\(\int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} (\frac {3 (-4 x-x^2)}{e}+\frac {12 x \log (x)}{e}+(-32-16 x-2 x^2) \log ^3(x))}{(16 x+8 x^2+x^3) \log ^2(x)} \, dx\) [670]

Optimal result

Integrand size = 94, antiderivative size = 25 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{25+\frac {3 x}{e (4+x) \log (x)}-\log ^2(x)} \]

[Out]

exp(exp(ln(3)-1)/ln(x)/(4+x)*x-ln(x)^2+25)

Rubi [A] (verified)

Time = 2.60 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1608, 27, 6820, 6838} \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=\exp \left (-\log ^2(x)+\frac {3 x}{e x \log (x)+4 e \log (x)}+25\right ) \]

[In]

Int[(E^(((3*x)/E + (100 + 25*x)*Log[x] + (-4 - x)*Log[x]^3)/((4 + x)*Log[x]))*((3*(-4*x - x^2))/E + (12*x*Log[
x])/E + (-32 - 16*x - 2*x^2)*Log[x]^3))/((16*x + 8*x^2 + x^3)*Log[x]^2),x]

[Out]

E^(25 - Log[x]^2 + (3*x)/(4*E*Log[x] + E*x*Log[x]))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 6820

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}\right ) \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{x \left (16+8 x+x^2\right ) \log ^2(x)} \, dx \\ & = \int \frac {\exp \left (\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}\right ) \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{x (4+x)^2 \log ^2(x)} \, dx \\ & = \int \frac {e^{24-\log ^2(x)+\frac {3 x}{4 e \log (x)+e x \log (x)}} \left (-3 x (4+x)+12 x \log (x)-2 e (4+x)^2 \log ^3(x)\right )}{x (4+x)^2 \log ^2(x)} \, dx \\ & = e^{25-\log ^2(x)+\frac {3 x}{4 e \log (x)+e x \log (x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{25+\frac {3 x}{e (4+x) \log (x)}-\log ^2(x)} \]

[In]

Integrate[(E^(((3*x)/E + (100 + 25*x)*Log[x] + (-4 - x)*Log[x]^3)/((4 + x)*Log[x]))*((3*(-4*x - x^2))/E + (12*
x*Log[x])/E + (-32 - 16*x - 2*x^2)*Log[x]^3))/((16*x + 8*x^2 + x^3)*Log[x]^2),x]

[Out]

E^(25 + (3*x)/(E*(4 + x)*Log[x]) - Log[x]^2)

Maple [A] (verified)

Time = 3.86 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\left (-4-x \right ) \ln \left (x \right )^{3}+\left (25 x +100\right ) \ln \left (x \right )+x \,{\mathrm e}^{\ln \left (3\right )-1}}{\left (4+x \right ) \ln \left (x \right )}}\)	\(38\)
risch	\({\mathrm e}^{\frac {-x \ln \left (x \right )^{3}-4 \ln \left (x \right )^{3}+25 x \ln \left (x \right )+3 \,{\mathrm e}^{-1} x +100 \ln \left (x \right )}{\left (4+x \right ) \ln \left (x \right )}}\)	\(40\)

[In]

int(((-2*x^2-16*x-32)*ln(x)^3+4*x*exp(ln(3)-1)*ln(x)+(-x^2-4*x)*exp(ln(3)-1))*exp(((-4-x)*ln(x)^3+(25*x+100)*l
n(x)+x*exp(ln(3)-1))/(4+x)/ln(x))/(x^3+8*x^2+16*x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

exp(((-4-x)*ln(x)^3+(25*x+100)*ln(x)+x*exp(ln(3)-1))/(4+x)/ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{\left (-\frac {{\left (x + 4\right )} \log \left (x\right )^{3} - x e^{\left (\log \left (3\right ) - 1\right )} - 25 \, {\left (x + 4\right )} \log \left (x\right )}{{\left (x + 4\right )} \log \left (x\right )}\right )} \]

[In]

integrate(((-2*x^2-16*x-32)*log(x)^3+4*x*exp(log(3)-1)*log(x)+(-x^2-4*x)*exp(log(3)-1))*exp(((-4-x)*log(x)^3+(
25*x+100)*log(x)+x*exp(log(3)-1))/(4+x)/log(x))/(x^3+8*x^2+16*x)/log(x)^2,x, algorithm="fricas")

[Out]

e^(-((x + 4)*log(x)^3 - x*e^(log(3) - 1) - 25*(x + 4)*log(x))/((x + 4)*log(x)))

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{\frac {\frac {3 x}{e} + \left (- x - 4\right ) \log {\left (x \right )}^{3} + \left (25 x + 100\right ) \log {\left (x \right )}}{\left (x + 4\right ) \log {\left (x \right )}}} \]

[In]

integrate(((-2*x**2-16*x-32)*ln(x)**3+4*x*exp(ln(3)-1)*ln(x)+(-x**2-4*x)*exp(ln(3)-1))*exp(((-4-x)*ln(x)**3+(2
5*x+100)*ln(x)+x*exp(ln(3)-1))/(4+x)/ln(x))/(x**3+8*x**2+16*x)/ln(x)**2,x)

[Out]

exp((3*x*exp(-1) + (-x - 4)*log(x)**3 + (25*x + 100)*log(x))/((x + 4)*log(x)))

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{\left (-\log \left (x\right )^{2} + \frac {3 \, e^{\left (-1\right )}}{\log \left (x\right )} - \frac {12}{{\left (x e + 4 \, e\right )} \log \left (x\right )} + 25\right )} \]

[In]

integrate(((-2*x^2-16*x-32)*log(x)^3+4*x*exp(log(3)-1)*log(x)+(-x^2-4*x)*exp(log(3)-1))*exp(((-4-x)*log(x)^3+(
25*x+100)*log(x)+x*exp(log(3)-1))/(4+x)/log(x))/(x^3+8*x^2+16*x)/log(x)^2,x, algorithm="maxima")

[Out]

e^(-log(x)^2 + 3*e^(-1)/log(x) - 12/((x*e + 4*e)*log(x)) + 25)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (25) = 50\).

Time = 0.51 (sec) , antiderivative size = 86, normalized size of antiderivative = 3.44 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{\left (-\frac {x \log \left (x\right )^{3}}{x \log \left (x\right ) + 4 \, \log \left (x\right )} - \frac {4 \, \log \left (x\right )^{3}}{x \log \left (x\right ) + 4 \, \log \left (x\right )} + \frac {x e^{\left (\log \left (3\right ) - 1\right )}}{x \log \left (x\right ) + 4 \, \log \left (x\right )} + \frac {25 \, x \log \left (x\right )}{x \log \left (x\right ) + 4 \, \log \left (x\right )} + \frac {100 \, \log \left (x\right )}{x \log \left (x\right ) + 4 \, \log \left (x\right )}\right )} \]

[In]

integrate(((-2*x^2-16*x-32)*log(x)^3+4*x*exp(log(3)-1)*log(x)+(-x^2-4*x)*exp(log(3)-1))*exp(((-4-x)*log(x)^3+(
25*x+100)*log(x)+x*exp(log(3)-1))/(4+x)/log(x))/(x^3+8*x^2+16*x)/log(x)^2,x, algorithm="giac")

[Out]

e^(-x*log(x)^3/(x*log(x) + 4*log(x)) - 4*log(x)^3/(x*log(x) + 4*log(x)) + x*e^(log(3) - 1)/(x*log(x) + 4*log(x
)) + 25*x*log(x)/(x*log(x) + 4*log(x)) + 100*log(x)/(x*log(x) + 4*log(x)))

Mupad [B] (verification not implemented)

Time = 8.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=x^{\frac {25}{\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {4\,{\ln \left (x\right )}^2}{x+4}}\,{\mathrm {e}}^{-\frac {x\,{\ln \left (x\right )}^2}{x+4}}\,{\mathrm {e}}^{\frac {3\,x}{4\,\mathrm {e}\,\ln \left (x\right )+x\,\mathrm {e}\,\ln \left (x\right )}} \]

[In]

int(-(exp((x*exp(log(3) - 1) + log(x)*(25*x + 100) - log(x)^3*(x + 4))/(log(x)*(x + 4)))*(log(x)^3*(16*x + 2*x
^2 + 32) + exp(log(3) - 1)*(4*x + x^2) - 4*x*exp(log(3) - 1)*log(x)))/(log(x)^2*(16*x + 8*x^2 + x^3)),x)

[Out]

x^(25/log(x))*exp(-(4*log(x)^2)/(x + 4))*exp(-(x*log(x)^2)/(x + 4))*exp((3*x)/(4*exp(1)*log(x) + x*exp(1)*log(
x)))