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

Optimal. Leaf size=24 \[ x-x^2 \log \left (\frac {x^2}{-5 e^{-4 x}+\log (5)}\right ) \]

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Rubi [A]  time = 0.82, antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 16, number of rules used = 8, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.110, Rules used = {6742, 2184, 2190, 2531, 2282, 6589, 2551, 12} \begin {gather*} x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right ) \end {gather*}

Antiderivative was successfully verified.

[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 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 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 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 2531

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 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 6589

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 \text {Li}_2\left (\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{2} \int \text {Li}_2\left (\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 \text {Li}_2\left (\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\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 \text {Li}_2\left (\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{8} \text {Li}_3\left (\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 \text {Li}_2\left (\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{8} \text {Li}_3\left (\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} \text {Li}_3\left (\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{2} \int \text {Li}_2\left (\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} \text {Li}_3\left (\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{8} \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 31, normalized size = 1.29 \begin {gather*} x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right ) \end {gather*}

Antiderivative was successfully verified.

[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]))]

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fricas [A]  time = 0.92, size = 27, normalized size = 1.12 \begin {gather*} -x^{2} \log \left (\frac {x^{2} e^{\left (4 \, x\right )}}{e^{\left (4 \, x\right )} \log \relax (5) - 5}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.26, size = 30, normalized size = 1.25 \begin {gather*} -4 \, x^{3} - x^{2} \log \left (x^{2}\right ) + x^{2} \log \left (e^{\left (4 \, x\right )} \log \relax (5) - 5\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.29, size = 28, normalized size = 1.17

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.72, size = 45, normalized size = 1.88 \begin {gather*} -4 \, x^{3} - 2 \, x^{2} \log \relax (x) + \frac {1}{4} \, {\left (4 \, x^{2} + 1\right )} \log \left (e^{\left (4 \, x\right )} \log \relax (5) - 5\right ) + x - \frac {1}{4} \, \log \left (e^{\left (4 \, x\right )} \log \relax (5) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {10\,x+\ln \left (\frac {x^2\,{\mathrm {e}}^{4\,x}}{{\mathrm {e}}^{4\,x}\,\ln \relax (5)-5}\right )\,\left (10\,x-2\,x\,{\mathrm {e}}^{4\,x}\,\ln \relax (5)\right )+20\,x^2-{\mathrm {e}}^{4\,x}\,\ln \relax (5)\,\left (2\,x-1\right )-5}{{\mathrm {e}}^{4\,x}\,\ln \relax (5)-5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.39, size = 24, normalized size = 1.00 \begin {gather*} - x^{2} \log {\left (\frac {x^{2} e^{4 x}}{e^{4 x} \log {\relax (5 )} - 5} \right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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