3.52.51 \(\int \frac {120 x+2 e x-8 x^2+(64+e-4 x) \log (4096+e^2+e (128-8 x)-512 x+16 x^2)}{768+12 e-48 x} \, dx\)

Optimal. Leaf size=19 \[ 3+\frac {1}{12} x \left (x+\log \left ((e-4 (-16+x))^2\right )\right ) \]

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Rubi [B]  time = 0.10, antiderivative size = 41, normalized size of antiderivative = 2.16, number of steps used = 8, number of rules used = 6, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6, 6688, 12, 77, 2389, 2295} \begin {gather*} \frac {x^2}{12}+\frac {1}{24} (64+e) \log (-4 x+e+64)-\frac {1}{48} (-4 x+e+64) \log \left ((-4 x+e+64)^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(120*x + 2*E*x - 8*x^2 + (64 + E - 4*x)*Log[4096 + E^2 + E*(128 - 8*x) - 512*x + 16*x^2])/(768 + 12*E - 48
*x),x]

[Out]

x^2/12 + ((64 + E)*Log[64 + E - 4*x])/24 - ((64 + E - 4*x)*Log[(64 + E - 4*x)^2])/48

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 12

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

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 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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(120+2 e) x-8 x^2+(64+e-4 x) \log \left (4096+e^2+e (128-8 x)-512 x+16 x^2\right )}{768+12 e-48 x} \, dx\\ &=\int \frac {1}{12} \left (\frac {2 (60+e-4 x) x}{64+e-4 x}+\log \left ((64+e-4 x)^2\right )\right ) \, dx\\ &=\frac {1}{12} \int \left (\frac {2 (60+e-4 x) x}{64+e-4 x}+\log \left ((64+e-4 x)^2\right )\right ) \, dx\\ &=\frac {1}{12} \int \log \left ((64+e-4 x)^2\right ) \, dx+\frac {1}{6} \int \frac {(60+e-4 x) x}{64+e-4 x} \, dx\\ &=-\left (\frac {1}{48} \operatorname {Subst}\left (\int \log \left (x^2\right ) \, dx,x,64+e-4 x\right )\right )+\frac {1}{6} \int \left (1+\frac {-64-e}{64+e-4 x}+x\right ) \, dx\\ &=\frac {x^2}{12}+\frac {1}{24} (64+e) \log (64+e-4 x)-\frac {1}{48} (64+e-4 x) \log \left ((64+e-4 x)^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 66, normalized size = 3.47 \begin {gather*} \frac {1}{12} \left (-\frac {1}{8} (68+e) (64+e-4 x)+\frac {1}{16} (64+e-4 x)^2-2 x+\frac {1}{2} (64+e) \log (64+e-4 x)-\frac {1}{4} (64+e-4 x) \log \left ((64+e-4 x)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(120*x + 2*E*x - 8*x^2 + (64 + E - 4*x)*Log[4096 + E^2 + E*(128 - 8*x) - 512*x + 16*x^2])/(768 + 12*
E - 48*x),x]

[Out]

(-1/8*((68 + E)*(64 + E - 4*x)) + (64 + E - 4*x)^2/16 - 2*x + ((64 + E)*Log[64 + E - 4*x])/2 - ((64 + E - 4*x)
*Log[(64 + E - 4*x)^2])/4)/12

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fricas [A]  time = 0.65, size = 29, normalized size = 1.53 \begin {gather*} \frac {1}{12} \, x^{2} + \frac {1}{12} \, x \log \left (16 \, x^{2} - 8 \, {\left (x - 16\right )} e - 512 \, x + e^{2} + 4096\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)-4*x+64)*log(exp(1)^2+(-8*x+128)*exp(1)+16*x^2-512*x+4096)+2*x*exp(1)-8*x^2+120*x)/(12*exp(1
)-48*x+768),x, algorithm="fricas")

[Out]

1/12*x^2 + 1/12*x*log(16*x^2 - 8*(x - 16)*e - 512*x + e^2 + 4096)

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giac [A]  time = 0.24, size = 31, normalized size = 1.63 \begin {gather*} \frac {1}{12} \, x^{2} + \frac {1}{12} \, x \log \left (16 \, x^{2} - 8 \, x e - 512 \, x + e^{2} + 128 \, e + 4096\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)-4*x+64)*log(exp(1)^2+(-8*x+128)*exp(1)+16*x^2-512*x+4096)+2*x*exp(1)-8*x^2+120*x)/(12*exp(1
)-48*x+768),x, algorithm="giac")

[Out]

1/12*x^2 + 1/12*x*log(16*x^2 - 8*x*e - 512*x + e^2 + 128*e + 4096)

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maple [A]  time = 0.18, size = 31, normalized size = 1.63

method	result	size
risch	\(\frac {x^{2}}{12}+\frac {x \ln \left ({\mathrm e}^{2}+\left (-8 x +128\right ) {\mathrm e}+16 x^{2}-512 x +4096\right )}{12}\)	\(31\)
norman	\(\frac {x^{2}}{12}+\frac {x \ln \left ({\mathrm e}^{2}+\left (-8 x +128\right ) {\mathrm e}+16 x^{2}-512 x +4096\right )}{12}\)	\(33\)
default	\(\frac {x^{2}}{12}+\frac {\ln \left ({\mathrm e}^{2}-8 x \,{\mathrm e}+16 x^{2}+128 \,{\mathrm e}-512 x +4096\right ) x}{12}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(1)-4*x+64)*ln(exp(1)^2+(-8*x+128)*exp(1)+16*x^2-512*x+4096)+2*x*exp(1)-8*x^2+120*x)/(12*exp(1)-48*x+
768),x,method=_RETURNVERBOSE)

[Out]

1/12*x^2+1/12*x*ln(exp(2)+(-8*x+128)*exp(1)+16*x^2-512*x+4096)

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maxima [B]  time = 0.37, size = 238, normalized size = 12.53 \begin {gather*} -\frac {1}{48} \, e \log \left (16 \, x^{2} - 8 \, x e - 512 \, x + e^{2} + 128 \, e + 4096\right ) \log \left (4 \, x - e - 64\right ) - \frac {1}{48} \, {\left (e + 64\right )} \log \left (4 \, x - e - 64\right )^{2} + \frac {1}{12} \, x^{2} + \frac {1}{24} \, x {\left (e + 64\right )} - \frac {1}{96} \, {\left ({\left (e + 64\right )} \log \left (4 \, x - e - 64\right ) + 4 \, x\right )} e + \frac {1}{48} \, {\left (\log \left (16 \, x^{2} - 8 \, x e - 512 \, x + e^{2} + 128 \, e + 4096\right ) \log \left (4 \, x - e - 64\right ) - \log \left (4 \, x - e - 64\right )^{2}\right )} e + \frac {1}{48} \, {\left ({\left (e + 64\right )} \log \left (4 \, x - e - 64\right ) + 4 \, x\right )} \log \left (16 \, x^{2} - 8 \, x e - 512 \, x + e^{2} + 128 \, e + 4096\right ) + \frac {1}{96} \, {\left (e^{2} + 128 \, e + 4096\right )} \log \left (4 \, x - e - 64\right ) - \frac {2}{3} \, {\left (e + 64\right )} \log \left (4 \, x - e - 64\right ) - \frac {4}{3} \, \log \left (4 \, x - e - 64\right )^{2} - \frac {8}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)-4*x+64)*log(exp(1)^2+(-8*x+128)*exp(1)+16*x^2-512*x+4096)+2*x*exp(1)-8*x^2+120*x)/(12*exp(1
)-48*x+768),x, algorithm="maxima")

[Out]

-1/48*e*log(16*x^2 - 8*x*e - 512*x + e^2 + 128*e + 4096)*log(4*x - e - 64) - 1/48*(e + 64)*log(4*x - e - 64)^2
+ 1/12*x^2 + 1/24*x*(e + 64) - 1/96*((e + 64)*log(4*x - e - 64) + 4*x)*e + 1/48*(log(16*x^2 - 8*x*e - 512*x +
e^2 + 128*e + 4096)*log(4*x - e - 64) - log(4*x - e - 64)^2)*e + 1/48*((e + 64)*log(4*x - e - 64) + 4*x)*log(
16*x^2 - 8*x*e - 512*x + e^2 + 128*e + 4096) + 1/96*(e^2 + 128*e + 4096)*log(4*x - e - 64) - 2/3*(e + 64)*log(
4*x - e - 64) - 4/3*log(4*x - e - 64)^2 - 8/3*x

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mupad [B]  time = 0.68, size = 27, normalized size = 1.42 \begin {gather*} \frac {x\,\left (x+\ln \left (128\,\mathrm {e}-512\,x+{\mathrm {e}}^2-8\,x\,\mathrm {e}+16\,x^2+4096\right )\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((120*x + log(exp(2) - 512*x + 16*x^2 - exp(1)*(8*x - 128) + 4096)*(exp(1) - 4*x + 64) + 2*x*exp(1) - 8*x^2
)/(12*exp(1) - 48*x + 768),x)

[Out]

(x*(x + log(128*exp(1) - 512*x + exp(2) - 8*x*exp(1) + 16*x^2 + 4096)))/12

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sympy [A]  time = 0.16, size = 31, normalized size = 1.63 \begin {gather*} \frac {x^{2}}{12} + \frac {x \log {\left (16 x^{2} - 512 x + e \left (128 - 8 x\right ) + e^{2} + 4096 \right )}}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)-4*x+64)*ln(exp(1)**2+(-8*x+128)*exp(1)+16*x**2-512*x+4096)+2*x*exp(1)-8*x**2+120*x)/(12*exp
(1)-48*x+768),x)

[Out]

x**2/12 + x*log(16*x**2 - 512*x + E*(128 - 8*x) + exp(2) + 4096)/12

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