3.34.10 \(\int \frac {e^{98} (-3-x)+4 x+x^2+(-e^{98}+x) \log (\frac {e^{-x} x}{2 e^{98}-2 x})}{e^{98} x^2-x^3} \, dx\)

Optimal. Leaf size=27 \[ \frac {4+x+\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 4, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {1593, 6742, 893, 2551} \begin {gather*} \frac {4}{x}+\frac {\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^98*(-3 - x) + 4*x + x^2 + (-E^98 + x)*Log[x/(E^x*(2*E^98 - 2*x))])/(E^98*x^2 - x^3),x]

[Out]

4/x + Log[x/(2*E^x*(E^98 - x))]/x

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{98} (-3-x)+4 x+x^2+\left (-e^{98}+x\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{\left (e^{98}-x\right ) x^2} \, dx\\ &=\int \left (\frac {-3 e^{98}+\left (4-e^{98}\right ) x+x^2}{\left (e^{98}-x\right ) x^2}-\frac {\log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{x^2}\right ) \, dx\\ &=\int \frac {-3 e^{98}+\left (4-e^{98}\right ) x+x^2}{\left (e^{98}-x\right ) x^2} \, dx-\int \frac {\log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{x^2} \, dx\\ &=\frac {\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x}+\int \left (\frac {1}{e^{98} \left (e^{98}-x\right )}-\frac {3}{x^2}+\frac {1-e^{98}}{e^{98} x}\right ) \, dx-\int \frac {e^{98}-e^{98} x+x^2}{\left (e^{98}-x\right ) x^2} \, dx\\ &=\frac {3}{x}-\frac {\log \left (e^{98}-x\right )}{e^{98}}-\left (1-\frac {1}{e^{98}}\right ) \log (x)+\frac {\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x}-\int \left (\frac {1}{e^{98} \left (e^{98}-x\right )}+\frac {1}{x^2}+\frac {1-e^{98}}{e^{98} x}\right ) \, dx\\ &=\frac {4}{x}+\frac {\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 29, normalized size = 1.07 \begin {gather*} \frac {4}{x}+\frac {\log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^98*(-3 - x) + 4*x + x^2 + (-E^98 + x)*Log[x/(E^x*(2*E^98 - 2*x))])/(E^98*x^2 - x^3),x]

[Out]

4/x + Log[x/(E^x*(2*E^98 - 2*x))]/x

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 22, normalized size = 0.81 \begin {gather*} \frac {\log \left (-\frac {x e^{\left (-x\right )}}{2 \, {\left (x - e^{98}\right )}}\right ) + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(49)^2+x)*log(x/(2*exp(49)^2-2*x)/exp(x))+(-3-x)*exp(49)^2+x^2+4*x)/(x^2*exp(49)^2-x^3),x, alg
orithm="fricas")

[Out]

(log(-1/2*x*e^(-x)/(x - e^98)) + 4)/x

________________________________________________________________________________________

giac [A]  time = 0.28, size = 18, normalized size = 0.67 \begin {gather*} \frac {\log \left (-\frac {x}{2 \, {\left (x - e^{98}\right )}}\right ) + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(49)^2+x)*log(x/(2*exp(49)^2-2*x)/exp(x))+(-3-x)*exp(49)^2+x^2+4*x)/(x^2*exp(49)^2-x^3),x, alg
orithm="giac")

[Out]

(log(-1/2*x/(x - e^98)) + 4)/x

________________________________________________________________________________________

maple [A]  time = 0.66, size = 26, normalized size = 0.96

method	result	size
norman	\(\frac {4+\ln \left (\frac {x \,{\mathrm e}^{-x}}{2 \,{\mathrm e}^{98}-2 x}\right )}{x}\)	\(26\)
default	\(\frac {\ln \left (\frac {x \,{\mathrm e}^{-x}}{2 \,{\mathrm e}^{98}-2 x}\right )}{x}+\ln \left (2 \,{\mathrm e}^{98}-2 x \right )-2 \,{\mathrm e}^{98} {\mathrm e}^{-196} \ln \relax (x )+\frac {\left ({\mathrm e}^{98}\right )^{2} {\mathrm e}^{-196}}{x}+2 \,{\mathrm e}^{98} {\mathrm e}^{-196} \ln \left (-{\mathrm e}^{98}+x \right )+\frac {1}{{\mathrm e}^{98}-x}+2 \,{\mathrm e}^{-98} \ln \relax (x )-2 \,{\mathrm e}^{-98} \ln \left (-{\mathrm e}^{98}+x \right )+\frac {\left ({\mathrm e}^{98}\right )^{2} {\mathrm e}^{-196}}{-{\mathrm e}^{98}+x}-{\mathrm e}^{98} {\mathrm e}^{-98} \ln \left (-{\mathrm e}^{98}+x \right )+\frac {3 \,{\mathrm e}^{-98} {\mathrm e}^{98}}{x}\)	\(197\)
risch	\(-\frac {\ln \left ({\mathrm e}^{x}\right )}{x}-\frac {i \pi \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{3}-i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right ) \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{98}-x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2}-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right ) \mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )+i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{3}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{98}-x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )-8+2 \ln \relax (2)-2 \ln \relax (x )+2 \ln \left ({\mathrm e}^{98}-x \right )}{2 x}\)	\(292\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(49)^2+x)*ln(x/(2*exp(49)^2-2*x)/exp(x))+(-3-x)*exp(49)^2+x^2+4*x)/(x^2*exp(49)^2-x^3),x,method=_RET
URNVERBOSE)

[Out]

(4+ln(x/(2*exp(49)^2-2*x)/exp(x)))/x

________________________________________________________________________________________

maxima [B]  time = 0.51, size = 264, normalized size = 9.78 \begin {gather*} {\left (e^{\left (-196\right )} \log \left (x - e^{98}\right ) - e^{\left (-196\right )} \log \relax (x) + \frac {e^{\left (-98\right )}}{x}\right )} e^{98} \log \left (-\frac {x}{2 \, {\left (x e^{x} - e^{\left (x + 98\right )}\right )}}\right ) + {\left (e^{\left (-98\right )} \log \left (x - e^{98}\right ) - e^{\left (-98\right )} \log \relax (x)\right )} e^{98} + 3 \, {\left (e^{\left (-196\right )} \log \left (x - e^{98}\right ) - e^{\left (-196\right )} \log \relax (x) + \frac {e^{\left (-98\right )}}{x}\right )} e^{98} - \frac {1}{2} \, {\left (2 \, {\left (x - e^{98} - \log \relax (x)\right )} \log \left (x - e^{98}\right ) + \log \left (x - e^{98}\right )^{2} - 2 \, x \log \relax (x) + \log \relax (x)^{2}\right )} e^{\left (-98\right )} - 4 \, e^{\left (-98\right )} \log \left (x - e^{98}\right ) + 4 \, e^{\left (-98\right )} \log \relax (x) - {\left (e^{\left (-98\right )} \log \left (x - e^{98}\right ) - e^{\left (-98\right )} \log \relax (x)\right )} \log \left (-\frac {x}{2 \, {\left (x e^{x} - e^{\left (x + 98\right )}\right )}}\right ) + \frac {{\left (x \log \left (x - e^{98}\right )^{2} + x \log \relax (x)^{2} + 2 \, {\left (x^{2} - x {\left (e^{98} - 1\right )} - x \log \relax (x)\right )} \log \left (x - e^{98}\right ) - 2 \, {\left (x^{2} - x {\left (e^{98} - 1\right )}\right )} \log \relax (x) + 2 \, e^{98}\right )} e^{\left (-98\right )}}{2 \, x} - \log \left (x - e^{98}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(49)^2+x)*log(x/(2*exp(49)^2-2*x)/exp(x))+(-3-x)*exp(49)^2+x^2+4*x)/(x^2*exp(49)^2-x^3),x, alg
orithm="maxima")

[Out]

(e^(-196)*log(x - e^98) - e^(-196)*log(x) + e^(-98)/x)*e^98*log(-1/2*x/(x*e^x - e^(x + 98))) + (e^(-98)*log(x
- e^98) - e^(-98)*log(x))*e^98 + 3*(e^(-196)*log(x - e^98) - e^(-196)*log(x) + e^(-98)/x)*e^98 - 1/2*(2*(x - e
^98 - log(x))*log(x - e^98) + log(x - e^98)^2 - 2*x*log(x) + log(x)^2)*e^(-98) - 4*e^(-98)*log(x - e^98) + 4*e
^(-98)*log(x) - (e^(-98)*log(x - e^98) - e^(-98)*log(x))*log(-1/2*x/(x*e^x - e^(x + 98))) + 1/2*(x*log(x - e^9
8)^2 + x*log(x)^2 + 2*(x^2 - x*(e^98 - 1) - x*log(x))*log(x - e^98) - 2*(x^2 - x*(e^98 - 1))*log(x) + 2*e^98)*
e^(-98)/x - log(x - e^98)

________________________________________________________________________________________

mupad [B]  time = 2.46, size = 24, normalized size = 0.89 \begin {gather*} \frac {\ln \left (-\frac {x\,{\mathrm {e}}^{-x}}{2\,\left (x-{\mathrm {e}}^{98}\right )}\right )+4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - exp(98)*(x + 3) + x^2 + log(-(x*exp(-x))/(2*x - 2*exp(98)))*(x - exp(98)))/(x^2*exp(98) - x^3),x)

[Out]

(log(-(x*exp(-x))/(2*(x - exp(98)))) + 4)/x

________________________________________________________________________________________

sympy [A]  time = 0.25, size = 19, normalized size = 0.70 \begin {gather*} \frac {\log {\left (\frac {x e^{- x}}{- 2 x + 2 e^{98}} \right )}}{x} + \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(49)**2+x)*ln(x/(2*exp(49)**2-2*x)/exp(x))+(-3-x)*exp(49)**2+x**2+4*x)/(x**2*exp(49)**2-x**3),
x)

[Out]

log(x*exp(-x)/(-2*x + 2*exp(98)))/x + 4/x

________________________________________________________________________________________