3.71.50 \(\int \frac {2 x+2 x^2+(2+2 x) \log (-\frac {x}{5 e^5})}{x} \, dx\)

Optimal. Leaf size=13 \[ \left (x+\log \left (-\frac {x}{5 e^5}\right )\right )^2 \]

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Rubi [B]  time = 0.04, antiderivative size = 32, normalized size of antiderivative = 2.46, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {14, 2346, 2301, 2295} \begin {gather*} (x+1)^2-2 x+\log ^2\left (-\frac {x}{5 e^5}\right )+2 x \log \left (-\frac {x}{5 e^5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x + 2*x^2 + (2 + 2*x)*Log[-1/5*x/E^5])/x,x]

[Out]

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

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]

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]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rubi steps

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

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Mathematica [B]  time = 0.00, size = 28, normalized size = 2.15 \begin {gather*} -10 x+x^2+2 x \log \left (-\frac {x}{5}\right )+\log ^2\left (-\frac {x}{5}\right )-10 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x + 2*x^2 + (2 + 2*x)*Log[-1/5*x/E^5])/x,x]

[Out]

-10*x + x^2 + 2*x*Log[-1/5*x] + Log[-1/5*x]^2 - 10*Log[x]

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fricas [B]  time = 0.54, size = 21, normalized size = 1.62 \begin {gather*} x^{2} + 2 \, x \log \left (-\frac {1}{5} \, x e^{\left (-5\right )}\right ) + \log \left (-\frac {1}{5} \, x e^{\left (-5\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.22, size = 24, normalized size = 1.85 \begin {gather*} x^{2} + 2 \, x \log \left (-\frac {1}{5} \, x\right ) + \log \left (-\frac {1}{5} \, x\right )^{2} - 10 \, x - 10 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 22, normalized size = 1.69

method	result	size
risch	\(x^{2}+\ln \left (-\frac {x \,{\mathrm e}^{-5}}{5}\right )^{2}+2 \ln \left (-\frac {x \,{\mathrm e}^{-5}}{5}\right ) x\)	\(22\)
norman	\(x^{2}+\ln \left (-\frac {x \,{\mathrm e}^{-5}}{5}\right )^{2}+2 \ln \left (-\frac {x \,{\mathrm e}^{-5}}{5}\right ) x\)	\(26\)
derivativedivides	\(x^{2}-10 \,{\mathrm e}^{5} \left (-\frac {x \,{\mathrm e}^{-5} \ln \left (-\frac {x \,{\mathrm e}^{-5}}{5}\right )}{5}+\frac {x \,{\mathrm e}^{-5}}{5}\right )+2 x +\ln \left (-\frac {x \,{\mathrm e}^{-5}}{5}\right )^{2}\)	\(45\)
default	\(x^{2}-10 \,{\mathrm e}^{5} \left (-\frac {x \,{\mathrm e}^{-5} \ln \left (-\frac {x \,{\mathrm e}^{-5}}{5}\right )}{5}+\frac {x \,{\mathrm e}^{-5}}{5}\right )+2 x +\ln \left (-\frac {x \,{\mathrm e}^{-5}}{5}\right )^{2}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.38, size = 35, normalized size = 2.69 \begin {gather*} x^{2} + 2 \, {\left (x e^{\left (-5\right )} \log \left (-\frac {1}{5} \, x e^{\left (-5\right )}\right ) - x e^{\left (-5\right )}\right )} e^{5} + \log \left (-\frac {1}{5} \, x e^{\left (-5\right )}\right )^{2} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.14, size = 14, normalized size = 1.08 \begin {gather*} \left (x+\ln \left (-\frac {x}{5}\right )\right )\,\left (x+\ln \left (-\frac {x}{5}\right )-10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + log(-x/5))*(x + log(-x/5) - 10)

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sympy [B]  time = 0.11, size = 27, normalized size = 2.08 \begin {gather*} x^{2} + 2 x \log {\left (- \frac {x}{5 e^{5}} \right )} + \log {\left (- \frac {x}{5 e^{5}} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+2)*ln(-1/5*x/exp(5))+2*x**2+2*x)/x,x)

[Out]

x**2 + 2*x*log(-x*exp(-5)/5) + log(-x*exp(-5)/5)**2

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