3.16.9 \(\int \frac {1+e^5 (2-16 x)-6 x-16 x^2-32 x \log (1+2 e^5+2 x)+(-8-16 e^5-16 x) \log ^2(1+2 e^5+2 x)-32 \log ^3(1+2 e^5+2 x)}{1+2 e^5+2 x} \, dx\)

Optimal. Leaf size=21 \[ x-4 \left (x+\log ^2\left (3+2 \left (-1+e^5+x\right )\right )\right )^2 \]

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Rubi [B]  time = 0.27, antiderivative size = 69, normalized size of antiderivative = 3.29, number of steps used = 12, number of rules used = 10, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6742, 2411, 2346, 2301, 2295, 2389, 2296, 2390, 2302, 30} \begin {gather*} -4 x^2+x-4 \log ^4\left (2 x+2 e^5+1\right )-4 \left (2 x+2 e^5+1\right ) \log ^2\left (2 x+2 e^5+1\right )+4 \left (1+2 e^5\right ) \log ^2\left (2 x+2 e^5+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^5*(2 - 16*x) - 6*x - 16*x^2 - 32*x*Log[1 + 2*E^5 + 2*x] + (-8 - 16*E^5 - 16*x)*Log[1 + 2*E^5 + 2*x]
^2 - 32*Log[1 + 2*E^5 + 2*x]^3)/(1 + 2*E^5 + 2*x),x]

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

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 2302

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

Rule 2346

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

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

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Mathematica [B]  time = 0.03, size = 51, normalized size = 2.43 \begin {gather*} \frac {3}{2}+5 e^5+4 e^{10}+x-4 x^2-8 x \log ^2\left (1+2 e^5+2 x\right )-4 \log ^4\left (1+2 e^5+2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^5*(2 - 16*x) - 6*x - 16*x^2 - 32*x*Log[1 + 2*E^5 + 2*x] + (-8 - 16*E^5 - 16*x)*Log[1 + 2*E^5
+ 2*x]^2 - 32*Log[1 + 2*E^5 + 2*x]^3)/(1 + 2*E^5 + 2*x),x]

[Out]

3/2 + 5*E^5 + 4*E^10 + x - 4*x^2 - 8*x*Log[1 + 2*E^5 + 2*x]^2 - 4*Log[1 + 2*E^5 + 2*x]^4

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fricas [A]  time = 0.61, size = 36, normalized size = 1.71 \begin {gather*} -4 \, \log \left (2 \, x + 2 \, e^{5} + 1\right )^{4} - 8 \, x \log \left (2 \, x + 2 \, e^{5} + 1\right )^{2} - 4 \, x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(2*exp(5)+2*x+1)^3+(-16*exp(5)-16*x-8)*log(2*exp(5)+2*x+1)^2-32*x*log(2*exp(5)+2*x+1)+(-16*x
+2)*exp(5)-16*x^2-6*x+1)/(2*exp(5)+2*x+1),x, algorithm="fricas")

[Out]

-4*log(2*x + 2*e^5 + 1)^4 - 8*x*log(2*x + 2*e^5 + 1)^2 - 4*x^2 + x

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giac [A]  time = 0.26, size = 36, normalized size = 1.71 \begin {gather*} -4 \, \log \left (2 \, x + 2 \, e^{5} + 1\right )^{4} - 8 \, x \log \left (2 \, x + 2 \, e^{5} + 1\right )^{2} - 4 \, x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(2*exp(5)+2*x+1)^3+(-16*exp(5)-16*x-8)*log(2*exp(5)+2*x+1)^2-32*x*log(2*exp(5)+2*x+1)+(-16*x
+2)*exp(5)-16*x^2-6*x+1)/(2*exp(5)+2*x+1),x, algorithm="giac")

[Out]

-4*log(2*x + 2*e^5 + 1)^4 - 8*x*log(2*x + 2*e^5 + 1)^2 - 4*x^2 + x

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maple [A]  time = 0.24, size = 37, normalized size = 1.76

method	result	size
norman	\(x -4 x^{2}-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{4}-8 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} x\)	\(37\)
risch	\(x -4 x^{2}-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{4}-8 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} x\)	\(37\)
derivativedivides	\(-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{4}-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} \left (2 \,{\mathrm e}^{5}+2 x +1\right )+5 \,{\mathrm e}^{5}+5 x +\frac {5}{2}+8 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} {\mathrm e}^{5}+4 \,{\mathrm e}^{5} \left (2 \,{\mathrm e}^{5}+2 x +1\right )-\left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2}+4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2}\)	\(103\)
default	\(-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{4}-4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} \left (2 \,{\mathrm e}^{5}+2 x +1\right )+5 \,{\mathrm e}^{5}+5 x +\frac {5}{2}+8 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2} {\mathrm e}^{5}+4 \,{\mathrm e}^{5} \left (2 \,{\mathrm e}^{5}+2 x +1\right )-\left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2}+4 \ln \left (2 \,{\mathrm e}^{5}+2 x +1\right )^{2}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-32*ln(2*exp(5)+2*x+1)^3+(-16*exp(5)-16*x-8)*ln(2*exp(5)+2*x+1)^2-32*x*ln(2*exp(5)+2*x+1)+(-16*x+2)*exp(5
)-16*x^2-6*x+1)/(2*exp(5)+2*x+1),x,method=_RETURNVERBOSE)

[Out]

x-4*x^2-4*ln(2*exp(5)+2*x+1)^4-8*ln(2*exp(5)+2*x+1)^2*x

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maxima [B]  time = 0.39, size = 262, normalized size = 12.48 \begin {gather*} \frac {4}{3} \, {\left (2 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right )^{3} - \frac {8}{3} \, e^{5} \log \left (2 \, x + 2 \, e^{5} + 1\right )^{3} - 4 \, \log \left (2 \, x + 2 \, e^{5} + 1\right )^{4} - 4 \, {\left (2 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right )^{2} - \frac {4}{3} \, \log \left (2 \, x + 2 \, e^{5} + 1\right )^{3} - 4 \, {\left (\log \left (2 \, x + 2 \, e^{5} + 1\right )^{2} - 2 \, \log \left (2 \, x + 2 \, e^{5} + 1\right ) + 2\right )} {\left (2 \, x + 2 \, e^{5} + 1\right )} - 4 \, x^{2} + 4 \, x {\left (2 \, e^{5} + 1\right )} + 4 \, {\left ({\left (2 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right ) - 2 \, x\right )} e^{5} + 8 \, {\left ({\left (2 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right ) - 2 \, x\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right ) - 2 \, {\left (4 \, e^{10} + 4 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right ) - \frac {13}{2} \, {\left (2 \, e^{5} + 1\right )} \log \left (2 \, x + 2 \, e^{5} + 1\right ) + e^{5} \log \left (2 \, x + 2 \, e^{5} + 1\right ) + 13 \, x + \frac {1}{2} \, \log \left (2 \, x + 2 \, e^{5} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(2*exp(5)+2*x+1)^3+(-16*exp(5)-16*x-8)*log(2*exp(5)+2*x+1)^2-32*x*log(2*exp(5)+2*x+1)+(-16*x
+2)*exp(5)-16*x^2-6*x+1)/(2*exp(5)+2*x+1),x, algorithm="maxima")

[Out]

4/3*(2*e^5 + 1)*log(2*x + 2*e^5 + 1)^3 - 8/3*e^5*log(2*x + 2*e^5 + 1)^3 - 4*log(2*x + 2*e^5 + 1)^4 - 4*(2*e^5
+ 1)*log(2*x + 2*e^5 + 1)^2 - 4/3*log(2*x + 2*e^5 + 1)^3 - 4*(log(2*x + 2*e^5 + 1)^2 - 2*log(2*x + 2*e^5 + 1)
+ 2)*(2*x + 2*e^5 + 1) - 4*x^2 + 4*x*(2*e^5 + 1) + 4*((2*e^5 + 1)*log(2*x + 2*e^5 + 1) - 2*x)*e^5 + 8*((2*e^5
+ 1)*log(2*x + 2*e^5 + 1) - 2*x)*log(2*x + 2*e^5 + 1) - 2*(4*e^10 + 4*e^5 + 1)*log(2*x + 2*e^5 + 1) - 13/2*(2*
e^5 + 1)*log(2*x + 2*e^5 + 1) + e^5*log(2*x + 2*e^5 + 1) + 13*x + 1/2*log(2*x + 2*e^5 + 1)

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mupad [B]  time = 1.14, size = 36, normalized size = 1.71 \begin {gather*} -4\,x^2-8\,x\,{\ln \left (2\,x+2\,{\mathrm {e}}^5+1\right )}^2+x-4\,{\ln \left (2\,x+2\,{\mathrm {e}}^5+1\right )}^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x + log(2*x + 2*exp(5) + 1)^2*(16*x + 16*exp(5) + 8) + 32*x*log(2*x + 2*exp(5) + 1) + 32*log(2*x + 2*e
xp(5) + 1)^3 + 16*x^2 + exp(5)*(16*x - 2) - 1)/(2*x + 2*exp(5) + 1),x)

[Out]

x - 8*x*log(2*x + 2*exp(5) + 1)^2 - 4*log(2*x + 2*exp(5) + 1)^4 - 4*x^2

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sympy [A]  time = 0.18, size = 37, normalized size = 1.76 \begin {gather*} - 4 x^{2} - 8 x \log {\left (2 x + 1 + 2 e^{5} \right )}^{2} + x - 4 \log {\left (2 x + 1 + 2 e^{5} \right )}^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*ln(2*exp(5)+2*x+1)**3+(-16*exp(5)-16*x-8)*ln(2*exp(5)+2*x+1)**2-32*x*ln(2*exp(5)+2*x+1)+(-16*x+
2)*exp(5)-16*x**2-6*x+1)/(2*exp(5)+2*x+1),x)

[Out]

-4*x**2 - 8*x*log(2*x + 1 + 2*exp(5))**2 + x - 4*log(2*x + 1 + 2*exp(5))**4

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