3.89.87 $\int \frac{8e^{2x}x+12x^2+e^x(-8x-8x^2)+(-10+6x^2+e^{2x}(-2+8x)+e^x(-4x-8x^2))\log (x)+(-5+x^2-2e^x{x}^2+e^{2x}(-1+2x)){\log }^2(x)}{x^2}dx$

Optimal. Leaf size=28 $$4x+\frac{x+(5+{(e^x-x)}^2)(2+\log (x){)}^2}{x}$$

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Rubi [B]  time = 0.28, antiderivative size = 90, normalized size of antiderivative = 3.21, number of steps used = 14, number of rules used = 8, integrand size = 94, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.085, Rules used = {14, 2288, 2334, 2353, 2296, 2295, 2305, 2304} $$\begin{array}{c}\frac{e^{2x}(\log (x)+2)(2x+x\log (x))}{x^2}+8x+\frac{20}{x}+x{\log }^2(x)+\frac{5{\log }^2(x)}{x}-2x\log (x)+2(3x+\frac{5}{x})\log (x)+\frac{10\log (x)}{x}-\frac{2e^x(\log (x)+2)(2x+x\log (x))}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(8*E^(2*x)*x + 12*x^2 + E^x*(-8*x - 8*x^2) + (-10 + 6*x^2 + E^(2*x)*(-2 + 8*x) + E^x*(-4*x - 8*x^2))*Log[x
] + (-5 + x^2 - 2*E^x*x^2 + E^(2*x)*(-1 + 2*x))*Log[x]^2)/x^2,x]

[Out]

20/x + 8*x + (10*Log[x])/x - 2*x*Log[x] + 2*(5/x + 3*x)*Log[x] + (5*Log[x]^2)/x + x*Log[x]^2 + (E^(2*x)*(2 + L
og[x])*(2*x + x*Log[x]))/x^2 - (2*E^x*(2 + Log[x])*(2*x + x*Log[x]))/x

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-\frac{2e^x(2+\log (x))(2+2x+x\log (x))}{x}+\frac{e^{2x}(2+\log (x))(4x-\log (x)+2x\log (x))}{x^2}+\frac{12x^2-10\log (x)+6x^2\log (x)-5{\log }^2(x)+x^2{\log }^2(x)}{x^2})dx\\ & =-(2\int \frac{e^x(2+\log (x))(2+2x+x\log (x))}{x}dx)+\int \frac{e^{2x}(2+\log (x))(4x-\log (x)+2x\log (x))}{x^2}dx+\int \frac{12x^2-10\log (x)+6x^2\log (x)-5{\log }^2(x)+x^2{\log }^2(x)}{x^2}dx\\ & =\frac{e^{2x}(2+\log (x))(2x+x\log (x))}{x^2}-\frac{2e^x(2+\log (x))(2x+x\log (x))}{x}+\int (12+\frac{2(-5+3x^2)\log (x)}{x^2}+\frac{(-5+x^2){\log }^2(x)}{x^2})dx\\ & =12x+\frac{e^{2x}(2+\log (x))(2x+x\log (x))}{x^2}-\frac{2e^x(2+\log (x))(2x+x\log (x))}{x}+2\int \frac{(-5+3x^2)\log (x)}{x^2}dx+\int \frac{(-5+x^2){\log }^2(x)}{x^2}dx\\ & =12x+2(\frac{5}{x}+3x)\log (x)+\frac{e^{2x}(2+\log (x))(2x+x\log (x))}{x^2}-\frac{2e^x(2+\log (x))(2x+x\log (x))}{x}-2\int (3+\frac{5}{x^2})dx+\int ({\log }^2(x)-\frac{5{\log }^2(x)}{x^2})dx\\ & =\frac{10}{x}+6x+2(\frac{5}{x}+3x)\log (x)+\frac{e^{2x}(2+\log (x))(2x+x\log (x))}{x^2}-\frac{2e^x(2+\log (x))(2x+x\log (x))}{x}-5\int \frac{{\log }^2(x)}{x^2}dx+\int {\log }^2(x)dx\\ & =\frac{10}{x}+6x+2(\frac{5}{x}+3x)\log (x)+\frac{5{\log }^2(x)}{x}+x{\log }^2(x)+\frac{e^{2x}(2+\log (x))(2x+x\log (x))}{x^2}-\frac{2e^x(2+\log (x))(2x+x\log (x))}{x}-2\int \log (x)dx-10\int \frac{\log (x)}{x^2}dx\\ & =\frac{20}{x}+8x+\frac{10\log (x)}{x}-2x\log (x)+2(\frac{5}{x}+3x)\log (x)+\frac{5{\log }^2(x)}{x}+x{\log }^2(x)+\frac{e^{2x}(2+\log (x))(2x+x\log (x))}{x^2}-\frac{2e^x(2+\log (x))(2x+x\log (x))}{x}\end{array}\end{array}$$

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Mathematica [B]  time = 0.09, size = 66, normalized size = 2.36 $$\begin{array}{c}\frac{4(5+e^{2x}-2e^{x}x+2x^2)+4(5+e^{2x}-2e^{x}x+x^2)\log (x)+(5+e^{2x}-2e^{x}x+x^2){\log }^2(x)}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(8*E^(2*x)*x + 12*x^2 + E^x*(-8*x - 8*x^2) + (-10 + 6*x^2 + E^(2*x)*(-2 + 8*x) + E^x*(-4*x - 8*x^2))
*Log[x] + (-5 + x^2 - 2*E^x*x^2 + E^(2*x)*(-1 + 2*x))*Log[x]^2)/x^2,x]

[Out]

(4*(5 + E^(2*x) - 2*E^x*x + 2*x^2) + 4*(5 + E^(2*x) - 2*E^x*x + x^2)*Log[x] + (5 + E^(2*x) - 2*E^x*x + x^2)*Lo
g[x]^2)/x

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fricas [B]  time = 0.46, size = 59, normalized size = 2.11 $$\begin{array}{c}\frac{(x^2-2x{e}^x+e^{\left(2x\right)}+5)\log (x{)}^2+8x^2-8x{e}^x+4(x^2-2x{e}^x+e^{\left(2x\right)}+5)\log (x)+4e^{\left(2x\right)}+20}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-1)*exp(x)^2-2*exp(x)*x^2+x^2-5)*log(x)^2+((8*x-2)*exp(x)^2+(-8*x^2-4*x)*exp(x)+6*x^2-10)*log(
x)+8*x*exp(x)^2+(-8*x^2-8*x)*exp(x)+12*x^2)/x^2,x, algorithm="fricas")

[Out]

((x^2 - 2*x*e^x + e^(2*x) + 5)*log(x)^2 + 8*x^2 - 8*x*e^x + 4*(x^2 - 2*x*e^x + e^(2*x) + 5)*log(x) + 4*e^(2*x)
+ 20)/x

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giac [B]  time = 0.12, size = 80, normalized size = 2.86 $$\begin{array}{c}\frac{x^2\log (x{)}^2-2x{e}^x\log (x{)}^2+4x^2\log (x)-8x{e}^x\log (x)+e^{\left(2x\right)}\log (x{)}^2+8x^2-8x{e}^x+4e^{\left(2x\right)}\log (x)+5\log (x{)}^2+4e^{\left(2x\right)}+20\log (x)+20}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-1)*exp(x)^2-2*exp(x)*x^2+x^2-5)*log(x)^2+((8*x-2)*exp(x)^2+(-8*x^2-4*x)*exp(x)+6*x^2-10)*log(
x)+8*x*exp(x)^2+(-8*x^2-8*x)*exp(x)+12*x^2)/x^2,x, algorithm="giac")

[Out]

(x^2*log(x)^2 - 2*x*e^x*log(x)^2 + 4*x^2*log(x) - 8*x*e^x*log(x) + e^(2*x)*log(x)^2 + 8*x^2 - 8*x*e^x + 4*e^(2
*x)*log(x) + 5*log(x)^2 + 4*e^(2*x) + 20*log(x) + 20)/x

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maple [B]  time = 0.10, size = 66, normalized size = 2.36

method	result	size
risch	$\frac{(x^2-2{\mathrm{e}}^{x}x+{\mathrm{e}}^{2x}+5)\ln (x{)}^2}{x}+\frac{4(x^2-2{\mathrm{e}}^{x}x+{\mathrm{e}}^{2x}+5)\ln (x)}{x}+\frac{8x^2-8{\mathrm{e}}^{x}x+4{\mathrm{e}}^{2x}+20}{x}$	$66$
default	$8x-8{\mathrm{e}}^x\ln (x)-2{\mathrm{e}}^x\ln (x{)}^2-8{\mathrm{e}}^x+\frac{\ln (x{)}^2{\mathrm{e}}^{2x}+4\ln (x){\mathrm{e}}^{2x}+4{\mathrm{e}}^{2x}}{x}+x\ln (x{)}^2+4x\ln (x)+\frac{5\ln (x{)}^2}{x}+\frac{20\ln (x)}{x}+\frac{20}{x}$	$83$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x-1)*exp(x)^2-2*exp(x)*x^2+x^2-5)*ln(x)^2+((8*x-2)*exp(x)^2+(-8*x^2-4*x)*exp(x)+6*x^2-10)*ln(x)+8*x*e
xp(x)^2+(-8*x^2-8*x)*exp(x)+12*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}6x\log (x)-8e^x\log (x)+6x-\frac{2x{e}^x\log (x{)}^2-(x^2+5)\log (x{)}^2-2x^2-(\log (x{)}^2+4\log (x))e^{\left(2x\right)}+2(x^2-5)\log (x)-10}{x}+\frac{10\log (x)}{x}+\frac{10}{x}+8\mathrm{E}\mathrm{i}\left(2x\right)-8e^x-4\int \frac{e^{\left(2x\right)}}{x^2}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-1)*exp(x)^2-2*exp(x)*x^2+x^2-5)*log(x)^2+((8*x-2)*exp(x)^2+(-8*x^2-4*x)*exp(x)+6*x^2-10)*log(
x)+8*x*exp(x)^2+(-8*x^2-8*x)*exp(x)+12*x^2)/x^2,x, algorithm="maxima")

[Out]

6*x*log(x) - 8*e^x*log(x) + 6*x - (2*x*e^x*log(x)^2 - (x^2 + 5)*log(x)^2 - 2*x^2 - (log(x)^2 + 4*log(x))*e^(2*
x) + 2*(x^2 - 5)*log(x) - 10)/x + 10*log(x)/x + 10/x + 8*Ei(2*x) - 8*e^x - 4*integrate(e^(2*x)/x^2, x)

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mupad [B]  time = 5.35, size = 70, normalized size = 2.50 $$\begin{array}{c}\frac{4{\mathrm{e}}^{2x}+20\ln (x)+5{\ln (x)}^2+4{\mathrm{e}}^{2x}\ln (x)+{\mathrm{e}}^{2x}{\ln (x)}^2+20}{x}-8{\mathrm{e}}^x\ln (x)-8{\mathrm{e}}^x-2{\mathrm{e}}^x{\ln (x)}^2+x({\ln (x)}^2+4\ln (x)+8)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^2*(2*x^2*exp(x) - exp(2*x)*(2*x - 1) - x^2 + 5) - 8*x*exp(2*x) + log(x)*(exp(x)*(4*x + 8*x^2) - e
xp(2*x)*(8*x - 2) - 6*x^2 + 10) + exp(x)*(8*x + 8*x^2) - 12*x^2)/x^2,x)

[Out]

(4*exp(2*x) + 20*log(x) + 5*log(x)^2 + 4*exp(2*x)*log(x) + exp(2*x)*log(x)^2 + 20)/x - 8*exp(x)*log(x) - 8*exp
(x) - 2*exp(x)*log(x)^2 + x*(4*log(x) + log(x)^2 + 8)

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sympy [B]  time = 0.44, size = 71, normalized size = 2.54 $$\begin{array}{c}8x+\frac{(x^2+5)\log {(x)}^2}{x}+\frac{(4x^2+20)\log (x)}{x}+\frac{(-2x\log {(x)}^2-8x\log (x)-8x)e^x+(\log {(x)}^2+4\log (x)+4)e^{2x}}{x}+\frac{20}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-1)*exp(x)**2-2*exp(x)*x**2+x**2-5)*ln(x)**2+((8*x-2)*exp(x)**2+(-8*x**2-4*x)*exp(x)+6*x**2-10
)*ln(x)+8*x*exp(x)**2+(-8*x**2-8*x)*exp(x)+12*x**2)/x**2,x)

[Out]

8*x + (x**2 + 5)*log(x)**2/x + (4*x**2 + 20)*log(x)/x + ((-2*x*log(x)**2 - 8*x*log(x) - 8*x)*exp(x) + (log(x)*
*2 + 4*log(x) + 4)*exp(2*x))/x + 20/x

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