3.3.42 \(\int \frac {x^3}{1+x+\frac {x^2}{2}+\frac {x^3}{6}} \, dx\) [242]

Optimal. Leaf size=24 \[ 6 x-6 \log \left (1+x+\frac {x^2}{2}+\frac {x^3}{6}\right ) \]

[Out]

6*x-6*ln(1+x+1/2*x^2+1/6*x^3)

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(193\) vs. \(2(24)=48\).
time = 0.43, antiderivative size = 193, normalized size of antiderivative = 8.04, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2106, 2104, 1642, 642} \begin {gather*} 6 x-6 \log \left (\sqrt [3]{\sqrt {2}-1} (x+1)-\left (\sqrt {2}-1\right )^{2/3}+1\right )-\frac {12 \left (1-\sqrt {2}\right ) \left (1-\sqrt [3]{\sqrt {2}-1}-\left (\sqrt {2}-1\right )^{2/3}\right ) \log \left (\left (\sqrt {2}-1\right )^{2/3} (x+1)^2-\left (1-\sqrt {2}+\sqrt [3]{\sqrt {2}-1}\right ) (x+1)+\left (\sqrt {2}-1\right )^{4/3}+\left (\sqrt {2}-1\right )^{2/3}+1\right )}{\left (1-\sqrt {2}+\sqrt [3]{\sqrt {2}-1}\right ) \left (1-\left (\sqrt {2}-1\right )^{2/3}+\left (\sqrt {2}-1\right )^{4/3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(1 + x + x^2/2 + x^3/6),x]

[Out]

6*x - 6*Log[1 - (-1 + Sqrt[2])^(2/3) + (-1 + Sqrt[2])^(1/3)*(1 + x)] - (12*(1 - Sqrt[2])*(1 - (-1 + Sqrt[2])^(
1/3) - (-1 + Sqrt[2])^(2/3))*Log[1 + (-1 + Sqrt[2])^(2/3) + (-1 + Sqrt[2])^(4/3) - (1 - Sqrt[2] + (-1 + Sqrt[2
])^(1/3))*(1 + x) + (-1 + Sqrt[2])^(2/3)*(1 + x)^2])/((1 - Sqrt[2] + (-1 + Sqrt[2])^(1/3))*(1 - (-1 + Sqrt[2])
^(2/3) + (-1 + Sqrt[2])^(4/3)))

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2104

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> With[{r = Rt[-9*a*d^2 + S
qrt[3]*d*Sqrt[4*b^3*d + 27*a^2*d^2], 3]}, Dist[1/d^(2*p), Int[(e + f*x)^m*Simp[18^(1/3)*b*(d/(3*r)) - r/18^(1/
3) + d*x, x]^p*Simp[b*(d/3) + 12^(1/3)*b^2*(d^2/(3*r^2)) + r^2/(3*12^(1/3)) - d*(2^(1/3)*b*(d/(3^(1/3)*r)) - r
/18^(1/3))*x + d^2*x^2, x]^p, x], x]] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[4*b^3 + 27*a^2*d, 0] && ILtQ[p, 0
]

Rule 2106

Int[(P3_)^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = C
oeff[P3, x, 2], d = Coeff[P3, x, 3]}, Subst[Int[((3*d*e - c*f)/(3*d) + f*x)^m*Simp[(2*c^3 - 9*b*c*d + 27*a*d^2
)/(27*d^2) - (c^2 - 3*b*d)*(x/(3*d)) + d*x^3, x]^p, x], x, x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[{e, f, m, p}, x
] && PolyQ[P3, x, 3]

Rubi steps

\begin {gather*} \begin {aligned} \text {Integral} &=\text {Subst}\left (\int \frac {(-1+x)^3}{\frac {1}{3}+\frac {x}{2}+\frac {x^3}{6}} \, dx,x,1+x\right )\\ &=\frac {1}{36} \text {Subst}\left (\int \frac {(-1+x)^3}{\left (\frac {1-\left (-1+\sqrt {2}\right )^{2/3}}{6 \sqrt [3]{-1+\sqrt {2}}}+\frac {x}{6}\right ) \left (\frac {1}{36} \left (1+\frac {1}{\left (-1+\sqrt {2}\right )^{2/3}}+\left (-1+\sqrt {2}\right )^{2/3}\right )-\frac {\left (1-\left (-1+\sqrt {2}\right )^{2/3}\right ) x}{36 \sqrt [3]{-1+\sqrt {2}}}+\frac {x^2}{36}\right )} \, dx,x,1+x\right )\\ &=\frac {1}{36} \text {Subst}\left (\int \left (216+\frac {216 \sqrt [3]{-1+\sqrt {2}}}{-1+\left (-1+\sqrt {2}\right )^{2/3}-\sqrt [3]{-1+\sqrt {2}} x}+\frac {432 \left (\left (1-\sqrt {2}\right ) \left (1-\sqrt [3]{-1+\sqrt {2}}-\left (-1+\sqrt {2}\right )^{2/3}\right )-\left (3-2 \sqrt {2}+\left (-1+\sqrt {2}\right )^{2/3}-\left (-1+\sqrt {2}\right )^{4/3}\right ) x\right )}{\left (1-\left (-1+\sqrt {2}\right )^{2/3}+\left (-1+\sqrt {2}\right )^{4/3}\right ) \left (1+\left (-1+\sqrt {2}\right )^{2/3}+\left (-1+\sqrt {2}\right )^{4/3}-\left (1-\sqrt {2}+\sqrt [3]{-1+\sqrt {2}}\right ) x+\left (-1+\sqrt {2}\right )^{2/3} x^2\right )}\right ) \, dx,x,1+x\right )\\ &=6 x-6 \log \left (1-\left (-1+\sqrt {2}\right )^{2/3}+\sqrt [3]{-1+\sqrt {2}} (1+x)\right )+\frac {12 \text {Subst}\left (\int \frac {\left (1-\sqrt {2}\right ) \left (1-\sqrt [3]{-1+\sqrt {2}}-\left (-1+\sqrt {2}\right )^{2/3}\right )-\left (3-2 \sqrt {2}+\left (-1+\sqrt {2}\right )^{2/3}-\left (-1+\sqrt {2}\right )^{4/3}\right ) x}{1+\left (-1+\sqrt {2}\right )^{2/3}+\left (-1+\sqrt {2}\right )^{4/3}+\left (-1+\sqrt {2}-\sqrt [3]{-1+\sqrt {2}}\right ) x+\left (-1+\sqrt {2}\right )^{2/3} x^2} \, dx,x,1+x\right )}{1-\left (-1+\sqrt {2}\right )^{2/3}+\left (-1+\sqrt {2}\right )^{4/3}}\\ &=6 x-6 \log \left (1-\left (-1+\sqrt {2}\right )^{2/3}+\sqrt [3]{-1+\sqrt {2}} (1+x)\right )-\frac {12 \left (1-\sqrt {2}\right ) \left (1-\sqrt [3]{-1+\sqrt {2}}-\left (-1+\sqrt {2}\right )^{2/3}\right ) \log \left (1+\left (-1+\sqrt {2}\right )^{2/3}+\left (-1+\sqrt {2}\right )^{4/3}-\left (1-\sqrt {2}+\sqrt [3]{-1+\sqrt {2}}\right ) (1+x)+\left (-1+\sqrt {2}\right )^{2/3} (1+x)^2\right )}{\left (1-\sqrt {2}+\sqrt [3]{-1+\sqrt {2}}\right ) \left (1-\left (-1+\sqrt {2}\right )^{2/3}+\left (-1+\sqrt {2}\right )^{4/3}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 20, normalized size = 0.83 \begin {gather*} 6 \left (x-\log \left (6+6 x+3 x^2+x^3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(1 + x + x^2/2 + x^3/6),x]

[Out]

6*(x - Log[6 + 6*x + 3*x^2 + x^3])

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Maple [A]
time = 0.02, size = 21, normalized size = 0.88

method	result	size
default	\(6 x -6 \ln \left (x^{3}+3 x^{2}+6 x +6\right )\)	\(21\)
norman	\(6 x -6 \ln \left (x^{3}+3 x^{2}+6 x +6\right )\)	\(21\)
risch	\(6 x -6 \ln \left (x^{3}+3 x^{2}+6 x +6\right )\)	\(21\)
parallelrisch	\(6 x -6 \ln \left (x^{3}+3 x^{2}+6 x +6\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(1+x+1/2*x^2+1/6*x^3),x,method=_RETURNVERBOSE)

[Out]

6*x-6*ln(x^3+3*x^2+6*x+6)

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Maxima [A]
time = 0.42, size = 20, normalized size = 0.83 \begin {gather*} 6 \, x - 6 \, \log \left (x^{3} + 3 \, x^{2} + 6 \, x + 6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(1+x+1/2*x^2+1/6*x^3),x, algorithm="maxima")

[Out]

6*x - 6*log(x^3 + 3*x^2 + 6*x + 6)

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Fricas [A]
time = 0.56, size = 20, normalized size = 0.83 \begin {gather*} 6 \, x - 6 \, \log \left (x^{3} + 3 \, x^{2} + 6 \, x + 6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(1+x+1/2*x^2+1/6*x^3),x, algorithm="fricas")

[Out]

6*x - 6*log(x^3 + 3*x^2 + 6*x + 6)

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Sympy [A]
time = 0.03, size = 19, normalized size = 0.79 \begin {gather*} 6 x - 6 \log {\left (x^{3} + 3 x^{2} + 6 x + 6 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(1+x+1/2*x**2+1/6*x**3),x)

[Out]

6*x - 6*log(x**3 + 3*x**2 + 6*x + 6)

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Giac [A]
time = 0.41, size = 21, normalized size = 0.88 \begin {gather*} 6 \, x - 6 \, \log \left ({\left | x^{3} + 3 \, x^{2} + 6 \, x + 6 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(1+x+1/2*x^2+1/6*x^3),x, algorithm="giac")

[Out]

6*x - 6*log(abs(x^3 + 3*x^2 + 6*x + 6))

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Mupad [B]
time = 0.12, size = 20, normalized size = 0.83 \begin {gather*} 6\,x-6\,\ln \left (x^3+3\,x^2+6\,x+6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x + x^2/2 + x^3/6 + 1),x)

[Out]

6*x - 6*log(6*x + 3*x^2 + x^3 + 6)

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Chatgpt [F] Failed to verify
time = 1.00, size = 33, normalized size = 1.38 \begin {gather*} -6 x^{2}+12 x -24 \ln \left (2 x^{3}+3 x^{2}+3 x +2\right )+36 \ln \left (x +1\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

int(x^3/(1+x+1/2*x^2+1/6*x^3),x)

[Out]

-6*x^2+12*x-24*ln(2*x^3+3*x^2+3*x+2)+36*ln(x+1)

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