3.72.44 \(\int \frac {-10+2 x^3+(3 x^2-4 x^3) \log (2)-x^2 \log (5)}{2 x^2} \, dx\)

Optimal. Leaf size=29 \[ e^9+\frac {5}{x}-\frac {1}{2} x (-x-(3-2 x) \log (2)+\log (5)) \]

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Rubi [A]  time = 0.02, antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {12, 14} \begin {gather*} \frac {1}{2} x^2 (1-\log (4))+\frac {5}{x}+\frac {1}{2} x \log \left (\frac {8}{5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 + 2*x^3 + (3*x^2 - 4*x^3)*Log[2] - x^2*Log[5])/(2*x^2),x]

[Out]

5/x + (x*Log[8/5])/2 + (x^2*(1 - Log[4]))/2

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 {1}{2} \int \frac {-10+2 x^3+\left (3 x^2-4 x^3\right ) \log (2)-x^2 \log (5)}{x^2} \, dx\\ &=\frac {1}{2} \int \left (-\frac {10}{x^2}+2 x (1-\log (4))-\log (5) \left (1-\frac {\log (8)}{\log (5)}\right )\right ) \, dx\\ &=\frac {5}{x}+\frac {1}{2} x \log \left (\frac {8}{5}\right )+\frac {1}{2} x^2 (1-\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 1.07 \begin {gather*} \frac {5}{x}+\frac {x^2}{2}+\frac {1}{2} x \log \left (\frac {8}{5}\right )-\frac {1}{2} x^2 \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + 2*x^3 + (3*x^2 - 4*x^3)*Log[2] - x^2*Log[5])/(2*x^2),x]

[Out]

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

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fricas [A]  time = 0.81, size = 32, normalized size = 1.10 \begin {gather*} \frac {x^{3} - x^{2} \log \relax (5) - {\left (2 \, x^{3} - 3 \, x^{2}\right )} \log \relax (2) + 10}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-x^2*log(5)+(-4*x^3+3*x^2)*log(2)+2*x^3-10)/x^2,x, algorithm="fricas")

[Out]

1/2*(x^3 - x^2*log(5) - (2*x^3 - 3*x^2)*log(2) + 10)/x

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giac [A]  time = 2.89, size = 28, normalized size = 0.97 \begin {gather*} -x^{2} \log \relax (2) + \frac {1}{2} \, x^{2} - \frac {1}{2} \, x \log \relax (5) + \frac {3}{2} \, x \log \relax (2) + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-x^2*log(5)+(-4*x^3+3*x^2)*log(2)+2*x^3-10)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 29, normalized size = 1.00

method	result	size
default	\(-x^{2} \ln \relax (2)-\frac {x \ln \relax (5)}{2}+\frac {3 x \ln \relax (2)}{2}+\frac {x^{2}}{2}+\frac {5}{x}\)	\(29\)
risch	\(-x^{2} \ln \relax (2)-\frac {x \ln \relax (5)}{2}+\frac {3 x \ln \relax (2)}{2}+\frac {x^{2}}{2}+\frac {5}{x}\)	\(29\)
norman	\(\frac {5+\left (-\ln \relax (2)+\frac {1}{2}\right ) x^{3}+\left (-\frac {\ln \relax (5)}{2}+\frac {3 \ln \relax (2)}{2}\right ) x^{2}}{x}\)	\(30\)
gosper	\(-\frac {2 x^{3} \ln \relax (2)+x^{2} \ln \relax (5)-3 x^{2} \ln \relax (2)-x^{3}-10}{2 x}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-x^2*ln(5)+(-4*x^3+3*x^2)*ln(2)+2*x^3-10)/x^2,x,method=_RETURNVERBOSE)

[Out]

-x^2*ln(2)-1/2*x*ln(5)+3/2*x*ln(2)+1/2*x^2+5/x

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maxima [A]  time = 0.36, size = 27, normalized size = 0.93 \begin {gather*} -\frac {1}{2} \, x^{2} {\left (2 \, \log \relax (2) - 1\right )} - \frac {1}{2} \, x {\left (\log \relax (5) - 3 \, \log \relax (2)\right )} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-x^2*log(5)+(-4*x^3+3*x^2)*log(2)+2*x^3-10)/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.13, size = 22, normalized size = 0.76 \begin {gather*} \frac {5}{x}-x^2\,\left (\frac {\ln \left (16\right )}{4}-\frac {1}{2}\right )-\frac {x\,\ln \left (\frac {5}{8}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/x - x^2*(log(16)/4 - 1/2) - (x*log(5/8))/2

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sympy [A]  time = 0.12, size = 26, normalized size = 0.90 \begin {gather*} \frac {x^{2} \left (1 - 2 \log {\relax (2 )}\right )}{2} + \frac {x \left (- \log {\relax (5 )} + 3 \log {\relax (2 )}\right )}{2} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-x**2*ln(5)+(-4*x**3+3*x**2)*ln(2)+2*x**3-10)/x**2,x)

[Out]

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

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