3.97.77 \(\int \frac {8 x^3+4 e x^3+4 x^4+e^2 x^4+e^2 (8 x^3+4 e x^3+4 x^4) \log (e^3 (2+e+x))}{2+e+x} \, dx\) [9677]

Optimal. Leaf size=19 \[ x^4 \left (1+e^2 \log \left (e^3 (2+e+x)\right )\right ) \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 37, normalized size of antiderivative = 1.95, number of steps used = 9, number of rules used = 5, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {6, 6874, 78, 2442, 45} \begin {gather*} \frac {1}{4} \left (4+e^2\right ) x^4-\frac {e^2 x^4}{4}+e^2 x^4 (\log (x+e+2)+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*x^3 + 4*E*x^3 + 4*x^4 + E^2*x^4 + E^2*(8*x^3 + 4*E*x^3 + 4*x^4)*Log[E^3*(2 + E + x)])/(2 + E + x),x]

[Out]

-1/4*(E^2*x^4) + ((4 + E^2)*x^4)/4 + E^2*x^4*(3 + Log[2 + E + x])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2442

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

Rule 6874

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

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 20, normalized size = 1.05 \begin {gather*} x^4 \left (1+3 e^2+e^2 \log (2+e+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x^3 + 4*E*x^3 + 4*x^4 + E^2*x^4 + E^2*(8*x^3 + 4*E*x^3 + 4*x^4)*Log[E^3*(2 + E + x)])/(2 + E + x)
,x]

[Out]

x^4*(1 + 3*E^2 + E^2*Log[2 + E + x])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1038\) vs. \(2(20)=40\).
time = 0.07, size = 1039, normalized size = 54.68

method	result	size
norman	\(x^{4}+x^{4} {\mathrm e}^{2} \ln \left (\left (2+x +{\mathrm e}\right ) {\mathrm e}^{3}\right )\)	\(20\)
risch	\(x^{4}+x^{4} {\mathrm e}^{2} \ln \left (\left (2+x +{\mathrm e}\right ) {\mathrm e}^{3}\right )\)	\(20\)
derivativedivides	\(\text {Expression too large to display}\)	\(1039\)
default	\(\text {Expression too large to display}\)	\(1039\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3*exp(1)+4*x^4+8*x^3)*exp(2)*ln((2+x+exp(1))*exp(3))+x^4*exp(2)+4*x^3*exp(1)+4*x^4+8*x^3)/(2+x+exp(1
)),x,method=_RETURNVERBOSE)

[Out]

1/exp(3)*(-32*x*exp(3)-32*exp(2)*((x*exp(3)+(exp(1)+2)*exp(3))*ln(x*exp(3)+(exp(1)+2)*exp(3))-x*exp(3)-(exp(1)
+2)*exp(3))+24/exp(3)*(x*exp(3)+(exp(1)+2)*exp(3))^2-8/exp(3)^2*(x*exp(3)+(exp(1)+2)*exp(3))^3-4*exp(1)^3*(x*e
xp(3)+(exp(1)+2)*exp(3))-24*exp(1)^2*(x*exp(3)+(exp(1)+2)*exp(3))-48*exp(1)*(x*exp(3)+(exp(1)+2)*exp(3))-32*ex
p(2)*(x*exp(3)+(exp(1)+2)*exp(3))+1/exp(3)^3*(x*exp(3)+(exp(1)+2)*exp(3))^4-32*(exp(1)+2)*exp(3)+12/exp(3)*exp
(1)*exp(2)*(x*exp(3)+(exp(1)+2)*exp(3))^2-4/3/exp(3)^2*exp(1)*exp(2)*(x*exp(3)+(exp(1)+2)*exp(3))^3+32*exp(2)*
exp(3)*exp(1)*ln(x*exp(3)+(exp(1)+2)*exp(3))+8*exp(3)*exp(1)^3*exp(2)*ln(x*exp(3)+(exp(1)+2)*exp(3))+24*exp(3)
*exp(1)^2*exp(2)*ln(x*exp(3)+(exp(1)+2)*exp(3))-12/exp(3)^2*exp(1)*exp(2)*(1/3*(x*exp(3)+(exp(1)+2)*exp(3))^3*
ln(x*exp(3)+(exp(1)+2)*exp(3))-1/9*(x*exp(3)+(exp(1)+2)*exp(3))^3)+3/exp(3)*exp(1)^2*exp(2)*(x*exp(3)+(exp(1)+
2)*exp(3))^2+exp(3)*exp(1)^4*exp(2)*ln(x*exp(3)+(exp(1)+2)*exp(3))+12/exp(3)*exp(1)^2*exp(2)*(1/2*(x*exp(3)+(e
xp(1)+2)*exp(3))^2*ln(x*exp(3)+(exp(1)+2)*exp(3))-1/4*(x*exp(3)+(exp(1)+2)*exp(3))^2)+48/exp(3)*exp(1)*exp(2)*
(1/2*(x*exp(3)+(exp(1)+2)*exp(3))^2*ln(x*exp(3)+(exp(1)+2)*exp(3))-1/4*(x*exp(3)+(exp(1)+2)*exp(3))^2)+24/exp(
3)*exp(1)*(x*exp(3)+(exp(1)+2)*exp(3))^2-4/exp(3)^2*exp(1)*(x*exp(3)+(exp(1)+2)*exp(3))^3+12/exp(3)*exp(2)*(x*
exp(3)+(exp(1)+2)*exp(3))^2-8/3/exp(3)^2*exp(2)*(x*exp(3)+(exp(1)+2)*exp(3))^3+4/exp(3)^3*exp(2)*(1/4*(x*exp(3
)+(exp(1)+2)*exp(3))^4*ln(x*exp(3)+(exp(1)+2)*exp(3))-1/16*(x*exp(3)+(exp(1)+2)*exp(3))^4)+48/exp(3)*exp(2)*(1
/2*(x*exp(3)+(exp(1)+2)*exp(3))^2*ln(x*exp(3)+(exp(1)+2)*exp(3))-1/4*(x*exp(3)+(exp(1)+2)*exp(3))^2)-24/exp(3)
^2*exp(2)*(1/3*(x*exp(3)+(exp(1)+2)*exp(3))^3*ln(x*exp(3)+(exp(1)+2)*exp(3))-1/9*(x*exp(3)+(exp(1)+2)*exp(3))^
3)-4*exp(1)^3*exp(2)*(x*exp(3)+(exp(1)+2)*exp(3))+16*exp(2)*exp(3)*ln(x*exp(3)+(exp(1)+2)*exp(3))+1/4/exp(3)^3
*exp(2)*(x*exp(3)+(exp(1)+2)*exp(3))^4-48*exp(1)*exp(2)*(x*exp(3)+(exp(1)+2)*exp(3))-4*exp(1)^3*exp(2)*((x*exp
(3)+(exp(1)+2)*exp(3))*ln(x*exp(3)+(exp(1)+2)*exp(3))-x*exp(3)-(exp(1)+2)*exp(3))-24*exp(2)*exp(1)^2*((x*exp(3
)+(exp(1)+2)*exp(3))*ln(x*exp(3)+(exp(1)+2)*exp(3))-x*exp(3)-(exp(1)+2)*exp(3))-48*exp(1)*exp(2)*((x*exp(3)+(e
xp(1)+2)*exp(3))*ln(x*exp(3)+(exp(1)+2)*exp(3))-x*exp(3)-(exp(1)+2)*exp(3))-24*exp(2)*exp(1)^2*(x*exp(3)+(exp(
1)+2)*exp(3))+6/exp(3)*exp(1)^2*(x*exp(3)+(exp(1)+2)*exp(3))^2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 679 vs. \(2 (18) = 36\).
time = 0.28, size = 679, normalized size = 35.74 \begin {gather*} x^{4} - \frac {4}{3} \, x^{3} {\left (e + 2\right )} + \frac {8}{3} \, x^{3} + 2 \, x^{2} {\left (e^{2} + 4 \, e + 4\right )} - 4 \, x^{2} {\left (e + 2\right )} + \frac {2}{3} \, {\left (2 \, x^{3} - 3 \, x^{2} {\left (e + 2\right )} + 6 \, x {\left (e^{2} + 4 \, e + 4\right )} - 6 \, {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} \log \left (x + e + 2\right )\right )} e^{3} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right ) + \frac {1}{3} \, {\left (3 \, x^{4} - 4 \, x^{3} {\left (e + 2\right )} + 6 \, x^{2} {\left (e^{2} + 4 \, e + 4\right )} - 12 \, x {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} + 12 \, {\left (e^{4} + 8 \, e^{3} + 24 \, e^{2} + 32 \, e + 16\right )} \log \left (x + e + 2\right )\right )} e^{2} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right ) + \frac {4}{3} \, {\left (2 \, x^{3} - 3 \, x^{2} {\left (e + 2\right )} + 6 \, x {\left (e^{2} + 4 \, e + 4\right )} - 6 \, {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} \log \left (x + e + 2\right )\right )} e^{2} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right ) - 4 \, x {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} + 8 \, x {\left (e^{2} + 4 \, e + 4\right )} - \frac {1}{9} \, {\left (4 \, x^{3} - 15 \, x^{2} {\left (e + 2\right )} - 18 \, {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} \log \left (x + e + 2\right )^{2} + 66 \, x {\left (e^{2} + 4 \, e + 4\right )} - 66 \, {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} \log \left (x + e + 2\right )\right )} e^{3} - \frac {1}{36} \, {\left (9 \, x^{4} - 28 \, x^{3} {\left (e + 2\right )} + 78 \, x^{2} {\left (e^{2} + 4 \, e + 4\right )} + 72 \, {\left (e^{4} + 8 \, e^{3} + 24 \, e^{2} + 32 \, e + 16\right )} \log \left (x + e + 2\right )^{2} - 300 \, x {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} + 300 \, {\left (e^{4} + 8 \, e^{3} + 24 \, e^{2} + 32 \, e + 16\right )} \log \left (x + e + 2\right )\right )} e^{2} + \frac {1}{12} \, {\left (3 \, x^{4} - 4 \, x^{3} {\left (e + 2\right )} + 6 \, x^{2} {\left (e^{2} + 4 \, e + 4\right )} - 12 \, x {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} + 12 \, {\left (e^{4} + 8 \, e^{3} + 24 \, e^{2} + 32 \, e + 16\right )} \log \left (x + e + 2\right )\right )} e^{2} - \frac {2}{9} \, {\left (4 \, x^{3} - 15 \, x^{2} {\left (e + 2\right )} - 18 \, {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} \log \left (x + e + 2\right )^{2} + 66 \, x {\left (e^{2} + 4 \, e + 4\right )} - 66 \, {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} \log \left (x + e + 2\right )\right )} e^{2} + \frac {2}{3} \, {\left (2 \, x^{3} - 3 \, x^{2} {\left (e + 2\right )} + 6 \, x {\left (e^{2} + 4 \, e + 4\right )} - 6 \, {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} \log \left (x + e + 2\right )\right )} e + 4 \, {\left (e^{4} + 8 \, e^{3} + 24 \, e^{2} + 32 \, e + 16\right )} \log \left (x + e + 2\right ) - 8 \, {\left (e^{3} + 6 \, e^{2} + 12 \, e + 8\right )} \log \left (x + e + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3*exp(1)+4*x^4+8*x^3)*exp(2)*log((2+x+exp(1))*exp(3))+x^4*exp(2)+4*x^3*exp(1)+4*x^4+8*x^3)/(2+
x+exp(1)),x, algorithm="maxima")

[Out]

x^4 - 4/3*x^3*(e + 2) + 8/3*x^3 + 2*x^2*(e^2 + 4*e + 4) - 4*x^2*(e + 2) + 2/3*(2*x^3 - 3*x^2*(e + 2) + 6*x*(e^
2 + 4*e + 4) - 6*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2))*e^3*log(x*e^3 + e^4 + 2*e^3) + 1/3*(3*x^4 - 4*x^3*(e
+ 2) + 6*x^2*(e^2 + 4*e + 4) - 12*x*(e^3 + 6*e^2 + 12*e + 8) + 12*(e^4 + 8*e^3 + 24*e^2 + 32*e + 16)*log(x +
e + 2))*e^2*log(x*e^3 + e^4 + 2*e^3) + 4/3*(2*x^3 - 3*x^2*(e + 2) + 6*x*(e^2 + 4*e + 4) - 6*(e^3 + 6*e^2 + 12*
e + 8)*log(x + e + 2))*e^2*log(x*e^3 + e^4 + 2*e^3) - 4*x*(e^3 + 6*e^2 + 12*e + 8) + 8*x*(e^2 + 4*e + 4) - 1/9
*(4*x^3 - 15*x^2*(e + 2) - 18*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2)^2 + 66*x*(e^2 + 4*e + 4) - 66*(e^3 + 6*e
^2 + 12*e + 8)*log(x + e + 2))*e^3 - 1/36*(9*x^4 - 28*x^3*(e + 2) + 78*x^2*(e^2 + 4*e + 4) + 72*(e^4 + 8*e^3 +
24*e^2 + 32*e + 16)*log(x + e + 2)^2 - 300*x*(e^3 + 6*e^2 + 12*e + 8) + 300*(e^4 + 8*e^3 + 24*e^2 + 32*e + 16
)*log(x + e + 2))*e^2 + 1/12*(3*x^4 - 4*x^3*(e + 2) + 6*x^2*(e^2 + 4*e + 4) - 12*x*(e^3 + 6*e^2 + 12*e + 8) +
12*(e^4 + 8*e^3 + 24*e^2 + 32*e + 16)*log(x + e + 2))*e^2 - 2/9*(4*x^3 - 15*x^2*(e + 2) - 18*(e^3 + 6*e^2 + 12
*e + 8)*log(x + e + 2)^2 + 66*x*(e^2 + 4*e + 4) - 66*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2))*e^2 + 2/3*(2*x^3
- 3*x^2*(e + 2) + 6*x*(e^2 + 4*e + 4) - 6*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2))*e + 4*(e^4 + 8*e^3 + 24*e^
2 + 32*e + 16)*log(x + e + 2) - 8*(e^3 + 6*e^2 + 12*e + 8)*log(x + e + 2)

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Fricas [A]
time = 0.38, size = 20, normalized size = 1.05 \begin {gather*} x^{4} e^{2} \log \left ({\left (x + 2\right )} e^{3} + e^{4}\right ) + x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3*exp(1)+4*x^4+8*x^3)*exp(2)*log((2+x+exp(1))*exp(3))+x^4*exp(2)+4*x^3*exp(1)+4*x^4+8*x^3)/(2+
x+exp(1)),x, algorithm="fricas")

[Out]

x^4*e^2*log((x + 2)*e^3 + e^4) + x^4

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Sympy [A]
time = 0.11, size = 20, normalized size = 1.05 \begin {gather*} x^{4} e^{2} \log {\left (\left (x + 2 + e\right ) e^{3} \right )} + x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3*exp(1)+4*x**4+8*x**3)*exp(2)*ln((2+x+exp(1))*exp(3))+x**4*exp(2)+4*x**3*exp(1)+4*x**4+8*x**
3)/(2+x+exp(1)),x)

[Out]

x**4*exp(2)*log((x + 2 + E)*exp(3)) + x**4

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Giac [A]
time = 0.43, size = 22, normalized size = 1.16 \begin {gather*} x^{4} e^{2} \log \left (x e^{3} + e^{4} + 2 \, e^{3}\right ) + x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3*exp(1)+4*x^4+8*x^3)*exp(2)*log((2+x+exp(1))*exp(3))+x^4*exp(2)+4*x^3*exp(1)+4*x^4+8*x^3)/(2+
x+exp(1)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 6.42, size = 18, normalized size = 0.95 \begin {gather*} x^4\,\left ({\mathrm {e}}^2\,\ln \left ({\mathrm {e}}^3\,\left (x+\mathrm {e}+2\right )\right )+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^3*exp(1) + x^4*exp(2) + 8*x^3 + 4*x^4 + exp(2)*log(exp(3)*(x + exp(1) + 2))*(4*x^3*exp(1) + 8*x^3 + 4
*x^4))/(x + exp(1) + 2),x)

[Out]

x^4*(exp(2)*log(exp(3)*(x + exp(1) + 2)) + 1)

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