3.50.90 \(\int \frac {6 x^3+3 x^2 \log (2)+(9 x^3+9 x^2 \log (2)) \log (\frac {1}{4} (x^2+x \log (2)))}{10 x+10 \log (2)} \, dx\)

Optimal. Leaf size=17 \[ \frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right ) \]

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Rubi [A]  time = 0.27, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.137, Rules used = {6688, 12, 6742, 77, 2495, 30, 43} \begin {gather*} \frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*x^3 + 3*x^2*Log[2] + (9*x^3 + 9*x^2*Log[2])*Log[(x^2 + x*Log[2])/4])/(10*x + 10*Log[2]),x]

[Out]

(3*x^3*Log[(x*(x + Log[2]))/4])/10

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2495

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(h*(m + 1)), x] + (-Dist[(b*p*r)/(
h*(m + 1)), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[(d*q*r)/(h*(m + 1)), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 x^2 \left (2 x+\log (2)+3 (x+\log (2)) \log \left (\frac {1}{4} x (x+\log (2))\right )\right )}{10 (x+\log (2))} \, dx\\ &=\frac {3}{10} \int \frac {x^2 \left (2 x+\log (2)+3 (x+\log (2)) \log \left (\frac {1}{4} x (x+\log (2))\right )\right )}{x+\log (2)} \, dx\\ &=\frac {3}{10} \int \left (\frac {x^2 (2 x+\log (2))}{x+\log (2)}+3 x^2 \log \left (\frac {1}{4} x (x+\log (2))\right )\right ) \, dx\\ &=\frac {3}{10} \int \frac {x^2 (2 x+\log (2))}{x+\log (2)} \, dx+\frac {9}{10} \int x^2 \log \left (\frac {1}{4} x (x+\log (2))\right ) \, dx\\ &=\frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right )-\frac {3 \int x^2 \, dx}{10}-\frac {3}{10} \int \frac {x^3}{x+\log (2)} \, dx+\frac {3}{10} \int \left (2 x^2-x \log (2)+\log ^2(2)-\frac {\log ^3(2)}{x+\log (2)}\right ) \, dx\\ &=\frac {x^3}{10}-\frac {3}{20} x^2 \log (2)+\frac {3}{10} x \log ^2(2)-\frac {3}{10} \log ^3(2) \log (x+\log (2))+\frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right )-\frac {3}{10} \int \left (x^2-x \log (2)+\log ^2(2)-\frac {\log ^3(2)}{x+\log (2)}\right ) \, dx\\ &=\frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 17, normalized size = 1.00 \begin {gather*} \frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*x^3 + 3*x^2*Log[2] + (9*x^3 + 9*x^2*Log[2])*Log[(x^2 + x*Log[2])/4])/(10*x + 10*Log[2]),x]

[Out]

(3*x^3*Log[(x*(x + Log[2]))/4])/10

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fricas [A]  time = 0.80, size = 17, normalized size = 1.00 \begin {gather*} \frac {3}{10} \, x^{3} \log \left (\frac {1}{4} \, x^{2} + \frac {1}{4} \, x \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x^2*log(2)+9*x^3)*log(1/4*x*log(2)+1/4*x^2)+3*x^2*log(2)+6*x^3)/(10*log(2)+10*x),x, algorithm="f
ricas")

[Out]

3/10*x^3*log(1/4*x^2 + 1/4*x*log(2))

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giac [A]  time = 0.15, size = 22, normalized size = 1.29 \begin {gather*} -\frac {3}{5} \, x^{3} \log \relax (2) + \frac {3}{10} \, x^{3} \log \left (x^{2} + x \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x^2*log(2)+9*x^3)*log(1/4*x*log(2)+1/4*x^2)+3*x^2*log(2)+6*x^3)/(10*log(2)+10*x),x, algorithm="g
iac")

[Out]

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

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maple [A]  time = 0.12, size = 18, normalized size = 1.06

method	result	size
norman	\(\frac {3 x^{3} \ln \left (\frac {x \ln \relax (2)}{4}+\frac {x^{2}}{4}\right )}{10}\)	\(18\)
risch	\(\frac {3 x^{3} \ln \left (\frac {x \ln \relax (2)}{4}+\frac {x^{2}}{4}\right )}{10}\)	\(18\)
default	\(\frac {3 x^{3} \ln \left (x \ln \relax (2)+x^{2}\right )}{10}-\frac {3 x^{3} \ln \relax (2)}{5}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((9*x^2*ln(2)+9*x^3)*ln(1/4*x*ln(2)+1/4*x^2)+3*x^2*ln(2)+6*x^3)/(10*ln(2)+10*x),x,method=_RETURNVERBOSE)

[Out]

3/10*x^3*ln(1/4*x*ln(2)+1/4*x^2)

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maxima [B]  time = 0.49, size = 88, normalized size = 5.18 \begin {gather*} -\frac {1}{5} \, x^{3} {\left (3 \, \log \relax (2) + 1\right )} - \frac {3}{5} \, \log \relax (2)^{3} \log \left (x + \log \relax (2)\right ) + \frac {3}{10} \, x^{3} \log \relax (x) + \frac {1}{5} \, x^{3} - \frac {3}{20} \, x^{2} \log \relax (2) + \frac {3}{10} \, x \log \relax (2)^{2} + \frac {3}{20} \, {\left (2 \, \log \relax (2)^{2} \log \left (x + \log \relax (2)\right ) + x^{2} - 2 \, x \log \relax (2)\right )} \log \relax (2) + \frac {3}{10} \, {\left (x^{3} + \log \relax (2)^{3}\right )} \log \left (x + \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x^2*log(2)+9*x^3)*log(1/4*x*log(2)+1/4*x^2)+3*x^2*log(2)+6*x^3)/(10*log(2)+10*x),x, algorithm="m
axima")

[Out]

-1/5*x^3*(3*log(2) + 1) - 3/5*log(2)^3*log(x + log(2)) + 3/10*x^3*log(x) + 1/5*x^3 - 3/20*x^2*log(2) + 3/10*x*
log(2)^2 + 3/20*(2*log(2)^2*log(x + log(2)) + x^2 - 2*x*log(2))*log(2) + 3/10*(x^3 + log(2)^3)*log(x + log(2))

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mupad [B]  time = 3.63, size = 19, normalized size = 1.12 \begin {gather*} -\frac {3\,x^3\,\left (\ln \relax (4)-\ln \left (x^2+\ln \relax (2)\,x\right )\right )}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2*log(2) + 6*x^3 + log((x*log(2))/4 + x^2/4)*(9*x^2*log(2) + 9*x^3))/(10*x + 10*log(2)),x)

[Out]

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

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sympy [A]  time = 0.16, size = 19, normalized size = 1.12 \begin {gather*} \frac {3 x^{3} \log {\left (\frac {x^{2}}{4} + \frac {x \log {\relax (2 )}}{4} \right )}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x**2*ln(2)+9*x**3)*ln(1/4*x*ln(2)+1/4*x**2)+3*x**2*ln(2)+6*x**3)/(10*ln(2)+10*x),x)

[Out]

3*x**3*log(x**2/4 + x*log(2)/4)/10

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