3.102.18 \(\int \frac {10 x^2+2 e^3 x^3+2 e^3 \log (x)-e^3 \log ^2(x)}{2 e^3 x^2 \log (2)} \, dx\)

Optimal. Leaf size=29 \[ e+\frac {1+\left (\frac {5}{e^3}+x\right )^2+\frac {\log ^2(x)}{x}}{2 \log (2)} \]

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Rubi [A]  time = 0.05, antiderivative size = 36, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 4, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 14, 2304, 2305} \begin {gather*} \frac {\log ^2(x)}{2 x \log (2)}+\frac {\left (e^3 x+5\right )^2}{2 e^6 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x^2 + 2*E^3*x^3 + 2*E^3*Log[x] - E^3*Log[x]^2)/(2*E^3*x^2*Log[2]),x]

[Out]

(5 + E^3*x)^2/(2*E^6*Log[2]) + Log[x]^2/(2*x*Log[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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} &=\frac {\int \frac {10 x^2+2 e^3 x^3+2 e^3 \log (x)-e^3 \log ^2(x)}{x^2} \, dx}{2 e^3 \log (2)}\\ &=\frac {\int \left (2 \left (5+e^3 x\right )+\frac {2 e^3 \log (x)}{x^2}-\frac {e^3 \log ^2(x)}{x^2}\right ) \, dx}{2 e^3 \log (2)}\\ &=\frac {\left (5+e^3 x\right )^2}{2 e^6 \log (2)}-\frac {\int \frac {\log ^2(x)}{x^2} \, dx}{2 \log (2)}+\frac {\int \frac {\log (x)}{x^2} \, dx}{\log (2)}\\ &=-\frac {1}{x \log (2)}+\frac {\left (5+e^3 x\right )^2}{2 e^6 \log (2)}-\frac {\log (x)}{x \log (2)}+\frac {\log ^2(x)}{2 x \log (2)}-\frac {\int \frac {\log (x)}{x^2} \, dx}{\log (2)}\\ &=\frac {\left (5+e^3 x\right )^2}{2 e^6 \log (2)}+\frac {\log ^2(x)}{2 x \log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 30, normalized size = 1.03 \begin {gather*} \frac {10 x+e^3 x^2+\frac {e^3 \log ^2(x)}{x}}{e^3 \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x^2 + 2*E^3*x^3 + 2*E^3*Log[x] - E^3*Log[x]^2)/(2*E^3*x^2*Log[2]),x]

[Out]

(10*x + E^3*x^2 + (E^3*Log[x]^2)/x)/(E^3*Log[4])

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fricas [A]  time = 0.67, size = 30, normalized size = 1.03 \begin {gather*} \frac {{\left (x^{3} e^{3} + e^{3} \log \relax (x)^{2} + 10 \, x^{2}\right )} e^{\left (-3\right )}}{2 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(3)*log(x)^2+2*log(x)*exp(3)+2*x^3*exp(3)+10*x^2)/x^2/exp(3)/log(2),x, algorithm="fricas")

[Out]

1/2*(x^3*e^3 + e^3*log(x)^2 + 10*x^2)*e^(-3)/(x*log(2))

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giac [A]  time = 0.35, size = 30, normalized size = 1.03 \begin {gather*} \frac {{\left (x^{3} e^{3} + e^{3} \log \relax (x)^{2} + 10 \, x^{2}\right )} e^{\left (-3\right )}}{2 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(3)*log(x)^2+2*log(x)*exp(3)+2*x^3*exp(3)+10*x^2)/x^2/exp(3)/log(2),x, algorithm="giac")

[Out]

1/2*(x^3*e^3 + e^3*log(x)^2 + 10*x^2)*e^(-3)/(x*log(2))

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maple [A]  time = 0.06, size = 33, normalized size = 1.14

method	result	size
risch	\(\frac {\ln \relax (x )^{2}}{2 \ln \relax (2) x}+\frac {x^{2}}{2 \ln \relax (2)}+\frac {5 \,{\mathrm e}^{-3} x}{\ln \relax (2)}\)	\(33\)
norman	\(\frac {\frac {x^{3}}{2 \ln \relax (2)}+\frac {\ln \relax (x )^{2}}{2 \ln \relax (2)}+\frac {5 \,{\mathrm e}^{-3} x^{2}}{\ln \relax (2)}}{x}\)	\(38\)
default	\(\frac {{\mathrm e}^{-3} \left (x^{2} {\mathrm e}^{3}-{\mathrm e}^{3} \left (-\frac {\ln \relax (x )^{2}}{x}-\frac {2 \ln \relax (x )}{x}-\frac {2}{x}\right )+2 \,{\mathrm e}^{3} \left (-\frac {\ln \relax (x )}{x}-\frac {1}{x}\right )+10 x \right )}{2 \ln \relax (2)}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-exp(3)*ln(x)^2+2*ln(x)*exp(3)+2*x^3*exp(3)+10*x^2)/x^2/exp(3)/ln(2),x,method=_RETURNVERBOSE)

[Out]

1/2/ln(2)/x*ln(x)^2+1/2*x^2/ln(2)+5*exp(-3)/ln(2)*x

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maxima [A]  time = 0.35, size = 48, normalized size = 1.66 \begin {gather*} \frac {{\left (x^{2} e^{3} - 2 \, {\left (\frac {\log \relax (x)}{x} + \frac {1}{x}\right )} e^{3} + 10 \, x + \frac {{\left (\log \relax (x)^{2} + 2 \, \log \relax (x) + 2\right )} e^{3}}{x}\right )} e^{\left (-3\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(3)*log(x)^2+2*log(x)*exp(3)+2*x^3*exp(3)+10*x^2)/x^2/exp(3)/log(2),x, algorithm="maxima")

[Out]

1/2*(x^2*e^3 - 2*(log(x)/x + 1/x)*e^3 + 10*x + (log(x)^2 + 2*log(x) + 2)*e^3/x)*e^(-3)/log(2)

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mupad [B]  time = 6.67, size = 29, normalized size = 1.00 \begin {gather*} \frac {{\ln \relax (x)}^2}{2\,x\,\ln \relax (2)}+\frac {x\,{\mathrm {e}}^{-3}\,\left (x\,{\mathrm {e}}^3+10\right )}{2\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3)*(x^3*exp(3) - (exp(3)*log(x)^2)/2 + exp(3)*log(x) + 5*x^2))/(x^2*log(2)),x)

[Out]

log(x)^2/(2*x*log(2)) + (x*exp(-3)*(x*exp(3) + 10))/(2*log(2))

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sympy [A]  time = 0.14, size = 29, normalized size = 1.00 \begin {gather*} \frac {x^{2}}{2 \log {\relax (2 )}} + \frac {5 x}{e^{3} \log {\relax (2 )}} + \frac {\log {\relax (x )}^{2}}{2 x \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(3)*ln(x)**2+2*ln(x)*exp(3)+2*x**3*exp(3)+10*x**2)/x**2/exp(3)/ln(2),x)

[Out]

x**2/(2*log(2)) + 5*x*exp(-3)/log(2) + log(x)**2/(2*x*log(2))

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