3.46.62 \(\int \frac {-10 x^2+2 x^3-9 x^5+4 x^6-40 x^8+10 x^9+8 x^{12}+e^{5/2} (2 x^2+2 x^5+8 x^8)+(2+20 x^2-4 e^{5/2} x^2-2 x^3+4 x^6) \log (x)-4 \log ^2(x)}{x^5} \, dx\)

Optimal. Leaf size=22 \[ -5+x+\left (-5+e^{5/2}+x+x^4+\frac {\log (x)}{x^2}\right )^2 \]

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Rubi [B]  time = 0.20, antiderivative size = 76, normalized size of antiderivative = 3.45, number of steps used = 12, number of rules used = 4, integrand size = 98, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {14, 2357, 2304, 2305} \begin {gather*} x^8+2 x^5-2 \left (5-e^{5/2}\right ) x^4+\frac {\log ^2(x)}{x^4}+x^2+2 x^2 \log (x)-\frac {2 \left (5-e^{5/2}\right ) \log (x)}{x^2}-\left (9-2 e^{5/2}\right ) x+\frac {2 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*x^2 + 2*x^3 - 9*x^5 + 4*x^6 - 40*x^8 + 10*x^9 + 8*x^12 + E^(5/2)*(2*x^2 + 2*x^5 + 8*x^8) + (2 + 20*x^
2 - 4*E^(5/2)*x^2 - 2*x^3 + 4*x^6)*Log[x] - 4*Log[x]^2)/x^5,x]

[Out]

-((9 - 2*E^(5/2))*x) + x^2 - 2*(5 - E^(5/2))*x^4 + 2*x^5 + x^8 - (2*(5 - E^(5/2))*Log[x])/x^2 + (2*Log[x])/x +
2*x^2*Log[x] + Log[x]^2/x^4

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 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 2305

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

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-10 \left (1-\frac {e^{5/2}}{5}\right )+2 x-9 \left (1-\frac {2 e^{5/2}}{9}\right ) x^3+4 x^4-40 \left (1-\frac {e^{5/2}}{5}\right ) x^6+10 x^7+8 x^{10}}{x^3}+\frac {2 \left (1+10 \left (1-\frac {e^{5/2}}{5}\right ) x^2-x^3+2 x^6\right ) \log (x)}{x^5}-\frac {4 \log ^2(x)}{x^5}\right ) \, dx\\ &=2 \int \frac {\left (1+10 \left (1-\frac {e^{5/2}}{5}\right ) x^2-x^3+2 x^6\right ) \log (x)}{x^5} \, dx-4 \int \frac {\log ^2(x)}{x^5} \, dx+\int \frac {-10 \left (1-\frac {e^{5/2}}{5}\right )+2 x-9 \left (1-\frac {2 e^{5/2}}{9}\right ) x^3+4 x^4-40 \left (1-\frac {e^{5/2}}{5}\right ) x^6+10 x^7+8 x^{10}}{x^3} \, dx\\ &=\frac {\log ^2(x)}{x^4}-2 \int \frac {\log (x)}{x^5} \, dx+2 \int \left (\frac {\log (x)}{x^5}-\frac {2 \left (-5+e^{5/2}\right ) \log (x)}{x^3}-\frac {\log (x)}{x^2}+2 x \log (x)\right ) \, dx+\int \left (-9 \left (1-\frac {2 e^{5/2}}{9}\right )+\frac {2 \left (-5+e^{5/2}\right )}{x^3}+\frac {2}{x^2}+4 x+8 \left (-5+e^{5/2}\right ) x^3+10 x^4+8 x^7\right ) \, dx\\ &=\frac {1}{8 x^4}+\frac {5-e^{5/2}}{x^2}-\frac {2}{x}-\left (9-2 e^{5/2}\right ) x+2 x^2-2 \left (5-e^{5/2}\right ) x^4+2 x^5+x^8+\frac {\log (x)}{2 x^4}+\frac {\log ^2(x)}{x^4}+2 \int \frac {\log (x)}{x^5} \, dx-2 \int \frac {\log (x)}{x^2} \, dx+4 \int x \log (x) \, dx+\left (4 \left (5-e^{5/2}\right )\right ) \int \frac {\log (x)}{x^3} \, dx\\ &=-\left (\left (9-2 e^{5/2}\right ) x\right )+x^2-2 \left (5-e^{5/2}\right ) x^4+2 x^5+x^8-\frac {2 \left (5-e^{5/2}\right ) \log (x)}{x^2}+\frac {2 \log (x)}{x}+2 x^2 \log (x)+\frac {\log ^2(x)}{x^4}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.07, size = 57, normalized size = 2.59 \begin {gather*} x \left (-9+x-10 x^3+2 x^4+x^7+2 e^{5/2} \left (1+x^3\right )\right )+\frac {2 \left (-5+e^{5/2}+x+x^4\right ) \log (x)}{x^2}+\frac {\log ^2(x)}{x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x^2 + 2*x^3 - 9*x^5 + 4*x^6 - 40*x^8 + 10*x^9 + 8*x^12 + E^(5/2)*(2*x^2 + 2*x^5 + 8*x^8) + (2 +
20*x^2 - 4*E^(5/2)*x^2 - 2*x^3 + 4*x^6)*Log[x] - 4*Log[x]^2)/x^5,x]

[Out]

x*(-9 + x - 10*x^3 + 2*x^4 + x^7 + 2*E^(5/2)*(1 + x^3)) + (2*(-5 + E^(5/2) + x + x^4)*Log[x])/x^2 + Log[x]^2/x
^4

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fricas [B]  time = 0.60, size = 63, normalized size = 2.86 \begin {gather*} \frac {x^{12} + 2 \, x^{9} - 10 \, x^{8} + x^{6} - 9 \, x^{5} + 2 \, {\left (x^{8} + x^{5}\right )} e^{\frac {5}{2}} + 2 \, {\left (x^{6} + x^{3} + x^{2} e^{\frac {5}{2}} - 5 \, x^{2}\right )} \log \relax (x) + \log \relax (x)^{2}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x)^2+(-4*x^2*exp(5/2)+4*x^6-2*x^3+20*x^2+2)*log(x)+(8*x^8+2*x^5+2*x^2)*exp(5/2)+8*x^12+10*x^
9-40*x^8+4*x^6-9*x^5+2*x^3-10*x^2)/x^5,x, algorithm="fricas")

[Out]

(x^12 + 2*x^9 - 10*x^8 + x^6 - 9*x^5 + 2*(x^8 + x^5)*e^(5/2) + 2*(x^6 + x^3 + x^2*e^(5/2) - 5*x^2)*log(x) + lo
g(x)^2)/x^4

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giac [B]  time = 0.17, size = 74, normalized size = 3.36 \begin {gather*} \frac {x^{12} + 2 \, x^{9} + 2 \, x^{8} e^{\frac {5}{2}} - 10 \, x^{8} + 2 \, x^{6} \log \relax (x) + x^{6} + 2 \, x^{5} e^{\frac {5}{2}} - 9 \, x^{5} + 2 \, x^{3} \log \relax (x) + 2 \, x^{2} e^{\frac {5}{2}} \log \relax (x) - 10 \, x^{2} \log \relax (x) + \log \relax (x)^{2}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x)^2+(-4*x^2*exp(5/2)+4*x^6-2*x^3+20*x^2+2)*log(x)+(8*x^8+2*x^5+2*x^2)*exp(5/2)+8*x^12+10*x^
9-40*x^8+4*x^6-9*x^5+2*x^3-10*x^2)/x^5,x, algorithm="giac")

[Out]

(x^12 + 2*x^9 + 2*x^8*e^(5/2) - 10*x^8 + 2*x^6*log(x) + x^6 + 2*x^5*e^(5/2) - 9*x^5 + 2*x^3*log(x) + 2*x^2*e^(
5/2)*log(x) - 10*x^2*log(x) + log(x)^2)/x^4

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maple [B]  time = 0.02, size = 56, normalized size = 2.55

method	result	size
risch	\(\frac {\ln \relax (x )^{2}}{x^{4}}+\frac {2 \left (x^{4}+{\mathrm e}^{\frac {5}{2}}+x -5\right ) \ln \relax (x )}{x^{2}}+x^{8}+2 \,{\mathrm e}^{\frac {5}{2}} x^{4}+2 x^{5}-10 x^{4}+2 x \,{\mathrm e}^{\frac {5}{2}}+x^{2}-9 x\)	\(56\)
default	\(x^{8}+2 \,{\mathrm e}^{\frac {5}{2}} x^{4}+2 x^{5}-10 x^{4}+2 x^{2} \ln \relax (x )+x^{2}+2 x \,{\mathrm e}^{\frac {5}{2}}-9 x -4 \,{\mathrm e}^{\frac {5}{2}} \left (-\frac {\ln \relax (x )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )+\frac {2 \ln \relax (x )}{x}-\frac {10 \ln \relax (x )}{x^{2}}-\frac {{\mathrm e}^{\frac {5}{2}}}{x^{2}}+\frac {\ln \relax (x )^{2}}{x^{4}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*ln(x)^2+(-4*x^2*exp(5/2)+4*x^6-2*x^3+20*x^2+2)*ln(x)+(8*x^8+2*x^5+2*x^2)*exp(5/2)+8*x^12+10*x^9-40*x^8
+4*x^6-9*x^5+2*x^3-10*x^2)/x^5,x,method=_RETURNVERBOSE)

[Out]

1/x^4*ln(x)^2+2*(x^4+exp(5/2)+x-5)/x^2*ln(x)+x^8+2*exp(5/2)*x^4+2*x^5-10*x^4+2*x*exp(5/2)+x^2-9*x

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maxima [B]  time = 0.37, size = 103, normalized size = 4.68 \begin {gather*} x^{8} + 2 \, x^{5} + 2 \, x^{4} e^{\frac {5}{2}} - 10 \, x^{4} + 2 \, x^{2} \log \relax (x) + x^{2} + 2 \, x e^{\frac {5}{2}} + {\left (\frac {2 \, \log \relax (x)}{x^{2}} + \frac {1}{x^{2}}\right )} e^{\frac {5}{2}} - 9 \, x + \frac {2 \, \log \relax (x)}{x} - \frac {e^{\frac {5}{2}}}{x^{2}} - \frac {10 \, \log \relax (x)}{x^{2}} + \frac {8 \, \log \relax (x)^{2} + 4 \, \log \relax (x) + 1}{8 \, x^{4}} - \frac {\log \relax (x)}{2 \, x^{4}} - \frac {1}{8 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x)^2+(-4*x^2*exp(5/2)+4*x^6-2*x^3+20*x^2+2)*log(x)+(8*x^8+2*x^5+2*x^2)*exp(5/2)+8*x^12+10*x^
9-40*x^8+4*x^6-9*x^5+2*x^3-10*x^2)/x^5,x, algorithm="maxima")

[Out]

x^8 + 2*x^5 + 2*x^4*e^(5/2) - 10*x^4 + 2*x^2*log(x) + x^2 + 2*x*e^(5/2) + (2*log(x)/x^2 + 1/x^2)*e^(5/2) - 9*x
+ 2*log(x)/x - e^(5/2)/x^2 - 10*log(x)/x^2 + 1/8*(8*log(x)^2 + 4*log(x) + 1)/x^4 - 1/2*log(x)/x^4 - 1/8/x^4

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mupad [B]  time = 3.38, size = 68, normalized size = 3.09 \begin {gather*} \frac {x^{13}+2\,x^{10}+\left (2\,{\mathrm {e}}^{5/2}-10\right )\,x^9+2\,x^7\,\ln \relax (x)+x^7+\left (2\,{\mathrm {e}}^{5/2}-9\right )\,x^6+2\,x^4\,\ln \relax (x)+\left (2\,{\mathrm {e}}^{5/2}-10\right )\,x^3\,\ln \relax (x)+x\,{\ln \relax (x)}^2}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5/2)*(2*x^2 + 2*x^5 + 8*x^8) - 4*log(x)^2 - 10*x^2 + 2*x^3 - 9*x^5 + 4*x^6 - 40*x^8 + 10*x^9 + 8*x^12
+ log(x)*(20*x^2 - 4*x^2*exp(5/2) - 2*x^3 + 4*x^6 + 2))/x^5,x)

[Out]

(x*log(x)^2 + 2*x^4*log(x) + 2*x^7*log(x) + x^6*(2*exp(5/2) - 9) + x^9*(2*exp(5/2) - 10) + x^7 + 2*x^10 + x^13
+ x^3*log(x)*(2*exp(5/2) - 10))/x^5

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sympy [B]  time = 0.22, size = 65, normalized size = 2.95 \begin {gather*} x^{8} + 2 x^{5} + x^{4} \left (-10 + 2 e^{\frac {5}{2}}\right ) + x^{2} + x \left (-9 + 2 e^{\frac {5}{2}}\right ) + \frac {\left (2 x^{4} + 2 x - 10 + 2 e^{\frac {5}{2}}\right ) \log {\relax (x )}}{x^{2}} + \frac {\log {\relax (x )}^{2}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*ln(x)**2+(-4*x**2*exp(5/2)+4*x**6-2*x**3+20*x**2+2)*ln(x)+(8*x**8+2*x**5+2*x**2)*exp(5/2)+8*x**1
2+10*x**9-40*x**8+4*x**6-9*x**5+2*x**3-10*x**2)/x**5,x)

[Out]

x**8 + 2*x**5 + x**4*(-10 + 2*exp(5/2)) + x**2 + x*(-9 + 2*exp(5/2)) + (2*x**4 + 2*x - 10 + 2*exp(5/2))*log(x)
/x**2 + log(x)**2/x**4

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