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

Optimal. Leaf size=25 \[ e^{25+\frac {3 x}{e (4+x) \log (x)}-\log ^2(x)} \]

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Rubi [A]  time = 3.93, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1594, 27, 6688, 6706} \begin {gather*} \exp \left (-\log ^2(x)+\frac {3 x}{e x \log (x)+4 e \log (x)}+25\right ) \end {gather*}

Antiderivative was successfully verified.

[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 1594

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 6688

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

Rule 6706

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 25, normalized size = 1.00 \begin {gather*} e^{25+\frac {3 x}{e (4+x) \log (x)}-\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 1.02, size = 36, normalized size = 1.44 \begin {gather*} e^{\left (-\frac {{\left (x + 4\right )} \log \relax (x)^{3} - x e^{\left (\log \relax (3) - 1\right )} - 25 \, {\left (x + 4\right )} \log \relax (x)}{{\left (x + 4\right )} \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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(((-x-4)*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)))

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giac [B]  time = 0.95, size = 86, normalized size = 3.44 \begin {gather*} e^{\left (-\frac {x \log \relax (x)^{3}}{x \log \relax (x) + 4 \, \log \relax (x)} - \frac {4 \, \log \relax (x)^{3}}{x \log \relax (x) + 4 \, \log \relax (x)} + \frac {x e^{\left (\log \relax (3) - 1\right )}}{x \log \relax (x) + 4 \, \log \relax (x)} + \frac {25 \, x \log \relax (x)}{x \log \relax (x) + 4 \, \log \relax (x)} + \frac {100 \, \log \relax (x)}{x \log \relax (x) + 4 \, \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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(((-x-4)*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)))

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maple [A]  time = 0.04, size = 40, normalized size = 1.60

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

Verification of antiderivative is not currently implemented for this CAS.

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

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maxima [A]  time = 0.90, size = 34, normalized size = 1.36 \begin {gather*} e^{\left (-\log \relax (x)^{2} + \frac {3 \, e^{\left (-1\right )}}{\log \relax (x)} - \frac {12}{{\left (x e + 4 \, e\right )} \log \relax (x)} + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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(((-x-4)*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)

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mupad [B]  time = 0.92, size = 53, normalized size = 2.12 \begin {gather*} x^{\frac {25}{\ln \relax (x)}}\,{\mathrm {e}}^{-\frac {4\,{\ln \relax (x)}^2}{x+4}}\,{\mathrm {e}}^{-\frac {x\,{\ln \relax (x)}^2}{x+4}}\,{\mathrm {e}}^{\frac {3\,x}{4\,\mathrm {e}\,\ln \relax (x)+x\,\mathrm {e}\,\ln \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 0.98, size = 32, normalized size = 1.28 \begin {gather*} e^{\frac {\frac {3 x}{e} + \left (- x - 4\right ) \log {\relax (x )}^{3} + \left (25 x + 100\right ) \log {\relax (x )}}{\left (x + 4\right ) \log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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(((-x-4)*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)))

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