3.1.52 \(\int \frac {-160 x^2-240 x^3 \log (5)+(2-80 x^4) \log ^2(5)}{x \log ^2(5)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (e^{-20 x^2 \left (x+\frac {2}{\log (5)}\right )^2} x^2\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.27, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {12, 14} \begin {gather*} -20 x^4-\frac {80 x^3}{\log (5)}-\frac {80 x^2}{\log ^2(5)}+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-160*x^2 - 240*x^3*Log[5] + (2 - 80*x^4)*Log[5]^2)/(x*Log[5]^2),x]

[Out]

-20*x^4 - (80*x^2)/Log[5]^2 - (80*x^3)/Log[5] + 2*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 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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-160 x^2-240 x^3 \log (5)+\left (2-80 x^4\right ) \log ^2(5)}{x} \, dx}{\log ^2(5)}\\ &=\frac {\int \left (-160 x-240 x^2 \log (5)+\frac {2 \log ^2(5)}{x}-80 x^3 \log ^2(5)\right ) \, dx}{\log ^2(5)}\\ &=-20 x^4-\frac {80 x^2}{\log ^2(5)}-\frac {80 x^3}{\log (5)}+2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.27 \begin {gather*} -20 x^4-\frac {80 x^2}{\log ^2(5)}-\frac {80 x^3}{\log (5)}+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-160*x^2 - 240*x^3*Log[5] + (2 - 80*x^4)*Log[5]^2)/(x*Log[5]^2),x]

[Out]

-20*x^4 - (80*x^2)/Log[5]^2 - (80*x^3)/Log[5] + 2*Log[x]

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fricas [A]  time = 0.57, size = 36, normalized size = 1.64 \begin {gather*} -\frac {2 \, {\left (10 \, x^{4} \log \relax (5)^{2} + 40 \, x^{3} \log \relax (5) - \log \relax (5)^{2} \log \relax (x) + 40 \, x^{2}\right )}}{\log \relax (5)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-80*x^4+2)*log(5)^2-240*x^3*log(5)-160*x^2)/x/log(5)^2,x, algorithm="fricas")

[Out]

-2*(10*x^4*log(5)^2 + 40*x^3*log(5) - log(5)^2*log(x) + 40*x^2)/log(5)^2

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giac [A]  time = 0.36, size = 37, normalized size = 1.68 \begin {gather*} -\frac {2 \, {\left (10 \, x^{4} \log \relax (5)^{2} + 40 \, x^{3} \log \relax (5) - \log \relax (5)^{2} \log \left ({\left | x \right |}\right ) + 40 \, x^{2}\right )}}{\log \relax (5)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-80*x^4+2)*log(5)^2-240*x^3*log(5)-160*x^2)/x/log(5)^2,x, algorithm="giac")

[Out]

-2*(10*x^4*log(5)^2 + 40*x^3*log(5) - log(5)^2*log(abs(x)) + 40*x^2)/log(5)^2

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maple [A]  time = 0.05, size = 29, normalized size = 1.32

method	result	size
risch	\(-20 x^{4}-\frac {80 x^{3}}{\ln \relax (5)}-\frac {80 x^{2}}{\ln \relax (5)^{2}}+2 \ln \relax (x )\)	\(29\)
norman	\(\frac {-80 x^{3}-\frac {80 x^{2}}{\ln \relax (5)}-20 x^{4} \ln \relax (5)}{\ln \relax (5)}+2 \ln \relax (x )\)	\(33\)
default	\(\frac {-20 x^{4} \ln \relax (5)^{2}-80 x^{3} \ln \relax (5)-80 x^{2}+2 \ln \relax (x ) \ln \relax (5)^{2}}{\ln \relax (5)^{2}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-80*x^4+2)*ln(5)^2-240*x^3*ln(5)-160*x^2)/x/ln(5)^2,x,method=_RETURNVERBOSE)

[Out]

-20*x^4-80/ln(5)*x^3-80*x^2/ln(5)^2+2*ln(x)

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maxima [A]  time = 0.66, size = 36, normalized size = 1.64 \begin {gather*} -\frac {2 \, {\left (10 \, x^{4} \log \relax (5)^{2} + 40 \, x^{3} \log \relax (5) - \log \relax (5)^{2} \log \relax (x) + 40 \, x^{2}\right )}}{\log \relax (5)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-80*x^4+2)*log(5)^2-240*x^3*log(5)-160*x^2)/x/log(5)^2,x, algorithm="maxima")

[Out]

-2*(10*x^4*log(5)^2 + 40*x^3*log(5) - log(5)^2*log(x) + 40*x^2)/log(5)^2

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mupad [B]  time = 0.06, size = 28, normalized size = 1.27 \begin {gather*} 2\,\ln \relax (x)-\frac {80\,x^2}{{\ln \relax (5)}^2}-\frac {80\,x^3}{\ln \relax (5)}-20\,x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(240*x^3*log(5) + log(5)^2*(80*x^4 - 2) + 160*x^2)/(x*log(5)^2),x)

[Out]

2*log(x) - (80*x^2)/log(5)^2 - (80*x^3)/log(5) - 20*x^4

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sympy [A]  time = 0.10, size = 37, normalized size = 1.68 \begin {gather*} \frac {- 20 x^{4} \log {\relax (5 )}^{2} - 80 x^{3} \log {\relax (5 )} - 80 x^{2} + 2 \log {\relax (5 )}^{2} \log {\relax (x )}}{\log {\relax (5 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-80*x**4+2)*ln(5)**2-240*x**3*ln(5)-160*x**2)/x/ln(5)**2,x)

[Out]

(-20*x**4*log(5)**2 - 80*x**3*log(5) - 80*x**2 + 2*log(5)**2*log(x))/log(5)**2

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