\(\int ((40-8 e^4-120 x+8 \log (5)) \log (x)+(20-4 e^4-120 x+4 \log (5)) \log ^2(x)) \, dx\) [7631]

Optimal result

Integrand size = 37, antiderivative size = 19 \[ \int \left (\left (40-8 e^4-120 x+8 \log (5)\right ) \log (x)+\left (20-4 e^4-120 x+4 \log (5)\right ) \log ^2(x)\right ) \, dx=4 x \left (5-e^4-15 x+\log (5)\right ) \log ^2(x) \]

[Out]

4*(ln(5)-15*x+5-exp(4))*x*ln(x)^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {2350, 2367, 2342, 2341, 2333, 2332} \[ \int \left (\left (40-8 e^4-120 x+8 \log (5)\right ) \log (x)+\left (20-4 e^4-120 x+4 \log (5)\right ) \log ^2(x)\right ) \, dx=4 x \left (5-e^4+\log (5)\right ) \log ^2(x)-60 x^2 \log ^2(x) \]

[In]

Int[(40 - 8*E^4 - 120*x + 8*Log[5])*Log[x] + (20 - 4*E^4 - 120*x + 4*Log[5])*Log[x]^2,x]

[Out]

-60*x^2*Log[x]^2 + 4*x*(5 - E^4 + Log[5])*Log[x]^2

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 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

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 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

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 2367

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

Rubi steps \begin{align*} \text {integral}& = \int \left (40-8 e^4-120 x+8 \log (5)\right ) \log (x) \, dx+\int \left (20-4 e^4-120 x+4 \log (5)\right ) \log ^2(x) \, dx \\ & = -60 x^2 \log (x)+8 x \left (5-e^4+\log (5)\right ) \log (x)-\int \left (-8 e^4-60 x+8 (5+\log (5))\right ) \, dx+\int \left (-120 x \log ^2(x)+20 \left (1+\frac {1}{5} \left (-e^4+\log (5)\right )\right ) \log ^2(x)\right ) \, dx \\ & = 30 x^2-8 x \left (5-e^4+\log (5)\right )-60 x^2 \log (x)+8 x \left (5-e^4+\log (5)\right ) \log (x)-120 \int x \log ^2(x) \, dx+\left (4 \left (5-e^4+\log (5)\right )\right ) \int \log ^2(x) \, dx \\ & = 30 x^2-8 x \left (5-e^4+\log (5)\right )-60 x^2 \log (x)+8 x \left (5-e^4+\log (5)\right ) \log (x)-60 x^2 \log ^2(x)+4 x \left (5-e^4+\log (5)\right ) \log ^2(x)+120 \int x \log (x) \, dx-\left (8 \left (5-e^4+\log (5)\right )\right ) \int \log (x) \, dx \\ & = -60 x^2 \log ^2(x)+4 x \left (5-e^4+\log (5)\right ) \log ^2(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89 \[ \int \left (\left (40-8 e^4-120 x+8 \log (5)\right ) \log (x)+\left (20-4 e^4-120 x+4 \log (5)\right ) \log ^2(x)\right ) \, dx=20 x \log ^2(x)-4 e^4 x \log ^2(x)-60 x^2 \log ^2(x)+4 x \log (5) \log ^2(x) \]

[In]

Integrate[(40 - 8*E^4 - 120*x + 8*Log[5])*Log[x] + (20 - 4*E^4 - 120*x + 4*Log[5])*Log[x]^2,x]

[Out]

20*x*Log[x]^2 - 4*E^4*x*Log[x]^2 - 60*x^2*Log[x]^2 + 4*x*Log[5]*Log[x]^2

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42

method	result	size
norman	\(\left (4 \ln \left (5\right )-4 \,{\mathrm e}^{4}+20\right ) x \ln \left (x \right )^{2}-60 x^{2} \ln \left (x \right )^{2}\)	\(27\)
risch	\(4 \ln \left (x \right )^{2} \ln \left (5\right ) x -4 x \,{\mathrm e}^{4} \ln \left (x \right )^{2}-60 x^{2} \ln \left (x \right )^{2}+20 x \ln \left (x \right )^{2}\)	\(36\)
parallelrisch	\(4 \ln \left (x \right )^{2} \ln \left (5\right ) x -4 x \,{\mathrm e}^{4} \ln \left (x \right )^{2}-60 x^{2} \ln \left (x \right )^{2}+20 x \ln \left (x \right )^{2}\)	\(36\)
default	\(4 \ln \left (5\right ) \left (x \ln \left (x \right )^{2}-2 x \ln \left (x \right )+2 x \right )-4 \,{\mathrm e}^{4} \left (x \ln \left (x \right )^{2}-2 x \ln \left (x \right )+2 x \right )-60 x^{2} \ln \left (x \right )^{2}+20 x \ln \left (x \right )^{2}+8 \ln \left (5\right ) \left (x \ln \left (x \right )-x \right )-8 \,{\mathrm e}^{4} \left (x \ln \left (x \right )-x \right )\)	\(80\)
parts	\(4 \ln \left (5\right ) \left (x \ln \left (x \right )^{2}-2 x \ln \left (x \right )+2 x \right )-4 \,{\mathrm e}^{4} \left (x \ln \left (x \right )^{2}-2 x \ln \left (x \right )+2 x \right )-60 x^{2} \ln \left (x \right )^{2}+20 x \ln \left (x \right )^{2}+8 \ln \left (5\right ) \left (x \ln \left (x \right )-x \right )-8 \,{\mathrm e}^{4} \left (x \ln \left (x \right )-x \right )\)	\(80\)

[In]

int((4*ln(5)-4*exp(4)-120*x+20)*ln(x)^2+(8*ln(5)-8*exp(4)-120*x+40)*ln(x),x,method=_RETURNVERBOSE)

[Out]

(4*ln(5)-4*exp(4)+20)*x*ln(x)^2-60*x^2*ln(x)^2

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \left (\left (40-8 e^4-120 x+8 \log (5)\right ) \log (x)+\left (20-4 e^4-120 x+4 \log (5)\right ) \log ^2(x)\right ) \, dx=-4 \, {\left (15 \, x^{2} + x e^{4} - x \log \left (5\right ) - 5 \, x\right )} \log \left (x\right )^{2} \]

[In]

integrate((4*log(5)-4*exp(4)-120*x+20)*log(x)^2+(8*log(5)-8*exp(4)-120*x+40)*log(x),x, algorithm="fricas")

[Out]

-4*(15*x^2 + x*e^4 - x*log(5) - 5*x)*log(x)^2

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \left (\left (40-8 e^4-120 x+8 \log (5)\right ) \log (x)+\left (20-4 e^4-120 x+4 \log (5)\right ) \log ^2(x)\right ) \, dx=\left (- 60 x^{2} - 4 x e^{4} + 4 x \log {\left (5 \right )} + 20 x\right ) \log {\left (x \right )}^{2} \]

[In]

integrate((4*ln(5)-4*exp(4)-120*x+20)*ln(x)**2+(8*ln(5)-8*exp(4)-120*x+40)*ln(x),x)

[Out]

(-60*x**2 - 4*x*exp(4) + 4*x*log(5) + 20*x)*log(x)**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (18) = 36\).

Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 5.00 \[ \int \left (\left (40-8 e^4-120 x+8 \log (5)\right ) \log (x)+\left (20-4 e^4-120 x+4 \log (5)\right ) \log ^2(x)\right ) \, dx=-30 \, {\left (2 \, \log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 1\right )} x^{2} - 4 \, {\left ({\left (e^{4} - \log \left (5\right ) - 5\right )} \log \left (x\right )^{2} - 2 \, {\left (e^{4} - \log \left (5\right ) - 5\right )} \log \left (x\right ) + 2 \, e^{4} - 2 \, \log \left (5\right ) - 10\right )} x + 30 \, x^{2} + 8 \, x {\left (e^{4} - \log \left (5\right ) - 5\right )} - 4 \, {\left (15 \, x^{2} + 2 \, x e^{4} - 2 \, x \log \left (5\right ) - 10 \, x\right )} \log \left (x\right ) \]

[In]

integrate((4*log(5)-4*exp(4)-120*x+20)*log(x)^2+(8*log(5)-8*exp(4)-120*x+40)*log(x),x, algorithm="maxima")

[Out]

-30*(2*log(x)^2 - 2*log(x) + 1)*x^2 - 4*((e^4 - log(5) - 5)*log(x)^2 - 2*(e^4 - log(5) - 5)*log(x) + 2*e^4 - 2
*log(5) - 10)*x + 30*x^2 + 8*x*(e^4 - log(5) - 5) - 4*(15*x^2 + 2*x*e^4 - 2*x*log(5) - 10*x)*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84 \[ \int \left (\left (40-8 e^4-120 x+8 \log (5)\right ) \log (x)+\left (20-4 e^4-120 x+4 \log (5)\right ) \log ^2(x)\right ) \, dx=-60 \, x^{2} \log \left (x\right )^{2} - 4 \, x e^{4} \log \left (x\right )^{2} + 4 \, x \log \left (5\right ) \log \left (x\right )^{2} + 20 \, x \log \left (x\right )^{2} \]

[In]

integrate((4*log(5)-4*exp(4)-120*x+20)*log(x)^2+(8*log(5)-8*exp(4)-120*x+40)*log(x),x, algorithm="giac")

[Out]

-60*x^2*log(x)^2 - 4*x*e^4*log(x)^2 + 4*x*log(5)*log(x)^2 + 20*x*log(x)^2

Mupad [B] (verification not implemented)

Time = 12.97 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \left (\left (40-8 e^4-120 x+8 \log (5)\right ) \log (x)+\left (20-4 e^4-120 x+4 \log (5)\right ) \log ^2(x)\right ) \, dx=-x\,{\ln \left (x\right )}^2\,\left (60\,x+4\,{\mathrm {e}}^4-\ln \left (625\right )-20\right ) \]

[In]

int(- log(x)*(120*x + 8*exp(4) - 8*log(5) - 40) - log(x)^2*(120*x + 4*exp(4) - 4*log(5) - 20),x)

[Out]

-x*log(x)^2*(60*x + 4*exp(4) - log(625) - 20)