3.55.69 \(\int \frac {-10 x^2-10 \log (e^{-2-x} (1+x) \log (4))}{x^2} \, dx\)

Optimal. Leaf size=22 \[ \frac {10 (1+x) \log \left (e^{-2-x} (1+x) \log (4)\right )}{x} \]

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Rubi [A]  time = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.32, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {14, 2551, 31} \begin {gather*} -10 x+10 \log (x+1)+\frac {10 \log \left (e^{-x-2} (x+1) \log (4)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*x^2 - 10*Log[E^(-2 - x)*(1 + x)*Log[4]])/x^2,x]

[Out]

-10*x + 10*Log[1 + x] + (10*Log[E^(-2 - x)*(1 + x)*Log[4]])/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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-10-\frac {10 \log \left (e^{-2-x} (1+x) \log (4)\right )}{x^2}\right ) \, dx\\ &=-10 x-10 \int \frac {\log \left (e^{-2-x} (1+x) \log (4)\right )}{x^2} \, dx\\ &=-10 x+\frac {10 \log \left (e^{-2-x} (1+x) \log (4)\right )}{x}-10 \int \frac {1}{-1-x} \, dx\\ &=-10 x+10 \log (1+x)+\frac {10 \log \left (e^{-2-x} (1+x) \log (4)\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.32 \begin {gather*} -10 x+10 \log (1+x)+\frac {10 \log \left (e^{-2-x} (1+x) \log (4)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x^2 - 10*Log[E^(-2 - x)*(1 + x)*Log[4]])/x^2,x]

[Out]

-10*x + 10*Log[1 + x] + (10*Log[E^(-2 - x)*(1 + x)*Log[4]])/x

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fricas [A]  time = 1.57, size = 22, normalized size = 1.00 \begin {gather*} \frac {10 \, {\left (x + 1\right )} \log \left (2 \, {\left (x + 1\right )} e^{\left (-x - 2\right )} \log \relax (2)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 31, normalized size = 1.41 \begin {gather*} -10 \, x + \frac {10 \, {\left (\log \relax (2) + \log \left (\log \relax (2)\right ) - 2\right )}}{x} + \frac {10 \, \log \left (x + 1\right )}{x} + 10 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*x + 10*(log(2) + log(log(2)) - 2)/x + 10*log(x + 1)/x + 10*log(x + 1)

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maple [A]  time = 0.20, size = 39, normalized size = 1.77

method	result	size
norman	\(\frac {10 \ln \left (2 \left (x +1\right ) \ln \relax (2) {\mathrm e}^{-x -2}\right ) x +10 \ln \left (2 \left (x +1\right ) \ln \relax (2) {\mathrm e}^{-x -2}\right )}{x}\)	\(39\)
default	\(-10 x +\frac {10 \ln \relax (2)}{x}+\frac {10 \ln \left (\ln \relax (2)\right )}{x}+\frac {10 \ln \left (\left (x +1\right ) {\mathrm e}^{-x -2}\right )}{x}+10 \ln \left (-x -1\right )\)	\(44\)
risch	\(-\frac {10 \ln \left ({\mathrm e}^{2+x}\right )}{x}+\frac {-5 i \pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x -2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (x +1\right )\right )+5 i \pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (x +1\right )\right )^{2}+5 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x -2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (x +1\right )\right )^{2}-5 i \pi \mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (x +1\right )\right )^{3}+10 \ln \left (x +1\right ) x -10 x^{2}+10 \ln \relax (2)+10 \ln \left (\ln \relax (2)\right )+10 \ln \left (x +1\right )}{x}\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(10*ln(2*(x+1)*ln(2)/exp(2+x))*x+10*ln(2*(x+1)*ln(2)/exp(2+x)))/x

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maxima [A]  time = 0.46, size = 38, normalized size = 1.73 \begin {gather*} -10 \, x + \frac {10 \, \log \left (2 \, x e^{\left (-x - 2\right )} \log \relax (2) + 2 \, e^{\left (-x - 2\right )} \log \relax (2)\right )}{x} + 10 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.19, size = 22, normalized size = 1.00 \begin {gather*} \frac {10\,\left (x+1\right )\,\left (\ln \left (x+1\right )-x+\ln \relax (2)+\ln \left (\ln \relax (2)\right )-2\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*log(2*exp(- x - 2)*log(2)*(x + 1)) + 10*x^2)/x^2,x)

[Out]

(10*(x + 1)*(log(x + 1) - x + log(2) + log(log(2)) - 2))/x

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sympy [A]  time = 0.16, size = 29, normalized size = 1.32 \begin {gather*} - 10 x + 10 \log {\left (x + 1 \right )} + \frac {10 \log {\left (\left (2 x + 2\right ) e^{- x - 2} \log {\relax (2 )} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*ln(2*(x+1)*ln(2)/exp(2+x))-10*x**2)/x**2,x)

[Out]

-10*x + 10*log(x + 1) + 10*log((2*x + 2)*exp(-x - 2)*log(2))/x

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