3.29.55 \(\int \frac {e^5 x+4 x^2+(4-e^4+e^5 x+2 x^2) \log (4-e^4+e^5 x+2 x^2)}{4-e^4+e^5 x+2 x^2} \, dx\)

Optimal. Leaf size=20 \[ x \log \left (4-e^4+x^2+x \left (e^5+x\right )\right ) \]

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Rubi [A]  time = 0.19, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {6688, 773, 634, 618, 206, 628, 2523} \begin {gather*} x \log \left (2 x^2+e^5 x-e^4+4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5*x + 4*x^2 + (4 - E^4 + E^5*x + 2*x^2)*Log[4 - E^4 + E^5*x + 2*x^2])/(4 - E^4 + E^5*x + 2*x^2),x]

[Out]

x*Log[4 - E^4 + E^5*x + 2*x^2]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*g)*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2523

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[(x*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

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 \left (\frac {x \left (e^5+4 x\right )}{4-e^4+e^5 x+2 x^2}+\log \left (4-e^4+e^5 x+2 x^2\right )\right ) \, dx\\ &=\int \frac {x \left (e^5+4 x\right )}{4-e^4+e^5 x+2 x^2} \, dx+\int \log \left (4-e^4+e^5 x+2 x^2\right ) \, dx\\ &=2 x+x \log \left (4-e^4+e^5 x+2 x^2\right )+\frac {1}{2} \int \frac {-4 \left (4-e^4\right )-2 e^5 x}{4-e^4+e^5 x+2 x^2} \, dx-\int \frac {x \left (e^5+4 x\right )}{4-e^4+e^5 x+2 x^2} \, dx\\ &=x \log \left (4-e^4+e^5 x+2 x^2\right )-\frac {1}{2} \int \frac {-4 \left (4-e^4\right )-2 e^5 x}{4-e^4+e^5 x+2 x^2} \, dx-\frac {1}{4} e^5 \int \frac {e^5+4 x}{4-e^4+e^5 x+2 x^2} \, dx+\frac {1}{4} \left (-32+8 e^4+e^{10}\right ) \int \frac {1}{4-e^4+e^5 x+2 x^2} \, dx\\ &=-\frac {1}{4} e^5 \log \left (4-e^4+e^5 x+2 x^2\right )+x \log \left (4-e^4+e^5 x+2 x^2\right )+\frac {1}{4} e^5 \int \frac {e^5+4 x}{4-e^4+e^5 x+2 x^2} \, dx+\frac {1}{2} \left (32-8 e^4-e^{10}\right ) \operatorname {Subst}\left (\int \frac {1}{-32+8 e^4+e^{10}-x^2} \, dx,x,e^5+4 x\right )-\frac {1}{4} \left (-32+8 e^4+e^{10}\right ) \int \frac {1}{4-e^4+e^5 x+2 x^2} \, dx\\ &=-\frac {1}{2} \sqrt {-32+8 e^4+e^{10}} \tanh ^{-1}\left (\frac {e^5+4 x}{\sqrt {-32+8 e^4+e^{10}}}\right )+x \log \left (4-e^4+e^5 x+2 x^2\right )-\frac {1}{2} \left (32-8 e^4-e^{10}\right ) \operatorname {Subst}\left (\int \frac {1}{-32+8 e^4+e^{10}-x^2} \, dx,x,e^5+4 x\right )\\ &=x \log \left (4-e^4+e^5 x+2 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 1.00 \begin {gather*} x \log \left (4-e^4+e^5 x+2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*x + 4*x^2 + (4 - E^4 + E^5*x + 2*x^2)*Log[4 - E^4 + E^5*x + 2*x^2])/(4 - E^4 + E^5*x + 2*x^2),x
]

[Out]

x*Log[4 - E^4 + E^5*x + 2*x^2]

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fricas [A]  time = 0.91, size = 18, normalized size = 0.90 \begin {gather*} x \log \left (2 \, x^{2} + x e^{5} - e^{4} + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(5)-exp(4)+2*x^2+4)*log(x*exp(5)-exp(4)+2*x^2+4)+x*exp(5)+4*x^2)/(x*exp(5)-exp(4)+2*x^2+4),x,
algorithm="fricas")

[Out]

x*log(2*x^2 + x*e^5 - e^4 + 4)

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giac [A]  time = 0.32, size = 18, normalized size = 0.90 \begin {gather*} x \log \left (2 \, x^{2} + x e^{5} - e^{4} + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(5)-exp(4)+2*x^2+4)*log(x*exp(5)-exp(4)+2*x^2+4)+x*exp(5)+4*x^2)/(x*exp(5)-exp(4)+2*x^2+4),x,
algorithm="giac")

[Out]

x*log(2*x^2 + x*e^5 - e^4 + 4)

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maple [A]  time = 0.61, size = 19, normalized size = 0.95

method	result	size
default	\(x \ln \left (x \,{\mathrm e}^{5}-{\mathrm e}^{4}+2 x^{2}+4\right )\)	\(19\)
norman	\(x \ln \left (x \,{\mathrm e}^{5}-{\mathrm e}^{4}+2 x^{2}+4\right )\)	\(19\)
risch	\(x \ln \left (x \,{\mathrm e}^{5}-{\mathrm e}^{4}+2 x^{2}+4\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(5)-exp(4)+2*x^2+4)*ln(x*exp(5)-exp(4)+2*x^2+4)+x*exp(5)+4*x^2)/(x*exp(5)-exp(4)+2*x^2+4),x,method=
_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.84, size = 227, normalized size = 11.35 \begin {gather*} -\frac {1}{4} \, {\left (\frac {e^{5} \log \left (\frac {4 \, x - \sqrt {e^{10} + 8 \, e^{4} - 32} + e^{5}}{4 \, x + \sqrt {e^{10} + 8 \, e^{4} - 32} + e^{5}}\right )}{\sqrt {e^{10} + 8 \, e^{4} - 32}} - \log \left (2 \, x^{2} + x e^{5} - e^{4} + 4\right )\right )} e^{5} + \frac {1}{4} \, {\left (4 \, x + e^{5}\right )} \log \left (2 \, x^{2} + x e^{5} - e^{4} + 4\right ) - \frac {1}{2} \, e^{5} \log \left (2 \, x^{2} + x e^{5} - e^{4} + 4\right ) - \frac {1}{4} \, \sqrt {e^{10} + 8 \, e^{4} - 32} \log \left (\frac {4 \, x - \sqrt {e^{10} + 8 \, e^{4} - 32} + e^{5}}{4 \, x + \sqrt {e^{10} + 8 \, e^{4} - 32} + e^{5}}\right ) + \frac {{\left (e^{10} + 4 \, e^{4} - 16\right )} \log \left (\frac {4 \, x - \sqrt {e^{10} + 8 \, e^{4} - 32} + e^{5}}{4 \, x + \sqrt {e^{10} + 8 \, e^{4} - 32} + e^{5}}\right )}{2 \, \sqrt {e^{10} + 8 \, e^{4} - 32}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(5)-exp(4)+2*x^2+4)*log(x*exp(5)-exp(4)+2*x^2+4)+x*exp(5)+4*x^2)/(x*exp(5)-exp(4)+2*x^2+4),x,
algorithm="maxima")

[Out]

-1/4*(e^5*log((4*x - sqrt(e^10 + 8*e^4 - 32) + e^5)/(4*x + sqrt(e^10 + 8*e^4 - 32) + e^5))/sqrt(e^10 + 8*e^4 -
32) - log(2*x^2 + x*e^5 - e^4 + 4))*e^5 + 1/4*(4*x + e^5)*log(2*x^2 + x*e^5 - e^4 + 4) - 1/2*e^5*log(2*x^2 +
x*e^5 - e^4 + 4) - 1/4*sqrt(e^10 + 8*e^4 - 32)*log((4*x - sqrt(e^10 + 8*e^4 - 32) + e^5)/(4*x + sqrt(e^10 + 8*
e^4 - 32) + e^5)) + 1/2*(e^10 + 4*e^4 - 16)*log((4*x - sqrt(e^10 + 8*e^4 - 32) + e^5)/(4*x + sqrt(e^10 + 8*e^4
- 32) + e^5))/sqrt(e^10 + 8*e^4 - 32)

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mupad [B]  time = 1.88, size = 18, normalized size = 0.90 \begin {gather*} x\,\ln \left (2\,x^2+{\mathrm {e}}^5\,x-{\mathrm {e}}^4+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(5) + 4*x^2 + log(x*exp(5) - exp(4) + 2*x^2 + 4)*(x*exp(5) - exp(4) + 2*x^2 + 4))/(x*exp(5) - exp(4)
+ 2*x^2 + 4),x)

[Out]

x*log(x*exp(5) - exp(4) + 2*x^2 + 4)

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sympy [A]  time = 0.15, size = 17, normalized size = 0.85 \begin {gather*} x \log {\left (2 x^{2} + x e^{5} - e^{4} + 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(5)-exp(4)+2*x**2+4)*ln(x*exp(5)-exp(4)+2*x**2+4)+x*exp(5)+4*x**2)/(x*exp(5)-exp(4)+2*x**2+4)
,x)

[Out]

x*log(2*x**2 + x*exp(5) - exp(4) + 4)

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