3.86.27 \(\int \frac {10 x+(3 x+2 e^2 x+3 x^2) \log (5)-2 x \log (5) \log (x)}{2 \log (5)} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{2} x^2 \left (2+e^2+x+\frac {5}{\log (5)}-\log (x)\right ) \]

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Rubi [B]  time = 0.02, antiderivative size = 49, normalized size of antiderivative = 2.13, number of steps used = 5, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {12, 6, 2304} \begin {gather*} \frac {x^3}{2}+\frac {1}{4} \left (3+2 e^2\right ) x^2+\frac {x^2}{4}-\frac {1}{2} x^2 \log (x)+\frac {5 x^2}{2 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x + (3*x + 2*E^2*x + 3*x^2)*Log[5] - 2*x*Log[5]*Log[x])/(2*Log[5]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (10 x+\left (3 x+2 e^2 x+3 x^2\right ) \log (5)-2 x \log (5) \log (x)\right ) \, dx}{2 \log (5)}\\ &=\frac {5 x^2}{2 \log (5)}+\frac {1}{2} \int \left (3 x+2 e^2 x+3 x^2\right ) \, dx-\int x \log (x) \, dx\\ &=\frac {x^2}{4}+\frac {5 x^2}{2 \log (5)}-\frac {1}{2} x^2 \log (x)+\frac {1}{2} \int \left (\left (3+2 e^2\right ) x+3 x^2\right ) \, dx\\ &=\frac {x^2}{4}+\frac {1}{4} \left (3+2 e^2\right ) x^2+\frac {x^3}{2}+\frac {5 x^2}{2 \log (5)}-\frac {1}{2} x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 1.83 \begin {gather*} \frac {5 x^2+2 x^2 \log (5)+e^2 x^2 \log (5)+x^3 \log (5)-x^2 \log (5) \log (x)}{\log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x + (3*x + 2*E^2*x + 3*x^2)*Log[5] - 2*x*Log[5]*Log[x])/(2*Log[5]),x]

[Out]

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

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fricas [A]  time = 0.81, size = 39, normalized size = 1.70 \begin {gather*} -\frac {x^{2} \log \relax (5) \log \relax (x) - 5 \, x^{2} - {\left (x^{3} + x^{2} e^{2} + 2 \, x^{2}\right )} \log \relax (5)}{2 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*x*log(5)*log(x)+(2*exp(2)*x+3*x^2+3*x)*log(5)+10*x)/log(5),x, algorithm="fricas")

[Out]

-1/2*(x^2*log(5)*log(x) - 5*x^2 - (x^3 + x^2*e^2 + 2*x^2)*log(5))/log(5)

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giac [B]  time = 0.13, size = 50, normalized size = 2.17 \begin {gather*} \frac {10 \, x^{2} + {\left (2 \, x^{3} + 2 \, x^{2} e^{2} + 3 \, x^{2}\right )} \log \relax (5) - {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} \log \relax (5)}{4 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*x*log(5)*log(x)+(2*exp(2)*x+3*x^2+3*x)*log(5)+10*x)/log(5),x, algorithm="giac")

[Out]

1/4*(10*x^2 + (2*x^3 + 2*x^2*e^2 + 3*x^2)*log(5) - (2*x^2*log(x) - x^2)*log(5))/log(5)

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maple [A]  time = 0.04, size = 33, normalized size = 1.43

method	result	size
risch	\(\frac {x^{2} {\mathrm e}^{2}}{2}+\frac {x^{3}}{2}+x^{2}+\frac {5 x^{2}}{2 \ln \relax (5)}-\frac {x^{2} \ln \relax (x )}{2}\)	\(33\)
norman	\(\frac {x^{3}}{2}-\frac {x^{2} \ln \relax (x )}{2}+\frac {\left ({\mathrm e}^{2} \ln \relax (5)+2 \ln \relax (5)+5\right ) x^{2}}{2 \ln \relax (5)}\)	\(34\)
default	\(\frac {\ln \relax (5) x^{2} {\mathrm e}^{2}+x^{3} \ln \relax (5)+2 x^{2} \ln \relax (5)+5 x^{2}-x^{2} \ln \relax (5) \ln \relax (x )}{2 \ln \relax (5)}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-2*x*ln(5)*ln(x)+(2*exp(2)*x+3*x^2+3*x)*ln(5)+10*x)/ln(5),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.37, size = 50, normalized size = 2.17 \begin {gather*} \frac {10 \, x^{2} + {\left (2 \, x^{3} + 2 \, x^{2} e^{2} + 3 \, x^{2}\right )} \log \relax (5) - {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} \log \relax (5)}{4 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*x*log(5)*log(x)+(2*exp(2)*x+3*x^2+3*x)*log(5)+10*x)/log(5),x, algorithm="maxima")

[Out]

1/4*(10*x^2 + (2*x^3 + 2*x^2*e^2 + 3*x^2)*log(5) - (2*x^2*log(x) - x^2)*log(5))/log(5)

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mupad [B]  time = 5.19, size = 30, normalized size = 1.30 \begin {gather*} \frac {x^2\,\left (2\,\ln \relax (5)+{\mathrm {e}}^2\,\ln \relax (5)+x\,\ln \relax (5)-\ln \relax (5)\,\ln \relax (x)+5\right )}{2\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.13, size = 34, normalized size = 1.48 \begin {gather*} \frac {x^{3}}{2} - \frac {x^{2} \log {\relax (x )}}{2} + \frac {x^{2} \left (2 \log {\relax (5 )} + 5 + e^{2} \log {\relax (5 )}\right )}{2 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*x*ln(5)*ln(x)+(2*exp(2)*x+3*x**2+3*x)*ln(5)+10*x)/ln(5),x)

[Out]

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

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