3.60.93 \(\int \frac {-3-3 e^5-5 x-3 x^2+3 x^3+(-5 x-6 x^2+9 x^3) \log (x)}{3 x} \, dx\)

Optimal. Leaf size=23 \[ \left (-1-e^5-\frac {5 x}{3}-x^2+x^3\right ) \log (x) \]

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Rubi [A]  time = 0.05, antiderivative size = 30, normalized size of antiderivative = 1.30, number of steps used = 10, number of rules used = 5, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 14, 2356, 2295, 2304} \begin {gather*} x^3 \log (x)-x^2 \log (x)-\frac {5}{3} x \log (x)-\left (1+e^5\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 - 3*E^5 - 5*x - 3*x^2 + 3*x^3 + (-5*x - 6*x^2 + 9*x^3)*Log[x])/(3*x),x]

[Out]

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

Rule 12

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

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2304

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

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {-3-3 e^5-5 x-3 x^2+3 x^3+\left (-5 x-6 x^2+9 x^3\right ) \log (x)}{x} \, dx\\ &=\frac {1}{3} \int \left (\frac {-3 \left (1+e^5\right )-5 x-3 x^2+3 x^3}{x}+\left (-5-6 x+9 x^2\right ) \log (x)\right ) \, dx\\ &=\frac {1}{3} \int \frac {-3 \left (1+e^5\right )-5 x-3 x^2+3 x^3}{x} \, dx+\frac {1}{3} \int \left (-5-6 x+9 x^2\right ) \log (x) \, dx\\ &=\frac {1}{3} \int \left (-5-\frac {3 \left (1+e^5\right )}{x}-3 x+3 x^2\right ) \, dx+\frac {1}{3} \int \left (-5 \log (x)-6 x \log (x)+9 x^2 \log (x)\right ) \, dx\\ &=-\frac {5 x}{3}-\frac {x^2}{2}+\frac {x^3}{3}-\left (1+e^5\right ) \log (x)-\frac {5}{3} \int \log (x) \, dx-2 \int x \log (x) \, dx+3 \int x^2 \log (x) \, dx\\ &=-\left (\left (1+e^5\right ) \log (x)\right )-\frac {5}{3} x \log (x)-x^2 \log (x)+x^3 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.39 \begin {gather*} -\log (x)-e^5 \log (x)-\frac {5}{3} x \log (x)-x^2 \log (x)+x^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 - 3*E^5 - 5*x - 3*x^2 + 3*x^3 + (-5*x - 6*x^2 + 9*x^3)*Log[x])/(3*x),x]

[Out]

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

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fricas [A]  time = 0.64, size = 23, normalized size = 1.00 \begin {gather*} \frac {1}{3} \, {\left (3 \, x^{3} - 3 \, x^{2} - 5 \, x - 3 \, e^{5} - 3\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x^3-6*x^2-5*x)*log(x)-3*exp(5)+3*x^3-3*x^2-5*x-3)/x,x, algorithm="fricas")

[Out]

1/3*(3*x^3 - 3*x^2 - 5*x - 3*e^5 - 3)*log(x)

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giac [A]  time = 0.11, size = 29, normalized size = 1.26 \begin {gather*} x^{3} \log \relax (x) - x^{2} \log \relax (x) - \frac {5}{3} \, x \log \relax (x) - e^{5} \log \relax (x) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x^3-6*x^2-5*x)*log(x)-3*exp(5)+3*x^3-3*x^2-5*x-3)/x,x, algorithm="giac")

[Out]

x^3*log(x) - x^2*log(x) - 5/3*x*log(x) - e^5*log(x) - log(x)

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maple [A]  time = 0.04, size = 29, normalized size = 1.26

method	result	size
norman	\(x^{3} \ln \relax (x )+\left (-{\mathrm e}^{5}-1\right ) \ln \relax (x )-\frac {5 x \ln \relax (x )}{3}-x^{2} \ln \relax (x )\)	\(29\)
default	\(x^{3} \ln \relax (x )-x^{2} \ln \relax (x )-\frac {5 x \ln \relax (x )}{3}-{\mathrm e}^{5} \ln \relax (x )-\ln \relax (x )\)	\(30\)
risch	\(\frac {\left (3 x^{3}-3 x^{2}-5 x \right ) \ln \relax (x )}{3}-{\mathrm e}^{5} \ln \relax (x )-\ln \relax (x )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((9*x^3-6*x^2-5*x)*ln(x)-3*exp(5)+3*x^3-3*x^2-5*x-3)/x,x,method=_RETURNVERBOSE)

[Out]

x^3*ln(x)+(-exp(5)-1)*ln(x)-5/3*x*ln(x)-x^2*ln(x)

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maxima [A]  time = 0.34, size = 29, normalized size = 1.26 \begin {gather*} x^{3} \log \relax (x) - x^{2} \log \relax (x) - \frac {5}{3} \, x \log \relax (x) - e^{5} \log \relax (x) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x^3-6*x^2-5*x)*log(x)-3*exp(5)+3*x^3-3*x^2-5*x-3)/x,x, algorithm="maxima")

[Out]

x^3*log(x) - x^2*log(x) - 5/3*x*log(x) - e^5*log(x) - log(x)

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mupad [B]  time = 4.43, size = 23, normalized size = 1.00 \begin {gather*} -\frac {\ln \relax (x)\,\left (-3\,x^3+3\,x^2+5\,x+3\,{\mathrm {e}}^5+3\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((5*x)/3 + exp(5) + x^2 - x^3 + (log(x)*(5*x + 6*x^2 - 9*x^3))/3 + 1)/x,x)

[Out]

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

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sympy [A]  time = 0.12, size = 24, normalized size = 1.04 \begin {gather*} \left (x^{3} - x^{2} - \frac {5 x}{3}\right ) \log {\relax (x )} + \left (- e^{5} - 1\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x**3-6*x**2-5*x)*ln(x)-3*exp(5)+3*x**3-3*x**2-5*x-3)/x,x)

[Out]

(x**3 - x**2 - 5*x/3)*log(x) + (-exp(5) - 1)*log(x)

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