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

Optimal. Leaf size=33 \[ 3-x-\left (5-x-2 x^2\right ) \log \left (x+5 \left (-x+\frac {x \log (3)}{e}\right )\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 29, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14, 2313} \begin {gather*} \left (2 x^2+x\right ) \log \left (-x \left (4-\frac {5 \log (3)}{e}\right )\right )-x-5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 39, normalized size = 1.18 \begin {gather*} -x-5 \log (x)+x \log \left (x \left (-4+\frac {5 \log (3)}{e}\right )\right )+2 x^2 \log \left (x \left (-4+\frac {5 \log (3)}{e}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.75, size = 29, normalized size = 0.88 \begin {gather*} {\left (2 \, x^{2} + x - 5\right )} \log \left (-{\left (4 \, x e - 5 \, x \log \relax (3)\right )} e^{\left (-1\right )}\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 44, normalized size = 1.33 \begin {gather*} 2 \, x^{2} \log \left (-4 \, x e + 5 \, x \log \relax (3)\right ) - 2 \, x^{2} + x \log \left (-4 \, x e + 5 \, x \log \relax (3)\right ) - 2 \, x - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 32, normalized size = 0.97

method	result	size
risch	\(\left (2 x^{2}+x \right ) \ln \left (\left (5 x \ln \relax (3)-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right )-x -5 \ln \relax (x )\)	\(32\)
norman	\(x \ln \left (\left (5 x \ln \relax (3)-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right )-5 \ln \left (\left (5 x \ln \relax (3)-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right )-x +2 x^{2} \ln \left (\left (5 x \ln \relax (3)-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right )\)	\(65\)
derivativedivides	\(\frac {4 \,{\mathrm e}^{2} \left (\frac {x^{2} \left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2} {\mathrm e}^{-2} \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{2}-\frac {x^{2} \left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2} {\mathrm e}^{-2}}{4}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}-\frac {4 \,{\mathrm e}^{2} \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1} \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )+x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}+x^{2}+\frac {5 \,{\mathrm e} \ln \relax (3) \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1} \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )+x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}-\frac {80 \,{\mathrm e}^{2} \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}+\frac {200 \,{\mathrm e} \ln \relax (3) \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}-\frac {125 \ln \relax (3)^{2} \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}\)	\(314\)
default	\(\frac {4 \,{\mathrm e}^{2} \left (\frac {x^{2} \left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2} {\mathrm e}^{-2} \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{2}-\frac {x^{2} \left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2} {\mathrm e}^{-2}}{4}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}-\frac {4 \,{\mathrm e}^{2} \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1} \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )+x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}+x^{2}+\frac {5 \,{\mathrm e} \ln \relax (3) \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1} \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )+x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}-\frac {80 \,{\mathrm e}^{2} \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}+\frac {200 \,{\mathrm e} \ln \relax (3) \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}-\frac {125 \ln \relax (3)^{2} \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \relax (3)\right ) {\mathrm e}^{-1}\right )}{\left (4 \,{\mathrm e}-5 \ln \relax (3)\right )^{2}}\)	\(314\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x^2+x)*ln((5*x*ln(3)-4*x*exp(1))*exp(-1))-x-5*ln(x)

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maxima [B]  time = 0.37, size = 98, normalized size = 2.97 \begin {gather*} \frac {{\left (5 \, e^{\left (-1\right )} \log \relax (3) - 4\right )} x^{2} e}{4 \, e - 5 \, \log \relax (3)} + 2 \, x^{2} \log \left (5 \, x e^{\left (-1\right )} \log \relax (3) - 4 \, x\right ) + x^{2} - \frac {5 \, x e^{\left (-1\right )} \log \relax (3) - {\left (5 \, x e^{\left (-1\right )} \log \relax (3) - 4 \, x\right )} \log \left (5 \, x e^{\left (-1\right )} \log \relax (3) - 4 \, x\right ) - 4 \, x}{5 \, e^{\left (-1\right )} \log \relax (3) - 4} - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.18, size = 39, normalized size = 1.18 \begin {gather*} x\,\ln \left (5\,x\,{\mathrm {e}}^{-1}\,\ln \relax (3)-4\,x\right )-5\,\ln \relax (x)-x+2\,x^2\,\ln \left (5\,x\,{\mathrm {e}}^{-1}\,\ln \relax (3)-4\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-exp(-1)*(4*x*exp(1) - 5*x*log(3)))*(x + 4*x^2) + 2*x^2 - 5)/x,x)

[Out]

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

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sympy [A]  time = 0.13, size = 31, normalized size = 0.94 \begin {gather*} - x + \left (2 x^{2} + x\right ) \log {\left (\frac {- 4 e x + 5 x \log {\relax (3 )}}{e} \right )} - 5 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + (2*x**2 + x)*log((-4*E*x + 5*x*log(3))*exp(-1)) - 5*log(x)

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