3.52.82 \(\int \frac {(-8-6 x) \log ^2(5) \log (2 x-x \log (3))-3 x \log ^2(5) \log ^2(2 x-x \log (3))}{x} \, dx\)

Optimal. Leaf size=21 \[ (-4-3 x) \log ^2(5) \log ^2(x (2-\log (3))) \]

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Rubi [A]  time = 0.07, antiderivative size = 36, normalized size of antiderivative = 1.71, number of steps used = 7, number of rules used = 5, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {14, 2346, 2301, 2295, 2296} \begin {gather*} -3 x \log ^2(5) \log ^2(x (2-\log (3)))-4 \log ^2(5) \log ^2(x (2-\log (3))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

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 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 (-4-3 x) \log ^2(5) \log (x (2-\log (3)))}{x}-3 \log ^2(5) \log ^2(x (2-\log (3)))\right ) \, dx\\ &=\left (2 \log ^2(5)\right ) \int \frac {(-4-3 x) \log (x (2-\log (3)))}{x} \, dx-\left (3 \log ^2(5)\right ) \int \log ^2(x (2-\log (3))) \, dx\\ &=-3 x \log ^2(5) \log ^2(x (2-\log (3)))-\left (8 \log ^2(5)\right ) \int \frac {\log (x (2-\log (3)))}{x} \, dx\\ &=-4 \log ^2(5) \log ^2(x (2-\log (3)))-3 x \log ^2(5) \log ^2(x (2-\log (3)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 1.71 \begin {gather*} -4 \log ^2(5) \log ^2(x (2-\log (3)))-3 x \log ^2(5) \log ^2(x (2-\log (3))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 23, normalized size = 1.10 \begin {gather*} -{\left (3 \, x + 4\right )} \log \relax (5)^{2} \log \left (-x \log \relax (3) + 2 \, x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 28, normalized size = 1.33 \begin {gather*} -{\left (3 \, x \log \relax (5)^{2} + 4 \, \log \relax (5)^{2}\right )} \log \left (-x \log \relax (3) + 2 \, x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 28, normalized size = 1.33

method	result	size
risch	\(\left (-3 x \ln \relax (5)^{2}-4 \ln \relax (5)^{2}\right ) \ln \left (-x \ln \relax (3)+2 x \right )^{2}\)	\(28\)
norman	\(-4 \ln \relax (5)^{2} \ln \left (-x \ln \relax (3)+2 x \right )^{2}-3 x \ln \relax (5)^{2} \ln \left (-x \ln \relax (3)+2 x \right )^{2}\)	\(39\)
derivativedivides	\(\frac {3 \ln \relax (5)^{2} \left (x \left (2-\ln \relax (3)\right ) \ln \left (x \left (2-\ln \relax (3)\right )\right )^{2}-2 x \left (2-\ln \relax (3)\right ) \ln \left (x \left (2-\ln \relax (3)\right )\right )+2 x \left (2-\ln \relax (3)\right )\right )}{\ln \relax (3)-2}-\frac {4 \ln \relax (5)^{2} \ln \relax (3) \ln \left (x \left (2-\ln \relax (3)\right )\right )^{2}}{\ln \relax (3)-2}+\frac {6 \ln \relax (5)^{2} \left (x \left (2-\ln \relax (3)\right ) \ln \left (x \left (2-\ln \relax (3)\right )\right )-x \left (2-\ln \relax (3)\right )\right )}{\ln \relax (3)-2}+\frac {8 \ln \relax (5)^{2} \ln \left (x \left (2-\ln \relax (3)\right )\right )^{2}}{\ln \relax (3)-2}\)	\(148\)
default	\(\frac {3 \ln \relax (5)^{2} \left (x \left (2-\ln \relax (3)\right ) \ln \left (x \left (2-\ln \relax (3)\right )\right )^{2}-2 x \left (2-\ln \relax (3)\right ) \ln \left (x \left (2-\ln \relax (3)\right )\right )+2 x \left (2-\ln \relax (3)\right )\right )}{\ln \relax (3)-2}-\frac {4 \ln \relax (5)^{2} \ln \relax (3) \ln \left (x \left (2-\ln \relax (3)\right )\right )^{2}}{\ln \relax (3)-2}+\frac {6 \ln \relax (5)^{2} \left (x \left (2-\ln \relax (3)\right ) \ln \left (x \left (2-\ln \relax (3)\right )\right )-x \left (2-\ln \relax (3)\right )\right )}{\ln \relax (3)-2}+\frac {8 \ln \relax (5)^{2} \ln \left (x \left (2-\ln \relax (3)\right )\right )^{2}}{\ln \relax (3)-2}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.44, size = 143, normalized size = 6.81 \begin {gather*} -8 \, \log \relax (5)^{2} \log \left (-x \log \relax (3) + 2 \, x\right ) \log \relax (x) + 4 \, {\left (2 \, \log \left (-x \log \relax (3) + 2 \, x\right ) \log \relax (x) - \log \relax (x)^{2} - 2 \, \log \relax (x) \log \left (-\log \relax (3) + 2\right )\right )} \log \relax (5)^{2} - \frac {3 \, {\left (x \log \relax (3) - 2 \, x\right )} {\left (\log \left (-x \log \relax (3) + 2 \, x\right )^{2} - 2 \, \log \left (-x \log \relax (3) + 2 \, x\right ) + 2\right )} \log \relax (5)^{2}}{\log \relax (3) - 2} + \frac {6 \, {\left (x \log \relax (3) - {\left (x \log \relax (3) - 2 \, x\right )} \log \left (-x \log \relax (3) + 2 \, x\right ) - 2 \, x\right )} \log \relax (5)^{2}}{\log \relax (3) - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.29, size = 23, normalized size = 1.10 \begin {gather*} -{\ln \relax (5)}^2\,{\ln \left (2\,x-x\,\ln \relax (3)\right )}^2\,\left (3\,x+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.14, size = 27, normalized size = 1.29 \begin {gather*} \left (- 3 x \log {\relax (5 )}^{2} - 4 \log {\relax (5 )}^{2}\right ) \log {\left (- x \log {\relax (3 )} + 2 x \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*ln(5)**2*ln(-x*ln(3)+2*x)**2+(-6*x-8)*ln(5)**2*ln(-x*ln(3)+2*x))/x,x)

[Out]

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

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