3.73.60 \(\int \frac {-e^{4+2 e^3} x-10 x^2-2 x^3-20 x^4-5 x^5+e^{2+e^3} (-10-2 x-20 x^2-6 x^3)+(-10-2 x-2 e^{2+e^3} x-20 x^2-6 x^3) \log (2 x)-x \log ^2(2 x)}{x} \, dx\)

Optimal. Leaf size=23 \[ (-5-x) \left (e^{2+e^3}+x^2+\log (2 x)\right )^2 \]

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Rubi [B]  time = 0.25, antiderivative size = 160, normalized size of antiderivative = 6.96, number of steps used = 12, number of rules used = 6, integrand size = 99, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {14, 2357, 2295, 2301, 2304, 2296} \begin {gather*} -x^5-5 x^4-\frac {2}{3} \left (1+3 e^{2+e^3}\right ) x^3+\frac {2 x^3}{3}-2 x^3 \log (2 x)-5 \left (1+2 e^{2+e^3}\right ) x^2+5 x^2-10 x^2 \log (2 x)-e^{2+e^3} \left (2+e^{2+e^3}\right ) x+2 \left (1+e^{2+e^3}\right ) x-2 x-x \log ^2(2 x)-5 \log ^2(2 x)-2 \left (1+e^{2+e^3}\right ) x \log (2 x)+2 x \log (2 x)-10 e^{2+e^3} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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 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 e^{2+e^3}-2 e^{2+e^3} \left (1+\frac {e^{2+e^3}}{2}\right ) x-10 \left (1+2 e^{2+e^3}\right ) x^2-2 \left (1+3 e^{2+e^3}\right ) x^3-20 x^4-5 x^5}{x}+\frac {2 \left (-5-\left (1+e^{2+e^3}\right ) x-10 x^2-3 x^3\right ) \log (2 x)}{x}-\log ^2(2 x)\right ) \, dx\\ &=2 \int \frac {\left (-5-\left (1+e^{2+e^3}\right ) x-10 x^2-3 x^3\right ) \log (2 x)}{x} \, dx+\int \frac {-10 e^{2+e^3}-2 e^{2+e^3} \left (1+\frac {e^{2+e^3}}{2}\right ) x-10 \left (1+2 e^{2+e^3}\right ) x^2-2 \left (1+3 e^{2+e^3}\right ) x^3-20 x^4-5 x^5}{x} \, dx-\int \log ^2(2 x) \, dx\\ &=-x \log ^2(2 x)+2 \int \log (2 x) \, dx+2 \int \left (\left (-1-e^{2+e^3}\right ) \log (2 x)-\frac {5 \log (2 x)}{x}-10 x \log (2 x)-3 x^2 \log (2 x)\right ) \, dx+\int \left (-e^{2+e^3} \left (2+e^{2+e^3}\right )-\frac {10 e^{2+e^3}}{x}-10 \left (1+2 e^{2+e^3}\right ) x-2 \left (1+3 e^{2+e^3}\right ) x^2-20 x^3-5 x^4\right ) \, dx\\ &=-2 x-e^{2+e^3} \left (2+e^{2+e^3}\right ) x-5 \left (1+2 e^{2+e^3}\right ) x^2-\frac {2}{3} \left (1+3 e^{2+e^3}\right ) x^3-5 x^4-x^5-10 e^{2+e^3} \log (x)+2 x \log (2 x)-x \log ^2(2 x)-6 \int x^2 \log (2 x) \, dx-10 \int \frac {\log (2 x)}{x} \, dx-20 \int x \log (2 x) \, dx-\left (2 \left (1+e^{2+e^3}\right )\right ) \int \log (2 x) \, dx\\ &=-2 x+2 \left (1+e^{2+e^3}\right ) x-e^{2+e^3} \left (2+e^{2+e^3}\right ) x+5 x^2-5 \left (1+2 e^{2+e^3}\right ) x^2+\frac {2 x^3}{3}-\frac {2}{3} \left (1+3 e^{2+e^3}\right ) x^3-5 x^4-x^5-10 e^{2+e^3} \log (x)+2 x \log (2 x)-2 \left (1+e^{2+e^3}\right ) x \log (2 x)-10 x^2 \log (2 x)-2 x^3 \log (2 x)-5 \log ^2(2 x)-x \log ^2(2 x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.13, size = 76, normalized size = 3.30 \begin {gather*} -x \left (e^{4+2 e^3}+2 e^{2+e^3} x (5+x)+x^3 (5+x)\right )-10 e^{2+e^3} \log (x)-2 x \left (e^{2+e^3}+x (5+x)\right ) \log (2 x)-(5+x) \log ^2(2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.81, size = 72, normalized size = 3.13 \begin {gather*} -x^{5} - 5 \, x^{4} - {\left (x + 5\right )} \log \left (2 \, x\right )^{2} - x e^{\left (2 \, e^{3} + 4\right )} - 2 \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (e^{3} + 2\right )} - 2 \, {\left (x^{3} + 5 \, x^{2} + {\left (x + 5\right )} e^{\left (e^{3} + 2\right )}\right )} \log \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^5 - 5*x^4 - (x + 5)*log(2*x)^2 - x*e^(2*e^3 + 4) - 2*(x^3 + 5*x^2)*e^(e^3 + 2) - 2*(x^3 + 5*x^2 + (x + 5)*e
^(e^3 + 2))*log(2*x)

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giac [B]  time = 0.27, size = 97, normalized size = 4.22 \begin {gather*} -x^{5} - 5 \, x^{4} - 2 \, x^{3} e^{\left (e^{3} + 2\right )} - 2 \, x^{3} \log \left (2 \, x\right ) - 10 \, x^{2} e^{\left (e^{3} + 2\right )} - 10 \, x^{2} \log \left (2 \, x\right ) - 2 \, x e^{\left (e^{3} + 2\right )} \log \left (2 \, x\right ) - x \log \left (2 \, x\right )^{2} - x e^{\left (2 \, e^{3} + 4\right )} - 5 \, \log \left (2 \, x\right )^{2} - 10 \, e^{\left (e^{3} + 2\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^5 - 5*x^4 - 2*x^3*e^(e^3 + 2) - 2*x^3*log(2*x) - 10*x^2*e^(e^3 + 2) - 10*x^2*log(2*x) - 2*x*e^(e^3 + 2)*log
(2*x) - x*log(2*x)^2 - x*e^(2*e^3 + 4) - 5*log(2*x)^2 - 10*e^(e^3 + 2)*log(x)

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maple [B]  time = 0.06, size = 87, normalized size = 3.78

method	result	size
risch	\(\left (-x -5\right ) \ln \left (2 x \right )^{2}+\left (-2 x \,{\mathrm e}^{2+{\mathrm e}^{3}}-2 x^{3}-10 x^{2}\right ) \ln \left (2 x \right )-x \,{\mathrm e}^{4+2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{2+{\mathrm e}^{3}} x^{3}-10 \,{\mathrm e}^{2+{\mathrm e}^{3}} x^{2}-x^{5}-5 x^{4}-10 \ln \relax (x ) {\mathrm e}^{2+{\mathrm e}^{3}}\)	\(87\)
norman	\(-10 \ln \left (2 x \right ) {\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}}-5 x^{4}-x^{5}-5 \ln \left (2 x \right )^{2}-x \ln \left (2 x \right )^{2}-10 x^{2} \ln \left (2 x \right )-2 x^{3} \ln \left (2 x \right )-x \,{\mathrm e}^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}}-10 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{2}-2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{3}-2 x \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} \ln \left (2 x \right )\)	\(102\)
derivativedivides	\(-x^{5}-2 x^{3} \ln \left (2 x \right )-2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{3}-5 x^{4}-x \ln \left (2 x \right )^{2}-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} \left (2 x \ln \left (2 x \right )-2 x \right )-10 x^{2} \ln \left (2 x \right )-x \,{\mathrm e}^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}}-10 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{2}-2 x \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}}-5 \ln \left (2 x \right )^{2}-10 \ln \left (2 x \right ) {\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}}\)	\(116\)
default	\(-x^{5}-2 x^{3} \ln \left (2 x \right )-2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{3}-5 x^{4}-x \ln \left (2 x \right )^{2}-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} \left (2 x \ln \left (2 x \right )-2 x \right )-10 x^{2} \ln \left (2 x \right )-x \,{\mathrm e}^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}}-10 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{2}-2 x \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}}-5 \ln \left (2 x \right )^{2}-10 \ln \left (2 x \right ) {\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(2*x)^2+(-2*x*exp(2+exp(3))-6*x^3-20*x^2-2*x-10)*ln(2*x)-x*exp(2+exp(3))^2+(-6*x^3-20*x^2-2*x-10)*ex
p(2+exp(3))-5*x^5-20*x^4-2*x^3-10*x^2)/x,x,method=_RETURNVERBOSE)

[Out]

(-x-5)*ln(2*x)^2+(-2*x*exp(2+exp(3))-2*x^3-10*x^2)*ln(2*x)-x*exp(4+2*exp(3))-2*exp(2+exp(3))*x^3-10*exp(2+exp(
3))*x^2-x^5-5*x^4-10*ln(x)*exp(2+exp(3))

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maxima [B]  time = 0.36, size = 128, normalized size = 5.57 \begin {gather*} -x^{5} - 5 \, x^{4} - 2 \, x^{3} e^{\left (e^{3} + 2\right )} - 2 \, x^{3} \log \left (2 \, x\right ) - 10 \, x^{2} e^{\left (e^{3} + 2\right )} - 10 \, x^{2} \log \left (2 \, x\right ) - {\left (\log \left (2 \, x\right )^{2} - 2 \, \log \left (2 \, x\right ) + 2\right )} x - x e^{\left (2 \, e^{3} + 4\right )} - 2 \, {\left (x \log \left (2 \, x\right ) - x\right )} e^{\left (e^{3} + 2\right )} - 2 \, x e^{\left (e^{3} + 2\right )} - 2 \, x \log \left (2 \, x\right ) - 5 \, \log \left (2 \, x\right )^{2} - 10 \, e^{\left (e^{3} + 2\right )} \log \relax (x) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^5 - 5*x^4 - 2*x^3*e^(e^3 + 2) - 2*x^3*log(2*x) - 10*x^2*e^(e^3 + 2) - 10*x^2*log(2*x) - (log(2*x)^2 - 2*log
(2*x) + 2)*x - x*e^(2*e^3 + 4) - 2*(x*log(2*x) - x)*e^(e^3 + 2) - 2*x*e^(e^3 + 2) - 2*x*log(2*x) - 5*log(2*x)^
2 - 10*e^(e^3 + 2)*log(x) + 2*x

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mupad [B]  time = 4.43, size = 94, normalized size = 4.09 \begin {gather*} -x\,\left ({\ln \left (2\,x\right )}^2+2\,{\mathrm {e}}^{{\mathrm {e}}^3+2}\,\ln \left (2\,x\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^3+4}\right )-10\,{\mathrm {e}}^{{\mathrm {e}}^3+2}\,\ln \relax (x)-5\,{\ln \left (2\,x\right )}^2-x^3\,\left (2\,{\mathrm {e}}^{{\mathrm {e}}^3+2}+2\,\ln \left (2\,x\right )\right )-x^2\,\left (10\,{\mathrm {e}}^{{\mathrm {e}}^3+2}+10\,\ln \left (2\,x\right )\right )-5\,x^4-x^5 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.31, size = 100, normalized size = 4.35 \begin {gather*} - x^{5} - 5 x^{4} - 2 x^{3} e^{2} e^{e^{3}} - 10 x^{2} e^{2} e^{e^{3}} - x e^{4} e^{2 e^{3}} + \left (- x - 5\right ) \log {\left (2 x \right )}^{2} + \left (- 2 x^{3} - 10 x^{2} - 2 x e^{2} e^{e^{3}}\right ) \log {\left (2 x \right )} - 10 e^{2} e^{e^{3}} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**5 - 5*x**4 - 2*x**3*exp(2)*exp(exp(3)) - 10*x**2*exp(2)*exp(exp(3)) - x*exp(4)*exp(2*exp(3)) + (-x - 5)*lo
g(2*x)**2 + (-2*x**3 - 10*x**2 - 2*x*exp(2)*exp(exp(3)))*log(2*x) - 10*exp(2)*exp(exp(3))*log(x)

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