3.101.28 \(\int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx\)

Optimal. Leaf size=26 \[ 4+\frac {10 x^2}{-x+4 x \left (-e^{2 x}+\log (3)\right )} \]

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Rubi [C]  time = 0.62, antiderivative size = 250, normalized size of antiderivative = 9.62, number of steps used = 18, number of rules used = 13, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.255, Rules used = {6688, 12, 6742, 2184, 2190, 2279, 2391, 2185, 2191, 2282, 36, 29, 31} \begin {gather*} \frac {5 (2-\log (6561)) \text {Li}_2\left (-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{2 (1-\log (81))^2}-\frac {5 \text {Li}_2\left (-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}-\frac {5 x^2 (2-\log (6561))}{(1-\log (81))^2}+\frac {5 (1-2 x)^2}{2 (1-4 \log (3))}+\frac {5 (1-2 x) \log \left (\frac {4 e^{2 x}}{1-4 \log (3)}+1\right )}{1-4 \log (3)}+\frac {5 x (2-\log (6561)) \log \left (\frac {4 e^{2 x}}{1-4 \log (3)}+1\right )}{(1-\log (81))^2}-\frac {5 (2-\log (6561)) \log \left (4 e^{2 x}+1-4 \log (3)\right )}{2 (1-\log (81))^2}-\frac {5 x (2-\log (6561))}{(1-\log (81)) \left (4 e^{2 x}+1-4 \log (3)\right )}+\frac {5 x (2-\log (6561))}{(1-\log (81))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 + E^(2*x)*(-40 + 80*x) + 40*Log[3])/(1 + 16*E^(4*x) + E^(2*x)*(8 - 32*Log[3]) - 8*Log[3] + 16*Log[3]^
2),x]

[Out]

(5*(1 - 2*x)^2)/(2*(1 - 4*Log[3])) + (5*x*(2 - Log[6561]))/(1 - Log[81])^2 - (5*x^2*(2 - Log[6561]))/(1 - Log[
81])^2 - (5*x*(2 - Log[6561]))/((1 + 4*E^(2*x) - 4*Log[3])*(1 - Log[81])) + (5*(1 - 2*x)*Log[1 + (4*E^(2*x))/(
1 - 4*Log[3])])/(1 - 4*Log[3]) + (5*x*(2 - Log[6561])*Log[1 + (4*E^(2*x))/(1 - 4*Log[3])])/(1 - Log[81])^2 - (
5*(2 - Log[6561])*Log[1 + 4*E^(2*x) - 4*Log[3]])/(2*(1 - Log[81])^2) - (5*PolyLog[2, (-4*E^(2*x))/(1 - 4*Log[3
])])/(1 - 4*Log[3]) + (5*(2 - Log[6561])*PolyLog[2, (-4*E^(2*x))/(1 - 4*Log[3])])/(2*(1 - Log[81])^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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2185

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 0] && IGtQ[m, 0
]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2191

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 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 {10 \left (-1+e^{2 x} (-4+8 x)+\log (81)\right )}{\left (1+4 e^{2 x}-4 \log (3)\right )^2} \, dx\\ &=10 \int \frac {-1+e^{2 x} (-4+8 x)+\log (81)}{\left (1+4 e^{2 x}-4 \log (3)\right )^2} \, dx\\ &=10 \int \left (\frac {-1+2 x}{1+4 e^{2 x}-4 \log (3)}+\frac {x (-2+\log (6561))}{\left (1+4 e^{2 x}-4 \log (3)\right )^2}\right ) \, dx\\ &=10 \int \frac {-1+2 x}{1+4 e^{2 x}-4 \log (3)} \, dx-(10 (2-\log (6561))) \int \frac {x}{\left (1+4 e^{2 x}-4 \log (3)\right )^2} \, dx\\ &=\frac {5 (1-2 x)^2}{2 (1-4 \log (3))}-\frac {40 \int \frac {e^{2 x} (-1+2 x)}{1+4 e^{2 x}-4 \log (3)} \, dx}{1-4 \log (3)}-\frac {(10 (2-\log (6561))) \int \frac {x}{1+4 e^{2 x}-4 \log (3)} \, dx}{1-\log (81)}+\frac {(40 (2-\log (6561))) \int \frac {e^{2 x} x}{\left (1+4 e^{2 x}-4 \log (3)\right )^2} \, dx}{1-\log (81)}\\ &=\frac {5 (1-2 x)^2}{2 (1-4 \log (3))}-\frac {5 x^2 (2-\log (6561))}{(1-\log (81))^2}-\frac {5 x (2-\log (6561))}{\left (1+4 e^{2 x}-4 \log (3)\right ) (1-\log (81))}+\frac {5 (1-2 x) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {10 \int \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right ) \, dx}{1-4 \log (3)}+\frac {(40 (2-\log (6561))) \int \frac {e^{2 x} x}{1+4 e^{2 x}-4 \log (3)} \, dx}{(1-\log (81))^2}+\frac {(5 (2-\log (6561))) \int \frac {1}{1+4 e^{2 x}-4 \log (3)} \, dx}{1-\log (81)}\\ &=\frac {5 (1-2 x)^2}{2 (1-4 \log (3))}-\frac {5 x^2 (2-\log (6561))}{(1-\log (81))^2}-\frac {5 x (2-\log (6561))}{\left (1+4 e^{2 x}-4 \log (3)\right ) (1-\log (81))}+\frac {5 (1-2 x) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {5 x (2-\log (6561)) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{(1-\log (81))^2}+\frac {5 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {4 x}{1-4 \log (3)}\right )}{x} \, dx,x,e^{2 x}\right )}{1-4 \log (3)}-\frac {(5 (2-\log (6561))) \int \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right ) \, dx}{(1-\log (81))^2}+\frac {(5 (2-\log (6561))) \operatorname {Subst}\left (\int \frac {1}{x (1+4 x-4 \log (3))} \, dx,x,e^{2 x}\right )}{2 (1-\log (81))}\\ &=\frac {5 (1-2 x)^2}{2 (1-4 \log (3))}-\frac {5 x^2 (2-\log (6561))}{(1-\log (81))^2}-\frac {5 x (2-\log (6561))}{\left (1+4 e^{2 x}-4 \log (3)\right ) (1-\log (81))}+\frac {5 (1-2 x) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {5 x (2-\log (6561)) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{(1-\log (81))^2}-\frac {5 \text {Li}_2\left (-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {(5 (2-\log (6561))) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x}\right )}{2 (1-\log (81))^2}-\frac {(5 (2-\log (6561))) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {4 x}{1-4 \log (3)}\right )}{x} \, dx,x,e^{2 x}\right )}{2 (1-\log (81))^2}-\frac {(10 (2-\log (6561))) \operatorname {Subst}\left (\int \frac {1}{1+4 x-4 \log (3)} \, dx,x,e^{2 x}\right )}{(1-\log (81))^2}\\ &=\frac {5 (1-2 x)^2}{2 (1-4 \log (3))}+\frac {5 x (2-\log (6561))}{(1-\log (81))^2}-\frac {5 x^2 (2-\log (6561))}{(1-\log (81))^2}-\frac {5 x (2-\log (6561))}{\left (1+4 e^{2 x}-4 \log (3)\right ) (1-\log (81))}+\frac {5 (1-2 x) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {5 x (2-\log (6561)) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{(1-\log (81))^2}-\frac {5 (2-\log (6561)) \log \left (1+4 e^{2 x}-4 \log (3)\right )}{2 (1-\log (81))^2}-\frac {5 \text {Li}_2\left (-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {5 (2-\log (6561)) \text {Li}_2\left (-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{2 (1-\log (81))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 18, normalized size = 0.69 \begin {gather*} -\frac {10 x}{1+4 e^{2 x}-\log (81)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + E^(2*x)*(-40 + 80*x) + 40*Log[3])/(1 + 16*E^(4*x) + E^(2*x)*(8 - 32*Log[3]) - 8*Log[3] + 16*L
og[3]^2),x]

[Out]

(-10*x)/(1 + 4*E^(2*x) - Log[81])

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fricas [A]  time = 2.33, size = 17, normalized size = 0.65 \begin {gather*} -\frac {10 \, x}{4 \, e^{\left (2 \, x\right )} - 4 \, \log \relax (3) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((80*x-40)*exp(x)^2+40*log(3)-10)/(16*exp(x)^4+(-32*log(3)+8)*exp(x)^2+16*log(3)^2-8*log(3)+1),x, al
gorithm="fricas")

[Out]

-10*x/(4*e^(2*x) - 4*log(3) + 1)

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giac [B]  time = 0.17, size = 139, normalized size = 5.35 \begin {gather*} -\frac {5 \, {\left (8 \, x \log \relax (3) + 4 \, e^{\left (2 \, x\right )} \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \relax (3) + 1\right ) - 4 \, \log \relax (3) \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \relax (3) + 1\right ) - 4 \, e^{\left (2 \, x\right )} \log \left (-4 \, e^{\left (2 \, x\right )} + 4 \, \log \relax (3) - 1\right ) + 4 \, \log \relax (3) \log \left (-4 \, e^{\left (2 \, x\right )} + 4 \, \log \relax (3) - 1\right ) - 2 \, x + \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \relax (3) + 1\right ) - \log \left (-4 \, e^{\left (2 \, x\right )} + 4 \, \log \relax (3) - 1\right )\right )}}{16 \, e^{\left (2 \, x\right )} \log \relax (3) - 16 \, \log \relax (3)^{2} - 4 \, e^{\left (2 \, x\right )} + 8 \, \log \relax (3) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((80*x-40)*exp(x)^2+40*log(3)-10)/(16*exp(x)^4+(-32*log(3)+8)*exp(x)^2+16*log(3)^2-8*log(3)+1),x, al
gorithm="giac")

[Out]

-5*(8*x*log(3) + 4*e^(2*x)*log(4*e^(2*x) - 4*log(3) + 1) - 4*log(3)*log(4*e^(2*x) - 4*log(3) + 1) - 4*e^(2*x)*
log(-4*e^(2*x) + 4*log(3) - 1) + 4*log(3)*log(-4*e^(2*x) + 4*log(3) - 1) - 2*x + log(4*e^(2*x) - 4*log(3) + 1)
- log(-4*e^(2*x) + 4*log(3) - 1))/(16*e^(2*x)*log(3) - 16*log(3)^2 - 4*e^(2*x) + 8*log(3) - 1)

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maple [A]  time = 0.13, size = 18, normalized size = 0.69

method	result	size
norman	\(\frac {10 x}{-4 \,{\mathrm e}^{2 x}+4 \ln \relax (3)-1}\)	\(18\)
risch	\(\frac {10 x}{-4 \,{\mathrm e}^{2 x}+4 \ln \relax (3)-1}\)	\(18\)
default	\(-\frac {10 \ln \left ({\mathrm e}^{x}\right )}{\left (-1+4 \ln \relax (3)\right )^{2}}+\frac {5 \ln \left (4 \,{\mathrm e}^{2 x}-4 \ln \relax (3)+1\right )}{\left (-1+4 \ln \relax (3)\right )^{2}}+\frac {40 \ln \relax (3)}{\left (-1+4 \ln \relax (3)\right )^{2} \left (4 \,{\mathrm e}^{2 x}-4 \ln \relax (3)+1\right )}-\frac {5}{\left (-1+4 \ln \relax (3)\right )^{2} \left (4 \,{\mathrm e}^{2 x}-4 \ln \relax (3)+1\right )}+\frac {5}{4 \,{\mathrm e}^{2 x}-4 \ln \relax (3)+1}+\frac {40 \ln \relax (3) \ln \left ({\mathrm e}^{x}\right )}{\left (-1+4 \ln \relax (3)\right )^{2}}-\frac {20 \ln \relax (3) \ln \left (4 \,{\mathrm e}^{2 x}-4 \ln \relax (3)+1\right )}{\left (-1+4 \ln \relax (3)\right )^{2}}-\frac {80 \ln \relax (3)^{2}}{\left (-1+4 \ln \relax (3)\right )^{2} \left (4 \,{\mathrm e}^{2 x}-4 \ln \relax (3)+1\right )}+\frac {5 \ln \left (-4 \,{\mathrm e}^{2 x}+4 \ln \relax (3)-1\right )}{-1+4 \ln \relax (3)}+\frac {40 x \,{\mathrm e}^{2 x}}{\left (-1+4 \ln \relax (3)\right ) \left (-4 \,{\mathrm e}^{2 x}+4 \ln \relax (3)-1\right )}\)	\(224\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((80*x-40)*exp(x)^2+40*ln(3)-10)/(16*exp(x)^4+(-32*ln(3)+8)*exp(x)^2+16*ln(3)^2-8*ln(3)+1),x,method=_RETUR
NVERBOSE)

[Out]

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

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maxima [B]  time = 0.44, size = 224, normalized size = 8.62 \begin {gather*} 20 \, {\left (\frac {2 \, x}{16 \, \log \relax (3)^{2} - 8 \, \log \relax (3) + 1} - \frac {\log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \relax (3) + 1\right )}{16 \, \log \relax (3)^{2} - 8 \, \log \relax (3) + 1} - \frac {1}{4 \, {\left (4 \, \log \relax (3) - 1\right )} e^{\left (2 \, x\right )} - 16 \, \log \relax (3)^{2} + 8 \, \log \relax (3) - 1}\right )} \log \relax (3) - \frac {40 \, x e^{\left (2 \, x\right )}}{4 \, {\left (4 \, \log \relax (3) - 1\right )} e^{\left (2 \, x\right )} - 16 \, \log \relax (3)^{2} + 8 \, \log \relax (3) - 1} - \frac {10 \, x}{16 \, \log \relax (3)^{2} - 8 \, \log \relax (3) + 1} + \frac {5 \, \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \relax (3) + 1\right )}{16 \, \log \relax (3)^{2} - 8 \, \log \relax (3) + 1} + \frac {5 \, \log \left (e^{\left (2 \, x\right )} - \log \relax (3) + \frac {1}{4}\right )}{4 \, \log \relax (3) - 1} + \frac {5}{4 \, {\left (4 \, \log \relax (3) - 1\right )} e^{\left (2 \, x\right )} - 16 \, \log \relax (3)^{2} + 8 \, \log \relax (3) - 1} + \frac {5}{4 \, e^{\left (2 \, x\right )} - 4 \, \log \relax (3) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((80*x-40)*exp(x)^2+40*log(3)-10)/(16*exp(x)^4+(-32*log(3)+8)*exp(x)^2+16*log(3)^2-8*log(3)+1),x, al
gorithm="maxima")

[Out]

20*(2*x/(16*log(3)^2 - 8*log(3) + 1) - log(4*e^(2*x) - 4*log(3) + 1)/(16*log(3)^2 - 8*log(3) + 1) - 1/(4*(4*lo
g(3) - 1)*e^(2*x) - 16*log(3)^2 + 8*log(3) - 1))*log(3) - 40*x*e^(2*x)/(4*(4*log(3) - 1)*e^(2*x) - 16*log(3)^2
+ 8*log(3) - 1) - 10*x/(16*log(3)^2 - 8*log(3) + 1) + 5*log(4*e^(2*x) - 4*log(3) + 1)/(16*log(3)^2 - 8*log(3)
+ 1) + 5*log(e^(2*x) - log(3) + 1/4)/(4*log(3) - 1) + 5/(4*(4*log(3) - 1)*e^(2*x) - 16*log(3)^2 + 8*log(3) -
1) + 5/(4*e^(2*x) - 4*log(3) + 1)

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mupad [B]  time = 0.25, size = 17, normalized size = 0.65 \begin {gather*} -\frac {10\,x}{4\,{\mathrm {e}}^{2\,x}-\ln \left (81\right )+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((40*log(3) + exp(2*x)*(80*x - 40) - 10)/(16*exp(4*x) - 8*log(3) - exp(2*x)*(32*log(3) - 8) + 16*log(3)^2 +
1),x)

[Out]

-(10*x)/(4*exp(2*x) - log(81) + 1)

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sympy [A]  time = 0.13, size = 17, normalized size = 0.65 \begin {gather*} - \frac {10 x}{4 e^{2 x} - 4 \log {\relax (3 )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((80*x-40)*exp(x)**2+40*ln(3)-10)/(16*exp(x)**4+(-32*ln(3)+8)*exp(x)**2+16*ln(3)**2-8*ln(3)+1),x)

[Out]

-10*x/(4*exp(2*x) - 4*log(3) + 1)

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