3.20.24 \(\int \frac {32+16 x+4 x^2+(32+32 x+12 x^2) \log (x)}{5 \log (5)} \, dx\)

Optimal. Leaf size=29 \[ \frac {4 x^2 \left (4+\frac {4}{x}+\frac {4+x^2}{x}\right ) \log (x)}{5 \log (5)} \]

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Rubi [A]  time = 0.05, antiderivative size = 38, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {12, 2356, 2295, 2304} \begin {gather*} \frac {4 x^3 \log (x)}{5 \log (5)}+\frac {16 x^2 \log (x)}{5 \log (5)}+\frac {32 x \log (x)}{5 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32 + 16*x + 4*x^2 + (32 + 32*x + 12*x^2)*Log[x])/(5*Log[5]),x]

[Out]

(32*x*Log[x])/(5*Log[5]) + (16*x^2*Log[x])/(5*Log[5]) + (4*x^3*Log[x])/(5*Log[5])

Rule 12

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, 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 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (32+16 x+4 x^2+\left (32+32 x+12 x^2\right ) \log (x)\right ) \, dx}{5 \log (5)}\\ &=\frac {32 x}{5 \log (5)}+\frac {8 x^2}{5 \log (5)}+\frac {4 x^3}{15 \log (5)}+\frac {\int \left (32+32 x+12 x^2\right ) \log (x) \, dx}{5 \log (5)}\\ &=\frac {32 x}{5 \log (5)}+\frac {8 x^2}{5 \log (5)}+\frac {4 x^3}{15 \log (5)}+\frac {\int \left (32 \log (x)+32 x \log (x)+12 x^2 \log (x)\right ) \, dx}{5 \log (5)}\\ &=\frac {32 x}{5 \log (5)}+\frac {8 x^2}{5 \log (5)}+\frac {4 x^3}{15 \log (5)}+\frac {12 \int x^2 \log (x) \, dx}{5 \log (5)}+\frac {32 \int \log (x) \, dx}{5 \log (5)}+\frac {32 \int x \log (x) \, dx}{5 \log (5)}\\ &=\frac {32 x \log (x)}{5 \log (5)}+\frac {16 x^2 \log (x)}{5 \log (5)}+\frac {4 x^3 \log (x)}{5 \log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.93 \begin {gather*} \frac {4 \left (8 x \log (x)+4 x^2 \log (x)+x^3 \log (x)\right )}{5 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 + 16*x + 4*x^2 + (32 + 32*x + 12*x^2)*Log[x])/(5*Log[5]),x]

[Out]

(4*(8*x*Log[x] + 4*x^2*Log[x] + x^3*Log[x]))/(5*Log[5])

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fricas [A]  time = 0.71, size = 20, normalized size = 0.69 \begin {gather*} \frac {4 \, {\left (x^{3} + 4 \, x^{2} + 8 \, x\right )} \log \relax (x)}{5 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((12*x^2+32*x+32)*log(x)+4*x^2+16*x+32)/log(5),x, algorithm="fricas")

[Out]

4/5*(x^3 + 4*x^2 + 8*x)*log(x)/log(5)

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giac [A]  time = 0.80, size = 25, normalized size = 0.86 \begin {gather*} \frac {4 \, {\left (x^{3} \log \relax (x) + 4 \, x^{2} \log \relax (x) + 8 \, x \log \relax (x)\right )}}{5 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((12*x^2+32*x+32)*log(x)+4*x^2+16*x+32)/log(5),x, algorithm="giac")

[Out]

4/5*(x^3*log(x) + 4*x^2*log(x) + 8*x*log(x))/log(5)

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maple [A]  time = 0.03, size = 23, normalized size = 0.79

method	result	size
risch	\(\frac {\left (4 x^{3}+16 x^{2}+32 x \right ) \ln \relax (x )}{5 \ln \relax (5)}\)	\(23\)
default	\(\frac {4 x^{3} \ln \relax (x )+16 x^{2} \ln \relax (x )+32 x \ln \relax (x )}{5 \ln \relax (5)}\)	\(27\)
norman	\(\frac {32 x \ln \relax (x )}{5 \ln \relax (5)}+\frac {16 x^{2} \ln \relax (x )}{5 \ln \relax (5)}+\frac {4 x^{3} \ln \relax (x )}{5 \ln \relax (5)}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((12*x^2+32*x+32)*ln(x)+4*x^2+16*x+32)/ln(5),x,method=_RETURNVERBOSE)

[Out]

1/5/ln(5)*(4*x^3+16*x^2+32*x)*ln(x)

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maxima [A]  time = 0.45, size = 20, normalized size = 0.69 \begin {gather*} \frac {4 \, {\left (x^{3} + 4 \, x^{2} + 8 \, x\right )} \log \relax (x)}{5 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((12*x^2+32*x+32)*log(x)+4*x^2+16*x+32)/log(5),x, algorithm="maxima")

[Out]

4/5*(x^3 + 4*x^2 + 8*x)*log(x)/log(5)

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mupad [B]  time = 1.27, size = 31, normalized size = 1.07 \begin {gather*} \frac {32\,x^2\,\ln \relax (x)+16\,x^3\,\ln \relax (x)+4\,x^4\,\ln \relax (x)}{5\,x\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x)/5 + (log(x)*(32*x + 12*x^2 + 32))/5 + (4*x^2)/5 + 32/5)/log(5),x)

[Out]

(32*x^2*log(x) + 16*x^3*log(x) + 4*x^4*log(x))/(5*x*log(5))

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sympy [A]  time = 0.14, size = 20, normalized size = 0.69 \begin {gather*} \frac {\left (4 x^{3} + 16 x^{2} + 32 x\right ) \log {\relax (x )}}{5 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((12*x**2+32*x+32)*ln(x)+4*x**2+16*x+32)/ln(5),x)

[Out]

(4*x**3 + 16*x**2 + 32*x)*log(x)/(5*log(5))

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