3.73.48 \(\int \frac {e^{10}-2 e^5 x+x^2+(4 e^5 x-4 x^2) \log ^3(x)+e^5 x \log ^4(x)}{e^{10} x-2 e^5 x^2+x^3} \, dx\)

Optimal. Leaf size=23 \[ 5+e^{20}+\log (x)-\frac {x \log ^4(x)}{-e^5+x} \]

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Rubi [A]  time = 0.39, antiderivative size = 18, normalized size of antiderivative = 0.78, number of steps used = 13, number of rules used = 8, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1594, 27, 6688, 2317, 2374, 2383, 6589, 2318} \begin {gather*} \frac {x \log ^4(x)}{e^5-x}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] + (x*Log[x]^4)/(E^5 - 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 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2318

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_))^2, x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])
^p)/(d*(d + e*x)), x] - Dist[(b*n*p)/d, Int[(a + b*Log[c*x^n])^(p - 1)/(d + e*x), x], x] /; FreeQ[{a, b, c, d,
e, n, p}, x] && GtQ[p, 0]

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2383

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[(PolyL
og[k + 1, e*x^q]*(a + b*Log[c*x^n])^p)/q, x] - Dist[(b*n*p)/q, Int[(PolyLog[k + 1, e*x^q]*(a + b*Log[c*x^n])^(
p - 1))/x, x], x] /; FreeQ[{a, b, c, e, k, n, q}, x] && GtQ[p, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

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 {e^{10}-2 e^5 x+x^2+\left (4 e^5 x-4 x^2\right ) \log ^3(x)+e^5 x \log ^4(x)}{x \left (e^{10}-2 e^5 x+x^2\right )} \, dx\\ &=\int \frac {e^{10}-2 e^5 x+x^2+\left (4 e^5 x-4 x^2\right ) \log ^3(x)+e^5 x \log ^4(x)}{x \left (-e^5+x\right )^2} \, dx\\ &=\int \left (\frac {1}{x}+\frac {4 \log ^3(x)}{e^5-x}+\frac {e^5 \log ^4(x)}{\left (e^5-x\right )^2}\right ) \, dx\\ &=\log (x)+4 \int \frac {\log ^3(x)}{e^5-x} \, dx+e^5 \int \frac {\log ^4(x)}{\left (e^5-x\right )^2} \, dx\\ &=\log (x)+\frac {x \log ^4(x)}{e^5-x}-4 \log ^3(x) \log \left (1-\frac {x}{e^5}\right )-4 \int \frac {\log ^3(x)}{e^5-x} \, dx+12 \int \frac {\log ^2(x) \log \left (1-\frac {x}{e^5}\right )}{x} \, dx\\ &=\log (x)+\frac {x \log ^4(x)}{e^5-x}-12 \log ^2(x) \text {Li}_2\left (\frac {x}{e^5}\right )-12 \int \frac {\log ^2(x) \log \left (1-\frac {x}{e^5}\right )}{x} \, dx+24 \int \frac {\log (x) \text {Li}_2\left (\frac {x}{e^5}\right )}{x} \, dx\\ &=\log (x)+\frac {x \log ^4(x)}{e^5-x}+24 \log (x) \text {Li}_3\left (\frac {x}{e^5}\right )-24 \int \frac {\log (x) \text {Li}_2\left (\frac {x}{e^5}\right )}{x} \, dx-24 \int \frac {\text {Li}_3\left (\frac {x}{e^5}\right )}{x} \, dx\\ &=\log (x)+\frac {x \log ^4(x)}{e^5-x}-24 \text {Li}_4\left (\frac {x}{e^5}\right )+24 \int \frac {\text {Li}_3\left (\frac {x}{e^5}\right )}{x} \, dx\\ &=\log (x)+\frac {x \log ^4(x)}{e^5-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 25, normalized size = 1.09 \begin {gather*} \frac {\log (x) \left (e^5-x+x \log ^3(x)\right )}{e^5-x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 27, normalized size = 1.17 \begin {gather*} -\frac {x \log \relax (x)^{4} - {\left (x - e^{5}\right )} \log \relax (x)}{x - e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*log(x)^4 - (x - e^5)*log(x))/(x - e^5)

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giac [A]  time = 0.14, size = 27, normalized size = 1.17 \begin {gather*} -\frac {x \log \relax (x)^{4} - x \log \relax (x) + e^{5} \log \relax (x)}{x - e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 18, normalized size = 0.78

method	result	size
risch	\(\frac {x \ln \relax (x )^{4}}{{\mathrm e}^{5}-x}+\ln \relax (x )\)	\(18\)
norman	\(\frac {x \ln \relax (x )^{4}+{\mathrm e}^{5} \ln \relax (x )-x \ln \relax (x )}{{\mathrm e}^{5}-x}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(5)*ln(x)^4+(4*x*exp(5)-4*x^2)*ln(x)^3+exp(5)^2-2*x*exp(5)+x^2)/(x*exp(5)^2-2*x^2*exp(5)+x^3),x,meth
od=_RETURNVERBOSE)

[Out]

x/(exp(5)-x)*ln(x)^4+ln(x)

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maxima [B]  time = 0.43, size = 66, normalized size = 2.87 \begin {gather*} -\frac {x \log \relax (x)^{4}}{x - e^{5}} - {\left (e^{\left (-10\right )} \log \left (x - e^{5}\right ) - e^{\left (-10\right )} \log \relax (x) + \frac {1}{x e^{5} - e^{10}}\right )} e^{10} + \frac {e^{5}}{x - e^{5}} + \log \left (x - e^{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*log(x)^4/(x - e^5) - (e^(-10)*log(x - e^5) - e^(-10)*log(x) + 1/(x*e^5 - e^10))*e^10 + e^5/(x - e^5) + log(
x - e^5)

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mupad [B]  time = 4.56, size = 22, normalized size = 0.96 \begin {gather*} \ln \relax (x)-{\ln \relax (x)}^4\,\left (\frac {{\mathrm {e}}^5}{x-{\mathrm {e}}^5}+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(10) - 2*x*exp(5) + log(x)^3*(4*x*exp(5) - 4*x^2) + x^2 + x*exp(5)*log(x)^4)/(x*exp(10) - 2*x^2*exp(5)
+ x^3),x)

[Out]

log(x) - log(x)^4*(exp(5)/(x - exp(5)) + 1)

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sympy [A]  time = 0.18, size = 14, normalized size = 0.61 \begin {gather*} - \frac {x \log {\relax (x )}^{4}}{x - e^{5}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(5)*ln(x)**4+(4*x*exp(5)-4*x**2)*ln(x)**3+exp(5)**2-2*x*exp(5)+x**2)/(x*exp(5)**2-2*x**2*exp(5
)+x**3),x)

[Out]

-x*log(x)**4/(x - exp(5)) + log(x)

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