3.35.47 \(\int \frac {e^{-x} (20+40 x+20 x^2+e^x (-20 x-32 x^2-16 x^3)+(-20 x-40 x^2-20 x^3) \log (x)+(-5-10 x-5 x^2+e^x (5 x+8 x^2+4 x^3)+(5 x+10 x^2+5 x^3) \log (x)) \log (\log (2)))}{5 x+10 x^2+5 x^3} \, dx\)

Optimal. Leaf size=31 \[ \left (-\frac {4 x}{5}+\frac {1}{5 (1+x)}+e^{-x} \log (x)\right ) (4-\log (\log (2))) \]

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Rubi [A]  time = 2.50, antiderivative size = 44, normalized size of antiderivative = 1.42, number of steps used = 10, number of rules used = 7, integrand size = 116, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {1594, 27, 12, 6688, 6742, 683, 2288} \begin {gather*} -\frac {4}{5} x (4-\log (\log (2)))+e^{-x} (4-\log (\log (2))) \log (x)+\frac {4-\log (\log (2))}{5 (x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 + 40*x + 20*x^2 + E^x*(-20*x - 32*x^2 - 16*x^3) + (-20*x - 40*x^2 - 20*x^3)*Log[x] + (-5 - 10*x - 5*x^
2 + E^x*(5*x + 8*x^2 + 4*x^3) + (5*x + 10*x^2 + 5*x^3)*Log[x])*Log[Log[2]])/(E^x*(5*x + 10*x^2 + 5*x^3)),x]

[Out]

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

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{x \left (5+10 x+5 x^2\right )} \, dx\\ &=\int \frac {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{5 x (1+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{x (1+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{-x} \left (5-5 \left (-2+e^x\right ) x-\left (-5+8 e^x\right ) x^2-4 e^x x^3-5 x (1+x)^2 \log (x)\right ) (4-\log (\log (2)))}{x (1+x)^2} \, dx\\ &=\frac {1}{5} (4-\log (\log (2))) \int \frac {e^{-x} \left (5-5 \left (-2+e^x\right ) x-\left (-5+8 e^x\right ) x^2-4 e^x x^3-5 x (1+x)^2 \log (x)\right )}{x (1+x)^2} \, dx\\ &=\frac {1}{5} (4-\log (\log (2))) \int \left (-\frac {5+8 x+4 x^2}{(1+x)^2}-\frac {5 e^{-x} (-1+x \log (x))}{x}\right ) \, dx\\ &=\frac {1}{5} (-4+\log (\log (2))) \int \frac {5+8 x+4 x^2}{(1+x)^2} \, dx+(-4+\log (\log (2))) \int \frac {e^{-x} (-1+x \log (x))}{x} \, dx\\ &=e^{-x} \log (x) (4-\log (\log (2)))+\frac {1}{5} (-4+\log (\log (2))) \int \left (4+\frac {1}{(1+x)^2}\right ) \, dx\\ &=-\frac {4}{5} x (4-\log (\log (2)))+\frac {4-\log (\log (2))}{5 (1+x)}+e^{-x} \log (x) (4-\log (\log (2)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 29, normalized size = 0.94 \begin {gather*} \frac {1}{5} \left (4 x-\frac {1}{1+x}-5 e^{-x} \log (x)\right ) (-4+\log (\log (2))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + 40*x + 20*x^2 + E^x*(-20*x - 32*x^2 - 16*x^3) + (-20*x - 40*x^2 - 20*x^3)*Log[x] + (-5 - 10*x
- 5*x^2 + E^x*(5*x + 8*x^2 + 4*x^3) + (5*x + 10*x^2 + 5*x^3)*Log[x])*Log[Log[2]])/(E^x*(5*x + 10*x^2 + 5*x^3))
,x]

[Out]

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

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fricas [B]  time = 0.51, size = 59, normalized size = 1.90 \begin {gather*} -\frac {{\left (4 \, {\left (4 \, x^{2} + 4 \, x - 1\right )} e^{x} - 20 \, {\left (x + 1\right )} \log \relax (x) - {\left ({\left (4 \, x^{2} + 4 \, x - 1\right )} e^{x} - 5 \, {\left (x + 1\right )} \log \relax (x)\right )} \log \left (\log \relax (2)\right )\right )} e^{\left (-x\right )}}{5 \, {\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^3+10*x^2+5*x)*log(x)+(4*x^3+8*x^2+5*x)*exp(x)-5*x^2-10*x-5)*log(log(2))+(-20*x^3-40*x^2-20*x)
*log(x)+(-16*x^3-32*x^2-20*x)*exp(x)+20*x^2+40*x+20)/(5*x^3+10*x^2+5*x)/exp(x),x, algorithm="fricas")

[Out]

-1/5*(4*(4*x^2 + 4*x - 1)*e^x - 20*(x + 1)*log(x) - ((4*x^2 + 4*x - 1)*e^x - 5*(x + 1)*log(x))*log(log(2)))*e^
(-x)/(x + 1)

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giac [B]  time = 0.16, size = 74, normalized size = 2.39 \begin {gather*} -\frac {5 \, x e^{\left (-x\right )} \log \relax (x) \log \left (\log \relax (2)\right ) - 20 \, x e^{\left (-x\right )} \log \relax (x) - 4 \, x^{2} \log \left (\log \relax (2)\right ) + 5 \, e^{\left (-x\right )} \log \relax (x) \log \left (\log \relax (2)\right ) + 16 \, x^{2} - 20 \, e^{\left (-x\right )} \log \relax (x) - 4 \, x \log \left (\log \relax (2)\right ) + 16 \, x + \log \left (\log \relax (2)\right ) - 4}{5 \, {\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^3+10*x^2+5*x)*log(x)+(4*x^3+8*x^2+5*x)*exp(x)-5*x^2-10*x-5)*log(log(2))+(-20*x^3-40*x^2-20*x)
*log(x)+(-16*x^3-32*x^2-20*x)*exp(x)+20*x^2+40*x+20)/(5*x^3+10*x^2+5*x)/exp(x),x, algorithm="giac")

[Out]

-1/5*(5*x*e^(-x)*log(x)*log(log(2)) - 20*x*e^(-x)*log(x) - 4*x^2*log(log(2)) + 5*e^(-x)*log(x)*log(log(2)) + 1
6*x^2 - 20*e^(-x)*log(x) - 4*x*log(log(2)) + 16*x + log(log(2)) - 4)/(x + 1)

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maple [A]  time = 0.11, size = 43, normalized size = 1.39

method	result	size
default	\(\frac {\left (-5 \ln \left (\ln \relax (2)\right )+20\right ) \ln \relax (x ) {\mathrm e}^{-x}}{5}+\frac {4 x \ln \left (\ln \relax (2)\right )}{5}-\frac {\ln \left (\ln \relax (2)\right )}{5 \left (x +1\right )}-\frac {16 x}{5}+\frac {4}{5 \left (x +1\right )}\)	\(43\)
risch	\(-\left (\ln \left (\ln \relax (2)\right )-4\right ) {\mathrm e}^{-x} \ln \relax (x )+\frac {4 x^{2} \ln \left (\ln \relax (2)\right )+4 x \ln \left (\ln \relax (2)\right )-16 x^{2}-\ln \left (\ln \relax (2)\right )-16 x +4}{5 x +5}\)	\(51\)
norman	\(\frac {\left (\left (4-\ln \left (\ln \relax (2)\right )\right ) \ln \relax (x )+\left (4-\ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{x}+\left (4-\ln \left (\ln \relax (2)\right )\right ) x \ln \relax (x )+\left (-\frac {16}{5}+\frac {4 \ln \left (\ln \relax (2)\right )}{5}\right ) x^{2} {\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x +1}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*x^3+10*x^2+5*x)*ln(x)+(4*x^3+8*x^2+5*x)*exp(x)-5*x^2-10*x-5)*ln(ln(2))+(-20*x^3-40*x^2-20*x)*ln(x)+(-
16*x^3-32*x^2-20*x)*exp(x)+20*x^2+40*x+20)/(5*x^3+10*x^2+5*x)/exp(x),x,method=_RETURNVERBOSE)

[Out]

1/5*(-5*ln(ln(2))+20)*ln(x)/exp(x)+4/5*x*ln(ln(2))-1/5*ln(ln(2))/(x+1)-16/5*x+4/5/(x+1)

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maxima [B]  time = 0.65, size = 52, normalized size = 1.68 \begin {gather*} \frac {4 \, x^{2} {\left (\log \left (\log \relax (2)\right ) - 4\right )} - 5 \, {\left (x {\left (\log \left (\log \relax (2)\right ) - 4\right )} + \log \left (\log \relax (2)\right ) - 4\right )} e^{\left (-x\right )} \log \relax (x) + 4 \, x {\left (\log \left (\log \relax (2)\right ) - 4\right )} - \log \left (\log \relax (2)\right ) + 4}{5 \, {\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^3+10*x^2+5*x)*log(x)+(4*x^3+8*x^2+5*x)*exp(x)-5*x^2-10*x-5)*log(log(2))+(-20*x^3-40*x^2-20*x)
*log(x)+(-16*x^3-32*x^2-20*x)*exp(x)+20*x^2+40*x+20)/(5*x^3+10*x^2+5*x)/exp(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.23, size = 38, normalized size = 1.23 \begin {gather*} \frac {{\mathrm {e}}^{-x}\,\left (\ln \left (\ln \relax (2)\right )-4\right )\,\left ({\mathrm {e}}^x-5\,\ln \relax (x)+4\,x\,{\mathrm {e}}^x\right )}{5}-\frac {\frac {\ln \left (\ln \relax (2)\right )}{5}-\frac {4}{5}}{x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(40*x - log(log(2))*(10*x + 5*x^2 - exp(x)*(5*x + 8*x^2 + 4*x^3) - log(x)*(5*x + 10*x^2 + 5*x^3)
+ 5) + 20*x^2 - exp(x)*(20*x + 32*x^2 + 16*x^3) - log(x)*(20*x + 40*x^2 + 20*x^3) + 20))/(5*x + 10*x^2 + 5*x^3
),x)

[Out]

(exp(-x)*(log(log(2)) - 4)*(exp(x) - 5*log(x) + 4*x*exp(x)))/5 - (log(log(2))/5 - 4/5)/(x + 1)

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sympy [A]  time = 0.61, size = 41, normalized size = 1.32 \begin {gather*} - x \left (\frac {16}{5} - \frac {4 \log {\left (\log {\relax (2 )} \right )}}{5}\right ) + \left (- \log {\relax (x )} \log {\left (\log {\relax (2 )} \right )} + 4 \log {\relax (x )}\right ) e^{- x} - \frac {-4 + \log {\left (\log {\relax (2 )} \right )}}{5 x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x**3+10*x**2+5*x)*ln(x)+(4*x**3+8*x**2+5*x)*exp(x)-5*x**2-10*x-5)*ln(ln(2))+(-20*x**3-40*x**2-2
0*x)*ln(x)+(-16*x**3-32*x**2-20*x)*exp(x)+20*x**2+40*x+20)/(5*x**3+10*x**2+5*x)/exp(x),x)

[Out]

-x*(16/5 - 4*log(log(2))/5) + (-log(x)*log(log(2)) + 4*log(x))*exp(-x) - (-4 + log(log(2)))/(5*x + 5)

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