3.99.95 \(\int \frac {2 x+e^5 (-1-12 x-3 x^2)+e^5 \log (\frac {2-e}{x})}{e^5} \, dx\)

Optimal. Leaf size=24 \[ x \left (-x \left (6-\frac {1}{e^5}+x\right )+\log \left (\frac {2-e}{x}\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 2295} \begin {gather*} -x^3+\frac {x^2}{e^5}-6 x^2+x \log \left (\frac {2-e}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x + E^5*(-1 - 12*x - 3*x^2) + E^5*Log[(2 - E)/x])/E^5,x]

[Out]

-6*x^2 + x^2/E^5 - x^3 + x*Log[(2 - E)/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 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 {\int \left (2 x+e^5 \left (-1-12 x-3 x^2\right )+e^5 \log \left (\frac {2-e}{x}\right )\right ) \, dx}{e^5}\\ &=\frac {x^2}{e^5}+\int \left (-1-12 x-3 x^2\right ) \, dx+\int \log \left (\frac {2-e}{x}\right ) \, dx\\ &=-6 x^2+\frac {x^2}{e^5}-x^3+x \log \left (\frac {2-e}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 1.38 \begin {gather*} -\frac {\left (-1+6 e^5\right ) x^2}{e^5}-x^3+x \log \left (\frac {2-e}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x + E^5*(-1 - 12*x - 3*x^2) + E^5*Log[(2 - E)/x])/E^5,x]

[Out]

-(((-1 + 6*E^5)*x^2)/E^5) - x^3 + x*Log[(2 - E)/x]

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fricas [A]  time = 1.60, size = 34, normalized size = 1.42 \begin {gather*} {\left (x e^{5} \log \left (-\frac {e - 2}{x}\right ) + x^{2} - {\left (x^{3} + 6 \, x^{2}\right )} e^{5}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*log((2-exp(1))/x)+(-3*x^2-12*x-1)*exp(5)+2*x)/exp(5),x, algorithm="fricas")

[Out]

(x*e^5*log(-(e - 2)/x) + x^2 - (x^3 + 6*x^2)*e^5)*e^(-5)

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giac [B]  time = 0.43, size = 73, normalized size = 3.04 \begin {gather*} {\left (x^{2} - {\left (x^{3} + 6 \, x^{2} + x\right )} e^{5} + \frac {{\left (\frac {x {\left (e^{2} - 4 \, e + 4\right )} \log \left (-\frac {e - 2}{x}\right )}{e - 2} + \frac {x {\left (e^{2} - 4 \, e + 4\right )}}{e - 2}\right )} e^{5}}{e - 2}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*log((2-exp(1))/x)+(-3*x^2-12*x-1)*exp(5)+2*x)/exp(5),x, algorithm="giac")

[Out]

(x^2 - (x^3 + 6*x^2 + x)*e^5 + (x*(e^2 - 4*e + 4)*log(-(e - 2)/x)/(e - 2) + x*(e^2 - 4*e + 4)/(e - 2))*e^5/(e
- 2))*e^(-5)

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maple [A]  time = 0.07, size = 31, normalized size = 1.29

method	result	size
risch	\(-x^{3}-6 x^{2}+x \ln \left (\frac {2-{\mathrm e}}{x}\right )+{\mathrm e}^{-5} x^{2}\)	\(31\)
norman	\(x \ln \left (\frac {2-{\mathrm e}}{x}\right )-x^{3}-\left (6 \,{\mathrm e}^{5}-1\right ) {\mathrm e}^{-5} x^{2}\)	\(35\)
default	\({\mathrm e}^{-5} \left ({\mathrm e}^{5} \left (-x^{3}-6 x^{2}-x \right )+\frac {2 \,{\mathrm e}^{5} x \ln \left (\frac {2-{\mathrm e}}{x}\right )}{2-{\mathrm e}}+\frac {2 \,{\mathrm e}^{5} x}{2-{\mathrm e}}-\frac {{\mathrm e}^{5} {\mathrm e} x \ln \left (\frac {2-{\mathrm e}}{x}\right )}{2-{\mathrm e}}-\frac {{\mathrm e}^{5} {\mathrm e} x}{2-{\mathrm e}}+x^{2}\right )\)	\(105\)
derivativedivides	\(-{\mathrm e}^{-5} \left (2-{\mathrm e}\right ) \left ({\mathrm e}^{5} \left (-\frac {x \ln \left (\frac {2-{\mathrm e}}{x}\right )}{2-{\mathrm e}}-\frac {x}{2-{\mathrm e}}\right )+\frac {{\mathrm e}^{2} {\mathrm e}^{5} x^{3}}{\left (2-{\mathrm e}\right )^{3}}-\frac {6 \,{\mathrm e} \,{\mathrm e}^{5} x^{2}}{\left (2-{\mathrm e}\right )^{2}}+\frac {{\mathrm e}^{5} x}{2-{\mathrm e}}-\frac {4 x^{3} {\mathrm e}^{5} {\mathrm e}}{\left (2-{\mathrm e}\right )^{3}}+\frac {12 x^{2} {\mathrm e}^{5}}{\left (2-{\mathrm e}\right )^{2}}+\frac {{\mathrm e} x^{2}}{\left (2-{\mathrm e}\right )^{2}}+\frac {4 x^{3} {\mathrm e}^{5}}{\left (2-{\mathrm e}\right )^{3}}-\frac {2 x^{2}}{\left (2-{\mathrm e}\right )^{2}}\right )\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5)*ln((2-exp(1))/x)+(-3*x^2-12*x-1)*exp(5)+2*x)/exp(5),x,method=_RETURNVERBOSE)

[Out]

-x^3-6*x^2+x*ln((2-exp(1))/x)+exp(-5)*x^2

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maxima [A]  time = 0.36, size = 38, normalized size = 1.58 \begin {gather*} {\left (x^{2} - {\left (x^{3} + 6 \, x^{2} + x\right )} e^{5} + {\left (x \log \left (-\frac {e - 2}{x}\right ) + x\right )} e^{5}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*log((2-exp(1))/x)+(-3*x^2-12*x-1)*exp(5)+2*x)/exp(5),x, algorithm="maxima")

[Out]

(x^2 - (x^3 + 6*x^2 + x)*e^5 + (x*log(-(e - 2)/x) + x)*e^5)*e^(-5)

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mupad [B]  time = 5.54, size = 26, normalized size = 1.08 \begin {gather*} x\,\left (\ln \left (\mathrm {e}-2\right )-6\,x+\ln \left (-\frac {1}{x}\right )+x\,{\mathrm {e}}^{-5}-x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-5)*(2*x + exp(5)*log(-(exp(1) - 2)/x) - exp(5)*(12*x + 3*x^2 + 1)),x)

[Out]

x*(log(exp(1) - 2) - 6*x + log(-1/x) + x*exp(-5) - x^2)

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sympy [A]  time = 0.13, size = 26, normalized size = 1.08 \begin {gather*} - x^{3} + \frac {x^{2} \left (1 - 6 e^{5}\right )}{e^{5}} + x \log {\left (\frac {2 - e}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*ln((2-exp(1))/x)+(-3*x**2-12*x-1)*exp(5)+2*x)/exp(5),x)

[Out]

-x**3 + x**2*(1 - 6*exp(5))*exp(-5) + x*log((2 - E)/x)

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