3.36.15 \(\int \frac {-4+32 x+8 e^3 x-8 x \log (5)-8 x \log (x)}{x^2} \, dx\)

Optimal. Leaf size=24 \[ 5+\frac {4}{x}-4 \left (4+e^3-\log (5)-\log (x)\right )^2 \]

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6, 14, 43, 2301} \begin {gather*} \frac {4}{x}-4 \log ^2(x)+8 \left (4+e^3-\log (5)\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + 32*x + 8*E^3*x - 8*x*Log[5] - 8*x*Log[x])/x^2,x]

[Out]

4/x + 8*(4 + E^3 - Log[5])*Log[x] - 4*Log[x]^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4+\left (32+8 e^3\right ) x-8 x \log (5)-8 x \log (x)}{x^2} \, dx\\ &=\int \frac {-4+x \left (32+8 e^3-8 \log (5)\right )-8 x \log (x)}{x^2} \, dx\\ &=\int \left (\frac {4 \left (-1+2 x \left (4+e^3-\log (5)\right )\right )}{x^2}-\frac {8 \log (x)}{x}\right ) \, dx\\ &=4 \int \frac {-1+2 x \left (4+e^3-\log (5)\right )}{x^2} \, dx-8 \int \frac {\log (x)}{x} \, dx\\ &=-4 \log ^2(x)+4 \int \left (-\frac {1}{x^2}+\frac {2 \left (4+e^3-\log (5)\right )}{x}\right ) \, dx\\ &=\frac {4}{x}+8 \left (4+e^3-\log (5)\right ) \log (x)-4 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.21 \begin {gather*} \frac {4}{x}+32 \log (x)+8 e^3 \log (x)-8 \log (5) \log (x)-4 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 32*x + 8*E^3*x - 8*x*Log[5] - 8*x*Log[x])/x^2,x]

[Out]

4/x + 32*Log[x] + 8*E^3*Log[x] - 8*Log[5]*Log[x] - 4*Log[x]^2

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fricas [A]  time = 0.88, size = 30, normalized size = 1.25 \begin {gather*} -\frac {4 \, {\left (x \log \relax (x)^{2} - 2 \, {\left (x e^{3} - x \log \relax (5) + 4 \, x\right )} \log \relax (x) - 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*log(x)-8*x*log(5)+8*x*exp(3)+32*x-4)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 33, normalized size = 1.38 \begin {gather*} \frac {4 \, {\left (2 \, x e^{3} \log \relax (x) - 2 \, x \log \relax (5) \log \relax (x) - x \log \relax (x)^{2} + 8 \, x \log \relax (x) + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*log(x)-8*x*log(5)+8*x*exp(3)+32*x-4)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 28, normalized size = 1.17

method	result	size
norman	\(\frac {4+\left (-8 \ln \relax (5)+8 \,{\mathrm e}^{3}+32\right ) x \ln \relax (x )-4 x \ln \relax (x )^{2}}{x}\)	\(28\)
default	\(-4 \ln \relax (x )^{2}+8 \ln \relax (x ) {\mathrm e}^{3}-8 \ln \relax (5) \ln \relax (x )+32 \ln \relax (x )+\frac {4}{x}\)	\(29\)
risch	\(-4 \ln \relax (x )^{2}+\frac {8 x \,{\mathrm e}^{3} \ln \relax (x )-8 x \ln \relax (5) \ln \relax (x )+32 x \ln \relax (x )+4}{x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*x*ln(x)-8*x*ln(5)+8*x*exp(3)+32*x-4)/x^2,x,method=_RETURNVERBOSE)

[Out]

(4+(-8*ln(5)+8*exp(3)+32)*x*ln(x)-4*x*ln(x)^2)/x

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maxima [A]  time = 0.44, size = 28, normalized size = 1.17 \begin {gather*} 8 \, e^{3} \log \relax (x) - 8 \, \log \relax (5) \log \relax (x) - 4 \, \log \relax (x)^{2} + \frac {4}{x} + 32 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*log(x)-8*x*log(5)+8*x*exp(3)+32*x-4)/x^2,x, algorithm="maxima")

[Out]

8*e^3*log(x) - 8*log(5)*log(x) - 4*log(x)^2 + 4/x + 32*log(x)

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mupad [B]  time = 2.03, size = 25, normalized size = 1.04 \begin {gather*} \ln \relax (x)\,\left (8\,{\mathrm {e}}^3-8\,\ln \relax (5)+32\right )-4\,{\ln \relax (x)}^2+\frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x*log(5) - 8*x*exp(3) - 32*x + 8*x*log(x) + 4)/x^2,x)

[Out]

log(x)*(8*exp(3) - 8*log(5) + 32) - 4*log(x)^2 + 4/x

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sympy [A]  time = 0.18, size = 22, normalized size = 0.92 \begin {gather*} - 4 \log {\relax (x )}^{2} + 8 \left (- \log {\relax (5 )} + 4 + e^{3}\right ) \log {\relax (x )} + \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*ln(x)-8*x*ln(5)+8*x*exp(3)+32*x-4)/x**2,x)

[Out]

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

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