3.50.75 \(\int \frac {-4+e^x (-200 x+50 x^2)}{-4 x+x^2} \, dx\)

Optimal. Leaf size=18 \[ \log \left (\frac {e^{50 e^x} x}{6 (-4+x)}\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1593, 6688, 2194, 36, 31, 29} \begin {gather*} 50 e^x-\log (4-x)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + E^x*(-200*x + 50*x^2))/(-4*x + x^2),x]

[Out]

50*E^x - Log[4 - x] + Log[x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4+e^x \left (-200 x+50 x^2\right )}{(-4+x) x} \, dx\\ &=\int \left (50 e^x-\frac {4}{(-4+x) x}\right ) \, dx\\ &=-\left (4 \int \frac {1}{(-4+x) x} \, dx\right )+50 \int e^x \, dx\\ &=50 e^x-\int \frac {1}{-4+x} \, dx+\int \frac {1}{x} \, dx\\ &=50 e^x-\log (4-x)+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.89 \begin {gather*} 50 e^x-\log (4-x)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + E^x*(-200*x + 50*x^2))/(-4*x + x^2),x]

[Out]

50*E^x - Log[4 - x] + Log[x]

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fricas [A]  time = 0.83, size = 13, normalized size = 0.72 \begin {gather*} 50 \, e^{x} - \log \left (x - 4\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^2-200*x)*exp(x)-4)/(x^2-4*x),x, algorithm="fricas")

[Out]

50*e^x - log(x - 4) + log(x)

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giac [A]  time = 0.21, size = 13, normalized size = 0.72 \begin {gather*} 50 \, e^{x} - \log \left (x - 4\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^2-200*x)*exp(x)-4)/(x^2-4*x),x, algorithm="giac")

[Out]

50*e^x - log(x - 4) + log(x)

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maple [A]  time = 0.13, size = 14, normalized size = 0.78

method	result	size
default	\(-\ln \left (x -4\right )+\ln \relax (x )+50 \,{\mathrm e}^{x}\)	\(14\)
norman	\(-\ln \left (x -4\right )+\ln \relax (x )+50 \,{\mathrm e}^{x}\)	\(14\)
risch	\(-\ln \left (x -4\right )+\ln \relax (x )+50 \,{\mathrm e}^{x}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*x^2-200*x)*exp(x)-4)/(x^2-4*x),x,method=_RETURNVERBOSE)

[Out]

-ln(x-4)+ln(x)+50*exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 200 \, e^{4} E_{1}\left (-x + 4\right ) + \frac {50 \, x e^{x}}{x - 4} + 200 \, \int \frac {e^{x}}{x^{2} - 8 \, x + 16}\,{d x} - \log \left (x - 4\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^2-200*x)*exp(x)-4)/(x^2-4*x),x, algorithm="maxima")

[Out]

200*e^4*exp_integral_e(1, -x + 4) + 50*x*e^x/(x - 4) + 200*integrate(e^x/(x^2 - 8*x + 16), x) - log(x - 4) + l
og(x)

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mupad [B]  time = 0.08, size = 13, normalized size = 0.72 \begin {gather*} 50\,{\mathrm {e}}^x-\ln \left (x-4\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(200*x - 50*x^2) + 4)/(4*x - x^2),x)

[Out]

50*exp(x) - log(x - 4) + log(x)

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sympy [A]  time = 0.12, size = 12, normalized size = 0.67 \begin {gather*} 50 e^{x} + \log {\relax (x )} - \log {\left (x - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x**2-200*x)*exp(x)-4)/(x**2-4*x),x)

[Out]

50*exp(x) + log(x) - log(x - 4)

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