3.2.40 \(\int x^3 (a+b \text {ArcCos}(c x)) \, dx\) [140]

Optimal. Leaf size=76 \[ -\frac {3 b x \sqrt {1-c^2 x^2}}{32 c^3}-\frac {b x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {1}{4} x^4 (a+b \text {ArcCos}(c x))+\frac {3 b \text {ArcSin}(c x)}{32 c^4} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4724, 327, 222} \begin {gather*} \frac {1}{4} x^4 (a+b \text {ArcCos}(c x))+\frac {3 b \text {ArcSin}(c x)}{32 c^4}-\frac {b x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {3 b x \sqrt {1-c^2 x^2}}{32 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcCos[c*x]),x]

[Out]

(-3*b*x*Sqrt[1 - c^2*x^2])/(32*c^3) - (b*x^3*Sqrt[1 - c^2*x^2])/(16*c) + (x^4*(a + b*ArcCos[c*x]))/4 + (3*b*Ar
cSin[c*x])/(32*c^4)

Rule 222

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 327

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

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 68, normalized size = 0.89 \begin {gather*} \frac {a x^4}{4}+b \sqrt {1-c^2 x^2} \left (-\frac {3 x}{32 c^3}-\frac {x^3}{16 c}\right )+\frac {1}{4} b x^4 \text {ArcCos}(c x)+\frac {3 b \text {ArcSin}(c x)}{32 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcCos[c*x]),x]

[Out]

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

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Maple [A]
time = 0.10, size = 72, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {\frac {c^{4} x^{4} a}{4}+b \left (\frac {c^{4} x^{4} \arccos \left (c x \right )}{4}-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}-\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{32}+\frac {3 \arcsin \left (c x \right )}{32}\right )}{c^{4}}\)	\(72\)
default	\(\frac {\frac {c^{4} x^{4} a}{4}+b \left (\frac {c^{4} x^{4} \arccos \left (c x \right )}{4}-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}-\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{32}+\frac {3 \arcsin \left (c x \right )}{32}\right )}{c^{4}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arccos(c*x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.47, size = 71, normalized size = 0.93 \begin {gather*} \frac {1}{4} \, a x^{4} + \frac {1}{32} \, {\left (8 \, x^{4} \arccos \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccos(c*x)),x, algorithm="maxima")

[Out]

1/4*a*x^4 + 1/32*(8*x^4*arccos(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

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Fricas [A]
time = 3.89, size = 62, normalized size = 0.82 \begin {gather*} \frac {8 \, a c^{4} x^{4} + {\left (8 \, b c^{4} x^{4} - 3 \, b\right )} \arccos \left (c x\right ) - {\left (2 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt {-c^{2} x^{2} + 1}}{32 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccos(c*x)),x, algorithm="fricas")

[Out]

1/32*(8*a*c^4*x^4 + (8*b*c^4*x^4 - 3*b)*arccos(c*x) - (2*b*c^3*x^3 + 3*b*c*x)*sqrt(-c^2*x^2 + 1))/c^4

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Sympy [A]
time = 0.23, size = 85, normalized size = 1.12 \begin {gather*} \begin {cases} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {acos}{\left (c x \right )}}{4} - \frac {b x^{3} \sqrt {- c^{2} x^{2} + 1}}{16 c} - \frac {3 b x \sqrt {- c^{2} x^{2} + 1}}{32 c^{3}} - \frac {3 b \operatorname {acos}{\left (c x \right )}}{32 c^{4}} & \text {for}\: c \neq 0 \\\frac {x^{4} \left (a + \frac {\pi b}{2}\right )}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*acos(c*x)),x)

[Out]

Piecewise((a*x**4/4 + b*x**4*acos(c*x)/4 - b*x**3*sqrt(-c**2*x**2 + 1)/(16*c) - 3*b*x*sqrt(-c**2*x**2 + 1)/(32
*c**3) - 3*b*acos(c*x)/(32*c**4), Ne(c, 0)), (x**4*(a + pi*b/2)/4, True))

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Giac [A]
time = 0.42, size = 67, normalized size = 0.88 \begin {gather*} \frac {1}{4} \, b x^{4} \arccos \left (c x\right ) + \frac {1}{4} \, a x^{4} - \frac {\sqrt {-c^{2} x^{2} + 1} b x^{3}}{16 \, c} - \frac {3 \, \sqrt {-c^{2} x^{2} + 1} b x}{32 \, c^{3}} - \frac {3 \, b \arccos \left (c x\right )}{32 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccos(c*x)),x, algorithm="giac")

[Out]

1/4*b*x^4*arccos(c*x) + 1/4*a*x^4 - 1/16*sqrt(-c^2*x^2 + 1)*b*x^3/c - 3/32*sqrt(-c^2*x^2 + 1)*b*x/c^3 - 3/32*b
*arccos(c*x)/c^4

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\left (a+b\,\mathrm {acos}\left (c\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*acos(c*x)),x)

[Out]

int(x^3*(a + b*acos(c*x)), x)

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