\(\int \frac {1}{25} (588+168 x^2+12 x^4+(588+504 x^2+60 x^4) \log (x)) \, dx\) [5884]

Optimal result

Integrand size = 31, antiderivative size = 14 \[ \int \frac {1}{25} \left (588+168 x^2+12 x^4+\left (588+504 x^2+60 x^4\right ) \log (x)\right ) \, dx=\frac {12}{25} x \left (7+x^2\right )^2 \log (x) \]

[Out]

12/5*(x^2+7)*(1/5*x^2+7/5)*x*ln(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.86, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {12, 2403, 2332, 2341} \[ \int \frac {1}{25} \left (588+168 x^2+12 x^4+\left (588+504 x^2+60 x^4\right ) \log (x)\right ) \, dx=\frac {12}{25} x^5 \log (x)+\frac {168}{25} x^3 \log (x)+\frac {588}{25} x \log (x) \]

[In]

Int[(588 + 168*x^2 + 12*x^4 + (588 + 504*x^2 + 60*x^4)*Log[x])/25,x]

[Out]

(588*x*Log[x])/25 + (168*x^3*Log[x])/25 + (12*x^5*Log[x])/25

Rule 12

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

Rule 2332

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

Rule 2341

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 2403

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{align*} \text {integral}& = \frac {1}{25} \int \left (588+168 x^2+12 x^4+\left (588+504 x^2+60 x^4\right ) \log (x)\right ) \, dx \\ & = \frac {588 x}{25}+\frac {56 x^3}{25}+\frac {12 x^5}{125}+\frac {1}{25} \int \left (588+504 x^2+60 x^4\right ) \log (x) \, dx \\ & = \frac {588 x}{25}+\frac {56 x^3}{25}+\frac {12 x^5}{125}+\frac {1}{25} \int \left (588 \log (x)+504 x^2 \log (x)+60 x^4 \log (x)\right ) \, dx \\ & = \frac {588 x}{25}+\frac {56 x^3}{25}+\frac {12 x^5}{125}+\frac {12}{5} \int x^4 \log (x) \, dx+\frac {504}{25} \int x^2 \log (x) \, dx+\frac {588}{25} \int \log (x) \, dx \\ & = \frac {588}{25} x \log (x)+\frac {168}{25} x^3 \log (x)+\frac {12}{25} x^5 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.86 \[ \int \frac {1}{25} \left (588+168 x^2+12 x^4+\left (588+504 x^2+60 x^4\right ) \log (x)\right ) \, dx=\frac {588}{25} x \log (x)+\frac {168}{25} x^3 \log (x)+\frac {12}{25} x^5 \log (x) \]

[In]

Integrate[(588 + 168*x^2 + 12*x^4 + (588 + 504*x^2 + 60*x^4)*Log[x])/25,x]

[Out]

(588*x*Log[x])/25 + (168*x^3*Log[x])/25 + (12*x^5*Log[x])/25

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36

method	result	size
risch	\(\frac {\left (12 x^{5}+168 x^{3}+588 x \right ) \ln \left (x \right )}{25}\)	\(19\)
default	\(\frac {12 x^{5} \ln \left (x \right )}{25}+\frac {168 x^{3} \ln \left (x \right )}{25}+\frac {588 x \ln \left (x \right )}{25}\)	\(21\)
norman	\(\frac {12 x^{5} \ln \left (x \right )}{25}+\frac {168 x^{3} \ln \left (x \right )}{25}+\frac {588 x \ln \left (x \right )}{25}\)	\(21\)
parallelrisch	\(\frac {12 x^{5} \ln \left (x \right )}{25}+\frac {168 x^{3} \ln \left (x \right )}{25}+\frac {588 x \ln \left (x \right )}{25}\)	\(21\)
parts	\(\frac {12 x^{5} \ln \left (x \right )}{25}+\frac {168 x^{3} \ln \left (x \right )}{25}+\frac {588 x \ln \left (x \right )}{25}\)	\(21\)

[In]

int(1/25*(60*x^4+504*x^2+588)*ln(x)+12/25*x^4+168/25*x^2+588/25,x,method=_RETURNVERBOSE)

[Out]

1/25*(12*x^5+168*x^3+588*x)*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {1}{25} \left (588+168 x^2+12 x^4+\left (588+504 x^2+60 x^4\right ) \log (x)\right ) \, dx=\frac {12}{25} \, {\left (x^{5} + 14 \, x^{3} + 49 \, x\right )} \log \left (x\right ) \]

[In]

integrate(1/25*(60*x^4+504*x^2+588)*log(x)+12/25*x^4+168/25*x^2+588/25,x, algorithm="fricas")

[Out]

12/25*(x^5 + 14*x^3 + 49*x)*log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {1}{25} \left (588+168 x^2+12 x^4+\left (588+504 x^2+60 x^4\right ) \log (x)\right ) \, dx=\left (\frac {12 x^{5}}{25} + \frac {168 x^{3}}{25} + \frac {588 x}{25}\right ) \log {\left (x \right )} \]

[In]

integrate(1/25*(60*x**4+504*x**2+588)*ln(x)+12/25*x**4+168/25*x**2+588/25,x)

[Out]

(12*x**5/25 + 168*x**3/25 + 588*x/25)*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {1}{25} \left (588+168 x^2+12 x^4+\left (588+504 x^2+60 x^4\right ) \log (x)\right ) \, dx=\frac {12}{25} \, {\left (x^{5} + 14 \, x^{3} + 49 \, x\right )} \log \left (x\right ) \]

[In]

integrate(1/25*(60*x^4+504*x^2+588)*log(x)+12/25*x^4+168/25*x^2+588/25,x, algorithm="maxima")

[Out]

12/25*(x^5 + 14*x^3 + 49*x)*log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {1}{25} \left (588+168 x^2+12 x^4+\left (588+504 x^2+60 x^4\right ) \log (x)\right ) \, dx=\frac {12}{25} \, x^{5} \log \left (x\right ) + \frac {168}{25} \, x^{3} \log \left (x\right ) + \frac {588}{25} \, x \log \left (x\right ) \]

[In]

integrate(1/25*(60*x^4+504*x^2+588)*log(x)+12/25*x^4+168/25*x^2+588/25,x, algorithm="giac")

[Out]

12/25*x^5*log(x) + 168/25*x^3*log(x) + 588/25*x*log(x)

Mupad [B] (verification not implemented)

Time = 12.51 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1}{25} \left (588+168 x^2+12 x^4+\left (588+504 x^2+60 x^4\right ) \log (x)\right ) \, dx=\frac {12\,x\,\ln \left (x\right )\,{\left (x^2+7\right )}^2}{25} \]

[In]

int((log(x)*(504*x^2 + 60*x^4 + 588))/25 + (168*x^2)/25 + (12*x^4)/25 + 588/25,x)

[Out]

(12*x*log(x)*(x^2 + 7)^2)/25