {5 b^{2} d x \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{16 c} - \frac {5 b^{2} d \operatorname {asin}^{2}{\left (c x \right )}}{32 c^{2}} & \text {for}\: c \neq 0 \\\frac {a^{2} d x^{2}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*c**2*d*x**4/4 + a**2*d*x**2/2 - a*b*c**2*d*x**4*asin(c*x)/2 - a*b*c*d*x**3*sqrt(-c**2*x**2 +
1)/8 + a*b*d*x**2*asin(c*x) + 5*a*b*d*x*sqrt(-c**2*x**2 + 1)/(16*c) - 5*a*b*d*asin(c*x)/(16*c**2) - b**2*c**2*
d*x**4*asin(c*x)**2/4 + b**2*c**2*d*x**4/32 - b**2*c*d*x**3*sqrt(-c**2*x**2 + 1)*asin(c*x)/8 + b**2*d*x**2*asi
n(c*x)**2/2 - 5*b**2*d*x**2/32 + 5*b**2*d*x*sqrt(-c**2*x**2 + 1)*asin(c*x)/(16*c) - 5*b**2*d*asin(c*x)**2/(32*
c**2), Ne(c, 0)), (a**2*d*x**2/2, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (121) = 242\).
time = 0.45, size = 248, normalized size = 1.80 \begin {gather*} -\frac {1}{4} \, a^{2} c^{2} d x^{4} + \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b^{2} d x \arcsin \left (c x\right )}{8 \, c} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b^{2} d \arcsin \left (c x\right )^{2}}{4 \, c^{2}} + \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} a b d x}{8 \, c} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} b^{2} d x \arcsin \left (c x\right )}{16 \, c} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} a b d \arcsin \left (c x\right )}{2 \, c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} a b d x}{16 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b^{2} d}{32 \, c^{2}} + \frac {3 \, b^{2} d \arcsin \left (c x\right )^{2}}{32 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )} a^{2} d}{2 \, c^{2}} - \frac {3 \, {\left (c^{2} x^{2} - 1\right )} b^{2} d}{32 \, c^{2}} + \frac {3 \, a b d \arcsin \left (c x\right )}{16 \, c^{2}} - \frac {15 \, b^{2} d}{256 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*c^2*d*x^4 + 1/8*(-c^2*x^2 + 1)^(3/2)*b^2*d*x*arcsin(c*x)/c - 1/4*(c^2*x^2 - 1)^2*b^2*d*arcsin(c*x)^2/
c^2 + 1/8*(-c^2*x^2 + 1)^(3/2)*a*b*d*x/c + 3/16*sqrt(-c^2*x^2 + 1)*b^2*d*x*arcsin(c*x)/c - 1/2*(c^2*x^2 - 1)^2
*a*b*d*arcsin(c*x)/c^2 + 3/16*sqrt(-c^2*x^2 + 1)*a*b*d*x/c + 1/32*(c^2*x^2 - 1)^2*b^2*d/c^2 + 3/32*b^2*d*arcsi
n(c*x)^2/c^2 + 1/2*(c^2*x^2 - 1)*a^2*d/c^2 - 3/32*(c^2*x^2 - 1)*b^2*d/c^2 + 3/16*a*b*d*arcsin(c*x)/c^2 - 15/25
6*b^2*d/c^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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