3.88.63 \(\int \frac {1}{2} (e (-18+9 x)+3 \log (\frac {5}{4})+(e (-6+6 x)+\log (\frac {5}{4})) \log (x^3)) \, dx\)

Optimal. Leaf size=20 \[ \frac {1}{2} x \left (3 e (-2+x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.40, number of steps used = 5, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 2332, 2313} \begin {gather*} \frac {1}{2} \left (3 e x^2-x \left (6 e-\log \left (\frac {5}{4}\right )\right )\right ) \log \left (x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E*(-18 + 9*x) + 3*Log[5/4] + (E*(-6 + 6*x) + Log[5/4])*Log[x^3])/2,x]

[Out]

((3*E*x^2 - x*(6*E - Log[5/4]))*Log[x^3])/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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rule 2332

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*(a + b*Log[c*
x^n])^p, x] /; FreeQ[{a, b, c, n, p, q}, x] && BinomialQ[u, x] &&  !BinomialMatchQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx\\ &=\frac {9}{4} e (2-x)^2+\frac {3}{2} x \log \left (\frac {5}{4}\right )+\frac {1}{2} \int \left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \, dx\\ &=\frac {9}{4} e (2-x)^2+\frac {3}{2} x \log \left (\frac {5}{4}\right )+\frac {1}{2} \int \left (-6 e+6 e x+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \, dx\\ &=\frac {9}{4} e (2-x)^2+\frac {3}{2} x \log \left (\frac {5}{4}\right )+\frac {1}{2} \left (3 e x^2-x \left (6 e-\log \left (\frac {5}{4}\right )\right )\right ) \log \left (x^3\right )-\frac {3}{2} \int \left (3 e (-2+x)+\log \left (\frac {5}{4}\right )\right ) \, dx\\ &=\frac {1}{2} \left (3 e x^2-x \left (6 e-\log \left (\frac {5}{4}\right )\right )\right ) \log \left (x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.70 \begin {gather*} -3 e x \log \left (x^3\right )+\frac {3}{2} e x^2 \log \left (x^3\right )+\frac {1}{2} x \log \left (\frac {5}{4}\right ) \log \left (x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E*(-18 + 9*x) + 3*Log[5/4] + (E*(-6 + 6*x) + Log[5/4])*Log[x^3])/2,x]

[Out]

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

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fricas [A]  time = 0.71, size = 22, normalized size = 1.10 \begin {gather*} \frac {1}{2} \, {\left (3 \, {\left (x^{2} - 2 \, x\right )} e + x \log \left (\frac {5}{4}\right )\right )} \log \left (x^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(log(5/4)+(6*x-6)*exp(1))*log(x^3)+3/2*log(5/4)+1/2*(9*x-18)*exp(1),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.12, size = 46, normalized size = 2.30 \begin {gather*} -\frac {9}{4} \, x^{2} e + \frac {9}{4} \, {\left (x^{2} - 4 \, x\right )} e + 9 \, x e + \frac {1}{2} \, {\left (3 \, {\left (x^{2} - 2 \, x\right )} e + x \log \left (\frac {5}{4}\right )\right )} \log \left (x^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(log(5/4)+(6*x-6)*exp(1))*log(x^3)+3/2*log(5/4)+1/2*(9*x-18)*exp(1),x, algorithm="giac")

[Out]

-9/4*x^2*e + 9/4*(x^2 - 4*x)*e + 9*x*e + 1/2*(3*(x^2 - 2*x)*e + x*log(5/4))*log(x^3)

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maple [A]  time = 0.05, size = 32, normalized size = 1.60

method	result	size
norman	\(\left (-3 \,{\mathrm e}+\frac {\ln \relax (5)}{2}-\ln \relax (2)\right ) x \ln \left (x^{3}\right )+\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}\)	\(32\)
risch	\(\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}+\frac {\ln \relax (5) \ln \left (x^{3}\right ) x}{2}-3 \,{\mathrm e} \ln \left (x^{3}\right ) x -\ln \relax (2) \ln \left (x^{3}\right ) x\)	\(40\)
default	\(\frac {{\mathrm e} \left (\frac {9}{2} x^{2}-18 x \right )}{2}+\frac {\ln \relax (5) \ln \left (x^{3}\right ) x}{2}-\frac {3 x \ln \relax (5)}{2}-3 \,{\mathrm e} \ln \left (x^{3}\right ) x +9 x \,{\mathrm e}-\ln \relax (2) \ln \left (x^{3}\right ) x +3 x \ln \relax (2)+\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}-\frac {9 x^{2} {\mathrm e}}{4}+\frac {3 \ln \left (\frac {5}{4}\right ) x}{2}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(ln(5/4)+(6*x-6)*exp(1))*ln(x^3)+3/2*ln(5/4)+1/2*(9*x-18)*exp(1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.35, size = 58, normalized size = 2.90 \begin {gather*} -\frac {9}{4} \, x^{2} e + \frac {3}{2} \, x {\left (6 \, e - \log \left (\frac {5}{4}\right )\right )} + \frac {9}{4} \, {\left (x^{2} - 4 \, x\right )} e + \frac {3}{2} \, x \log \left (\frac {5}{4}\right ) + \frac {1}{2} \, {\left (3 \, {\left (x^{2} - 2 \, x\right )} e + x \log \left (\frac {5}{4}\right )\right )} \log \left (x^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(log(5/4)+(6*x-6)*exp(1))*log(x^3)+3/2*log(5/4)+1/2*(9*x-18)*exp(1),x, algorithm="maxima")

[Out]

-9/4*x^2*e + 3/2*x*(6*e - log(5/4)) + 9/4*(x^2 - 4*x)*e + 3/2*x*log(5/4) + 1/2*(3*(x^2 - 2*x)*e + x*log(5/4))*
log(x^3)

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mupad [B]  time = 5.59, size = 19, normalized size = 0.95 \begin {gather*} \frac {x\,\ln \left (x^3\right )\,\left (\ln \left (\frac {5}{4}\right )-6\,\mathrm {e}+3\,x\,\mathrm {e}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*log(5/4))/2 + (log(x^3)*(log(5/4) + exp(1)*(6*x - 6)))/2 + (exp(1)*(9*x - 18))/2,x)

[Out]

(x*log(x^3)*(log(5/4) - 6*exp(1) + 3*x*exp(1)))/2

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sympy [A]  time = 0.15, size = 32, normalized size = 1.60 \begin {gather*} \left (\frac {3 e x^{2}}{2} - 3 e x - x \log {\relax (2 )} + \frac {x \log {\relax (5 )}}{2}\right ) \log {\left (x^{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(ln(5/4)+(6*x-6)*exp(1))*ln(x**3)+3/2*ln(5/4)+1/2*(9*x-18)*exp(1),x)

[Out]

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

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