3.2.10 \(\int \frac {-1062882+1023516 x+38880 x^2+484 x^3+2 x^4+e^{2 x} (-324+316 x+4 x^2)+(2125764+78732 x+972 x^2+4 x^3) \log (x)}{531441 x^3+19683 x^4+243 x^5+x^6} \, dx\)

Optimal. Leaf size=24 \[ \frac {2 \left (-x+\frac {e^{2 x}}{(81+x)^2}-\log (x)\right )}{x^2} \]

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Rubi [B]  time = 0.64, antiderivative size = 76, normalized size of antiderivative = 3.17, number of steps used = 22, number of rules used = 8, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6688, 12, 14, 6742, 2177, 2178, 37, 2304} \begin {gather*} \frac {(1-x)^2}{x^2}+\frac {2 e^{2 x}}{6561 x^2}-\frac {1}{x^2}-\frac {2 \log (x)}{x^2}-\frac {4 e^{2 x}}{531441 x}+\frac {4 e^{2 x}}{531441 (x+81)}+\frac {2 e^{2 x}}{6561 (x+81)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1062882 + 1023516*x + 38880*x^2 + 484*x^3 + 2*x^4 + E^(2*x)*(-324 + 316*x + 4*x^2) + (2125764 + 78732*x
+ 972*x^2 + 4*x^3)*Log[x])/(531441*x^3 + 19683*x^4 + 243*x^5 + x^6),x]

[Out]

-x^(-2) + (2*E^(2*x))/(6561*x^2) + (1 - x)^2/x^2 - (4*E^(2*x))/(531441*x) + (2*E^(2*x))/(6561*(81 + x)^2) + (4
*E^(2*x))/(531441*(81 + x)) - (2*Log[x])/x^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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2304

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 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 {2 \left (-1+x+\frac {2 e^{2 x} \left (-81+79 x+x^2\right )}{(81+x)^3}+2 \log (x)\right )}{x^3} \, dx\\ &=2 \int \frac {-1+x+\frac {2 e^{2 x} \left (-81+79 x+x^2\right )}{(81+x)^3}+2 \log (x)}{x^3} \, dx\\ &=2 \int \left (\frac {2 e^{2 x} \left (-81+79 x+x^2\right )}{x^3 (81+x)^3}+\frac {-1+x+2 \log (x)}{x^3}\right ) \, dx\\ &=2 \int \frac {-1+x+2 \log (x)}{x^3} \, dx+4 \int \frac {e^{2 x} \left (-81+79 x+x^2\right )}{x^3 (81+x)^3} \, dx\\ &=2 \int \left (\frac {-1+x}{x^3}+\frac {2 \log (x)}{x^3}\right ) \, dx+4 \int \left (-\frac {e^{2 x}}{6561 x^3}+\frac {82 e^{2 x}}{531441 x^2}-\frac {2 e^{2 x}}{531441 x}-\frac {e^{2 x}}{6561 (81+x)^3}+\frac {80 e^{2 x}}{531441 (81+x)^2}+\frac {2 e^{2 x}}{531441 (81+x)}\right ) \, dx\\ &=-\frac {8 \int \frac {e^{2 x}}{x} \, dx}{531441}+\frac {8 \int \frac {e^{2 x}}{81+x} \, dx}{531441}+\frac {320 \int \frac {e^{2 x}}{(81+x)^2} \, dx}{531441}-\frac {4 \int \frac {e^{2 x}}{x^3} \, dx}{6561}-\frac {4 \int \frac {e^{2 x}}{(81+x)^3} \, dx}{6561}+\frac {328 \int \frac {e^{2 x}}{x^2} \, dx}{531441}+2 \int \frac {-1+x}{x^3} \, dx+4 \int \frac {\log (x)}{x^3} \, dx\\ &=-\frac {1}{x^2}+\frac {2 e^{2 x}}{6561 x^2}+\frac {(1-x)^2}{x^2}-\frac {328 e^{2 x}}{531441 x}+\frac {2 e^{2 x}}{6561 (81+x)^2}-\frac {320 e^{2 x}}{531441 (81+x)}-\frac {8 \text {Ei}(2 x)}{531441}+\frac {8 \text {Ei}(2 (81+x))}{531441 e^{162}}-\frac {2 \log (x)}{x^2}-\frac {4 \int \frac {e^{2 x}}{x^2} \, dx}{6561}-\frac {4 \int \frac {e^{2 x}}{(81+x)^2} \, dx}{6561}+\frac {640 \int \frac {e^{2 x}}{81+x} \, dx}{531441}+\frac {656 \int \frac {e^{2 x}}{x} \, dx}{531441}\\ &=-\frac {1}{x^2}+\frac {2 e^{2 x}}{6561 x^2}+\frac {(1-x)^2}{x^2}-\frac {4 e^{2 x}}{531441 x}+\frac {2 e^{2 x}}{6561 (81+x)^2}+\frac {4 e^{2 x}}{531441 (81+x)}+\frac {8 \text {Ei}(2 x)}{6561}+\frac {8 \text {Ei}(2 (81+x))}{6561 e^{162}}-\frac {2 \log (x)}{x^2}-\frac {8 \int \frac {e^{2 x}}{x} \, dx}{6561}-\frac {8 \int \frac {e^{2 x}}{81+x} \, dx}{6561}\\ &=-\frac {1}{x^2}+\frac {2 e^{2 x}}{6561 x^2}+\frac {(1-x)^2}{x^2}-\frac {4 e^{2 x}}{531441 x}+\frac {2 e^{2 x}}{6561 (81+x)^2}+\frac {4 e^{2 x}}{531441 (81+x)}-\frac {2 \log (x)}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 33, normalized size = 1.38 \begin {gather*} \frac {2 \left (e^{2 x}-x (81+x)^2-(81+x)^2 \log (x)\right )}{x^2 (81+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1062882 + 1023516*x + 38880*x^2 + 484*x^3 + 2*x^4 + E^(2*x)*(-324 + 316*x + 4*x^2) + (2125764 + 78
732*x + 972*x^2 + 4*x^3)*Log[x])/(531441*x^3 + 19683*x^4 + 243*x^5 + x^6),x]

[Out]

(2*(E^(2*x) - x*(81 + x)^2 - (81 + x)^2*Log[x]))/(x^2*(81 + x)^2)

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fricas [B]  time = 0.54, size = 47, normalized size = 1.96 \begin {gather*} -\frac {2 \, {\left (x^{3} + 162 \, x^{2} + {\left (x^{2} + 162 \, x + 6561\right )} \log \relax (x) + 6561 \, x - e^{\left (2 \, x\right )}\right )}}{x^{4} + 162 \, x^{3} + 6561 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+972*x^2+78732*x+2125764)*log(x)+(4*x^2+316*x-324)*exp(2*x)+2*x^4+484*x^3+38880*x^2+1023516*x
-1062882)/(x^6+243*x^5+19683*x^4+531441*x^3),x, algorithm="fricas")

[Out]

-2*(x^3 + 162*x^2 + (x^2 + 162*x + 6561)*log(x) + 6561*x - e^(2*x))/(x^4 + 162*x^3 + 6561*x^2)

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giac [B]  time = 0.34, size = 51, normalized size = 2.12 \begin {gather*} -\frac {2 \, {\left (x^{3} + x^{2} \log \relax (x) + 162 \, x^{2} + 162 \, x \log \relax (x) + 6561 \, x - e^{\left (2 \, x\right )} + 6561 \, \log \relax (x)\right )}}{x^{4} + 162 \, x^{3} + 6561 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+972*x^2+78732*x+2125764)*log(x)+(4*x^2+316*x-324)*exp(2*x)+2*x^4+484*x^3+38880*x^2+1023516*x
-1062882)/(x^6+243*x^5+19683*x^4+531441*x^3),x, algorithm="giac")

[Out]

-2*(x^3 + x^2*log(x) + 162*x^2 + 162*x*log(x) + 6561*x - e^(2*x) + 6561*log(x))/(x^4 + 162*x^3 + 6561*x^2)

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maple [A]  time = 0.06, size = 37, normalized size = 1.54

method	result	size
risch	\(-\frac {2 \ln \relax (x )}{x^{2}}-\frac {2 \left (x^{3}+162 x^{2}+6561 x -{\mathrm e}^{2 x}\right )}{x^{2} \left (81+x \right )^{2}}\)	\(37\)
default	\(-\frac {2}{x}-\frac {2 \ln \relax (x )}{x^{2}}+\frac {8 \,{\mathrm e}^{2 x}}{6561 \left (2 x +162\right )^{2}}+\frac {8 \,{\mathrm e}^{2 x}}{531441 \left (2 x +162\right )}-\frac {4 \,{\mathrm e}^{2 x}}{531441 x}+\frac {2 \,{\mathrm e}^{2 x}}{6561 x^{2}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3+972*x^2+78732*x+2125764)*ln(x)+(4*x^2+316*x-324)*exp(2*x)+2*x^4+484*x^3+38880*x^2+1023516*x-106288
2)/(x^6+243*x^5+19683*x^4+531441*x^3),x,method=_RETURNVERBOSE)

[Out]

-2*ln(x)/x^2-2*(x^3+162*x^2+6561*x-exp(2*x))/x^2/(81+x)^2

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maxima [B]  time = 0.57, size = 150, normalized size = 6.25 \begin {gather*} -\frac {4 \, x^{3} + 486 \, x^{2} + 8748 \, x - 177147}{27 \, {\left (x^{4} + 162 \, x^{3} + 6561 \, x^{2}\right )}} - \frac {26 \, {\left (2 \, x^{2} + 243 \, x + 4374\right )}}{9 \, {\left (x^{3} + 162 \, x^{2} + 6561 \, x\right )}} - \frac {x^{2} + 2 \, {\left (x^{2} + 162 \, x + 6561\right )} \log \relax (x) + 162 \, x - 2 \, e^{\left (2 \, x\right )} + 6561}{x^{4} + 162 \, x^{3} + 6561 \, x^{2}} + \frac {80 \, {\left (2 \, x + 243\right )}}{27 \, {\left (x^{2} + 162 \, x + 6561\right )}} - \frac {2 \, x + 81}{x^{2} + 162 \, x + 6561} - \frac {242}{x^{2} + 162 \, x + 6561} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+972*x^2+78732*x+2125764)*log(x)+(4*x^2+316*x-324)*exp(2*x)+2*x^4+484*x^3+38880*x^2+1023516*x
-1062882)/(x^6+243*x^5+19683*x^4+531441*x^3),x, algorithm="maxima")

[Out]

-1/27*(4*x^3 + 486*x^2 + 8748*x - 177147)/(x^4 + 162*x^3 + 6561*x^2) - 26/9*(2*x^2 + 243*x + 4374)/(x^3 + 162*
x^2 + 6561*x) - (x^2 + 2*(x^2 + 162*x + 6561)*log(x) + 162*x - 2*e^(2*x) + 6561)/(x^4 + 162*x^3 + 6561*x^2) +
80/27*(2*x + 243)/(x^2 + 162*x + 6561) - (2*x + 81)/(x^2 + 162*x + 6561) - 242/(x^2 + 162*x + 6561)

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mupad [B]  time = 0.50, size = 35, normalized size = 1.46 \begin {gather*} \frac {2\,{\mathrm {e}}^{2\,x}}{x^4+162\,x^3+6561\,x^2}-\frac {2\,\ln \relax (x)}{x^2}-\frac {2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1023516*x + exp(2*x)*(316*x + 4*x^2 - 324) + 38880*x^2 + 484*x^3 + 2*x^4 + log(x)*(78732*x + 972*x^2 + 4*
x^3 + 2125764) - 1062882)/(531441*x^3 + 19683*x^4 + 243*x^5 + x^6),x)

[Out]

(2*exp(2*x))/(6561*x^2 + 162*x^3 + x^4) - (2*log(x))/x^2 - 2/x

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sympy [A]  time = 0.35, size = 31, normalized size = 1.29 \begin {gather*} \frac {2 e^{2 x}}{x^{4} + 162 x^{3} + 6561 x^{2}} - \frac {2}{x} - \frac {2 \log {\relax (x )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3+972*x**2+78732*x+2125764)*ln(x)+(4*x**2+316*x-324)*exp(2*x)+2*x**4+484*x**3+38880*x**2+1023
516*x-1062882)/(x**6+243*x**5+19683*x**4+531441*x**3),x)

[Out]

2*exp(2*x)/(x**4 + 162*x**3 + 6561*x**2) - 2/x - 2*log(x)/x**2

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