\(\int \frac {1}{2} (e (-18+9 x)+3 \log (\frac {5}{4})+(e (-6+6 x)+\log (\frac {5}{4})) \log (x^3)) \, dx\) [8763]

Optimal result

Integrand size = 35, antiderivative size = 20 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=\frac {1}{2} x \left (3 e (-2+x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 2369, 2350} \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=\frac {3}{2} e x^2 \log \left (x^3\right )-\frac {1}{2} x \left (6 e-\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \]

[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*Log[x^3])/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 2350

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

Rule 2369

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{align*} \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 {3}{2} e x^2 \log \left (x^3\right )-\frac {1}{2} x \left (6 e-\log \left (\frac {5}{4}\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 {3}{2} e x^2 \log \left (x^3\right )-\frac {1}{2} x \left (6 e-\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=-3 e x \log \left (x^3\right )+\frac {3}{2} e x^2 \log \left (x^3\right )+\frac {1}{2} x \log \left (\frac {5}{4}\right ) \log \left (x^3\right ) \]

[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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55

method	result	size
parallelrisch	\(\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}-3 \,{\mathrm e} \ln \left (x^{3}\right ) x +\frac {\ln \left (x^{3}\right ) x \ln \left (\frac {5}{4}\right )}{2}\)	\(31\)
norman	\(\left (-3 \,{\mathrm e}+\frac {\ln \left (5\right )}{2}-\ln \left (2\right )\right ) x \ln \left (x^{3}\right )+\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}\)	\(32\)
risch	\(\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}+\frac {\ln \left (5\right ) \ln \left (x^{3}\right ) x}{2}-3 \,{\mathrm e} \ln \left (x^{3}\right ) x -\ln \left (2\right ) \ln \left (x^{3}\right ) x\)	\(40\)
parts	\(\frac {\ln \left (5\right ) \left (x \ln \left (x^{3}\right )-3 x \right )}{2}-3 \,{\mathrm e} \left (x \ln \left (x^{3}\right )-3 x \right )-\ln \left (2\right ) \left (x \ln \left (x^{3}\right )-3 x \right )+\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}-9 x \,{\mathrm e}+\frac {3 x \ln \left (\frac {5}{4}\right )}{2}\)	\(65\)
default	\(\frac {9 \,{\mathrm e} \left (\frac {1}{2} x^{2}-2 x \right )}{2}+\frac {\ln \left (5\right ) \left (x \ln \left (x^{3}\right )-3 x \right )}{2}-3 \,{\mathrm e} \left (x \ln \left (x^{3}\right )-3 x \right )-\ln \left (2\right ) \left (x \ln \left (x^{3}\right )-3 x \right )+\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}-\frac {9 x^{2} {\mathrm e}}{4}+\frac {3 x \ln \left (\frac {5}{4}\right )}{2}\)	\(80\)

[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]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=\frac {1}{2} \, {\left (3 \, {\left (x^{2} - 2 \, x\right )} e + x \log \left (\frac {5}{4}\right )\right )} \log \left (x^{3}\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=\left (\frac {3 e x^{2}}{2} - 3 e x - x \log {\left (2 \right )} + \frac {x \log {\left (5 \right )}}{2}\right ) \log {\left (x^{3} \right )} \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (17) = 34\).

Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.90 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=-\frac {9}{4} \, x^{2} e + \frac {3}{2} \, x {\left (6 \, e - \log \left (\frac {5}{4}\right )\right )} + \frac {9}{4} \, {\left (x^{2} - 4 \, x\right )} e + \frac {3}{2} \, x \log \left (\frac {5}{4}\right ) + \frac {1}{2} \, {\left (3 \, {\left (x^{2} - 2 \, x\right )} e + x \log \left (\frac {5}{4}\right )\right )} \log \left (x^{3}\right ) \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (17) = 34\).

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.30 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=-\frac {9}{4} \, x^{2} e + \frac {9}{4} \, {\left (x^{2} - 4 \, x\right )} e + 9 \, x e + \frac {1}{2} \, {\left (3 \, {\left (x^{2} - 2 \, x\right )} e + x \log \left (\frac {5}{4}\right )\right )} \log \left (x^{3}\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 13.76 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=\frac {x\,\ln \left (x^3\right )\,\left (\ln \left (\frac {5}{4}\right )-6\,\mathrm {e}+3\,x\,\mathrm {e}\right )}{2} \]

[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