3.54.89 \(\int \frac {28 x^2-144 x^3+e^{20/x} (-60-20 x-10 x^2)+(-2 x^2-e^{20/x} x^2-16 x^3) \log (2+e^{20/x}+16 x)}{18 x^2+156 x^3+98 x^4+16 x^5+e^{20/x} (9 x^2+6 x^3+x^4)} \, dx\)

Optimal. Leaf size=21 \[ \frac {10+\log \left (2+e^{20/x}+16 x\right )}{3+x} \]

________________________________________________________________________________________

Rubi [A]  time = 2.79, antiderivative size = 27, normalized size of antiderivative = 1.29, number of steps used = 16, number of rules used = 5, integrand size = 112, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {6688, 6742, 893, 2551, 44} \begin {gather*} \frac {10}{x+3}+\frac {\log \left (16 x+e^{20/x}+2\right )}{x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(28*x^2 - 144*x^3 + E^(20/x)*(-60 - 20*x - 10*x^2) + (-2*x^2 - E^(20/x)*x^2 - 16*x^3)*Log[2 + E^(20/x) + 1
6*x])/(18*x^2 + 156*x^3 + 98*x^4 + 16*x^5 + E^(20/x)*(9*x^2 + 6*x^3 + x^4)),x]

[Out]

10/(3 + x) + Log[2 + E^(20/x) + 16*x]/(3 + x)

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4 x^2 (-7+36 x)-10 e^{20/x} \left (6+2 x+x^2\right )-x^2 \left (2+e^{20/x}+16 x\right ) \log \left (2+e^{20/x}+16 x\right )}{x^2 (3+x)^2 \left (2+e^{20/x}+16 x\right )} \, dx\\ &=\int \left (\frac {8 \left (5+40 x+2 x^2\right )}{x^2 (3+x) \left (2+e^{20/x}+16 x\right )}+\frac {-60-20 x-10 x^2-x^2 \log \left (2+e^{20/x}+16 x\right )}{x^2 (3+x)^2}\right ) \, dx\\ &=8 \int \frac {5+40 x+2 x^2}{x^2 (3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\int \frac {-60-20 x-10 x^2-x^2 \log \left (2+e^{20/x}+16 x\right )}{x^2 (3+x)^2} \, dx\\ &=8 \int \left (\frac {5}{3 x^2 \left (2+e^{20/x}+16 x\right )}+\frac {115}{9 x \left (2+e^{20/x}+16 x\right )}-\frac {97}{9 (3+x) \left (2+e^{20/x}+16 x\right )}\right ) \, dx+\int \left (-\frac {10 \left (6+2 x+x^2\right )}{x^2 (3+x)^2}-\frac {\log \left (2+e^{20/x}+16 x\right )}{(3+x)^2}\right ) \, dx\\ &=-\left (10 \int \frac {6+2 x+x^2}{x^2 (3+x)^2} \, dx\right )+\frac {40}{3} \int \frac {1}{x^2 \left (2+e^{20/x}+16 x\right )} \, dx-\frac {776}{9} \int \frac {1}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {920}{9} \int \frac {1}{x \left (2+e^{20/x}+16 x\right )} \, dx-\int \frac {\log \left (2+e^{20/x}+16 x\right )}{(3+x)^2} \, dx\\ &=\frac {\log \left (2+e^{20/x}+16 x\right )}{3+x}-10 \int \left (\frac {2}{3 x^2}-\frac {2}{9 x}+\frac {1}{(3+x)^2}+\frac {2}{9 (3+x)}\right ) \, dx+\frac {40}{3} \int \frac {1}{x^2 \left (2+e^{20/x}+16 x\right )} \, dx-\frac {776}{9} \int \frac {1}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {920}{9} \int \frac {1}{x \left (2+e^{20/x}+16 x\right )} \, dx-\int \frac {16-\frac {20 e^{20/x}}{x^2}}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx\\ &=\frac {20}{3 x}+\frac {10}{3+x}+\frac {20 \log (x)}{9}-\frac {20}{9} \log (3+x)+\frac {\log \left (2+e^{20/x}+16 x\right )}{3+x}+\frac {40}{3} \int \frac {1}{x^2 \left (2+e^{20/x}+16 x\right )} \, dx-\frac {776}{9} \int \frac {1}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {920}{9} \int \frac {1}{x \left (2+e^{20/x}+16 x\right )} \, dx-\int \left (-\frac {20}{x^2 (3+x)}+\frac {8 \left (5+40 x+2 x^2\right )}{x^2 (3+x) \left (2+e^{20/x}+16 x\right )}\right ) \, dx\\ &=\frac {20}{3 x}+\frac {10}{3+x}+\frac {20 \log (x)}{9}-\frac {20}{9} \log (3+x)+\frac {\log \left (2+e^{20/x}+16 x\right )}{3+x}-8 \int \frac {5+40 x+2 x^2}{x^2 (3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {40}{3} \int \frac {1}{x^2 \left (2+e^{20/x}+16 x\right )} \, dx+20 \int \frac {1}{x^2 (3+x)} \, dx-\frac {776}{9} \int \frac {1}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {920}{9} \int \frac {1}{x \left (2+e^{20/x}+16 x\right )} \, dx\\ &=\frac {20}{3 x}+\frac {10}{3+x}+\frac {20 \log (x)}{9}-\frac {20}{9} \log (3+x)+\frac {\log \left (2+e^{20/x}+16 x\right )}{3+x}-8 \int \left (\frac {5}{3 x^2 \left (2+e^{20/x}+16 x\right )}+\frac {115}{9 x \left (2+e^{20/x}+16 x\right )}-\frac {97}{9 (3+x) \left (2+e^{20/x}+16 x\right )}\right ) \, dx+\frac {40}{3} \int \frac {1}{x^2 \left (2+e^{20/x}+16 x\right )} \, dx+20 \int \left (\frac {1}{3 x^2}-\frac {1}{9 x}+\frac {1}{9 (3+x)}\right ) \, dx-\frac {776}{9} \int \frac {1}{(3+x) \left (2+e^{20/x}+16 x\right )} \, dx+\frac {920}{9} \int \frac {1}{x \left (2+e^{20/x}+16 x\right )} \, dx\\ &=\frac {10}{3+x}+\frac {\log \left (2+e^{20/x}+16 x\right )}{3+x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 21, normalized size = 1.00 \begin {gather*} \frac {10+\log \left (2+e^{20/x}+16 x\right )}{3+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(28*x^2 - 144*x^3 + E^(20/x)*(-60 - 20*x - 10*x^2) + (-2*x^2 - E^(20/x)*x^2 - 16*x^3)*Log[2 + E^(20/
x) + 16*x])/(18*x^2 + 156*x^3 + 98*x^4 + 16*x^5 + E^(20/x)*(9*x^2 + 6*x^3 + x^4)),x]

[Out]

(10 + Log[2 + E^(20/x) + 16*x])/(3 + x)

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 20, normalized size = 0.95 \begin {gather*} \frac {\log \left (16 \, x + e^{\frac {20}{x}} + 2\right ) + 10}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*exp(20/x)-16*x^3-2*x^2)*log(exp(20/x)+16*x+2)+(-10*x^2-20*x-60)*exp(20/x)-144*x^3+28*x^2)/((x
^4+6*x^3+9*x^2)*exp(20/x)+16*x^5+98*x^4+156*x^3+18*x^2),x, algorithm="fricas")

[Out]

(log(16*x + e^(20/x) + 2) + 10)/(x + 3)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 20, normalized size = 0.95 \begin {gather*} \frac {\log \left (16 \, x + e^{\frac {20}{x}} + 2\right ) + 10}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*exp(20/x)-16*x^3-2*x^2)*log(exp(20/x)+16*x+2)+(-10*x^2-20*x-60)*exp(20/x)-144*x^3+28*x^2)/((x
^4+6*x^3+9*x^2)*exp(20/x)+16*x^5+98*x^4+156*x^3+18*x^2),x, algorithm="giac")

[Out]

(log(16*x + e^(20/x) + 2) + 10)/(x + 3)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 27, normalized size = 1.29

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}^{\frac {20}{x}}+16 x +2\right )}{3+x}+\frac {10}{3+x}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2*exp(20/x)-16*x^3-2*x^2)*ln(exp(20/x)+16*x+2)+(-10*x^2-20*x-60)*exp(20/x)-144*x^3+28*x^2)/((x^4+6*x^
3+9*x^2)*exp(20/x)+16*x^5+98*x^4+156*x^3+18*x^2),x,method=_RETURNVERBOSE)

[Out]

1/(3+x)*ln(exp(20/x)+16*x+2)+10/(3+x)

________________________________________________________________________________________

maxima [A]  time = 0.39, size = 20, normalized size = 0.95 \begin {gather*} \frac {\log \left (16 \, x + e^{\frac {20}{x}} + 2\right ) + 10}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*exp(20/x)-16*x^3-2*x^2)*log(exp(20/x)+16*x+2)+(-10*x^2-20*x-60)*exp(20/x)-144*x^3+28*x^2)/((x
^4+6*x^3+9*x^2)*exp(20/x)+16*x^5+98*x^4+156*x^3+18*x^2),x, algorithm="maxima")

[Out]

(log(16*x + e^(20/x) + 2) + 10)/(x + 3)

________________________________________________________________________________________

mupad [B]  time = 3.87, size = 20, normalized size = 0.95 \begin {gather*} \frac {\ln \left (16\,x+{\mathrm {e}}^{20/x}+2\right )+10}{x+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(16*x + exp(20/x) + 2)*(x^2*exp(20/x) + 2*x^2 + 16*x^3) + exp(20/x)*(20*x + 10*x^2 + 60) - 28*x^2 + 1
44*x^3)/(exp(20/x)*(9*x^2 + 6*x^3 + x^4) + 18*x^2 + 156*x^3 + 98*x^4 + 16*x^5),x)

[Out]

(log(16*x + exp(20/x) + 2) + 10)/(x + 3)

________________________________________________________________________________________

sympy [A]  time = 0.44, size = 19, normalized size = 0.90 \begin {gather*} \frac {\log {\left (16 x + e^{\frac {20}{x}} + 2 \right )}}{x + 3} + \frac {10}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2*exp(20/x)-16*x**3-2*x**2)*ln(exp(20/x)+16*x+2)+(-10*x**2-20*x-60)*exp(20/x)-144*x**3+28*x**2
)/((x**4+6*x**3+9*x**2)*exp(20/x)+16*x**5+98*x**4+156*x**3+18*x**2),x)

[Out]

log(16*x + exp(20/x) + 2)/(x + 3) + 10/(x + 3)

________________________________________________________________________________________