3.99.2 \(\int \frac {e^8 (12288 e x^2+12800 x^3)+e^8 (1536 e x^2+1568 x^3) \log (e+x)+e^8 (48 e x^2+48 x^3) \log ^2(e+x)}{e+x} \, dx\)

Optimal. Leaf size=16 \[ 16 e^8 x^3 (16+\log (e+x))^2 \]

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Rubi [B]  time = 0.56, antiderivative size = 169, normalized size of antiderivative = 10.56, number of steps used = 23, number of rules used = 14, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.215, Rules used = {6688, 12, 6742, 77, 2418, 2389, 2295, 2395, 43, 2390, 2301, 2398, 2411, 2334} \begin {gather*} \frac {36832 e^8 x^3}{9}+16 e^8 x^3 \log ^2(x+e)+\frac {1568}{3} e^8 x^3 \log (x+e)+\frac {40 e^9 x^2}{3}-16 e^9 x^2 \log (x+e)+\frac {112 e^{10} x}{3}+\frac {32}{9} e^8 (x+e)^3-24 e^9 (x+e)^2-32 e^{11} \log ^2(x+e)+32 e^{10} (x+e) \log (x+e)+\frac {80}{3} e^{11} \log (x+e)-\frac {16}{3} e^8 \log (x+e) \left (2 (x+e)^3-9 e (x+e)^2+18 e^2 (x+e)-6 e^3 \log (x+e)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^8*(12288*E*x^2 + 12800*x^3) + E^8*(1536*E*x^2 + 1568*x^3)*Log[E + x] + E^8*(48*E*x^2 + 48*x^3)*Log[E +
x]^2)/(E + x),x]

[Out]

(112*E^10*x)/3 + (40*E^9*x^2)/3 + (36832*E^8*x^3)/9 - 24*E^9*(E + x)^2 + (32*E^8*(E + x)^3)/9 + (80*E^11*Log[E
+ x])/3 - 16*E^9*x^2*Log[E + x] + (1568*E^8*x^3*Log[E + x])/3 + 32*E^10*(E + x)*Log[E + x] - 32*E^11*Log[E +
x]^2 + 16*E^8*x^3*Log[E + x]^2 - (16*E^8*Log[E + x]*(18*E^2*(E + x) - 9*E*(E + x)^2 + 2*(E + x)^3 - 6*E^3*Log[
E + x]))/3

Rule 12

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

Rule 43

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 77

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, 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 2395

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 2398

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
f + g*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(g*(q + 1)), x] - Dist[(b*e*n*p)/(g*(q + 1)), Int[((f + g*x)^(q
+ 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*
f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && IntegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

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 {16 e^8 x^2 (16+\log (e+x)) (48 e+50 x+3 (e+x) \log (e+x))}{e+x} \, dx\\ &=\left (16 e^8\right ) \int \frac {x^2 (16+\log (e+x)) (48 e+50 x+3 (e+x) \log (e+x))}{e+x} \, dx\\ &=\left (16 e^8\right ) \int \left (\frac {32 x^2 (24 e+25 x)}{e+x}+\frac {2 x^2 (48 e+49 x) \log (e+x)}{e+x}+3 x^2 \log ^2(e+x)\right ) \, dx\\ &=\left (32 e^8\right ) \int \frac {x^2 (48 e+49 x) \log (e+x)}{e+x} \, dx+\left (48 e^8\right ) \int x^2 \log ^2(e+x) \, dx+\left (512 e^8\right ) \int \frac {x^2 (24 e+25 x)}{e+x} \, dx\\ &=16 e^8 x^3 \log ^2(e+x)-\left (32 e^8\right ) \int \frac {x^3 \log (e+x)}{e+x} \, dx+\left (32 e^8\right ) \int \left (e^2 \log (e+x)-e x \log (e+x)+49 x^2 \log (e+x)-\frac {e^3 \log (e+x)}{e+x}\right ) \, dx+\left (512 e^8\right ) \int \left (e^2-e x+25 x^2-\frac {e^3}{e+x}\right ) \, dx\\ &=512 e^{10} x-256 e^9 x^2+\frac {12800 e^8 x^3}{3}-512 e^{11} \log (e+x)+16 e^8 x^3 \log ^2(e+x)-\left (32 e^8\right ) \operatorname {Subst}\left (\int \frac {(-e+x)^3 \log (x)}{x} \, dx,x,e+x\right )+\left (1568 e^8\right ) \int x^2 \log (e+x) \, dx-\left (32 e^9\right ) \int x \log (e+x) \, dx+\left (32 e^{10}\right ) \int \log (e+x) \, dx-\left (32 e^{11}\right ) \int \frac {\log (e+x)}{e+x} \, dx\\ &=512 e^{10} x-256 e^9 x^2+\frac {12800 e^8 x^3}{3}-512 e^{11} \log (e+x)-16 e^9 x^2 \log (e+x)+\frac {1568}{3} e^8 x^3 \log (e+x)+16 e^8 x^3 \log ^2(e+x)-\frac {16}{3} e^8 \log (e+x) \left (18 e^2 (e+x)-9 e (e+x)^2+2 (e+x)^3-6 e^3 \log (e+x)\right )+\left (32 e^8\right ) \operatorname {Subst}\left (\int \left (3 e^2-\frac {3 e x}{2}+\frac {x^2}{3}-\frac {e^3 \log (x)}{x}\right ) \, dx,x,e+x\right )-\frac {1}{3} \left (1568 e^8\right ) \int \frac {x^3}{e+x} \, dx+\left (16 e^9\right ) \int \frac {x^2}{e+x} \, dx+\left (32 e^{10}\right ) \operatorname {Subst}(\int \log (x) \, dx,x,e+x)-\left (32 e^{11}\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,e+x\right )\\ &=576 e^{10} x-256 e^9 x^2+\frac {12800 e^8 x^3}{3}-24 e^9 (e+x)^2+\frac {32}{9} e^8 (e+x)^3-512 e^{11} \log (e+x)-16 e^9 x^2 \log (e+x)+\frac {1568}{3} e^8 x^3 \log (e+x)+32 e^{10} (e+x) \log (e+x)-16 e^{11} \log ^2(e+x)+16 e^8 x^3 \log ^2(e+x)-\frac {16}{3} e^8 \log (e+x) \left (18 e^2 (e+x)-9 e (e+x)^2+2 (e+x)^3-6 e^3 \log (e+x)\right )-\frac {1}{3} \left (1568 e^8\right ) \int \left (e^2-e x+x^2-\frac {e^3}{e+x}\right ) \, dx+\left (16 e^9\right ) \int \left (-e+x+\frac {e^2}{e+x}\right ) \, dx-\left (32 e^{11}\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,e+x\right )\\ &=\frac {112 e^{10} x}{3}+\frac {40 e^9 x^2}{3}+\frac {36832 e^8 x^3}{9}-24 e^9 (e+x)^2+\frac {32}{9} e^8 (e+x)^3+\frac {80}{3} e^{11} \log (e+x)-16 e^9 x^2 \log (e+x)+\frac {1568}{3} e^8 x^3 \log (e+x)+32 e^{10} (e+x) \log (e+x)-32 e^{11} \log ^2(e+x)+16 e^8 x^3 \log ^2(e+x)-\frac {16}{3} e^8 \log (e+x) \left (18 e^2 (e+x)-9 e (e+x)^2+2 (e+x)^3-6 e^3 \log (e+x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 30, normalized size = 1.88 \begin {gather*} 16 e^8 \left (256 x^3+32 x^3 \log (e+x)+x^3 \log ^2(e+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^8*(12288*E*x^2 + 12800*x^3) + E^8*(1536*E*x^2 + 1568*x^3)*Log[E + x] + E^8*(48*E*x^2 + 48*x^3)*Lo
g[E + x]^2)/(E + x),x]

[Out]

16*E^8*(256*x^3 + 32*x^3*Log[E + x] + x^3*Log[E + x]^2)

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fricas [B]  time = 0.67, size = 34, normalized size = 2.12 \begin {gather*} 16 \, x^{3} e^{8} \log \left (x + e\right )^{2} + 512 \, x^{3} e^{8} \log \left (x + e\right ) + 4096 \, x^{3} e^{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x^2*exp(1)+48*x^3)*exp(4)^2*log(x+exp(1))^2+(1536*x^2*exp(1)+1568*x^3)*exp(4)^2*log(x+exp(1))+(
12288*x^2*exp(1)+12800*x^3)*exp(4)^2)/(x+exp(1)),x, algorithm="fricas")

[Out]

16*x^3*e^8*log(x + e)^2 + 512*x^3*e^8*log(x + e) + 4096*x^3*e^8

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giac [B]  time = 0.17, size = 34, normalized size = 2.12 \begin {gather*} 16 \, x^{3} e^{8} \log \left (x + e\right )^{2} + 512 \, x^{3} e^{8} \log \left (x + e\right ) + 4096 \, x^{3} e^{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x^2*exp(1)+48*x^3)*exp(4)^2*log(x+exp(1))^2+(1536*x^2*exp(1)+1568*x^3)*exp(4)^2*log(x+exp(1))+(
12288*x^2*exp(1)+12800*x^3)*exp(4)^2)/(x+exp(1)),x, algorithm="giac")

[Out]

16*x^3*e^8*log(x + e)^2 + 512*x^3*e^8*log(x + e) + 4096*x^3*e^8

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maple [A]  time = 0.09, size = 35, normalized size = 2.19

method	result	size
risch	\(16 x^{3} {\mathrm e}^{8} \ln \left (x +{\mathrm e}\right )^{2}+512 x^{3} {\mathrm e}^{8} \ln \left (x +{\mathrm e}\right )+4096 x^{3} {\mathrm e}^{8}\)	\(35\)
norman	\(16 x^{3} {\mathrm e}^{8} \ln \left (x +{\mathrm e}\right )^{2}+512 x^{3} {\mathrm e}^{8} \ln \left (x +{\mathrm e}\right )+4096 x^{3} {\mathrm e}^{8}\)	\(41\)
derivativedivides	\(48 \,{\mathrm e}^{8} {\mathrm e}^{2} \left (\left (x +{\mathrm e}\right ) \ln \left (x +{\mathrm e}\right )^{2}-2 \left (x +{\mathrm e}\right ) \ln \left (x +{\mathrm e}\right )+2 x +2 \,{\mathrm e}\right )-96 \,{\mathrm e}^{8} {\mathrm e} \left (\frac {\left (x +{\mathrm e}\right )^{2} \ln \left (x +{\mathrm e}\right )^{2}}{2}-\frac {\left (x +{\mathrm e}\right )^{2} \ln \left (x +{\mathrm e}\right )}{2}+\frac {\left (x +{\mathrm e}\right )^{2}}{4}\right )+48 \,{\mathrm e}^{8} \left (\frac {\ln \left (x +{\mathrm e}\right )^{2} \left (x +{\mathrm e}\right )^{3}}{3}-\frac {2 \ln \left (x +{\mathrm e}\right ) \left (x +{\mathrm e}\right )^{3}}{9}+\frac {2 \left (x +{\mathrm e}\right )^{3}}{27}\right )-16 \,{\mathrm e}^{8} {\mathrm e}^{3} \ln \left (x +{\mathrm e}\right )^{2}+1632 \,{\mathrm e}^{8} {\mathrm e}^{2} \left (\left (x +{\mathrm e}\right ) \ln \left (x +{\mathrm e}\right )-x -{\mathrm e}\right )-3168 \,{\mathrm e}^{8} {\mathrm e} \left (\frac {\left (x +{\mathrm e}\right )^{2} \ln \left (x +{\mathrm e}\right )}{2}-\frac {\left (x +{\mathrm e}\right )^{2}}{4}\right )+1568 \,{\mathrm e}^{8} \left (\frac {\ln \left (x +{\mathrm e}\right ) \left (x +{\mathrm e}\right )^{3}}{3}-\frac {\left (x +{\mathrm e}\right )^{3}}{9}\right )-512 \,{\mathrm e}^{8} {\mathrm e}^{3} \ln \left (x +{\mathrm e}\right )+13824 \,{\mathrm e}^{8} {\mathrm e}^{2} \left (x +{\mathrm e}\right )-13056 \,{\mathrm e}^{8} {\mathrm e} \left (x +{\mathrm e}\right )^{2}+\frac {12800 \,{\mathrm e}^{8} \left (x +{\mathrm e}\right )^{3}}{3}\)	\(289\)
default	\(48 \,{\mathrm e}^{8} {\mathrm e}^{2} \left (\left (x +{\mathrm e}\right ) \ln \left (x +{\mathrm e}\right )^{2}-2 \left (x +{\mathrm e}\right ) \ln \left (x +{\mathrm e}\right )+2 x +2 \,{\mathrm e}\right )-96 \,{\mathrm e}^{8} {\mathrm e} \left (\frac {\left (x +{\mathrm e}\right )^{2} \ln \left (x +{\mathrm e}\right )^{2}}{2}-\frac {\left (x +{\mathrm e}\right )^{2} \ln \left (x +{\mathrm e}\right )}{2}+\frac {\left (x +{\mathrm e}\right )^{2}}{4}\right )+48 \,{\mathrm e}^{8} \left (\frac {\ln \left (x +{\mathrm e}\right )^{2} \left (x +{\mathrm e}\right )^{3}}{3}-\frac {2 \ln \left (x +{\mathrm e}\right ) \left (x +{\mathrm e}\right )^{3}}{9}+\frac {2 \left (x +{\mathrm e}\right )^{3}}{27}\right )-16 \,{\mathrm e}^{8} {\mathrm e}^{3} \ln \left (x +{\mathrm e}\right )^{2}+1632 \,{\mathrm e}^{8} {\mathrm e}^{2} \left (\left (x +{\mathrm e}\right ) \ln \left (x +{\mathrm e}\right )-x -{\mathrm e}\right )-3168 \,{\mathrm e}^{8} {\mathrm e} \left (\frac {\left (x +{\mathrm e}\right )^{2} \ln \left (x +{\mathrm e}\right )}{2}-\frac {\left (x +{\mathrm e}\right )^{2}}{4}\right )+1568 \,{\mathrm e}^{8} \left (\frac {\ln \left (x +{\mathrm e}\right ) \left (x +{\mathrm e}\right )^{3}}{3}-\frac {\left (x +{\mathrm e}\right )^{3}}{9}\right )-512 \,{\mathrm e}^{8} {\mathrm e}^{3} \ln \left (x +{\mathrm e}\right )+13824 \,{\mathrm e}^{8} {\mathrm e}^{2} \left (x +{\mathrm e}\right )-13056 \,{\mathrm e}^{8} {\mathrm e} \left (x +{\mathrm e}\right )^{2}+\frac {12800 \,{\mathrm e}^{8} \left (x +{\mathrm e}\right )^{3}}{3}\)	\(289\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((48*x^2*exp(1)+48*x^3)*exp(4)^2*ln(x+exp(1))^2+(1536*x^2*exp(1)+1568*x^3)*exp(4)^2*ln(x+exp(1))+(12288*x^
2*exp(1)+12800*x^3)*exp(4)^2)/(x+exp(1)),x,method=_RETURNVERBOSE)

[Out]

16*x^3*exp(8)*ln(x+exp(1))^2+512*x^3*exp(8)*ln(x+exp(1))+4096*x^3*exp(8)

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maxima [B]  time = 0.36, size = 367, normalized size = 22.94 \begin {gather*} 768 \, {\left (x^{2} - 2 \, x e + 2 \, e^{2} \log \left (x + e\right )\right )} e^{9} \log \left (x + e\right ) + \frac {784}{3} \, {\left (2 \, x^{3} - 3 \, x^{2} e + 6 \, x e^{2} - 6 \, e^{3} \log \left (x + e\right )\right )} e^{8} \log \left (x + e\right ) + 4 \, {\left (4 \, e^{2} \log \left (x + e\right )^{3} + 3 \, {\left (2 \, \log \left (x + e\right )^{2} - 2 \, \log \left (x + e\right ) + 1\right )} {\left (x + e\right )}^{2} - 24 \, {\left (e \log \left (x + e\right )^{2} - 2 \, e \log \left (x + e\right ) + 2 \, e\right )} {\left (x + e\right )}\right )} e^{9} - 384 \, {\left (2 \, e^{2} \log \left (x + e\right )^{2} + x^{2} - 6 \, x e + 6 \, e^{2} \log \left (x + e\right )\right )} e^{9} + 6144 \, {\left (x^{2} - 2 \, x e + 2 \, e^{2} \log \left (x + e\right )\right )} e^{9} + \frac {4}{9} \, {\left (4 \, {\left (9 \, \log \left (x + e\right )^{2} - 6 \, \log \left (x + e\right ) + 2\right )} {\left (x + e\right )}^{3} - 36 \, e^{3} \log \left (x + e\right )^{3} - 81 \, {\left (2 \, e \log \left (x + e\right )^{2} - 2 \, e \log \left (x + e\right ) + e\right )} {\left (x + e\right )}^{2} + 324 \, {\left (e^{2} \log \left (x + e\right )^{2} - 2 \, e^{2} \log \left (x + e\right ) + 2 \, e^{2}\right )} {\left (x + e\right )}\right )} e^{8} - \frac {392}{9} \, {\left (4 \, x^{3} - 15 \, x^{2} e - 18 \, e^{3} \log \left (x + e\right )^{2} + 66 \, x e^{2} - 66 \, e^{3} \log \left (x + e\right )\right )} e^{8} + \frac {6400}{3} \, {\left (2 \, x^{3} - 3 \, x^{2} e + 6 \, x e^{2} - 6 \, e^{3} \log \left (x + e\right )\right )} e^{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x^2*exp(1)+48*x^3)*exp(4)^2*log(x+exp(1))^2+(1536*x^2*exp(1)+1568*x^3)*exp(4)^2*log(x+exp(1))+(
12288*x^2*exp(1)+12800*x^3)*exp(4)^2)/(x+exp(1)),x, algorithm="maxima")

[Out]

768*(x^2 - 2*x*e + 2*e^2*log(x + e))*e^9*log(x + e) + 784/3*(2*x^3 - 3*x^2*e + 6*x*e^2 - 6*e^3*log(x + e))*e^8
*log(x + e) + 4*(4*e^2*log(x + e)^3 + 3*(2*log(x + e)^2 - 2*log(x + e) + 1)*(x + e)^2 - 24*(e*log(x + e)^2 - 2
*e*log(x + e) + 2*e)*(x + e))*e^9 - 384*(2*e^2*log(x + e)^2 + x^2 - 6*x*e + 6*e^2*log(x + e))*e^9 + 6144*(x^2
- 2*x*e + 2*e^2*log(x + e))*e^9 + 4/9*(4*(9*log(x + e)^2 - 6*log(x + e) + 2)*(x + e)^3 - 36*e^3*log(x + e)^3 -
81*(2*e*log(x + e)^2 - 2*e*log(x + e) + e)*(x + e)^2 + 324*(e^2*log(x + e)^2 - 2*e^2*log(x + e) + 2*e^2)*(x +
e))*e^8 - 392/9*(4*x^3 - 15*x^2*e - 18*e^3*log(x + e)^2 + 66*x*e^2 - 66*e^3*log(x + e))*e^8 + 6400/3*(2*x^3 -
3*x^2*e + 6*x*e^2 - 6*e^3*log(x + e))*e^8

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mupad [B]  time = 5.74, size = 16, normalized size = 1.00 \begin {gather*} 16\,x^3\,{\mathrm {e}}^8\,{\left (\ln \left (x+\mathrm {e}\right )+16\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(8)*(12288*x^2*exp(1) + 12800*x^3) + exp(8)*log(x + exp(1))^2*(48*x^2*exp(1) + 48*x^3) + exp(8)*log(x
+ exp(1))*(1536*x^2*exp(1) + 1568*x^3))/(x + exp(1)),x)

[Out]

16*x^3*exp(8)*(log(x + exp(1)) + 16)^2

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sympy [B]  time = 0.18, size = 39, normalized size = 2.44 \begin {gather*} 16 x^{3} e^{8} \log {\left (x + e \right )}^{2} + 512 x^{3} e^{8} \log {\left (x + e \right )} + 4096 x^{3} e^{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x**2*exp(1)+48*x**3)*exp(4)**2*ln(x+exp(1))**2+(1536*x**2*exp(1)+1568*x**3)*exp(4)**2*ln(x+exp(
1))+(12288*x**2*exp(1)+12800*x**3)*exp(4)**2)/(x+exp(1)),x)

[Out]

16*x**3*exp(8)*log(x + E)**2 + 512*x**3*exp(8)*log(x + E) + 4096*x**3*exp(8)

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