3.4.70 \(\int \frac {1}{5} (-55+6 x+60 x^2+(-60+12 x-90 x^2) \log (x)) \, dx\)

Optimal. Leaf size=28 \[ x-6 x^2 \left (-x+\left (-\frac {-2+\frac {x}{5}}{x}+x\right ) \log (x)\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {12, 2356, 2295, 2304} \begin {gather*} 6 x^3-6 x^3 \log (x)+\frac {6}{5} x^2 \log (x)+x-12 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-55 + 6*x + 60*x^2 + (-60 + 12*x - 90*x^2)*Log[x])/5,x]

[Out]

x + 6*x^3 - 12*x*Log[x] + (6*x^2*Log[x])/5 - 6*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 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}{5} \int \left (-55+6 x+60 x^2+\left (-60+12 x-90 x^2\right ) \log (x)\right ) \, dx\\ &=-11 x+\frac {3 x^2}{5}+4 x^3+\frac {1}{5} \int \left (-60+12 x-90 x^2\right ) \log (x) \, dx\\ &=-11 x+\frac {3 x^2}{5}+4 x^3+\frac {1}{5} \int \left (-60 \log (x)+12 x \log (x)-90 x^2 \log (x)\right ) \, dx\\ &=-11 x+\frac {3 x^2}{5}+4 x^3+\frac {12}{5} \int x \log (x) \, dx-12 \int \log (x) \, dx-18 \int x^2 \log (x) \, dx\\ &=x+6 x^3-12 x \log (x)+\frac {6}{5} x^2 \log (x)-6 x^3 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \begin {gather*} x+6 x^3-12 x \log (x)+\frac {6}{5} x^2 \log (x)-6 x^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-55 + 6*x + 60*x^2 + (-60 + 12*x - 90*x^2)*Log[x])/5,x]

[Out]

x + 6*x^3 - 12*x*Log[x] + (6*x^2*Log[x])/5 - 6*x^3*Log[x]

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fricas [A]  time = 0.73, size = 25, normalized size = 0.89 \begin {gather*} 6 \, x^{3} - \frac {6}{5} \, {\left (5 \, x^{3} - x^{2} + 10 \, x\right )} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-90*x^2+12*x-60)*log(x)+12*x^2+6/5*x-11,x, algorithm="fricas")

[Out]

6*x^3 - 6/5*(5*x^3 - x^2 + 10*x)*log(x) + x

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giac [A]  time = 0.28, size = 26, normalized size = 0.93 \begin {gather*} -6 \, x^{3} \log \relax (x) + 6 \, x^{3} + \frac {6}{5} \, x^{2} \log \relax (x) - 12 \, x \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-90*x^2+12*x-60)*log(x)+12*x^2+6/5*x-11,x, algorithm="giac")

[Out]

-6*x^3*log(x) + 6*x^3 + 6/5*x^2*log(x) - 12*x*log(x) + x

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maple [A]  time = 0.02, size = 26, normalized size = 0.93

method	result	size
risch	\(\frac {\left (-30 x^{3}+6 x^{2}-60 x \right ) \ln \relax (x )}{5}+6 x^{3}+x\)	\(26\)
default	\(x +6 x^{3}-6 x^{3} \ln \relax (x )+\frac {6 x^{2} \ln \relax (x )}{5}-12 x \ln \relax (x )\)	\(27\)
norman	\(x +6 x^{3}-6 x^{3} \ln \relax (x )+\frac {6 x^{2} \ln \relax (x )}{5}-12 x \ln \relax (x )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-90*x^2+12*x-60)*ln(x)+12*x^2+6/5*x-11,x,method=_RETURNVERBOSE)

[Out]

1/5*(-30*x^3+6*x^2-60*x)*ln(x)+6*x^3+x

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maxima [A]  time = 0.51, size = 25, normalized size = 0.89 \begin {gather*} 6 \, x^{3} - \frac {6}{5} \, {\left (5 \, x^{3} - x^{2} + 10 \, x\right )} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-90*x^2+12*x-60)*log(x)+12*x^2+6/5*x-11,x, algorithm="maxima")

[Out]

6*x^3 - 6/5*(5*x^3 - x^2 + 10*x)*log(x) + x

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mupad [B]  time = 0.45, size = 26, normalized size = 0.93 \begin {gather*} \frac {x\,\left (6\,x\,\ln \relax (x)-30\,x^2\,\ln \relax (x)-60\,\ln \relax (x)+30\,x^2+5\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x)/5 - (log(x)*(90*x^2 - 12*x + 60))/5 + 12*x^2 - 11,x)

[Out]

(x*(6*x*log(x) - 30*x^2*log(x) - 60*log(x) + 30*x^2 + 5))/5

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sympy [A]  time = 0.11, size = 24, normalized size = 0.86 \begin {gather*} 6 x^{3} + x + \left (- 6 x^{3} + \frac {6 x^{2}}{5} - 12 x\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-90*x**2+12*x-60)*ln(x)+12*x**2+6/5*x-11,x)

[Out]

6*x**3 + x + (-6*x**3 + 6*x**2/5 - 12*x)*log(x)

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