3.2.48 \(\int x^2 (a+b \text {ArcCos}(c x))^2 \, dx\) [148]

Optimal. Leaf size=102 \[ -\frac {4 b^2 x}{9 c^2}-\frac {2 b^2 x^3}{27}-\frac {4 b \sqrt {1-c^2 x^2} (a+b \text {ArcCos}(c x))}{9 c^3}-\frac {2 b x^2 \sqrt {1-c^2 x^2} (a+b \text {ArcCos}(c x))}{9 c}+\frac {1}{3} x^3 (a+b \text {ArcCos}(c x))^2 \]

[Out]

-4/9*b^2*x/c^2-2/27*b^2*x^3+1/3*x^3*(a+b*arccos(c*x))^2-4/9*b*(a+b*arccos(c*x))*(-c^2*x^2+1)^(1/2)/c^3-2/9*b*x
^2*(a+b*arccos(c*x))*(-c^2*x^2+1)^(1/2)/c

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Rubi [A]
time = 0.10, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4724, 4796, 4768, 8, 30} \begin {gather*} -\frac {2 b x^2 \sqrt {1-c^2 x^2} (a+b \text {ArcCos}(c x))}{9 c}-\frac {4 b \sqrt {1-c^2 x^2} (a+b \text {ArcCos}(c x))}{9 c^3}+\frac {1}{3} x^3 (a+b \text {ArcCos}(c x))^2-\frac {4 b^2 x}{9 c^2}-\frac {2}{27} b^2 x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcCos[c*x])^2,x]

[Out]

(-4*b^2*x)/(9*c^2) - (2*b^2*x^3)/27 - (4*b*Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x]))/(9*c^3) - (2*b*x^2*Sqrt[1 -
c^2*x^2]*(a + b*ArcCos[c*x]))/(9*c) + (x^3*(a + b*ArcCos[c*x])^2)/3

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

Rule 4724

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcCo
s[c*x])^n/(d*(m + 1))), x] + Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcCos[c*x])^(n - 1)/Sqrt[1 -
c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4768

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^(
p + 1)*((a + b*ArcCos[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*ArcCos[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]

Rule 4796

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

Rubi steps

\begin {align*} \int x^2 \left (a+b \cos ^{-1}(c x)\right )^2 \, dx &=\frac {1}{3} x^3 \left (a+b \cos ^{-1}(c x)\right )^2+\frac {1}{3} (2 b c) \int \frac {x^3 \left (a+b \cos ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {2 b x^2 \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )}{9 c}+\frac {1}{3} x^3 \left (a+b \cos ^{-1}(c x)\right )^2-\frac {1}{9} \left (2 b^2\right ) \int x^2 \, dx+\frac {(4 b) \int \frac {x \left (a+b \cos ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{9 c}\\ &=-\frac {2}{27} b^2 x^3-\frac {4 b \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )}{9 c^3}-\frac {2 b x^2 \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )}{9 c}+\frac {1}{3} x^3 \left (a+b \cos ^{-1}(c x)\right )^2-\frac {\left (4 b^2\right ) \int 1 \, dx}{9 c^2}\\ &=-\frac {4 b^2 x}{9 c^2}-\frac {2 b^2 x^3}{27}-\frac {4 b \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )}{9 c^3}-\frac {2 b x^2 \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )}{9 c}+\frac {1}{3} x^3 \left (a+b \cos ^{-1}(c x)\right )^2\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 121, normalized size = 1.19 \begin {gather*} \frac {9 a^2 c^3 x^3-6 a b \sqrt {1-c^2 x^2} \left (2+c^2 x^2\right )-2 b^2 c x \left (6+c^2 x^2\right )-6 b \left (-3 a c^3 x^3+b \sqrt {1-c^2 x^2} \left (2+c^2 x^2\right )\right ) \text {ArcCos}(c x)+9 b^2 c^3 x^3 \text {ArcCos}(c x)^2}{27 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcCos[c*x])^2,x]

[Out]

(9*a^2*c^3*x^3 - 6*a*b*Sqrt[1 - c^2*x^2]*(2 + c^2*x^2) - 2*b^2*c*x*(6 + c^2*x^2) - 6*b*(-3*a*c^3*x^3 + b*Sqrt[
1 - c^2*x^2]*(2 + c^2*x^2))*ArcCos[c*x] + 9*b^2*c^3*x^3*ArcCos[c*x]^2)/(27*c^3)

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Maple [A]
time = 0.07, size = 126, normalized size = 1.24

method	result	size
derivativedivides	\(\frac {\frac {c^{3} x^{3} a^{2}}{3}+b^{2} \left (\frac {c^{3} x^{3} \arccos \left (c x \right )^{2}}{3}-\frac {2 \arccos \left (c x \right ) \left (c^{2} x^{2}+2\right ) \sqrt {-c^{2} x^{2}+1}}{9}-\frac {2 c^{3} x^{3}}{27}-\frac {4 c x}{9}\right )+2 a b \left (\frac {c^{3} x^{3} \arccos \left (c x \right )}{3}-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{9}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{9}\right )}{c^{3}}\)	\(126\)
default	\(\frac {\frac {c^{3} x^{3} a^{2}}{3}+b^{2} \left (\frac {c^{3} x^{3} \arccos \left (c x \right )^{2}}{3}-\frac {2 \arccos \left (c x \right ) \left (c^{2} x^{2}+2\right ) \sqrt {-c^{2} x^{2}+1}}{9}-\frac {2 c^{3} x^{3}}{27}-\frac {4 c x}{9}\right )+2 a b \left (\frac {c^{3} x^{3} \arccos \left (c x \right )}{3}-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{9}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{9}\right )}{c^{3}}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arccos(c*x))^2,x,method=_RETURNVERBOSE)

[Out]

1/c^3*(1/3*c^3*x^3*a^2+b^2*(1/3*c^3*x^3*arccos(c*x)^2-2/9*arccos(c*x)*(c^2*x^2+2)*(-c^2*x^2+1)^(1/2)-2/27*c^3*
x^3-4/9*c*x)+2*a*b*(1/3*c^3*x^3*arccos(c*x)-1/9*c^2*x^2*(-c^2*x^2+1)^(1/2)-2/9*(-c^2*x^2+1)^(1/2)))

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Maxima [A]
time = 0.49, size = 142, normalized size = 1.39 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} \arccos \left (c x\right )^{2} + \frac {1}{3} \, a^{2} x^{3} + \frac {2}{9} \, {\left (3 \, x^{3} \arccos \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 )} a b - \frac {2}{27} \, {\left (3 \, c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )} \arccos \left (c x\right ) + \frac {c^{2} x^{3} + 6 \, x}{c^{2}}\right )} b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*x^3*arccos(c*x)^2 + 1/3*a^2*x^3 + 2/9*(3*x^3*arccos(c*x) - c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2
*x^2 + 1)/c^4))*a*b - 2/27*(3*c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4)*arccos(c*x) + (c^2*x^3
+ 6*x)/c^2)*b^2

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Fricas [A]
time = 1.93, size = 111, normalized size = 1.09 \begin {gather*} \frac {9 \, b^{2} c^{3} x^{3} \arccos \left (c x\right )^{2} + 18 \, a b c^{3} x^{3} \arccos \left (c x\right ) + {\left (9 \, a^{2} - 2 \, b^{2}\right )} c^{3} x^{3} - 12 \, b^{2} c x - 6 \, {\left (a b c^{2} x^{2} + 2 \, a b + {\left (b^{2} c^{2} x^{2} + 2 \, b^{2}\right )} \arccos \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{27 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*(9*b^2*c^3*x^3*arccos(c*x)^2 + 18*a*b*c^3*x^3*arccos(c*x) + (9*a^2 - 2*b^2)*c^3*x^3 - 12*b^2*c*x - 6*(a*b
*c^2*x^2 + 2*a*b + (b^2*c^2*x^2 + 2*b^2)*arccos(c*x))*sqrt(-c^2*x^2 + 1))/c^3

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Sympy [A]
time = 0.27, size = 175, normalized size = 1.72 \begin {gather*} \begin {cases} \frac {a^{2} x^{3}}{3} + \frac {2 a b x^{3} \operatorname {acos}{\left (c x \right )}}{3} - \frac {2 a b x^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c} - \frac {4 a b \sqrt {- c^{2} x^{2} + 1}}{9 c^{3}} + \frac {b^{2} x^{3} \operatorname {acos}^{2}{\left (c x \right )}}{3} - \frac {2 b^{2} x^{3}}{27} - \frac {2 b^{2} x^{2} \sqrt {- c^{2} x^{2} + 1} \operatorname {acos}{\left (c x \right )}}{9 c} - \frac {4 b^{2} x}{9 c^{2}} - \frac {4 b^{2} \sqrt {- c^{2} x^{2} + 1} \operatorname {acos}{\left (c x \right )}}{9 c^{3}} & \text {for}\: c \neq 0 \\\frac {x^{3} \left (a + \frac {\pi b}{2}\right )^{2}}{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*acos(c*x))**2,x)

[Out]

Piecewise((a**2*x**3/3 + 2*a*b*x**3*acos(c*x)/3 - 2*a*b*x**2*sqrt(-c**2*x**2 + 1)/(9*c) - 4*a*b*sqrt(-c**2*x**
2 + 1)/(9*c**3) + b**2*x**3*acos(c*x)**2/3 - 2*b**2*x**3/27 - 2*b**2*x**2*sqrt(-c**2*x**2 + 1)*acos(c*x)/(9*c)
- 4*b**2*x/(9*c**2) - 4*b**2*sqrt(-c**2*x**2 + 1)*acos(c*x)/(9*c**3), Ne(c, 0)), (x**3*(a + pi*b/2)**2/3, Tru
e))

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Giac [A]
time = 0.44, size = 143, normalized size = 1.40 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} \arccos \left (c x\right )^{2} + \frac {2}{3} \, a b x^{3} \arccos \left (c x\right ) + \frac {1}{3} \, a^{2} x^{3} - \frac {2}{27} \, b^{2} x^{3} - \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b^{2} x^{2} \arccos \left (c x\right )}{9 \, c} - \frac {2 \, \sqrt {-c^{2} x^{2} + 1} a b x^{2}}{9 \, c} - \frac {4 \, b^{2} x}{9 \, c^{2}} - \frac {4 \, \sqrt {-c^{2} x^{2} + 1} b^{2} \arccos \left (c x\right )}{9 \, c^{3}} - \frac {4 \, \sqrt {-c^{2} x^{2} + 1} a b}{9 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*x^3*arccos(c*x)^2 + 2/3*a*b*x^3*arccos(c*x) + 1/3*a^2*x^3 - 2/27*b^2*x^3 - 2/9*sqrt(-c^2*x^2 + 1)*b^2*
x^2*arccos(c*x)/c - 2/9*sqrt(-c^2*x^2 + 1)*a*b*x^2/c - 4/9*b^2*x/c^2 - 4/9*sqrt(-c^2*x^2 + 1)*b^2*arccos(c*x)/
c^3 - 4/9*sqrt(-c^2*x^2 + 1)*a*b/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*acos(c*x))^2,x)

[Out]

int(x^2*(a + b*acos(c*x))^2, x)

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