3.49.96 \(\int \frac {e^{-2 x} (576+36 x+e^{2 x} (288 x+19 x^2)+(576-1112 x-72 x^2) \log (x)+(64+4 x+e^{2 x} (32 x+2 x^2)+(64-124 x-8 x^2) \log (x)) \log (16+x))}{16+x} \, dx\)

Optimal. Leaf size=19 \[ x \left (x+4 e^{-2 x} \log (x)\right ) (9+\log (16+x)) \]

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Rubi [A]  time = 4.14, antiderivative size = 38, normalized size of antiderivative = 2.00, number of steps used = 45, number of rules used = 14, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {6742, 77, 2395, 43, 6688, 2194, 2199, 2176, 2178, 2554, 12, 2288, 2557, 14} \begin {gather*} 9 x^2+x^2 \log (x+16)+36 e^{-2 x} x \log (x)+4 e^{-2 x} x \log (x) \log (x+16) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(576 + 36*x + E^(2*x)*(288*x + 19*x^2) + (576 - 1112*x - 72*x^2)*Log[x] + (64 + 4*x + E^(2*x)*(32*x + 2*x^
2) + (64 - 124*x - 8*x^2)*Log[x])*Log[16 + x])/(E^(2*x)*(16 + x)),x]

[Out]

9*x^2 + (36*x*Log[x])/E^(2*x) + x^2*Log[16 + x] + (4*x*Log[x]*Log[16 + x])/E^(2*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2176

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

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

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

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 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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 2557

Int[Log[v_]*Log[w_]*(u_), x_Symbol] :> With[{z = IntHide[u, x]}, Dist[Log[v]*Log[w], z, x] + (-Int[SimplifyInt
egrand[(z*Log[w]*D[v, x])/v, x], x] - Int[SimplifyIntegrand[(z*Log[v]*D[w, x])/w, x], x]) /; InverseFunctionFr
eeQ[z, x]] /; InverseFunctionFreeQ[v, x] && InverseFunctionFreeQ[w, 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 \left (\frac {x (288+19 x+32 \log (16+x)+2 x \log (16+x))}{16+x}-\frac {4 e^{-2 x} \left (-144-9 x-144 \log (x)+278 x \log (x)+18 x^2 \log (x)-16 \log (16+x)-x \log (16+x)-16 \log (x) \log (16+x)+31 x \log (x) \log (16+x)+2 x^2 \log (x) \log (16+x)\right )}{16+x}\right ) \, dx\\ &=-\left (4 \int \frac {e^{-2 x} \left (-144-9 x-144 \log (x)+278 x \log (x)+18 x^2 \log (x)-16 \log (16+x)-x \log (16+x)-16 \log (x) \log (16+x)+31 x \log (x) \log (16+x)+2 x^2 \log (x) \log (16+x)\right )}{16+x} \, dx\right )+\int \frac {x (288+19 x+32 \log (16+x)+2 x \log (16+x))}{16+x} \, dx\\ &=-\left (4 \int \frac {e^{-2 x} \left (-((16+x) (9+\log (16+x)))+\log (x) \left (2 \left (-72+139 x+9 x^2\right )+\left (-16+31 x+2 x^2\right ) \log (16+x)\right )\right )}{16+x} \, dx\right )+\int \left (\frac {x (288+19 x)}{16+x}+2 x \log (16+x)\right ) \, dx\\ &=2 \int x \log (16+x) \, dx-4 \int \left (\frac {e^{-2 x} \left (-144-9 x-144 \log (x)+278 x \log (x)+18 x^2 \log (x)\right )}{16+x}+e^{-2 x} (-1-\log (x)+2 x \log (x)) \log (16+x)\right ) \, dx+\int \frac {x (288+19 x)}{16+x} \, dx\\ &=x^2 \log (16+x)-4 \int \frac {e^{-2 x} \left (-144-9 x-144 \log (x)+278 x \log (x)+18 x^2 \log (x)\right )}{16+x} \, dx-4 \int e^{-2 x} (-1-\log (x)+2 x \log (x)) \log (16+x) \, dx-\int \frac {x^2}{16+x} \, dx+\int \left (-16+19 x+\frac {256}{16+x}\right ) \, dx\\ &=-16 x+\frac {19 x^2}{2}+256 \log (16+x)+x^2 \log (16+x)-4 \int \left (-9 e^{-2 x}+\frac {2 e^{-2 x} \left (-72+139 x+9 x^2\right ) \log (x)}{16+x}\right ) \, dx-4 \int \left (-e^{-2 x} \log (16+x)-e^{-2 x} \log (x) \log (16+x)+2 e^{-2 x} x \log (x) \log (16+x)\right ) \, dx-\int \left (-16+x+\frac {256}{16+x}\right ) \, dx\\ &=9 x^2+x^2 \log (16+x)+4 \int e^{-2 x} \log (16+x) \, dx+4 \int e^{-2 x} \log (x) \log (16+x) \, dx-8 \int \frac {e^{-2 x} \left (-72+139 x+9 x^2\right ) \log (x)}{16+x} \, dx-8 \int e^{-2 x} x \log (x) \log (16+x) \, dx+36 \int e^{-2 x} \, dx\\ &=-18 e^{-2 x}+9 x^2-2 e^{-2 x} \log (x)+36 e^{-2 x} x \log (x)-64 e^{32} \text {Ei}(-2 (16+x)) \log (x)-2 e^{-2 x} \log (16+x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)-4 \int -\frac {e^{-2 x}}{2 (16+x)} \, dx-4 \int -\frac {e^{-2 x} \log (x)}{2 (16+x)} \, dx-4 \int -\frac {e^{-2 x} \log (16+x)}{2 x} \, dx+8 \int \frac {e^{-2 x} \left (1-18 x+32 e^{32+2 x} \text {Ei}(-2 (16+x))\right )}{4 x} \, dx+8 \int \frac {e^{-2 x} (-1-2 x) \log (x)}{4 (16+x)} \, dx+8 \int \frac {e^{-2 x} (-1-2 x) \log (16+x)}{4 x} \, dx\\ &=-18 e^{-2 x}+9 x^2-2 e^{-2 x} \log (x)+36 e^{-2 x} x \log (x)-64 e^{32} \text {Ei}(-2 (16+x)) \log (x)-2 e^{-2 x} \log (16+x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)+2 \int \frac {e^{-2 x}}{16+x} \, dx+2 \int \frac {e^{-2 x} \left (1-18 x+32 e^{32+2 x} \text {Ei}(-2 (16+x))\right )}{x} \, dx+2 \int \frac {e^{-2 x} \log (x)}{16+x} \, dx+2 \int \frac {e^{-2 x} (-1-2 x) \log (x)}{16+x} \, dx+2 \int \frac {e^{-2 x} \log (16+x)}{x} \, dx+2 \int \frac {e^{-2 x} (-1-2 x) \log (16+x)}{x} \, dx\\ &=-18 e^{-2 x}+9 x^2+2 e^{32} \text {Ei}(-2 (16+x))+36 e^{-2 x} x \log (x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)-2 \int \frac {e^{32} \text {Ei}(-32-2 x)}{x} \, dx+2 \int \left (\frac {e^{-2 x} (1-18 x)}{x}+\frac {32 e^{32} \text {Ei}(-32-2 x)}{x}\right ) \, dx-2 \int \frac {e^{-2 x}-\text {Ei}(-2 x)}{16+x} \, dx-2 \int \frac {\text {Ei}(-2 x)}{16+x} \, dx-2 \int \frac {e^{-2 x}+31 e^{32} \text {Ei}(-2 (16+x))}{x} \, dx\\ &=-18 e^{-2 x}+9 x^2+2 e^{32} \text {Ei}(-2 (16+x))+36 e^{-2 x} x \log (x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)+2 \int \frac {e^{-2 x} (1-18 x)}{x} \, dx-2 \int \left (\frac {e^{-2 x}}{x}+\frac {31 e^{32} \text {Ei}(-32-2 x)}{x}\right ) \, dx-2 \int \frac {\text {Ei}(-2 x)}{16+x} \, dx-2 \int \left (\frac {e^{-2 x}}{16+x}-\frac {\text {Ei}(-2 x)}{16+x}\right ) \, dx-\left (2 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx+\left (64 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx\\ &=-18 e^{-2 x}+9 x^2+2 e^{32} \text {Ei}(-2 (16+x))+36 e^{-2 x} x \log (x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)+2 \int \left (-18 e^{-2 x}+\frac {e^{-2 x}}{x}\right ) \, dx-2 \int \frac {e^{-2 x}}{x} \, dx-2 \int \frac {e^{-2 x}}{16+x} \, dx-\left (2 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx-\left (62 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx+\left (64 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx\\ &=-18 e^{-2 x}+9 x^2-2 \text {Ei}(-2 x)+36 e^{-2 x} x \log (x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)+2 \int \frac {e^{-2 x}}{x} \, dx-36 \int e^{-2 x} \, dx-\left (2 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx-\left (62 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx+\left (64 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx\\ &=9 x^2+36 e^{-2 x} x \log (x)+x^2 \log (16+x)+4 e^{-2 x} x \log (x) \log (16+x)-\left (2 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx-\left (62 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx+\left (64 e^{32}\right ) \int \frac {\text {Ei}(-32-2 x)}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 25, normalized size = 1.32 \begin {gather*} e^{-2 x} x \left (e^{2 x} x+4 \log (x)\right ) (9+\log (16+x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(576 + 36*x + E^(2*x)*(288*x + 19*x^2) + (576 - 1112*x - 72*x^2)*Log[x] + (64 + 4*x + E^(2*x)*(32*x
+ 2*x^2) + (64 - 124*x - 8*x^2)*Log[x])*Log[16 + x])/(E^(2*x)*(16 + x)),x]

[Out]

(x*(E^(2*x)*x + 4*Log[x])*(9 + Log[16 + x]))/E^(2*x)

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fricas [B]  time = 0.73, size = 39, normalized size = 2.05 \begin {gather*} {\left (9 \, x^{2} e^{\left (2 \, x\right )} + {\left (x^{2} e^{\left (2 \, x\right )} + 4 \, x \log \relax (x)\right )} \log \left (x + 16\right ) + 36 \, x \log \relax (x)\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x^2-124*x+64)*log(x)+(2*x^2+32*x)*exp(2*x)+4*x+64)*log(x+16)+(-72*x^2-1112*x+576)*log(x)+(19*x
^2+288*x)*exp(2*x)+36*x+576)/(x+16)/exp(2*x),x, algorithm="fricas")

[Out]

(9*x^2*e^(2*x) + (x^2*e^(2*x) + 4*x*log(x))*log(x + 16) + 36*x*log(x))*e^(-2*x)

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giac [A]  time = 0.15, size = 36, normalized size = 1.89 \begin {gather*} 4 \, x e^{\left (-2 \, x\right )} \log \left (x + 16\right ) \log \relax (x) + x^{2} \log \left (x + 16\right ) + 40 \, x e^{\left (-2 \, x\right )} \log \relax (x) + 9 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x^2-124*x+64)*log(x)+(2*x^2+32*x)*exp(2*x)+4*x+64)*log(x+16)+(-72*x^2-1112*x+576)*log(x)+(19*x
^2+288*x)*exp(2*x)+36*x+576)/(x+16)/exp(2*x),x, algorithm="giac")

[Out]

4*x*e^(-2*x)*log(x + 16)*log(x) + x^2*log(x + 16) + 40*x*e^(-2*x)*log(x) + 9*x^2

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maple [A]  time = 0.11, size = 41, normalized size = 2.16

method	result	size
risch	\(x \left (x \,{\mathrm e}^{2 x}+4 \ln \relax (x )\right ) {\mathrm e}^{-2 x} \ln \left (x +16\right )+9 x \left (x \,{\mathrm e}^{2 x}+4 \ln \relax (x )\right ) {\mathrm e}^{-2 x}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*x^2-124*x+64)*ln(x)+(2*x^2+32*x)*exp(2*x)+4*x+64)*ln(x+16)+(-72*x^2-1112*x+576)*ln(x)+(19*x^2+288*x)
*exp(2*x)+36*x+576)/(x+16)/exp(2*x),x,method=_RETURNVERBOSE)

[Out]

x*(x*exp(2*x)+4*ln(x))*exp(-2*x)*ln(x+16)+9*x*(x*exp(2*x)+4*ln(x))*exp(-2*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 36 \, x e^{\left (-2 \, x\right )} \log \relax (x) + 9 \, x^{2} - 576 \, e^{32} E_{1}\left (2 \, x + 32\right ) + {\left (4 \, x e^{\left (-2 \, x\right )} \log \relax (x) + x^{2}\right )} \log \left (x + 16\right ) - 576 \, \int \frac {e^{\left (-2 \, x\right )}}{x + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x^2-124*x+64)*log(x)+(2*x^2+32*x)*exp(2*x)+4*x+64)*log(x+16)+(-72*x^2-1112*x+576)*log(x)+(19*x
^2+288*x)*exp(2*x)+36*x+576)/(x+16)/exp(2*x),x, algorithm="maxima")

[Out]

36*x*e^(-2*x)*log(x) + 9*x^2 - 576*e^32*exp_integral_e(1, 2*x + 32) + (4*x*e^(-2*x)*log(x) + x^2)*log(x + 16)
- 576*integrate(e^(-2*x)/(x + 16), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{-2\,x}\,\left (36\,x+{\mathrm {e}}^{2\,x}\,\left (19\,x^2+288\,x\right )+\ln \left (x+16\right )\,\left (4\,x+{\mathrm {e}}^{2\,x}\,\left (2\,x^2+32\,x\right )-\ln \relax (x)\,\left (8\,x^2+124\,x-64\right )+64\right )-\ln \relax (x)\,\left (72\,x^2+1112\,x-576\right )+576\right )}{x+16} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*x)*(36*x + exp(2*x)*(288*x + 19*x^2) + log(x + 16)*(4*x + exp(2*x)*(32*x + 2*x^2) - log(x)*(124*x
+ 8*x^2 - 64) + 64) - log(x)*(1112*x + 72*x^2 - 576) + 576))/(x + 16),x)

[Out]

int((exp(-2*x)*(36*x + exp(2*x)*(288*x + 19*x^2) + log(x + 16)*(4*x + exp(2*x)*(32*x + 2*x^2) - log(x)*(124*x
+ 8*x^2 - 64) + 64) - log(x)*(1112*x + 72*x^2 - 576) + 576))/(x + 16), x)

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sympy [B]  time = 19.64, size = 48, normalized size = 2.53 \begin {gather*} 9 x^{2} + \left (x^{2} - \frac {256}{3}\right ) \log {\left (x + 16 \right )} + \left (4 x \log {\relax (x )} \log {\left (x + 16 \right )} + 36 x \log {\relax (x )}\right ) e^{- 2 x} + \frac {256 \log {\left (x + 16 \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x**2-124*x+64)*ln(x)+(2*x**2+32*x)*exp(2*x)+4*x+64)*ln(x+16)+(-72*x**2-1112*x+576)*ln(x)+(19*x
**2+288*x)*exp(2*x)+36*x+576)/(x+16)/exp(2*x),x)

[Out]

9*x**2 + (x**2 - 256/3)*log(x + 16) + (4*x*log(x)*log(x + 16) + 36*x*log(x))*exp(-2*x) + 256*log(x + 16)/3

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