3.67.28 \(\int \frac {-98 x+58 x^2+(-14-29 x) \log (2)-14 x \log (\frac {1}{x})}{14 x} \, dx\)

Optimal. Leaf size=18 \[ (-x+\log (2)) \left (8-\frac {29 x}{14}+\log \left (\frac {1}{x}\right )\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 35, normalized size of antiderivative = 1.94, number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 14, 2295} \begin {gather*} \frac {29 x^2}{14}-x-x \log \left (\frac {1}{x}\right )-\frac {1}{14} x (98+29 \log (2))-\log (2) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-98*x + 58*x^2 + (-14 - 29*x)*Log[2] - 14*x*Log[x^(-1)])/(14*x),x]

[Out]

-x + (29*x^2)/14 - (x*(98 + 29*Log[2]))/14 - x*Log[x^(-1)] - Log[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]]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{14} \int \frac {-98 x+58 x^2+(-14-29 x) \log (2)-14 x \log \left (\frac {1}{x}\right )}{x} \, dx\\ &=\frac {1}{14} \int \left (\frac {58 x^2-14 \log (2)-x (98+29 \log (2))}{x}-14 \log \left (\frac {1}{x}\right )\right ) \, dx\\ &=\frac {1}{14} \int \frac {58 x^2-14 \log (2)-x (98+29 \log (2))}{x} \, dx-\int \log \left (\frac {1}{x}\right ) \, dx\\ &=-x-x \log \left (\frac {1}{x}\right )+\frac {1}{14} \int \left (-98+58 x-29 \log (2)-\frac {14 \log (2)}{x}\right ) \, dx\\ &=-x+\frac {29 x^2}{14}-\frac {1}{14} x (98+29 \log (2))-x \log \left (\frac {1}{x}\right )-\log (2) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 1.72 \begin {gather*} -8 x+\frac {29 x^2}{14}-\frac {29}{14} x \log (2)-x \log \left (\frac {1}{x}\right )-\log (2) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-98*x + 58*x^2 + (-14 - 29*x)*Log[2] - 14*x*Log[x^(-1)])/(14*x),x]

[Out]

-8*x + (29*x^2)/14 - (29*x*Log[2])/14 - x*Log[x^(-1)] - Log[2]*Log[x]

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fricas [A]  time = 0.52, size = 26, normalized size = 1.44 \begin {gather*} \frac {29}{14} \, x^{2} - \frac {29}{14} \, x \log \relax (2) - {\left (x - \log \relax (2)\right )} \log \left (\frac {1}{x}\right ) - 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/14*(-14*x*log(1/x)+(-29*x-14)*log(2)+58*x^2-98*x)/x,x, algorithm="fricas")

[Out]

29/14*x^2 - 29/14*x*log(2) - (x - log(2))*log(1/x) - 8*x

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giac [A]  time = 0.27, size = 25, normalized size = 1.39 \begin {gather*} \frac {29}{14} \, x^{2} - \frac {1}{14} \, x {\left (29 \, \log \relax (2) + 112\right )} + x \log \relax (x) - \log \relax (2) \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/14*(-14*x*log(1/x)+(-29*x-14)*log(2)+58*x^2-98*x)/x,x, algorithm="giac")

[Out]

29/14*x^2 - 1/14*x*(29*log(2) + 112) + x*log(x) - log(2)*log(x)

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maple [A]  time = 0.05, size = 28, normalized size = 1.56

method	result	size
risch	\(-x \ln \left (\frac {1}{x}\right )-\frac {29 x \ln \relax (2)}{14}+\frac {29 x^{2}}{14}-8 x -\ln \relax (2) \ln \relax (x )\)	\(28\)
derivativedivides	\(\ln \relax (2) \ln \left (\frac {1}{x}\right )-\frac {29 x \ln \relax (2)}{14}-x \ln \left (\frac {1}{x}\right )-8 x +\frac {29 x^{2}}{14}\)	\(29\)
default	\(\ln \relax (2) \ln \left (\frac {1}{x}\right )-\frac {29 x \ln \relax (2)}{14}-x \ln \left (\frac {1}{x}\right )-8 x +\frac {29 x^{2}}{14}\)	\(29\)
norman	\(\left (-8-\frac {29 \ln \relax (2)}{14}\right ) x +\ln \relax (2) \ln \left (\frac {1}{x}\right )+\frac {29 x^{2}}{14}-x \ln \left (\frac {1}{x}\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/14*(-14*x*ln(1/x)+(-29*x-14)*ln(2)+58*x^2-98*x)/x,x,method=_RETURNVERBOSE)

[Out]

-x*ln(1/x)-29/14*x*ln(2)+29/14*x^2-8*x-ln(2)*ln(x)

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maxima [A]  time = 0.37, size = 24, normalized size = 1.33 \begin {gather*} \frac {29}{14} \, x^{2} - \frac {29}{14} \, x \log \relax (2) + x \log \relax (x) - \log \relax (2) \log \relax (x) - 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/14*(-14*x*log(1/x)+(-29*x-14)*log(2)+58*x^2-98*x)/x,x, algorithm="maxima")

[Out]

29/14*x^2 - 29/14*x*log(2) + x*log(x) - log(2)*log(x) - 8*x

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mupad [B]  time = 4.40, size = 28, normalized size = 1.56 \begin {gather*} \ln \left (\frac {1}{x}\right )\,\ln \relax (2)-8\,x-x\,\ln \left (\frac {1}{x}\right )-\frac {29\,x\,\ln \relax (2)}{14}+\frac {29\,x^2}{14} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(7*x + (log(2)*(29*x + 14))/14 + x*log(1/x) - (29*x^2)/7)/x,x)

[Out]

log(1/x)*log(2) - 8*x - x*log(1/x) - (29*x*log(2))/14 + (29*x^2)/14

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sympy [B]  time = 0.14, size = 31, normalized size = 1.72 \begin {gather*} \frac {29 x^{2}}{14} - x \log {\left (\frac {1}{x} \right )} + \frac {x \left (-112 - 29 \log {\relax (2 )}\right )}{14} - \log {\relax (2 )} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/14*(-14*x*ln(1/x)+(-29*x-14)*ln(2)+58*x**2-98*x)/x,x)

[Out]

29*x**2/14 - x*log(1/x) + x*(-112 - 29*log(2))/14 - log(2)*log(x)

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