3.98.73 \(\int \frac {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} (32 x^2-17 x^4+x^5)+(256-32 x+x^2) \log (x)}{256 x^2-32 x^3+x^4} \, dx\)

Optimal. Leaf size=25 \[ -2+x+\frac {e^{-10+x} (-2+x) x}{-16+x}-\frac {\log (x)}{x} \]

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Rubi [A]  time = 0.86, antiderivative size = 37, normalized size of antiderivative = 1.48, number of steps used = 13, number of rules used = 9, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1594, 27, 6688, 2199, 2194, 2177, 2178, 2176, 2304} \begin {gather*} e^{x-10} x+x+14 e^{x-10}-\frac {224 e^{x-10}}{16-x}-\frac {\log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-256 + 32*x + 255*x^2 - 32*x^3 + x^4 + E^(-10 + x)*(32*x^2 - 17*x^4 + x^5) + (256 - 32*x + x^2)*Log[x])/(
256*x^2 - 32*x^3 + x^4),x]

[Out]

14*E^(-10 + x) - (224*E^(-10 + x))/(16 - x) + x + E^(-10 + x)*x - Log[x]/x

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

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 {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} \left (32 x^2-17 x^4+x^5\right )+\left (256-32 x+x^2\right ) \log (x)}{x^2 \left (256-32 x+x^2\right )} \, dx\\ &=\int \frac {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} \left (32 x^2-17 x^4+x^5\right )+\left (256-32 x+x^2\right ) \log (x)}{(-16+x)^2 x^2} \, dx\\ &=\int \left (1-\frac {1}{x^2}+\frac {e^{-10+x} \left (32-17 x^2+x^3\right )}{(-16+x)^2}+\frac {\log (x)}{x^2}\right ) \, dx\\ &=\frac {1}{x}+x+\int \frac {e^{-10+x} \left (32-17 x^2+x^3\right )}{(-16+x)^2} \, dx+\int \frac {\log (x)}{x^2} \, dx\\ &=x-\frac {\log (x)}{x}+\int \left (15 e^{-10+x}-\frac {224 e^{-10+x}}{(-16+x)^2}+\frac {224 e^{-10+x}}{-16+x}+e^{-10+x} x\right ) \, dx\\ &=x-\frac {\log (x)}{x}+15 \int e^{-10+x} \, dx-224 \int \frac {e^{-10+x}}{(-16+x)^2} \, dx+224 \int \frac {e^{-10+x}}{-16+x} \, dx+\int e^{-10+x} x \, dx\\ &=15 e^{-10+x}-\frac {224 e^{-10+x}}{16-x}+x+e^{-10+x} x+224 e^6 \text {Ei}(-16+x)-\frac {\log (x)}{x}-224 \int \frac {e^{-10+x}}{-16+x} \, dx-\int e^{-10+x} \, dx\\ &=14 e^{-10+x}-\frac {224 e^{-10+x}}{16-x}+x+e^{-10+x} x-\frac {\log (x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 24, normalized size = 0.96 \begin {gather*} x+\frac {e^{-10+x} (-2+x) x}{-16+x}-\frac {\log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-256 + 32*x + 255*x^2 - 32*x^3 + x^4 + E^(-10 + x)*(32*x^2 - 17*x^4 + x^5) + (256 - 32*x + x^2)*Log
[x])/(256*x^2 - 32*x^3 + x^4),x]

[Out]

x + (E^(-10 + x)*(-2 + x)*x)/(-16 + x) - Log[x]/x

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fricas [A]  time = 0.79, size = 40, normalized size = 1.60 \begin {gather*} \frac {x^{3} - 16 \, x^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e^{\left (x - 10\right )} - {\left (x - 16\right )} \log \relax (x)}{x^{2} - 16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-32*x+256)*log(x)+(x^5-17*x^4+32*x^2)*exp(x-10)+x^4-32*x^3+255*x^2+32*x-256)/(x^4-32*x^3+256*x^
2),x, algorithm="fricas")

[Out]

(x^3 - 16*x^2 + (x^3 - 2*x^2)*e^(x - 10) - (x - 16)*log(x))/(x^2 - 16*x)

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giac [B]  time = 0.16, size = 55, normalized size = 2.20 \begin {gather*} \frac {x^{3} e^{10} + x^{3} e^{x} - 16 \, x^{2} e^{10} - 2 \, x^{2} e^{x} - x e^{10} \log \relax (x) + 16 \, e^{10} \log \relax (x)}{x^{2} e^{10} - 16 \, x e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-32*x+256)*log(x)+(x^5-17*x^4+32*x^2)*exp(x-10)+x^4-32*x^3+255*x^2+32*x-256)/(x^4-32*x^3+256*x^
2),x, algorithm="giac")

[Out]

(x^3*e^10 + x^3*e^x - 16*x^2*e^10 - 2*x^2*e^x - x*e^10*log(x) + 16*e^10*log(x))/(x^2*e^10 - 16*x*e^10)

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maple [A]  time = 0.04, size = 31, normalized size = 1.24

method	result	size
risch	\(-\frac {\ln \relax (x )}{x}+\frac {x \left (x \,{\mathrm e}^{x -10}+x -2 \,{\mathrm e}^{x -10}-16\right )}{x -16}\)	\(31\)
default	\(x -\frac {\ln \relax (x )}{x}+\left (x -10\right ) {\mathrm e}^{x -10}+24 \,{\mathrm e}^{x -10}+\frac {224 \,{\mathrm e}^{x -10}}{x -16}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2-32*x+256)*ln(x)+(x^5-17*x^4+32*x^2)*exp(x-10)+x^4-32*x^3+255*x^2+32*x-256)/(x^4-32*x^3+256*x^2),x,me
thod=_RETURNVERBOSE)

[Out]

-ln(x)/x+x*(x*exp(x-10)+x-2*exp(x-10)-16)/(x-16)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} x - \frac {32 \, e^{6} E_{2}\left (-x + 16\right )}{x - 16} + \frac {2 \, {\left (x - 8\right )}}{x^{2} - 16 \, x} - \frac {\log \relax (x) + 1}{x} - \frac {1}{x - 16} + \int \frac {{\left (x^{3} - 17 \, x^{2}\right )} e^{x}}{x^{2} e^{10} - 32 \, x e^{10} + 256 \, e^{10}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-32*x+256)*log(x)+(x^5-17*x^4+32*x^2)*exp(x-10)+x^4-32*x^3+255*x^2+32*x-256)/(x^4-32*x^3+256*x^
2),x, algorithm="maxima")

[Out]

x - 32*e^6*exp_integral_e(2, -x + 16)/(x - 16) + 2*(x - 8)/(x^2 - 16*x) - (log(x) + 1)/x - 1/(x - 16) + integr
ate((x^3 - 17*x^2)*e^x/(x^2*e^10 - 32*x*e^10 + 256*e^10), x)

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mupad [B]  time = 5.76, size = 29, normalized size = 1.16 \begin {gather*} x-\frac {\ln \relax (x)}{x}-\frac {{\mathrm {e}}^{x-10}\,\left (2\,x-x^2\right )}{x-16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x + exp(x - 10)*(32*x^2 - 17*x^4 + x^5) + log(x)*(x^2 - 32*x + 256) + 255*x^2 - 32*x^3 + x^4 - 256)/(2
56*x^2 - 32*x^3 + x^4),x)

[Out]

x - log(x)/x - (exp(x - 10)*(2*x - x^2))/(x - 16)

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sympy [A]  time = 0.36, size = 20, normalized size = 0.80 \begin {gather*} x + \frac {\left (x^{2} - 2 x\right ) e^{x - 10}}{x - 16} - \frac {\log {\relax (x )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2-32*x+256)*ln(x)+(x**5-17*x**4+32*x**2)*exp(x-10)+x**4-32*x**3+255*x**2+32*x-256)/(x**4-32*x**
3+256*x**2),x)

[Out]

x + (x**2 - 2*x)*exp(x - 10)/(x - 16) - log(x)/x

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