3.35.46 \(\int \frac {1}{5} (2 x+2 x \log (2)+e^{e^4} (x+x \log (2))+(4 x+4 x \log (2)+e^{e^4} (2 x+2 x \log (2))) \log (x)) \, dx\)

Optimal. Leaf size=22 \[ \frac {1}{5} x \left (2 x+e^{e^4} x\right ) (1+\log (2)) \log (x) \]

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Rubi [B]  time = 0.05, antiderivative size = 66, normalized size of antiderivative = 3.00, number of steps used = 7, number of rules used = 4, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6, 12, 2332, 2304} \begin {gather*} \frac {1}{5} \left (2+e^{e^4}\right ) x^2 (1+\log (2)) \log (x)-\frac {1}{10} \left (2+e^{e^4}\right ) x^2 (1+\log (2))+\frac {1}{10} e^{e^4} x^2 (1+\log (2))+\frac {1}{5} x^2 (1+\log (2)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2332

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*(a + b*Log[c*
x^n])^p, x] /; FreeQ[{a, b, c, n, p, q}, x] && BinomialQ[u, x] &&  !BinomialMatchQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 20, normalized size = 0.91 \begin {gather*} \frac {1}{5} \left (2+e^{e^4}\right ) x^2 (1+\log (2)) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 31, normalized size = 1.41 \begin {gather*} \frac {1}{5} \, {\left (2 \, x^{2} \log \relax (2) + 2 \, x^{2} + {\left (x^{2} \log \relax (2) + x^{2}\right )} e^{\left (e^{4}\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*log(2)+2*x)*exp(exp(4))+4*x*log(2)+4*x)*log(x)+1/5*(x*log(2)+x)*exp(exp(4))+2/5*x*log(2)+2
/5*x,x, algorithm="fricas")

[Out]

1/5*(2*x^2*log(2) + 2*x^2 + (x^2*log(2) + x^2)*e^(e^4))*log(x)

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giac [B]  time = 0.19, size = 72, normalized size = 3.27 \begin {gather*} \frac {1}{5} \, x^{2} e^{\left (e^{4}\right )} \log \relax (2) \log \relax (x) - \frac {1}{10} \, x^{2} e^{\left (e^{4}\right )} \log \relax (2) + \frac {1}{5} \, x^{2} e^{\left (e^{4}\right )} \log \relax (x) + \frac {2}{5} \, x^{2} \log \relax (2) \log \relax (x) - \frac {1}{10} \, x^{2} e^{\left (e^{4}\right )} + \frac {2}{5} \, x^{2} \log \relax (x) + \frac {1}{10} \, {\left (x^{2} \log \relax (2) + x^{2}\right )} e^{\left (e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*log(2)+2*x)*exp(exp(4))+4*x*log(2)+4*x)*log(x)+1/5*(x*log(2)+x)*exp(exp(4))+2/5*x*log(2)+2
/5*x,x, algorithm="giac")

[Out]

1/5*x^2*e^(e^4)*log(2)*log(x) - 1/10*x^2*e^(e^4)*log(2) + 1/5*x^2*e^(e^4)*log(x) + 2/5*x^2*log(2)*log(x) - 1/1
0*x^2*e^(e^4) + 2/5*x^2*log(x) + 1/10*(x^2*log(2) + x^2)*e^(e^4)

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maple [A]  time = 0.04, size = 25, normalized size = 1.14

method	result	size
norman	\(\left (\frac {{\mathrm e}^{{\mathrm e}^{4}} \ln \relax (2)}{5}+\frac {2 \ln \relax (2)}{5}+\frac {{\mathrm e}^{{\mathrm e}^{4}}}{5}+\frac {2}{5}\right ) x^{2} \ln \relax (x )\)	\(25\)
default	\(\frac {{\mathrm e}^{{\mathrm e}^{4}} \ln \relax (2) x^{2} \ln \relax (x )}{5}+\frac {2 x^{2} \ln \relax (2) \ln \relax (x )}{5}+\frac {{\mathrm e}^{{\mathrm e}^{4}} x^{2} \ln \relax (x )}{5}+\frac {2 x^{2} \ln \relax (x )}{5}\)	\(40\)
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{4}} \ln \relax (2) x^{2} \ln \relax (x )}{5}+\frac {2 x^{2} \ln \relax (2) \ln \relax (x )}{5}+\frac {{\mathrm e}^{{\mathrm e}^{4}} x^{2} \ln \relax (x )}{5}+\frac {2 x^{2} \ln \relax (x )}{5}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((2*x*ln(2)+2*x)*exp(exp(4))+4*x*ln(2)+4*x)*ln(x)+1/5*(x*ln(2)+x)*exp(exp(4))+2/5*x*ln(2)+2/5*x,x,meth
od=_RETURNVERBOSE)

[Out]

(1/5*exp(exp(4))*ln(2)+2/5*ln(2)+1/5*exp(exp(4))+2/5)*x^2*ln(x)

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maxima [B]  time = 0.40, size = 77, normalized size = 3.50 \begin {gather*} -\frac {1}{10} \, {\left ({\left (e^{\left (e^{4}\right )} + 2\right )} \log \relax (2) + e^{\left (e^{4}\right )} + 2\right )} x^{2} + \frac {1}{5} \, x^{2} \log \relax (2) + \frac {1}{5} \, x^{2} + \frac {1}{10} \, {\left (x^{2} \log \relax (2) + x^{2}\right )} e^{\left (e^{4}\right )} + \frac {1}{5} \, {\left (2 \, x^{2} \log \relax (2) + 2 \, x^{2} + {\left (x^{2} \log \relax (2) + x^{2}\right )} e^{\left (e^{4}\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*log(2)+2*x)*exp(exp(4))+4*x*log(2)+4*x)*log(x)+1/5*(x*log(2)+x)*exp(exp(4))+2/5*x*log(2)+2
/5*x,x, algorithm="maxima")

[Out]

-1/10*((e^(e^4) + 2)*log(2) + e^(e^4) + 2)*x^2 + 1/5*x^2*log(2) + 1/5*x^2 + 1/10*(x^2*log(2) + x^2)*e^(e^4) +
1/5*(2*x^2*log(2) + 2*x^2 + (x^2*log(2) + x^2)*e^(e^4))*log(x)

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mupad [B]  time = 2.31, size = 16, normalized size = 0.73 \begin {gather*} \frac {x^2\,\ln \relax (x)\,\left ({\mathrm {e}}^{{\mathrm {e}}^4}+2\right )\,\left (\ln \relax (2)+1\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x)/5 + (2*x*log(2))/5 + (exp(exp(4))*(x + x*log(2)))/5 + (log(x)*(4*x + 4*x*log(2) + exp(exp(4))*(2*x +
2*x*log(2))))/5,x)

[Out]

(x^2*log(x)*(exp(exp(4)) + 2)*(log(2) + 1))/5

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sympy [A]  time = 0.16, size = 42, normalized size = 1.91 \begin {gather*} \left (\frac {2 x^{2} \log {\relax (2 )}}{5} + \frac {2 x^{2}}{5} + \frac {x^{2} e^{e^{4}} \log {\relax (2 )}}{5} + \frac {x^{2} e^{e^{4}}}{5}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*ln(2)+2*x)*exp(exp(4))+4*x*ln(2)+4*x)*ln(x)+1/5*(x*ln(2)+x)*exp(exp(4))+2/5*x*ln(2)+2/5*x,
x)

[Out]

(2*x**2*log(2)/5 + 2*x**2/5 + x**2*exp(exp(4))*log(2)/5 + x**2*exp(exp(4))/5)*log(x)

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