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

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

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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} $$\begin{array}{c}-x^5-5x^4-\frac{2}{3}(1+3e^{2+e^3})x^3+\frac{2x^3}{3}-2x^3\log (2x)-5(1+2e^{2+e^3})x^2+5x^2-10x^2\log (2x)-e^{2+e^3}(2+e^{2+e^3})x+2(1+e^{2+e^3})x-2x-x{\log }^2(2x)-5{\log }^2(2x)-2(1+e^{2+e^3})x\log (2x)+2x\log (2x)-10e^{2+e^3}\log (x)\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{-10e^{2+e^3}-2e^{2+e^3}(1+\frac{e^{2+e^3}}{2})x-10(1+2e^{2+e^3})x^2-2(1+3e^{2+e^3})x^3-20x^4-5x^5}{x}+\frac{2(-5-(1+e^{2+e^3})x-10x^2-3x^3)\log (2x)}{x}-{\log }^2(2x))dx\\ & =2\int \frac{(-5-(1+e^{2+e^3})x-10x^2-3x^3)\log (2x)}{x}dx+\int \frac{-10e^{2+e^3}-2e^{2+e^3}(1+\frac{e^{2+e^3}}{2})x-10(1+2e^{2+e^3})x^2-2(1+3e^{2+e^3})x^3-20x^4-5x^5}{x}dx-\int {\log }^2(2x)dx\\ & =-x{\log }^2(2x)+2\int \log (2x)dx+2\int ((-1-e^{2+e^3})\log (2x)-\frac{5\log (2x)}{x}-10x\log (2x)-3x^2\log (2x))dx+\int (-e^{2+e^3}(2+e^{2+e^3})-\frac{10e^{2+e^3}}{x}-10(1+2e^{2+e^3})x-2(1+3e^{2+e^3})x^2-20x^3-5x^4)dx\\ & =-2x-e^{2+e^3}(2+e^{2+e^3})x-5(1+2e^{2+e^3})x^2-\frac{2}{3}(1+3e^{2+e^3})x^3-5x^4-x^5-10e^{2+e^3}\log (x)+2x\log (2x)-x{\log }^2(2x)-6\int x^2\log (2x)dx-10\int \frac{\log (2x)}{x}dx-20\int x\log (2x)dx-\left(2(1+e^{2+e^3})\right)\int \log (2x)dx\\ & =-2x+2(1+e^{2+e^3})x-e^{2+e^3}(2+e^{2+e^3})x+5x^2-5(1+2e^{2+e^3})x^2+\frac{2x^3}{3}-\frac{2}{3}(1+3e^{2+e^3})x^3-5x^4-x^5-10e^{2+e^3}\log (x)+2x\log (2x)-2(1+e^{2+e^3})x\log (2x)-10x^2\log (2x)-2x^3\log (2x)-5{\log }^2(2x)-x{\log }^2(2x)\end{array}\end{array}$$

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

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 $$\begin{array}{c}-x^5-5x^4-(x+5)\log {\left(2x\right)}^2-x{e}^{(2e^3+4)}-2(x^3+5x^2)e^{(e^3+2)}-2(x^3+5x^2+(x+5)e^{(e^3+2)})\log \left(2x\right)\end{array}$$

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 $$\begin{array}{c}-x^5-5x^4-2x^3{e}^{(e^3+2)}-2x^3\log \left(2x\right)-10x^2{e}^{(e^3+2)}-10x^2\log \left(2x\right)-2x{e}^{(e^3+2)}\log \left(2x\right)-x\log {\left(2x\right)}^2-x{e}^{(2e^3+4)}-5\log {\left(2x\right)}^2-10e^{(e^3+2)}\log (x)\end{array}$$

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	$(-x-5)\ln {\left(2x\right)}^2+(-2x{\mathrm{e}}^{2+{\mathrm{e}}^3}-2x^3-10x^2)\ln \left(2x\right)-x{\mathrm{e}}^{4+2{\mathrm{e}}^3}-2{\mathrm{e}}^{2+{\mathrm{e}}^3}{x}^3-10{\mathrm{e}}^{2+{\mathrm{e}}^3}{x}^2-x^5-5x^4-10\ln (x){\mathrm{e}}^{2+{\mathrm{e}}^3}$	$87$
norman	$-10\ln \left(2x\right){\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}-5x^4-x^5-5\ln {\left(2x\right)}^2-x\ln {\left(2x\right)}^2-10x^2\ln \left(2x\right)-2x^3\ln \left(2x\right)-x{\mathrm{e}}^4{\mathrm{e}}^{2{\mathrm{e}}^3}-10{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}{x}^2-2{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}{x}^3-2x{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}\ln \left(2x\right)$	$102$
derivativedivides	$-x^5-2x^3\ln \left(2x\right)-2{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}{x}^3-5x^4-x\ln {\left(2x\right)}^2-{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}(2x\ln \left(2x\right)-2x)-10x^2\ln \left(2x\right)-x{\mathrm{e}}^4{\mathrm{e}}^{2{\mathrm{e}}^3}-10{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}{x}^2-2x{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}-5\ln {\left(2x\right)}^2-10\ln \left(2x\right){\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}$	$116$
default	$-x^5-2x^3\ln \left(2x\right)-2{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}{x}^3-5x^4-x\ln {\left(2x\right)}^2-{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}(2x\ln \left(2x\right)-2x)-10x^2\ln \left(2x\right)-x{\mathrm{e}}^4{\mathrm{e}}^{2{\mathrm{e}}^3}-10{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}{x}^2-2x{\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}-5\ln {\left(2x\right)}^2-10\ln \left(2x\right){\mathrm{e}}^2{\mathrm{e}}^{{\mathrm{e}}^3}$	$116$

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 $$\begin{array}{c}-x^5-5x^4-2x^3{e}^{(e^3+2)}-2x^3\log \left(2x\right)-10x^2{e}^{(e^3+2)}-10x^2\log \left(2x\right)-(\log {\left(2x\right)}^2-2\log \left(2x\right)+2)x-x{e}^{(2e^3+4)}-2(x\log \left(2x\right)-x)e^{(e^3+2)}-2x{e}^{(e^3+2)}-2x\log \left(2x\right)-5\log {\left(2x\right)}^2-10e^{(e^3+2)}\log (x)+2x\end{array}$$

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 $$\begin{array}{c}-x({\ln \left(2x\right)}^2+2{\mathrm{e}}^{{\mathrm{e}}^3+2}\ln \left(2x\right)+{\mathrm{e}}^{2{\mathrm{e}}^3+4})-10{\mathrm{e}}^{{\mathrm{e}}^3+2}\ln (x)-5{\ln \left(2x\right)}^2-x^3(2{\mathrm{e}}^{{\mathrm{e}}^3+2}+2\ln \left(2x\right))-x^2(10{\mathrm{e}}^{{\mathrm{e}}^3+2}+10\ln \left(2x\right))-5x^4-x^5\end{array}$$

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 $$\begin{array}{c}-x^5-5x^4-2x^3{e}^2{e}^{e^3}-10x^2{e}^2{e}^{e^3}-x{e}^4{e}^{2e^3}+(-x-5)\log {\left(2x\right)}^2+(-2x^3-10x^2-2x{e}^2{e}^{e^3})\log \left(2x\right)-10e^2{e}^{e^3}\log (x)\end{array}$$

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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