3.71.92 \(\int \frac {-25+25 x+50 x^2-2 x^3-2 x^4+(4 x^2-6 x^3-8 x^4) \log (x)+4 x^2 \log ^2(x)}{5 x} \, dx\)

Optimal. Leaf size=21 \[ \left (x+x^2-\log (x)\right ) \left (5-\frac {2}{5} x^2 \log (x)\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 42, normalized size of antiderivative = 2.00, number of steps used = 12, number of rules used = 5, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 14, 2356, 2304, 2305} \begin {gather*} -\frac {2}{5} x^4 \log (x)-\frac {2}{5} x^3 \log (x)+5 x^2+\frac {2}{5} x^2 \log ^2(x)+5 x-5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 + 25*x + 50*x^2 - 2*x^3 - 2*x^4 + (4*x^2 - 6*x^3 - 8*x^4)*Log[x] + 4*x^2*Log[x]^2)/(5*x),x]

[Out]

5*x + 5*x^2 - 5*Log[x] - (2*x^3*Log[x])/5 - (2*x^4*Log[x])/5 + (2*x^2*Log[x]^2)/5

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 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 2305

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

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 \frac {-25+25 x+50 x^2-2 x^3-2 x^4+\left (4 x^2-6 x^3-8 x^4\right ) \log (x)+4 x^2 \log ^2(x)}{x} \, dx\\ &=\frac {1}{5} \int \left (\frac {-25+25 x+50 x^2-2 x^3-2 x^4}{x}-2 x \left (-2+3 x+4 x^2\right ) \log (x)+4 x \log ^2(x)\right ) \, dx\\ &=\frac {1}{5} \int \frac {-25+25 x+50 x^2-2 x^3-2 x^4}{x} \, dx-\frac {2}{5} \int x \left (-2+3 x+4 x^2\right ) \log (x) \, dx+\frac {4}{5} \int x \log ^2(x) \, dx\\ &=\frac {2}{5} x^2 \log ^2(x)+\frac {1}{5} \int \left (25-\frac {25}{x}+50 x-2 x^2-2 x^3\right ) \, dx-\frac {2}{5} \int \left (-2 x \log (x)+3 x^2 \log (x)+4 x^3 \log (x)\right ) \, dx-\frac {4}{5} \int x \log (x) \, dx\\ &=5 x+\frac {26 x^2}{5}-\frac {2 x^3}{15}-\frac {x^4}{10}-5 \log (x)-\frac {2}{5} x^2 \log (x)+\frac {2}{5} x^2 \log ^2(x)+\frac {4}{5} \int x \log (x) \, dx-\frac {6}{5} \int x^2 \log (x) \, dx-\frac {8}{5} \int x^3 \log (x) \, dx\\ &=5 x+5 x^2-5 \log (x)-\frac {2}{5} x^3 \log (x)-\frac {2}{5} x^4 \log (x)+\frac {2}{5} x^2 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 42, normalized size = 2.00 \begin {gather*} 5 x+5 x^2-5 \log (x)-\frac {2}{5} x^3 \log (x)-\frac {2}{5} x^4 \log (x)+\frac {2}{5} x^2 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 + 25*x + 50*x^2 - 2*x^3 - 2*x^4 + (4*x^2 - 6*x^3 - 8*x^4)*Log[x] + 4*x^2*Log[x]^2)/(5*x),x]

[Out]

5*x + 5*x^2 - 5*Log[x] - (2*x^3*Log[x])/5 - (2*x^4*Log[x])/5 + (2*x^2*Log[x]^2)/5

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fricas [A]  time = 1.11, size = 34, normalized size = 1.62 \begin {gather*} \frac {2}{5} \, x^{2} \log \relax (x)^{2} + 5 \, x^{2} - \frac {1}{5} \, {\left (2 \, x^{4} + 2 \, x^{3} + 25\right )} \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x^2*log(x)^2+(-8*x^4-6*x^3+4*x^2)*log(x)-2*x^4-2*x^3+50*x^2+25*x-25)/x,x, algorithm="fricas")

[Out]

2/5*x^2*log(x)^2 + 5*x^2 - 1/5*(2*x^4 + 2*x^3 + 25)*log(x) + 5*x

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giac [A]  time = 0.21, size = 33, normalized size = 1.57 \begin {gather*} \frac {2}{5} \, x^{2} \log \relax (x)^{2} + 5 \, x^{2} - \frac {2}{5} \, {\left (x^{4} + x^{3}\right )} \log \relax (x) + 5 \, x - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x^2*log(x)^2+(-8*x^4-6*x^3+4*x^2)*log(x)-2*x^4-2*x^3+50*x^2+25*x-25)/x,x, algorithm="giac")

[Out]

2/5*x^2*log(x)^2 + 5*x^2 - 2/5*(x^4 + x^3)*log(x) + 5*x - 5*log(x)

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maple [A]  time = 0.02, size = 37, normalized size = 1.76

method	result	size
default	\(-\frac {2 x^{4} \ln \relax (x )}{5}+\frac {2 x^{2} \ln \relax (x )^{2}}{5}+5 x^{2}-\frac {2 x^{3} \ln \relax (x )}{5}+5 x -5 \ln \relax (x )\)	\(37\)
norman	\(-\frac {2 x^{4} \ln \relax (x )}{5}+\frac {2 x^{2} \ln \relax (x )^{2}}{5}+5 x^{2}-\frac {2 x^{3} \ln \relax (x )}{5}+5 x -5 \ln \relax (x )\)	\(37\)
risch	\(\frac {2 x^{2} \ln \relax (x )^{2}}{5}+\frac {\left (-2 x^{4}-2 x^{3}\right ) \ln \relax (x )}{5}+5 x^{2}+5 x -5 \ln \relax (x )\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(4*x^2*ln(x)^2+(-8*x^4-6*x^3+4*x^2)*ln(x)-2*x^4-2*x^3+50*x^2+25*x-25)/x,x,method=_RETURNVERBOSE)

[Out]

-2/5*x^4*ln(x)+2/5*x^2*ln(x)^2+5*x^2-2/5*x^3*ln(x)+5*x-5*ln(x)

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maxima [B]  time = 0.37, size = 51, normalized size = 2.43 \begin {gather*} -\frac {2}{5} \, x^{4} \log \relax (x) - \frac {2}{5} \, x^{3} \log \relax (x) + \frac {1}{5} \, {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} + \frac {2}{5} \, x^{2} \log \relax (x) + \frac {24}{5} \, x^{2} + 5 \, x - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x^2*log(x)^2+(-8*x^4-6*x^3+4*x^2)*log(x)-2*x^4-2*x^3+50*x^2+25*x-25)/x,x, algorithm="maxima")

[Out]

-2/5*x^4*log(x) - 2/5*x^3*log(x) + 1/5*(2*log(x)^2 - 2*log(x) + 1)*x^2 + 2/5*x^2*log(x) + 24/5*x^2 + 5*x - 5*l
og(x)

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mupad [B]  time = 4.23, size = 20, normalized size = 0.95 \begin {gather*} -\frac {\left (2\,x^2\,\ln \relax (x)-25\right )\,\left (x-\ln \relax (x)+x^2\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((log(x)*(6*x^3 - 4*x^2 + 8*x^4))/5 - 5*x - (4*x^2*log(x)^2)/5 - 10*x^2 + (2*x^3)/5 + (2*x^4)/5 + 5)/x,x)

[Out]

-((2*x^2*log(x) - 25)*(x - log(x) + x^2))/5

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sympy [B]  time = 0.19, size = 42, normalized size = 2.00 \begin {gather*} \frac {2 x^{2} \log {\relax (x )}^{2}}{5} + 5 x^{2} + 5 x + \left (- \frac {2 x^{4}}{5} - \frac {2 x^{3}}{5}\right ) \log {\relax (x )} - 5 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x**2*ln(x)**2+(-8*x**4-6*x**3+4*x**2)*ln(x)-2*x**4-2*x**3+50*x**2+25*x-25)/x,x)

[Out]

2*x**2*log(x)**2/5 + 5*x**2 + 5*x + (-2*x**4/5 - 2*x**3/5)*log(x) - 5*log(x)

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