3.83.49 \(\int \frac {-4 e^{12+x}+4 e^{10+x} \log ^2(\frac {-1+e^{4+x}}{e^4})+(1-e^{4+x}) \log ^5(\frac {-1+e^{4+x}}{e^4})}{(-1+e^{4+x}) \log ^5(\frac {-1+e^{4+x}}{e^4})} \, dx\)

Optimal. Leaf size=27 \[ -x+\left (e^2-\frac {e^4}{\log ^2\left (-\frac {1}{e^4}+e^x\right )}\right )^2 \]

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Rubi [A]  time = 0.67, antiderivative size = 37, normalized size of antiderivative = 1.37, number of steps used = 11, number of rules used = 6, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {2282, 6742, 2390, 12, 2302, 30} \begin {gather*} -x+\frac {e^8}{\log ^4\left (e^x-\frac {1}{e^4}\right )}-\frac {2 e^6}{\log ^2\left (e^x-\frac {1}{e^4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*E^(12 + x) + 4*E^(10 + x)*Log[(-1 + E^(4 + x))/E^4]^2 + (1 - E^(4 + x))*Log[(-1 + E^(4 + x))/E^4]^5)/(
(-1 + E^(4 + x))*Log[(-1 + E^(4 + x))/E^4]^5),x]

[Out]

-x + E^8/Log[-E^(-4) + E^x]^4 - (2*E^6)/Log[-E^(-4) + E^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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {4 e^{12} x-4 e^{10} x \log ^2\left (-\frac {1}{e^4}+x\right )-\left (1-e^4 x\right ) \log ^5\left (-\frac {1}{e^4}+x\right )}{x \left (1-e^4 x\right ) \log ^5\left (-\frac {1}{e^4}+x\right )} \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \left (-\frac {1}{x}-\frac {4 e^{12}}{\left (-1+e^4 x\right ) \log ^5\left (-\frac {1}{e^4}+x\right )}+\frac {4 e^{10}}{\left (-1+e^4 x\right ) \log ^3\left (-\frac {1}{e^4}+x\right )}\right ) \, dx,x,e^x\right )\\ &=-x+\left (4 e^{10}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-1+e^4 x\right ) \log ^3\left (-\frac {1}{e^4}+x\right )} \, dx,x,e^x\right )-\left (4 e^{12}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-1+e^4 x\right ) \log ^5\left (-\frac {1}{e^4}+x\right )} \, dx,x,e^x\right )\\ &=-x+\left (4 e^{10}\right ) \operatorname {Subst}\left (\int \frac {1}{e^4 x \log ^3(x)} \, dx,x,-\frac {1}{e^4}+e^x\right )-\left (4 e^{12}\right ) \operatorname {Subst}\left (\int \frac {1}{e^4 x \log ^5(x)} \, dx,x,-\frac {1}{e^4}+e^x\right )\\ &=-x+\left (4 e^6\right ) \operatorname {Subst}\left (\int \frac {1}{x \log ^3(x)} \, dx,x,-\frac {1}{e^4}+e^x\right )-\left (4 e^8\right ) \operatorname {Subst}\left (\int \frac {1}{x \log ^5(x)} \, dx,x,-\frac {1}{e^4}+e^x\right )\\ &=-x+\left (4 e^6\right ) \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log \left (-\frac {1}{e^4}+e^x\right )\right )-\left (4 e^8\right ) \operatorname {Subst}\left (\int \frac {1}{x^5} \, dx,x,\log \left (-\frac {1}{e^4}+e^x\right )\right )\\ &=-x+\frac {e^8}{\log ^4\left (-\frac {1}{e^4}+e^x\right )}-\frac {2 e^6}{\log ^2\left (-\frac {1}{e^4}+e^x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 37, normalized size = 1.37 \begin {gather*} -x+\frac {e^8}{\left (-4+\log \left (-1+e^{4+x}\right )\right )^4}-\frac {2 e^6}{\left (-4+\log \left (-1+e^{4+x}\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E^(12 + x) + 4*E^(10 + x)*Log[(-1 + E^(4 + x))/E^4]^2 + (1 - E^(4 + x))*Log[(-1 + E^(4 + x))/E^4
]^5)/((-1 + E^(4 + x))*Log[(-1 + E^(4 + x))/E^4]^5),x]

[Out]

-x + E^8/(-4 + Log[-1 + E^(4 + x)])^4 - (2*E^6)/(-4 + Log[-1 + E^(4 + x)])^2

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fricas [B]  time = 0.79, size = 61, normalized size = 2.26 \begin {gather*} -\frac {x \log \left (-{\left (e^{8} - e^{\left (x + 12\right )}\right )} e^{\left (-12\right )}\right )^{4} + 2 \, e^{6} \log \left (-{\left (e^{8} - e^{\left (x + 12\right )}\right )} e^{\left (-12\right )}\right )^{2} - e^{8}}{\log \left (-{\left (e^{8} - e^{\left (x + 12\right )}\right )} e^{\left (-12\right )}\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(4)*exp(x)+1)*log((exp(4)*exp(x)-1)/exp(4))^5+4*exp(2)^3*exp(4)*exp(x)*log((exp(4)*exp(x)-1)/e
xp(4))^2-4*exp(2)^4*exp(4)*exp(x))/(exp(4)*exp(x)-1)/log((exp(4)*exp(x)-1)/exp(4))^5,x, algorithm="fricas")

[Out]

-(x*log(-(e^8 - e^(x + 12))*e^(-12))^4 + 2*e^6*log(-(e^8 - e^(x + 12))*e^(-12))^2 - e^8)/log(-(e^8 - e^(x + 12
))*e^(-12))^4

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giac [B]  time = 0.17, size = 147, normalized size = 5.44 \begin {gather*} -\frac {\log \left (e^{\left (x + 4\right )} - 1\right )^{4} \log \left (e^{\left (x + 4\right )}\right ) - 16 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{3} \log \left (e^{\left (x + 4\right )}\right ) + 2 \, e^{6} \log \left (e^{\left (x + 4\right )} - 1\right )^{2} + 96 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{2} \log \left (e^{\left (x + 4\right )}\right ) - 16 \, e^{6} \log \left (e^{\left (x + 4\right )} - 1\right ) - 256 \, \log \left (e^{\left (x + 4\right )} - 1\right ) \log \left (e^{\left (x + 4\right )}\right ) - e^{8} + 32 \, e^{6} + 256 \, \log \left (e^{\left (x + 4\right )}\right )}{\log \left (e^{\left (x + 4\right )} - 1\right )^{4} - 16 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{3} + 96 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{2} - 256 \, \log \left (e^{\left (x + 4\right )} - 1\right ) + 256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(4)*exp(x)+1)*log((exp(4)*exp(x)-1)/exp(4))^5+4*exp(2)^3*exp(4)*exp(x)*log((exp(4)*exp(x)-1)/e
xp(4))^2-4*exp(2)^4*exp(4)*exp(x))/(exp(4)*exp(x)-1)/log((exp(4)*exp(x)-1)/exp(4))^5,x, algorithm="giac")

[Out]

-(log(e^(x + 4) - 1)^4*log(e^(x + 4)) - 16*log(e^(x + 4) - 1)^3*log(e^(x + 4)) + 2*e^6*log(e^(x + 4) - 1)^2 +
96*log(e^(x + 4) - 1)^2*log(e^(x + 4)) - 16*e^6*log(e^(x + 4) - 1) - 256*log(e^(x + 4) - 1)*log(e^(x + 4)) - e
^8 + 32*e^6 + 256*log(e^(x + 4)))/(log(e^(x + 4) - 1)^4 - 16*log(e^(x + 4) - 1)^3 + 96*log(e^(x + 4) - 1)^2 -
256*log(e^(x + 4) - 1) + 256)

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maple [A]  time = 0.12, size = 37, normalized size = 1.37

method	result	size
risch	\(-x +\frac {{\mathrm e}^{6} \left (-2 \ln \left (\left ({\mathrm e}^{4+x}-1\right ) {\mathrm e}^{-4}\right )^{2}+{\mathrm e}^{2}\right )}{\ln \left (\left ({\mathrm e}^{4+x}-1\right ) {\mathrm e}^{-4}\right )^{4}}\)	\(37\)
derivativedivides	\(-\ln \left (\left ({\mathrm e}^{x}-{\mathrm e}^{-4}\right ) {\mathrm e}^{4}+1\right )-\frac {2 \,{\mathrm e}^{6}}{\ln \left ({\mathrm e}^{x}-{\mathrm e}^{-4}\right )^{2}}+\frac {{\mathrm e}^{8}}{\ln \left ({\mathrm e}^{x}-{\mathrm e}^{-4}\right )^{4}}\)	\(54\)
default	\(-\ln \left (\left ({\mathrm e}^{x}-{\mathrm e}^{-4}\right ) {\mathrm e}^{4}+1\right )-\frac {2 \,{\mathrm e}^{6}}{\ln \left ({\mathrm e}^{x}-{\mathrm e}^{-4}\right )^{2}}+\frac {{\mathrm e}^{8}}{\ln \left ({\mathrm e}^{x}-{\mathrm e}^{-4}\right )^{4}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(4)*exp(x)+1)*ln((exp(4)*exp(x)-1)/exp(4))^5+4*exp(2)^3*exp(4)*exp(x)*ln((exp(4)*exp(x)-1)/exp(4))^2
-4*exp(2)^4*exp(4)*exp(x))/(exp(4)*exp(x)-1)/ln((exp(4)*exp(x)-1)/exp(4))^5,x,method=_RETURNVERBOSE)

[Out]

-x+exp(6)*(-2*ln((exp(4+x)-1)*exp(-4))^2+exp(2))/ln((exp(4+x)-1)*exp(-4))^4

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maxima [B]  time = 0.44, size = 408, normalized size = 15.11 \begin {gather*} \frac {\log \left (-e^{\left (-4\right )} + e^{x}\right )^{5}}{4 \, {\left (\log \left (e^{\left (x + 4\right )} - 1\right )^{4} - 16 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{3} + 96 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{2} - 256 \, \log \left (e^{\left (x + 4\right )} - 1\right ) + 256\right )}} + \frac {5 \, \log \left (-e^{\left (-4\right )} + e^{x}\right )^{4}}{12 \, {\left (\log \left (e^{\left (x + 4\right )} - 1\right )^{3} - 12 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{2} + 48 \, \log \left (e^{\left (x + 4\right )} - 1\right ) - 64\right )}} - \frac {1}{3} \, {\left (\frac {2 \, \log \left (-e^{\left (-4\right )} + e^{x}\right )}{\log \left (e^{\left (x + 4\right )} - 1\right )^{3} - 12 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{2} + 48 \, \log \left (e^{\left (x + 4\right )} - 1\right ) - 64} + \frac {1}{\log \left (e^{\left (x + 4\right )} - 1\right )^{2} - 8 \, \log \left (e^{\left (x + 4\right )} - 1\right ) + 16}\right )} e^{6} + \frac {5}{6} \, {\left (\frac {e^{4} \log \left (-e^{\left (-4\right )} + e^{x}\right )^{3}}{\log \left (e^{\left (x + 4\right )} - 1\right )^{2} - 8 \, \log \left (e^{\left (x + 4\right )} - 1\right ) + 16} - 6 \, e^{4} \log \left (-e^{\left (-4\right )} + e^{x}\right ) \log \left (\log \left (e^{\left (x + 4\right )} - 1\right ) - 4\right ) + 6 \, {\left (\log \left (-e^{\left (-4\right )} + e^{x}\right ) \log \left (\log \left (e^{\left (x + 4\right )} - 1\right ) - 4\right ) - \log \left (e^{\left (x + 4\right )} - 1\right )\right )} e^{4} + \frac {3 \, e^{4} \log \left (-e^{\left (-4\right )} + e^{x}\right )^{2}}{\log \left (e^{\left (x + 4\right )} - 1\right ) - 4}\right )} e^{\left (-4\right )} - \frac {e^{6} \log \left (-e^{\left (-4\right )} + e^{x}\right )^{2}}{\log \left (e^{\left (x + 4\right )} - 1\right )^{4} - 16 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{3} + 96 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{2} - 256 \, \log \left (e^{\left (x + 4\right )} - 1\right ) + 256} - x + \frac {e^{8}}{\log \left (e^{\left (x + 4\right )} - 1\right )^{4} - 16 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{3} + 96 \, \log \left (e^{\left (x + 4\right )} - 1\right )^{2} - 256 \, \log \left (e^{\left (x + 4\right )} - 1\right ) + 256} + \log \left (e^{\left (x + 4\right )} - 1\right ) - 4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(4)*exp(x)+1)*log((exp(4)*exp(x)-1)/exp(4))^5+4*exp(2)^3*exp(4)*exp(x)*log((exp(4)*exp(x)-1)/e
xp(4))^2-4*exp(2)^4*exp(4)*exp(x))/(exp(4)*exp(x)-1)/log((exp(4)*exp(x)-1)/exp(4))^5,x, algorithm="maxima")

[Out]

1/4*log(-e^(-4) + e^x)^5/(log(e^(x + 4) - 1)^4 - 16*log(e^(x + 4) - 1)^3 + 96*log(e^(x + 4) - 1)^2 - 256*log(e
^(x + 4) - 1) + 256) + 5/12*log(-e^(-4) + e^x)^4/(log(e^(x + 4) - 1)^3 - 12*log(e^(x + 4) - 1)^2 + 48*log(e^(x
+ 4) - 1) - 64) - 1/3*(2*log(-e^(-4) + e^x)/(log(e^(x + 4) - 1)^3 - 12*log(e^(x + 4) - 1)^2 + 48*log(e^(x + 4
) - 1) - 64) + 1/(log(e^(x + 4) - 1)^2 - 8*log(e^(x + 4) - 1) + 16))*e^6 + 5/6*(e^4*log(-e^(-4) + e^x)^3/(log(
e^(x + 4) - 1)^2 - 8*log(e^(x + 4) - 1) + 16) - 6*e^4*log(-e^(-4) + e^x)*log(log(e^(x + 4) - 1) - 4) + 6*(log(
-e^(-4) + e^x)*log(log(e^(x + 4) - 1) - 4) - log(e^(x + 4) - 1))*e^4 + 3*e^4*log(-e^(-4) + e^x)^2/(log(e^(x +
4) - 1) - 4))*e^(-4) - e^6*log(-e^(-4) + e^x)^2/(log(e^(x + 4) - 1)^4 - 16*log(e^(x + 4) - 1)^3 + 96*log(e^(x
+ 4) - 1)^2 - 256*log(e^(x + 4) - 1) + 256) - x + e^8/(log(e^(x + 4) - 1)^4 - 16*log(e^(x + 4) - 1)^3 + 96*log
(e^(x + 4) - 1)^2 - 256*log(e^(x + 4) - 1) + 256) + log(e^(x + 4) - 1) - 4

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mupad [B]  time = 0.16, size = 43, normalized size = 1.59 \begin {gather*} -\frac {x\,{\ln \left ({\mathrm {e}}^x-{\mathrm {e}}^{-4}\right )}^4+2\,{\mathrm {e}}^6\,{\ln \left ({\mathrm {e}}^x-{\mathrm {e}}^{-4}\right )}^2-{\mathrm {e}}^8}{{\ln \left ({\mathrm {e}}^x-{\mathrm {e}}^{-4}\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(12)*exp(x) + log(exp(-4)*(exp(4)*exp(x) - 1))^5*(exp(4)*exp(x) - 1) - 4*log(exp(-4)*(exp(4)*exp(x)
- 1))^2*exp(10)*exp(x))/(log(exp(-4)*(exp(4)*exp(x) - 1))^5*(exp(4)*exp(x) - 1)),x)

[Out]

-(2*exp(6)*log(exp(x) - exp(-4))^2 - exp(8) + x*log(exp(x) - exp(-4))^4)/log(exp(x) - exp(-4))^4

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sympy [A]  time = 0.21, size = 39, normalized size = 1.44 \begin {gather*} - x + \frac {- 2 e^{6} \log {\left (\frac {e^{4} e^{x} - 1}{e^{4}} \right )}^{2} + e^{8}}{\log {\left (\frac {e^{4} e^{x} - 1}{e^{4}} \right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(4)*exp(x)+1)*ln((exp(4)*exp(x)-1)/exp(4))**5+4*exp(2)**3*exp(4)*exp(x)*ln((exp(4)*exp(x)-1)/e
xp(4))**2-4*exp(2)**4*exp(4)*exp(x))/(exp(4)*exp(x)-1)/ln((exp(4)*exp(x)-1)/exp(4))**5,x)

[Out]

-x + (-2*exp(6)*log((exp(4)*exp(x) - 1)*exp(-4))**2 + exp(8))/log((exp(4)*exp(x) - 1)*exp(-4))**4

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