3.26.17 \(\int \frac {28 x-32 x^2+4 x^3+(-20+20 x^2) \log (3)+(28 x-64 x^2+12 x^3+40 x^2 \log (3)) \log (x)}{x \log (3)} \, dx\)

Optimal. Leaf size=26 \[ 4 (-1+x) x \left (5+\frac {5}{x}-\frac {7-x}{\log (3)}\right ) \log (x) \]

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Rubi [A]  time = 0.10, antiderivative size = 48, normalized size of antiderivative = 1.85, number of steps used = 10, number of rules used = 6, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 14, 765, 2356, 2295, 2304} \begin {gather*} \frac {4 x^3 \log (x)}{\log (3)}-\frac {4 x^2 (8-\log (243)) \log (x)}{\log (3)}+\frac {28 x \log (x)}{\log (3)}-\frac {4 \log (243) \log (x)}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(28*x - 32*x^2 + 4*x^3 + (-20 + 20*x^2)*Log[3] + (28*x - 64*x^2 + 12*x^3 + 40*x^2*Log[3])*Log[x])/(x*Log[3
]),x]

[Out]

(28*x*Log[x])/Log[3] + (4*x^3*Log[x])/Log[3] - (4*x^2*(8 - Log[243])*Log[x])/Log[3] - (4*Log[243]*Log[x])/Log[
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 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 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

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 {\int \frac {28 x-32 x^2+4 x^3+\left (-20+20 x^2\right ) \log (3)+\left (28 x-64 x^2+12 x^3+40 x^2 \log (3)\right ) \log (x)}{x} \, dx}{\log (3)}\\ &=\frac {\int \left (\frac {4 (1-x) \left (-x^2+x (7-\log (243))-\log (243)\right )}{x}+4 \left (7+3 x^2-2 x (8-\log (243))\right ) \log (x)\right ) \, dx}{\log (3)}\\ &=\frac {4 \int \frac {(1-x) \left (-x^2+x (7-\log (243))-\log (243)\right )}{x} \, dx}{\log (3)}+\frac {4 \int \left (7+3 x^2-2 x (8-\log (243))\right ) \log (x) \, dx}{\log (3)}\\ &=\frac {4 \int \left (7+x^2-x (8-\log (243))-\frac {\log (243)}{x}\right ) \, dx}{\log (3)}+\frac {4 \int \left (7 \log (x)+3 x^2 \log (x)+2 x (-8+\log (243)) \log (x)\right ) \, dx}{\log (3)}\\ &=\frac {28 x}{\log (3)}+\frac {4 x^3}{3 \log (3)}-\frac {2 x^2 (8-\log (243))}{\log (3)}-\frac {4 \log (243) \log (x)}{\log (3)}+\frac {12 \int x^2 \log (x) \, dx}{\log (3)}+\frac {28 \int \log (x) \, dx}{\log (3)}-\frac {(8 (8-\log (243))) \int x \log (x) \, dx}{\log (3)}\\ &=\frac {28 x \log (x)}{\log (3)}+\frac {4 x^3 \log (x)}{\log (3)}-\frac {4 x^2 (8-\log (243)) \log (x)}{\log (3)}-\frac {4 \log (243) \log (x)}{\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 39, normalized size = 1.50 \begin {gather*} \frac {4 \left (7 x \log (x)-8 x^2 \log (x)+x^3 \log (x)-\log (243) \log (x)+x^2 \log (243) \log (x)\right )}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(28*x - 32*x^2 + 4*x^3 + (-20 + 20*x^2)*Log[3] + (28*x - 64*x^2 + 12*x^3 + 40*x^2*Log[3])*Log[x])/(x
*Log[3]),x]

[Out]

(4*(7*x*Log[x] - 8*x^2*Log[x] + x^3*Log[x] - Log[243]*Log[x] + x^2*Log[243]*Log[x]))/Log[3]

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fricas [A]  time = 0.59, size = 29, normalized size = 1.12 \begin {gather*} \frac {4 \, {\left (x^{3} - 8 \, x^{2} + 5 \, {\left (x^{2} - 1\right )} \log \relax (3) + 7 \, x\right )} \log \relax (x)}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^2*log(3)+12*x^3-64*x^2+28*x)*log(x)+(20*x^2-20)*log(3)+4*x^3-32*x^2+28*x)/x/log(3),x, algorit
hm="fricas")

[Out]

4*(x^3 - 8*x^2 + 5*(x^2 - 1)*log(3) + 7*x)*log(x)/log(3)

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giac [A]  time = 0.13, size = 33, normalized size = 1.27 \begin {gather*} \frac {4 \, {\left ({\left (x^{3} + x^{2} {\left (5 \, \log \relax (3) - 8\right )} + 7 \, x\right )} \log \relax (x) - 5 \, \log \relax (3) \log \relax (x)\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^2*log(3)+12*x^3-64*x^2+28*x)*log(x)+(20*x^2-20)*log(3)+4*x^3-32*x^2+28*x)/x/log(3),x, algorit
hm="giac")

[Out]

4*((x^3 + x^2*(5*log(3) - 8) + 7*x)*log(x) - 5*log(3)*log(x))/log(3)

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maple [A]  time = 0.04, size = 34, normalized size = 1.31

method	result	size
risch	\(\frac {\left (20 x^{2} \ln \relax (3)+4 x^{3}-32 x^{2}+28 x \right ) \ln \relax (x )}{\ln \relax (3)}-20 \ln \relax (x )\)	\(34\)
norman	\(-20 \ln \relax (x )+\frac {28 x \ln \relax (x )}{\ln \relax (3)}+\frac {4 x^{3} \ln \relax (x )}{\ln \relax (3)}+\frac {4 \left (5 \ln \relax (3)-8\right ) x^{2} \ln \relax (x )}{\ln \relax (3)}\)	\(43\)
default	\(\frac {40 \ln \relax (3) \left (\frac {x^{2} \ln \relax (x )}{2}-\frac {x^{2}}{4}\right )+4 x^{3} \ln \relax (x )-32 x^{2} \ln \relax (x )+10 x^{2} \ln \relax (3)+28 x \ln \relax (x )-20 \ln \relax (3) \ln \relax (x )}{\ln \relax (3)}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((40*x^2*ln(3)+12*x^3-64*x^2+28*x)*ln(x)+(20*x^2-20)*ln(3)+4*x^3-32*x^2+28*x)/x/ln(3),x,method=_RETURNVERB
OSE)

[Out]

1/ln(3)*(20*x^2*ln(3)+4*x^3-32*x^2+28*x)*ln(x)-20*ln(x)

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maxima [B]  time = 0.39, size = 56, normalized size = 2.15 \begin {gather*} \frac {2 \, {\left (2 \, x^{3} \log \relax (x) + 5 \, x^{2} \log \relax (3) - 16 \, x^{2} \log \relax (x) + 5 \, {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} \log \relax (3) + 14 \, x \log \relax (x) - 10 \, \log \relax (3) \log \relax (x)\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^2*log(3)+12*x^3-64*x^2+28*x)*log(x)+(20*x^2-20)*log(3)+4*x^3-32*x^2+28*x)/x/log(3),x, algorit
hm="maxima")

[Out]

2*(2*x^3*log(x) + 5*x^2*log(3) - 16*x^2*log(x) + 5*(2*x^2*log(x) - x^2)*log(3) + 14*x*log(x) - 10*log(3)*log(x
))/log(3)

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mupad [B]  time = 1.56, size = 41, normalized size = 1.58 \begin {gather*} \frac {28\,x\,\ln \relax (x)}{\ln \relax (3)}-20\,\ln \relax (x)+\frac {4\,x^3\,\ln \relax (x)}{\ln \relax (3)}+\frac {x^2\,\ln \relax (x)\,\left (20\,\ln \relax (3)-32\right )}{\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((28*x + log(3)*(20*x^2 - 20) + log(x)*(28*x + 40*x^2*log(3) - 64*x^2 + 12*x^3) - 32*x^2 + 4*x^3)/(x*log(3)
),x)

[Out]

(28*x*log(x))/log(3) - 20*log(x) + (4*x^3*log(x))/log(3) + (x^2*log(x)*(20*log(3) - 32))/log(3)

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sympy [A]  time = 0.16, size = 32, normalized size = 1.23 \begin {gather*} \frac {\left (4 x^{3} - 32 x^{2} + 20 x^{2} \log {\relax (3 )} + 28 x\right ) \log {\relax (x )}}{\log {\relax (3 )}} - 20 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x**2*ln(3)+12*x**3-64*x**2+28*x)*ln(x)+(20*x**2-20)*ln(3)+4*x**3-32*x**2+28*x)/x/ln(3),x)

[Out]

(4*x**3 - 32*x**2 + 20*x**2*log(3) + 28*x)*log(x)/log(3) - 20*log(x)

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