\(\int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+(10 x-2 e^{4 x} x \log (5)) \log (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)})}{-5+e^{4 x} \log (5)} \, dx\) [2157]

Optimal result

Integrand size = 73, antiderivative size = 24 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=x-x^2 \log \left (\frac {x^2}{-5 e^{-4 x}+\log (5)}\right ) \]

[Out]

x-x^2*ln(x^2/(ln(5)-5/exp(4*x)))

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 16, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.110, Rules used = {6874, 2215, 2221, 2611, 2320, 6724, 2631, 12} \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right ) \]

[In]

Int[(-5 + 10*x + 20*x^2 + E^(4*x)*(1 - 2*x)*Log[5] + (10*x - 2*E^(4*x)*x*Log[5])*Log[(E^(4*x)*x^2)/(-5 + E^(4*
x)*Log[5])])/(-5 + E^(4*x)*Log[5]),x]

[Out]

x - x^2*Log[-((E^(4*x)*x^2)/(5 - E^(4*x)*Log[5]))]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2215

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2631

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 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (1-2 x+\frac {20 x^2}{-5+e^{4 x} \log (5)}-2 x \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )\right ) \, dx \\ & = x-x^2-2 \int x \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right ) \, dx+20 \int \frac {x^2}{-5+e^{4 x} \log (5)} \, dx \\ & = x-x^2-\frac {4 x^3}{3}-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+(4 \log (5)) \int \frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)} \, dx+\int \frac {2 x \left (5+10 x-e^{4 x} \log (5)\right )}{5-e^{4 x} \log (5)} \, dx \\ & = x-x^2-\frac {4 x^3}{3}-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+x^2 \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right )+2 \int \frac {x \left (5+10 x-e^{4 x} \log (5)\right )}{5-e^{4 x} \log (5)} \, dx-2 \int x \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right ) \, dx \\ & = x-x^2-\frac {4 x^3}{3}-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+x^2 \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{2} x \operatorname {PolyLog}\left (2,\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{2} \int \operatorname {PolyLog}\left (2,\frac {1}{5} e^{4 x} \log (5)\right ) \, dx+2 \int \left (x-\frac {10 x^2}{-5+e^{4 x} \log (5)}\right ) \, dx \\ & = x-\frac {4 x^3}{3}-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+x^2 \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{2} x \operatorname {PolyLog}\left (2,\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{8} \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,\frac {1}{5} x \log (5)\right )}{x} \, dx,x,e^{4 x}\right )-20 \int \frac {x^2}{-5+e^{4 x} \log (5)} \, dx \\ & = x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+x^2 \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{2} x \operatorname {PolyLog}\left (2,\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{8} \operatorname {PolyLog}\left (3,\frac {1}{5} e^{4 x} \log (5)\right )-(4 \log (5)) \int \frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)} \, dx \\ & = x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+\frac {1}{2} x \operatorname {PolyLog}\left (2,\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{8} \operatorname {PolyLog}\left (3,\frac {1}{5} e^{4 x} \log (5)\right )+2 \int x \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right ) \, dx \\ & = x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )-\frac {1}{8} \operatorname {PolyLog}\left (3,\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{2} \int \operatorname {PolyLog}\left (2,\frac {1}{5} e^{4 x} \log (5)\right ) \, dx \\ & = x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )-\frac {1}{8} \operatorname {PolyLog}\left (3,\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{8} \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,\frac {1}{5} x \log (5)\right )}{x} \, dx,x,e^{4 x}\right ) \\ & = x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right ) \]

[In]

Integrate[(-5 + 10*x + 20*x^2 + E^(4*x)*(1 - 2*x)*Log[5] + (10*x - 2*E^(4*x)*x*Log[5])*Log[(E^(4*x)*x^2)/(-5 +
E^(4*x)*Log[5])])/(-5 + E^(4*x)*Log[5]),x]

[Out]

x - x^2*Log[-((E^(4*x)*x^2)/(5 - E^(4*x)*Log[5]))]

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17

method	result	size
norman	\(x -x^{2} \ln \left (\frac {x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )\)	\(28\)
parallelrisch	\(x -x^{2} \ln \left (\frac {x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )\)	\(28\)
risch	\(-x^{2} \ln \left ({\mathrm e}^{4 x}\right )+x^{2} \ln \left (\ln \left (5\right ) {\mathrm e}^{4 x}-5\right )-2 x^{2} \ln \left (x \right )+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}-i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right ) \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{2}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right ) \operatorname {csgn}\left (i {\mathrm e}^{4 x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{2}}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{4 x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{2}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{3}}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right ) \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{2}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{3}}{2}+x\)	\(422\)

[In]

int(((-2*x*ln(5)*exp(4*x)+10*x)*ln(x^2*exp(4*x)/(ln(5)*exp(4*x)-5))+(1-2*x)*ln(5)*exp(4*x)+20*x^2+10*x-5)/(ln(
5)*exp(4*x)-5),x,method=_RETURNVERBOSE)

[Out]

x-x^2*ln(x^2*exp(4*x)/(ln(5)*exp(4*x)-5))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=-x^{2} \log \left (\frac {x^{2} e^{\left (4 \, x\right )}}{e^{\left (4 \, x\right )} \log \left (5\right ) - 5}\right ) + x \]

[In]

integrate(((-2*x*log(5)*exp(4*x)+10*x)*log(x^2*exp(4*x)/(log(5)*exp(4*x)-5))+(1-2*x)*log(5)*exp(4*x)+20*x^2+10
*x-5)/(log(5)*exp(4*x)-5),x, algorithm="fricas")

[Out]

-x^2*log(x^2*e^(4*x)/(e^(4*x)*log(5) - 5)) + x

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=- x^{2} \log {\left (\frac {x^{2} e^{4 x}}{e^{4 x} \log {\left (5 \right )} - 5} \right )} + x \]

[In]

integrate(((-2*x*ln(5)*exp(4*x)+10*x)*ln(x**2*exp(4*x)/(ln(5)*exp(4*x)-5))+(1-2*x)*ln(5)*exp(4*x)+20*x**2+10*x
-5)/(ln(5)*exp(4*x)-5),x)

[Out]

-x**2*log(x**2*exp(4*x)/(exp(4*x)*log(5) - 5)) + x

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=-4 \, x^{3} - 2 \, x^{2} \log \left (x\right ) + \frac {1}{4} \, {\left (4 \, x^{2} + 1\right )} \log \left (e^{\left (4 \, x\right )} \log \left (5\right ) - 5\right ) + x - \frac {1}{4} \, \log \left (e^{\left (4 \, x\right )} \log \left (5\right ) - 5\right ) \]

[In]

integrate(((-2*x*log(5)*exp(4*x)+10*x)*log(x^2*exp(4*x)/(log(5)*exp(4*x)-5))+(1-2*x)*log(5)*exp(4*x)+20*x^2+10
*x-5)/(log(5)*exp(4*x)-5),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=-4 \, x^{3} - x^{2} \log \left (x^{2}\right ) + x^{2} \log \left (e^{\left (4 \, x\right )} \log \left (5\right ) - 5\right ) + x \]

[In]

integrate(((-2*x*log(5)*exp(4*x)+10*x)*log(x^2*exp(4*x)/(log(5)*exp(4*x)-5))+(1-2*x)*log(5)*exp(4*x)+20*x^2+10
*x-5)/(log(5)*exp(4*x)-5),x, algorithm="giac")

[Out]

-4*x^3 - x^2*log(x^2) + x^2*log(e^(4*x)*log(5) - 5) + x

Mupad [F(-1)]

Timed out. \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=\int \frac {10\,x+\ln \left (\frac {x^2\,{\mathrm {e}}^{4\,x}}{{\mathrm {e}}^{4\,x}\,\ln \left (5\right )-5}\right )\,\left (10\,x-2\,x\,{\mathrm {e}}^{4\,x}\,\ln \left (5\right )\right )+20\,x^2-{\mathrm {e}}^{4\,x}\,\ln \left (5\right )\,\left (2\,x-1\right )-5}{{\mathrm {e}}^{4\,x}\,\ln \left (5\right )-5} \,d x \]

[In]

int((10*x + log((x^2*exp(4*x))/(exp(4*x)*log(5) - 5))*(10*x - 2*x*exp(4*x)*log(5)) + 20*x^2 - exp(4*x)*log(5)*
(2*x - 1) - 5)/(exp(4*x)*log(5) - 5),x)

[Out]

int((10*x + log((x^2*exp(4*x))/(exp(4*x)*log(5) - 5))*(10*x - 2*x*exp(4*x)*log(5)) + 20*x^2 - exp(4*x)*log(5)*
(2*x - 1) - 5)/(exp(4*x)*log(5) - 5), x)