3.92.7 \(\int \frac {25+36 x-72 x^2+6 x^3-48 x^4-6 x^6+(-18+36 x+18 x^3) \log (x)}{25 x} \, dx\)

Optimal. Leaf size=25 \[ 2+\log (x)-\frac {1}{25} x^2 \left (6+x^2-\frac {3 \log (x)}{x}\right )^2 \]

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Rubi [A]  time = 0.07, antiderivative size = 48, normalized size of antiderivative = 1.92, number of steps used = 10, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 14, 2357, 2295, 2301, 2304} \begin {gather*} -\frac {x^6}{25}-\frac {12 x^4}{25}+\frac {6}{25} x^3 \log (x)-\frac {36 x^2}{25}-\frac {9 \log ^2(x)}{25}+\frac {36}{25} x \log (x)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 + 36*x - 72*x^2 + 6*x^3 - 48*x^4 - 6*x^6 + (-18 + 36*x + 18*x^3)*Log[x])/(25*x),x]

[Out]

(-36*x^2)/25 - (12*x^4)/25 - x^6/25 + Log[x] + (36*x*Log[x])/25 + (6*x^3*Log[x])/25 - (9*Log[x]^2)/25

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, 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 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \frac {25+36 x-72 x^2+6 x^3-48 x^4-6 x^6+\left (-18+36 x+18 x^3\right ) \log (x)}{x} \, dx\\ &=\frac {1}{25} \int \left (\frac {25+36 x-72 x^2+6 x^3-48 x^4-6 x^6}{x}+\frac {18 \left (-1+2 x+x^3\right ) \log (x)}{x}\right ) \, dx\\ &=\frac {1}{25} \int \frac {25+36 x-72 x^2+6 x^3-48 x^4-6 x^6}{x} \, dx+\frac {18}{25} \int \frac {\left (-1+2 x+x^3\right ) \log (x)}{x} \, dx\\ &=\frac {1}{25} \int \left (36+\frac {25}{x}-72 x+6 x^2-48 x^3-6 x^5\right ) \, dx+\frac {18}{25} \int \left (2 \log (x)-\frac {\log (x)}{x}+x^2 \log (x)\right ) \, dx\\ &=\frac {36 x}{25}-\frac {36 x^2}{25}+\frac {2 x^3}{25}-\frac {12 x^4}{25}-\frac {x^6}{25}+\log (x)-\frac {18}{25} \int \frac {\log (x)}{x} \, dx+\frac {18}{25} \int x^2 \log (x) \, dx+\frac {36}{25} \int \log (x) \, dx\\ &=-\frac {36 x^2}{25}-\frac {12 x^4}{25}-\frac {x^6}{25}+\log (x)+\frac {36}{25} x \log (x)+\frac {6}{25} x^3 \log (x)-\frac {9 \log ^2(x)}{25}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 48, normalized size = 1.92 \begin {gather*} -\frac {36 x^2}{25}-\frac {12 x^4}{25}-\frac {x^6}{25}+\log (x)+\frac {36}{25} x \log (x)+\frac {6}{25} x^3 \log (x)-\frac {9 \log ^2(x)}{25} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + 36*x - 72*x^2 + 6*x^3 - 48*x^4 - 6*x^6 + (-18 + 36*x + 18*x^3)*Log[x])/(25*x),x]

[Out]

(-36*x^2)/25 - (12*x^4)/25 - x^6/25 + Log[x] + (36*x*Log[x])/25 + (6*x^3*Log[x])/25 - (9*Log[x]^2)/25

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fricas [A]  time = 0.83, size = 36, normalized size = 1.44 \begin {gather*} -\frac {1}{25} \, x^{6} - \frac {12}{25} \, x^{4} - \frac {36}{25} \, x^{2} + \frac {1}{25} \, {\left (6 \, x^{3} + 36 \, x + 25\right )} \log \relax (x) - \frac {9}{25} \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((18*x^3+36*x-18)*log(x)-6*x^6-48*x^4+6*x^3-72*x^2+36*x+25)/x,x, algorithm="fricas")

[Out]

-1/25*x^6 - 12/25*x^4 - 36/25*x^2 + 1/25*(6*x^3 + 36*x + 25)*log(x) - 9/25*log(x)^2

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giac [A]  time = 0.21, size = 35, normalized size = 1.40 \begin {gather*} -\frac {1}{25} \, x^{6} - \frac {12}{25} \, x^{4} - \frac {36}{25} \, x^{2} + \frac {6}{25} \, {\left (x^{3} + 6 \, x\right )} \log \relax (x) - \frac {9}{25} \, \log \relax (x)^{2} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((18*x^3+36*x-18)*log(x)-6*x^6-48*x^4+6*x^3-72*x^2+36*x+25)/x,x, algorithm="giac")

[Out]

-1/25*x^6 - 12/25*x^4 - 36/25*x^2 + 6/25*(x^3 + 6*x)*log(x) - 9/25*log(x)^2 + log(x)

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maple [A]  time = 0.04, size = 37, normalized size = 1.48

method	result	size
default	\(-\frac {x^{6}}{25}+\frac {6 x^{3} \ln \relax (x )}{25}-\frac {12 x^{4}}{25}+\frac {36 x \ln \relax (x )}{25}-\frac {36 x^{2}}{25}-\frac {9 \ln \relax (x )^{2}}{25}+\ln \relax (x )\)	\(37\)
norman	\(-\frac {x^{6}}{25}+\frac {6 x^{3} \ln \relax (x )}{25}-\frac {12 x^{4}}{25}+\frac {36 x \ln \relax (x )}{25}-\frac {36 x^{2}}{25}-\frac {9 \ln \relax (x )^{2}}{25}+\ln \relax (x )\)	\(37\)
risch	\(-\frac {9 \ln \relax (x )^{2}}{25}+\frac {\left (6 x^{3}+36 x \right ) \ln \relax (x )}{25}-\frac {x^{6}}{25}-\frac {12 x^{4}}{25}-\frac {36 x^{2}}{25}+\ln \relax (x )\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*((18*x^3+36*x-18)*ln(x)-6*x^6-48*x^4+6*x^3-72*x^2+36*x+25)/x,x,method=_RETURNVERBOSE)

[Out]

-1/25*x^6+6/25*x^3*ln(x)-12/25*x^4+36/25*x*ln(x)-36/25*x^2-9/25*ln(x)^2+ln(x)

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maxima [A]  time = 0.39, size = 36, normalized size = 1.44 \begin {gather*} -\frac {1}{25} \, x^{6} - \frac {12}{25} \, x^{4} + \frac {6}{25} \, x^{3} \log \relax (x) - \frac {36}{25} \, x^{2} + \frac {36}{25} \, x \log \relax (x) - \frac {9}{25} \, \log \relax (x)^{2} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((18*x^3+36*x-18)*log(x)-6*x^6-48*x^4+6*x^3-72*x^2+36*x+25)/x,x, algorithm="maxima")

[Out]

-1/25*x^6 - 12/25*x^4 + 6/25*x^3*log(x) - 36/25*x^2 + 36/25*x*log(x) - 9/25*log(x)^2 + log(x)

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mupad [B]  time = 8.25, size = 36, normalized size = 1.44 \begin {gather*} -\frac {x^6}{25}-\frac {12\,x^4}{25}+\frac {6\,x^3\,\ln \relax (x)}{25}-\frac {36\,x^2}{25}+\frac {36\,x\,\ln \relax (x)}{25}-\frac {9\,{\ln \relax (x)}^2}{25}+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((36*x)/25 + (log(x)*(36*x + 18*x^3 - 18))/25 - (72*x^2)/25 + (6*x^3)/25 - (48*x^4)/25 - (6*x^6)/25 + 1)/x
,x)

[Out]

log(x) + (6*x^3*log(x))/25 - (9*log(x)^2)/25 + (36*x*log(x))/25 - (36*x^2)/25 - (12*x^4)/25 - x^6/25

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sympy [A]  time = 0.14, size = 44, normalized size = 1.76 \begin {gather*} - \frac {x^{6}}{25} - \frac {12 x^{4}}{25} - \frac {36 x^{2}}{25} + \left (\frac {6 x^{3}}{25} + \frac {36 x}{25}\right ) \log {\relax (x )} - \frac {9 \log {\relax (x )}^{2}}{25} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((18*x**3+36*x-18)*ln(x)-6*x**6-48*x**4+6*x**3-72*x**2+36*x+25)/x,x)

[Out]

-x**6/25 - 12*x**4/25 - 36*x**2/25 + (6*x**3/25 + 36*x/25)*log(x) - 9*log(x)**2/25 + log(x)

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