3.28.92 \(\int \frac {4-4 x-5 x^2+x^3+(-4+x^2) \log (\frac {3 e^x}{16 x-8 x^3+x^5})}{-4+x^2} \, dx\)

Optimal. Leaf size=18 \[ x \log \left (\frac {3 e^x}{x \left (-4+x^2\right )^2}\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 13, number of rules used = 9, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {6725, 207, 260, 321, 266, 43, 2548, 1810, 206} \begin {gather*} x \log \left (\frac {3 e^x}{x \left (4-x^2\right )^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 4*x - 5*x^2 + x^3 + (-4 + x^2)*Log[(3*E^x)/(16*x - 8*x^3 + x^5)])/(-4 + x^2),x]

[Out]

x*Log[(3*E^x)/(x*(4 - x^2)^2)]

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 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 207

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {4}{-4+x^2}-\frac {4 x}{-4+x^2}-\frac {5 x^2}{-4+x^2}+\frac {x^3}{-4+x^2}+\log \left (\frac {3 e^x}{x \left (-4+x^2\right )^2}\right )\right ) \, dx\\ &=4 \int \frac {1}{-4+x^2} \, dx-4 \int \frac {x}{-4+x^2} \, dx-5 \int \frac {x^2}{-4+x^2} \, dx+\int \frac {x^3}{-4+x^2} \, dx+\int \log \left (\frac {3 e^x}{x \left (-4+x^2\right )^2}\right ) \, dx\\ &=-5 x-2 \tanh ^{-1}\left (\frac {x}{2}\right )+x \log \left (\frac {3 e^x}{x \left (4-x^2\right )^2}\right )-2 \log \left (4-x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{-4+x} \, dx,x,x^2\right )-20 \int \frac {1}{-4+x^2} \, dx-\int \frac {-4+4 x+5 x^2-x^3}{4-x^2} \, dx\\ &=-5 x+8 \tanh ^{-1}\left (\frac {x}{2}\right )+x \log \left (\frac {3 e^x}{x \left (4-x^2\right )^2}\right )-2 \log \left (4-x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \left (1+\frac {4}{-4+x}\right ) \, dx,x,x^2\right )-\int \left (-5+x+\frac {16}{4-x^2}\right ) \, dx\\ &=8 \tanh ^{-1}\left (\frac {x}{2}\right )+x \log \left (\frac {3 e^x}{x \left (4-x^2\right )^2}\right )-16 \int \frac {1}{4-x^2} \, dx\\ &=x \log \left (\frac {3 e^x}{x \left (4-x^2\right )^2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 18, normalized size = 1.00 \begin {gather*} x \log \left (\frac {3 e^x}{x \left (-4+x^2\right )^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 4*x - 5*x^2 + x^3 + (-4 + x^2)*Log[(3*E^x)/(16*x - 8*x^3 + x^5)])/(-4 + x^2),x]

[Out]

x*Log[(3*E^x)/(x*(-4 + x^2)^2)]

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fricas [A]  time = 0.75, size = 21, normalized size = 1.17 \begin {gather*} x \log \left (\frac {3 \, e^{x}}{x^{5} - 8 \, x^{3} + 16 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-4)*log(3*exp(x)/(x^5-8*x^3+16*x))+x^3-5*x^2-4*x+4)/(x^2-4),x, algorithm="fricas")

[Out]

x*log(3*e^x/(x^5 - 8*x^3 + 16*x))

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giac [A]  time = 0.75, size = 23, normalized size = 1.28 \begin {gather*} x^{2} + x \log \left (\frac {3}{x^{5} - 8 \, x^{3} + 16 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-4)*log(3*exp(x)/(x^5-8*x^3+16*x))+x^3-5*x^2-4*x+4)/(x^2-4),x, algorithm="giac")

[Out]

x^2 + x*log(3/(x^5 - 8*x^3 + 16*x))

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maple [A]  time = 0.54, size = 22, normalized size = 1.22

method	result	size
default	\(x \ln \left (\frac {3 \,{\mathrm e}^{x}}{x^{5}-8 x^{3}+16 x}\right )\)	\(22\)
norman	\(x \ln \left (\frac {3 \,{\mathrm e}^{x}}{x^{5}-8 x^{3}+16 x}\right )\)	\(22\)
risch	\(x \ln \left ({\mathrm e}^{x}\right )-2 x \ln \left (x^{2}-4\right )-x \ln \relax (x )+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x \left (x^{2}-4\right )^{2}}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{\left (x^{2}-4\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{\left (x^{2}-4\right )^{2}}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{\left (x^{2}-4\right )^{2}}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{\left (x^{2}-4\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x \left (x^{2}-4\right )^{2}}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (i \left (x^{2}-4\right )^{2}\right )^{3}}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{\left (x^{2}-4\right )^{2}}\right )^{3}}{2}-i \pi x \,\mathrm {csgn}\left (i \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (i \left (x^{2}-4\right )^{2}\right )^{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{\left (x^{2}-4\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x \left (x^{2}-4\right )^{2}}\right )}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x \left (x^{2}-4\right )^{2}}\right )^{3}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i}{\left (x^{2}-4\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{\left (x^{2}-4\right )^{2}}\right )}{2}+\frac {i \pi x \mathrm {csgn}\left (i \left (x^{2}-4\right )\right )^{2} \mathrm {csgn}\left (i \left (x^{2}-4\right )^{2}\right )}{2}+x \ln \relax (3)\)	\(339\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2-4)*ln(3*exp(x)/(x^5-8*x^3+16*x))+x^3-5*x^2-4*x+4)/(x^2-4),x,method=_RETURNVERBOSE)

[Out]

x*ln(3*exp(x)/(x^5-8*x^3+16*x))

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maxima [B]  time = 0.76, size = 48, normalized size = 2.67 \begin {gather*} x^{2} + x {\left (\log \relax (3) + 5\right )} - 2 \, {\left (x + 2\right )} \log \left (x + 2\right ) - 2 \, {\left (x - 2\right )} \log \left (x - 2\right ) - x \log \relax (x) - 5 \, x + 4 \, \log \left (x + 2\right ) - 4 \, \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-4)*log(3*exp(x)/(x^5-8*x^3+16*x))+x^3-5*x^2-4*x+4)/(x^2-4),x, algorithm="maxima")

[Out]

x^2 + x*(log(3) + 5) - 2*(x + 2)*log(x + 2) - 2*(x - 2)*log(x - 2) - x*log(x) - 5*x + 4*log(x + 2) - 4*log(x -
2)

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mupad [B]  time = 2.04, size = 21, normalized size = 1.17 \begin {gather*} x\,\left (x+\ln \left (\frac {1}{x^5-8\,x^3+16\,x}\right )+\ln \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 - 5*x^2 - 4*x + log((3*exp(x))/(16*x - 8*x^3 + x^5))*(x^2 - 4) + 4)/(x^2 - 4),x)

[Out]

x*(x + log(1/(16*x - 8*x^3 + x^5)) + log(3))

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sympy [A]  time = 0.34, size = 19, normalized size = 1.06 \begin {gather*} x \log {\left (\frac {3 e^{x}}{x^{5} - 8 x^{3} + 16 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2-4)*ln(3*exp(x)/(x**5-8*x**3+16*x))+x**3-5*x**2-4*x+4)/(x**2-4),x)

[Out]

x*log(3*exp(x)/(x**5 - 8*x**3 + 16*x))

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