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

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

________________________________________________________________________________________

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 {gather*} \frac {x^2}{2}+x^2 \log ^2(2 x)+x^2 \left (3+\frac {8 \log (5)}{e^2}\right ) \log (2 x)-x^2 \log (2 x)+\frac {2 x^2 \left (e^2+\log (25)\right ) \left (e^2+\log (625)\right )}{e^4}-\frac {1}{2} x^2 \left (3+\frac {8 \log (5)}{e^2}\right ) \end {gather*}

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

________________________________________________________________________________________

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

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________