3.102.85 \(\int \frac {e^x (1-x)+e^2 (-20 x+11 x^2)+4 e^2 x^2 \log (x)}{e^2 x^2} \, dx\)

Optimal. Leaf size=26 \[ 3 (-3+x)-\frac {e^{-2+x}}{x}+(-5+x) (4+4 \log (x)) \]

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Rubi [A]  time = 0.06, antiderivative size = 23, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14, 2197, 43, 2295} \begin {gather*} 7 x-\frac {e^{x-2}}{x}+4 x \log (x)-20 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(1 - x) + E^2*(-20*x + 11*x^2) + 4*E^2*x^2*Log[x])/(E^2*x^2),x]

[Out]

-(E^(-2 + x)/x) + 7*x - 20*Log[x] + 4*x*Log[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 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, 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 \frac {e^x (1-x)+e^2 \left (-20 x+11 x^2\right )+4 e^2 x^2 \log (x)}{x^2} \, dx}{e^2}\\ &=\frac {\int \left (-\frac {e^x (-1+x)}{x^2}+\frac {e^2 (-20+11 x+4 x \log (x))}{x}\right ) \, dx}{e^2}\\ &=-\frac {\int \frac {e^x (-1+x)}{x^2} \, dx}{e^2}+\int \frac {-20+11 x+4 x \log (x)}{x} \, dx\\ &=-\frac {e^{-2+x}}{x}+\int \left (\frac {-20+11 x}{x}+4 \log (x)\right ) \, dx\\ &=-\frac {e^{-2+x}}{x}+4 \int \log (x) \, dx+\int \frac {-20+11 x}{x} \, dx\\ &=-\frac {e^{-2+x}}{x}-4 x+4 x \log (x)+\int \left (11-\frac {20}{x}\right ) \, dx\\ &=-\frac {e^{-2+x}}{x}+7 x-20 \log (x)+4 x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 23, normalized size = 0.88 \begin {gather*} -\frac {e^{-2+x}}{x}+7 x-20 \log (x)+4 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(1 - x) + E^2*(-20*x + 11*x^2) + 4*E^2*x^2*Log[x])/(E^2*x^2),x]

[Out]

-(E^(-2 + x)/x) + 7*x - 20*Log[x] + 4*x*Log[x]

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fricas [A]  time = 0.57, size = 31, normalized size = 1.19 \begin {gather*} \frac {{\left (7 \, x^{2} e^{2} + 4 \, {\left (x^{2} - 5 \, x\right )} e^{2} \log \relax (x) - e^{x}\right )} e^{\left (-2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*exp(2)*log(x)+(-x+1)*exp(x)+(11*x^2-20*x)*exp(2))/x^2/exp(2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.13, size = 34, normalized size = 1.31 \begin {gather*} \frac {{\left (4 \, x^{2} e^{2} \log \relax (x) + 7 \, x^{2} e^{2} - 20 \, x e^{2} \log \relax (x) - e^{x}\right )} e^{\left (-2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*exp(2)*log(x)+(-x+1)*exp(x)+(11*x^2-20*x)*exp(2))/x^2/exp(2),x, algorithm="giac")

[Out]

(4*x^2*e^2*log(x) + 7*x^2*e^2 - 20*x*e^2*log(x) - e^x)*e^(-2)/x

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maple [A]  time = 0.07, size = 31, normalized size = 1.19

method	result	size
norman	\(\frac {-20 x \ln \relax (x )+7 x^{2}+4 x^{2} \ln \relax (x )-{\mathrm e}^{x} {\mathrm e}^{-2}}{x}\)	\(31\)
risch	\(4 x \ln \relax (x )-\frac {{\mathrm e}^{-2} \left (20 x \,{\mathrm e}^{2} \ln \relax (x )-7 x^{2} {\mathrm e}^{2}+{\mathrm e}^{x}\right )}{x}\)	\(31\)
default	\({\mathrm e}^{-2} \left (-20 \,{\mathrm e}^{2} \ln \relax (x )+7 \,{\mathrm e}^{2} x +4 x \,{\mathrm e}^{2} \ln \relax (x )-\frac {{\mathrm e}^{x}}{x}\right )\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2*exp(2)*ln(x)+(1-x)*exp(x)+(11*x^2-20*x)*exp(2))/x^2/exp(2),x,method=_RETURNVERBOSE)

[Out]

(-20*x*ln(x)+7*x^2+4*x^2*ln(x)-exp(x)/exp(2))/x

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maxima [C]  time = 0.38, size = 36, normalized size = 1.38 \begin {gather*} {\left (4 \, {\left (x \log \relax (x) - x\right )} e^{2} + 11 \, x e^{2} - 20 \, e^{2} \log \relax (x) - {\rm Ei}\relax (x) + \Gamma \left (-1, -x\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*exp(2)*log(x)+(-x+1)*exp(x)+(11*x^2-20*x)*exp(2))/x^2/exp(2),x, algorithm="maxima")

[Out]

(4*(x*log(x) - x)*e^2 + 11*x*e^2 - 20*e^2*log(x) - Ei(x) + gamma(-1, -x))*e^(-2)

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mupad [B]  time = 6.98, size = 22, normalized size = 0.85 \begin {gather*} x\,\left (4\,\ln \relax (x)+7\right )-\frac {{\mathrm {e}}^{x-2}}{x}-20\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2)*(exp(x)*(x - 1) + exp(2)*(20*x - 11*x^2) - 4*x^2*exp(2)*log(x)))/x^2,x)

[Out]

x*(4*log(x) + 7) - exp(x - 2)/x - 20*log(x)

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sympy [A]  time = 0.30, size = 22, normalized size = 0.85 \begin {gather*} 4 x \log {\relax (x )} + 7 x - 20 \log {\relax (x )} - \frac {e^{x}}{x e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2*exp(2)*ln(x)+(-x+1)*exp(x)+(11*x**2-20*x)*exp(2))/x**2/exp(2),x)

[Out]

4*x*log(x) + 7*x - 20*log(x) - exp(-2)*exp(x)/x

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