3.1.79 \(\int \frac {1}{3} (-5-e^3-7 x^2+(-3-3 x^2) \log (x)) \, dx\)

Optimal. Leaf size=28 \[ x-x \log (x)-\frac {1}{3} x \left (5+e^3+x^2+x (x+x \log (x))\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2313} \begin {gather*} -\frac {2 x^3}{3}-\frac {1}{3} \left (x^3+3 x\right ) \log (x)-\frac {1}{3} \left (5+e^3\right ) x+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 - E^3 - 7*x^2 + (-3 - 3*x^2)*Log[x])/3,x]

[Out]

x - ((5 + E^3)*x)/3 - (2*x^3)/3 - ((3*x + x^3)*Log[x])/3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \left (-5-e^3-7 x^2+\left (-3-3 x^2\right ) \log (x)\right ) \, dx\\ &=-\frac {1}{3} \left (5+e^3\right ) x-\frac {7 x^3}{9}+\frac {1}{3} \int \left (-3-3 x^2\right ) \log (x) \, dx\\ &=-\frac {1}{3} \left (5+e^3\right ) x-\frac {7 x^3}{9}-\frac {1}{3} \left (3 x+x^3\right ) \log (x)-\frac {1}{3} \int \left (-3-x^2\right ) \, dx\\ &=x-\frac {1}{3} \left (5+e^3\right ) x-\frac {2 x^3}{3}-\frac {1}{3} \left (3 x+x^3\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 35, normalized size = 1.25 \begin {gather*} -\frac {2 x}{3}-\frac {e^3 x}{3}-\frac {2 x^3}{3}-x \log (x)-\frac {1}{3} x^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 - E^3 - 7*x^2 + (-3 - 3*x^2)*Log[x])/3,x]

[Out]

(-2*x)/3 - (E^3*x)/3 - (2*x^3)/3 - x*Log[x] - (x^3*Log[x])/3

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fricas [A]  time = 0.52, size = 25, normalized size = 0.89 \begin {gather*} -\frac {2}{3} \, x^{3} - \frac {1}{3} \, x e^{3} - \frac {1}{3} \, {\left (x^{3} + 3 \, x\right )} \log \relax (x) - \frac {2}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*x^2-3)*log(x)-1/3*exp(3)-7/3*x^2-5/3,x, algorithm="fricas")

[Out]

-2/3*x^3 - 1/3*x*e^3 - 1/3*(x^3 + 3*x)*log(x) - 2/3*x

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giac [A]  time = 0.28, size = 26, normalized size = 0.93 \begin {gather*} -\frac {1}{3} \, x^{3} \log \relax (x) - \frac {2}{3} \, x^{3} - \frac {1}{3} \, x e^{3} - x \log \relax (x) - \frac {2}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*x^2-3)*log(x)-1/3*exp(3)-7/3*x^2-5/3,x, algorithm="giac")

[Out]

-1/3*x^3*log(x) - 2/3*x^3 - 1/3*x*e^3 - x*log(x) - 2/3*x

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maple [A]  time = 0.03, size = 27, normalized size = 0.96

method	result	size
default	\(-\frac {2 x}{3}-\frac {2 x^{3}}{3}-\frac {x^{3} \ln \relax (x )}{3}-x \ln \relax (x )-\frac {x \,{\mathrm e}^{3}}{3}\)	\(27\)
norman	\(\left (-\frac {2}{3}-\frac {{\mathrm e}^{3}}{3}\right ) x -\frac {2 x^{3}}{3}-x \ln \relax (x )-\frac {x^{3} \ln \relax (x )}{3}\)	\(27\)
risch	\(\frac {\left (-x^{3}-3 x \right ) \ln \relax (x )}{3}-\frac {2 x^{3}}{3}-\frac {2 x}{3}-\frac {x \,{\mathrm e}^{3}}{3}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-3*x^2-3)*ln(x)-1/3*exp(3)-7/3*x^2-5/3,x,method=_RETURNVERBOSE)

[Out]

-2/3*x-2/3*x^3-1/3*x^3*ln(x)-x*ln(x)-1/3*x*exp(3)

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maxima [A]  time = 0.44, size = 25, normalized size = 0.89 \begin {gather*} -\frac {2}{3} \, x^{3} - \frac {1}{3} \, x e^{3} - \frac {1}{3} \, {\left (x^{3} + 3 \, x\right )} \log \relax (x) - \frac {2}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*x^2-3)*log(x)-1/3*exp(3)-7/3*x^2-5/3,x, algorithm="maxima")

[Out]

-2/3*x^3 - 1/3*x*e^3 - 1/3*(x^3 + 3*x)*log(x) - 2/3*x

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mupad [B]  time = 0.26, size = 22, normalized size = 0.79 \begin {gather*} -\frac {x\,\left ({\mathrm {e}}^3+3\,\ln \relax (x)+x^2\,\ln \relax (x)+2\,x^2+2\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(- exp(3)/3 - (7*x^2)/3 - (log(x)*(3*x^2 + 3))/3 - 5/3,x)

[Out]

-(x*(exp(3) + 3*log(x) + x^2*log(x) + 2*x^2 + 2))/3

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sympy [A]  time = 0.10, size = 29, normalized size = 1.04 \begin {gather*} - \frac {2 x^{3}}{3} + x \left (- \frac {e^{3}}{3} - \frac {2}{3}\right ) + \left (- \frac {x^{3}}{3} - x\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-3*x**2-3)*ln(x)-1/3*exp(3)-7/3*x**2-5/3,x)

[Out]

-2*x**3/3 + x*(-exp(3)/3 - 2/3) + (-x**3/3 - x)*log(x)

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