3.69.64 $\int \frac{4e^{4}x+24e^{2}x\log (5)+32x{\log }^2(5)+(6e^{4}x+16e^{2}x\log (5))\log (2x)+2e^{4}x{\log }^2(2x)}{e^4}dx$

Optimal. Leaf size=18 $${(x+\frac{4x\log (5)}{e^2}+x\log (2x))}^2$$

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Rubi [B]  time = 0.07, antiderivative size = 80, normalized size of antiderivative = 4.44, number of steps used = 9, number of rules used = 4, integrand size = 58, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.069, Rules used = {6, 12, 2304, 2305} $$\begin{array}{c}\frac{x^2}{2}+x^2{\log }^2(2x)+x^2(3+\frac{8\log (5)}{e^2})\log (2x)-x^2\log (2x)+\frac{2x^2(e^2+\log (25))(e^2+\log (625))}{e^4}-\frac{1}{2}{x}^2(3+\frac{8\log (5)}{e^2})\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(4*E^4*x + 24*E^2*x*Log[5] + 32*x*Log[5]^2 + (6*E^4*x + 16*E^2*x*Log[5])*Log[2*x] + 2*E^4*x*Log[2*x]^2)/E^
4,x]

[Out]

x^2/2 - (x^2*(3 + (8*Log[5])/E^2))/2 + (2*x^2*(E^2 + Log[25])*(E^2 + Log[625]))/E^4 - x^2*Log[2*x] + x^2*(3 +
(8*Log[5])/E^2)*Log[2*x] + x^2*Log[2*x]^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{32x{\log }^2(5)+x(4e^4+24e^2\log (5))+(6e^{4}x+16e^{2}x\log (5))\log (2x)+2e^{4}x{\log }^2(2x)}{e^4}dx\\ & =\int \frac{x(4e^4+24e^2\log (5)+32{\log }^2(5))+(6e^{4}x+16e^{2}x\log (5))\log (2x)+2e^{4}x{\log }^2(2x)}{e^4}dx\\ & =\frac{\int (x(4e^4+24e^2\log (5)+32{\log }^2(5))+(6e^{4}x+16e^{2}x\log (5))\log (2x)+2e^{4}x{\log }^2(2x))dx}{e^4}\\ & =\frac{2x^2(e^2+\log (25))(e^2+\log (625))}{e^4}+2\int x{\log }^2(2x)dx+\frac{\int (6e^{4}x+16e^{2}x\log (5))\log (2x)dx}{e^4}\\ & =\frac{2x^2(e^2+\log (25))(e^2+\log (625))}{e^4}+x^2{\log }^2(2x)-2\int x\log (2x)dx+\frac{\int x(6e^4+16e^2\log (5))\log (2x)dx}{e^4}\\ & =\frac{x^2}{2}+\frac{2x^2(e^2+\log (25))(e^2+\log (625))}{e^4}-x^2\log (2x)+x^2{\log }^2(2x)+\left(2(3+\frac{8\log (5)}{e^2})\right)\int x\log (2x)dx\\ & =\frac{x^2}{2}-\frac{1}{2}{x}^2(3+\frac{8\log (5)}{e^2})+\frac{2x^2(e^2+\log (25))(e^2+\log (625))}{e^4}-x^2\log (2x)+x^2(3+\frac{8\log (5)}{e^2})\log (2x)+x^2{\log }^2(2x)\end{array}\end{array}$$

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Mathematica [B]  time = 0.05, size = 76, normalized size = 4.22 $$\begin{array}{c}\frac{2(\frac{e^4{x}^2}{2}+4e^2{x}^2\log (5)+8x^2{\log }^2(5)+e^4{x}^2\log (2x)+4e^2{x}^2\log (5)\log (2x)+\frac{1}{2}{e}^4{x}^2{\log }^2(2x))}{e^4}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(4*E^4*x + 24*E^2*x*Log[5] + 32*x*Log[5]^2 + (6*E^4*x + 16*E^2*x*Log[5])*Log[2*x] + 2*E^4*x*Log[2*x]
^2)/E^4,x]

[Out]

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

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fricas [B]  time = 1.09, size = 62, normalized size = 3.44 $$\begin{array}{c}(x^2{e}^4\log {\left(2x\right)}^2+8x^2{e}^2\log (5)+16x^2\log (5{)}^2+x^2{e}^4+2(4x^2{e}^2\log (5)+x^2{e}^4)\log \left(2x\right))e^{(-4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(2)^2*log(2*x)^2+(16*x*exp(2)*log(5)+6*x*exp(2)^2)*log(2*x)+32*x*log(5)^2+24*x*exp(2)*log(5)
+4*x*exp(2)^2)/exp(2)^2,x, algorithm="fricas")

[Out]

(x^2*e^4*log(2*x)^2 + 8*x^2*e^2*log(5) + 16*x^2*log(5)^2 + x^2*e^4 + 2*(4*x^2*e^2*log(5) + x^2*e^4)*log(2*x))*
e^(-4)

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giac [B]  time = 0.20, size = 80, normalized size = 4.44 $$\begin{array}{c}\frac{1}{2}(16x^2{e}^2\log (5)\log \left(2x\right)+16x^2{e}^2\log (5)+32x^2\log (5{)}^2+6x^2{e}^4\log \left(2x\right)+x^2{e}^4+(2x^2\log {\left(2x\right)}^2-2x^2\log \left(2x\right)+x^2)e^4)e^{(-4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(2)^2*log(2*x)^2+(16*x*exp(2)*log(5)+6*x*exp(2)^2)*log(2*x)+32*x*log(5)^2+24*x*exp(2)*log(5)
+4*x*exp(2)^2)/exp(2)^2,x, algorithm="giac")

[Out]

1/2*(16*x^2*e^2*log(5)*log(2*x) + 16*x^2*e^2*log(5) + 32*x^2*log(5)^2 + 6*x^2*e^4*log(2*x) + x^2*e^4 + (2*x^2*
log(2*x)^2 - 2*x^2*log(2*x) + x^2)*e^4)*e^(-4)

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maple [B]  time = 0.04, size = 60, normalized size = 3.33

method	result	size
risch	$x^2\ln {\left(2x\right)}^2+2{\mathrm{e}}^{-2}{x}^2({\mathrm{e}}^2+4\ln (5))\ln \left(2x\right)+8{\mathrm{e}}^2\ln (5){\mathrm{e}}^{-4}{x}^2+16\ln (5{)}^2{\mathrm{e}}^{-4}{x}^2+{\mathrm{e}}^4{\mathrm{e}}^{-4}{x}^2$	$60$
norman	$(x^2{\mathrm{e}}^2\ln {\left(2x\right)}^2+(2{\mathrm{e}}^2+8\ln (5))x^2\ln \left(2x\right)+({\mathrm{e}}^4+8{\mathrm{e}}^2\ln (5)+16\ln (5{)}^2){\mathrm{e}}^{-2}{x}^2){\mathrm{e}}^{-2}$	$61$
default	${\mathrm{e}}^{-4}(2{\mathrm{e}}^4{x}^2\ln \left(2x\right)+x^2{\mathrm{e}}^4+8{\mathrm{e}}^2\ln (5)x^2\ln \left(2x\right)+8{\mathrm{e}}^2\ln (5)x^2+16x^2\ln (5{)}^2+{\mathrm{e}}^4{x}^2\ln {\left(2x\right)}^2)$	$73$
derivativedivides	$\frac{{\mathrm{e}}^{-4}(4{\mathrm{e}}^4{x}^2\ln \left(2x\right)+2x^2{\mathrm{e}}^4+16{\mathrm{e}}^2\ln (5)x^2\ln \left(2x\right)+16{\mathrm{e}}^2\ln (5)x^2+2{\mathrm{e}}^4{x}^2\ln {\left(2x\right)}^2+32x^2\ln (5{)}^2)}{2}$	$76$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(2)^2*ln(2*x)^2+(16*x*exp(2)*ln(5)+6*x*exp(2)^2)*ln(2*x)+32*x*ln(5)^2+24*x*exp(2)*ln(5)+4*x*exp(2)
^2)/exp(2)^2,x,method=_RETURNVERBOSE)

[Out]

x^2*ln(2*x)^2+2*exp(-2)*x^2*(exp(2)+4*ln(5))*ln(2*x)+8*exp(2)*ln(5)*exp(-4)*x^2+16*ln(5)^2*exp(-4)*x^2+exp(4)*
exp(-4)*x^2

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maxima [B]  time = 0.36, size = 91, normalized size = 5.06 $$\begin{array}{c}\frac{1}{2}((2\log {\left(2x\right)}^2-2\log \left(2x\right)+1)x^2{e}^4+24x^2{e}^2\log (5)+32x^2\log (5{)}^2-(8e^2\log (5)+3e^4)x^2+4x^2{e}^4+2(8x^2{e}^2\log (5)+3x^2{e}^4)\log \left(2x\right))e^{(-4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(2)^2*log(2*x)^2+(16*x*exp(2)*log(5)+6*x*exp(2)^2)*log(2*x)+32*x*log(5)^2+24*x*exp(2)*log(5)
+4*x*exp(2)^2)/exp(2)^2,x, algorithm="maxima")

[Out]

1/2*((2*log(2*x)^2 - 2*log(2*x) + 1)*x^2*e^4 + 24*x^2*e^2*log(5) + 32*x^2*log(5)^2 - (8*e^2*log(5) + 3*e^4)*x^
2 + 4*x^2*e^4 + 2*(8*x^2*e^2*log(5) + 3*x^2*e^4)*log(2*x))*e^(-4)

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mupad [B]  time = 4.24, size = 20, normalized size = 1.11 $$\begin{array}{c}{x}^2{\mathrm{e}}^{-4}{({\mathrm{e}}^2+\ln \left(625\right)+\ln \left(2x\right){\mathrm{e}}^2)}^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-4)*(4*x*exp(4) + 32*x*log(5)^2 + log(2*x)*(6*x*exp(4) + 16*x*exp(2)*log(5)) + 24*x*exp(2)*log(5) + 2*
x*log(2*x)^2*exp(4)),x)

[Out]

x^2*exp(-4)*(exp(2) + log(625) + log(2*x)*exp(2))^2

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sympy [B]  time = 0.21, size = 60, normalized size = 3.33 $$\begin{array}{c}{x}^2\log {\left(2x\right)}^2+\frac{x^2(16\log {(5)}^2+e^4+8e^2\log (5))}{e^4}+\frac{(8x^2\log (5)+2x^2{e}^2)\log \left(2x\right)}{e^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(2)**2*ln(2*x)**2+(16*x*exp(2)*ln(5)+6*x*exp(2)**2)*ln(2*x)+32*x*ln(5)**2+24*x*exp(2)*ln(5)+
4*x*exp(2)**2)/exp(2)**2,x)

[Out]

x**2*log(2*x)**2 + x**2*(16*log(5)**2 + exp(4) + 8*exp(2)*log(5))*exp(-4) + (8*x**2*log(5) + 2*x**2*exp(2))*ex
p(-2)*log(2*x)

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