3.50.83 \(\int \frac {3+5 x^2+2 x^3+(3+x-5 x^2-x^3) \log (\frac {3+x-5 x^2-x^3}{x})}{-3 x^2-x^3+5 x^4+x^5} \, dx\)

Optimal. Leaf size=27 \[ \frac {\log \left ((-1+x)^2+\frac {3}{x}-x-x (2+2 x)\right )}{x} \]

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Rubi [A]  time = 6.09, antiderivative size = 20, normalized size of antiderivative = 0.74, number of steps used = 26, number of rules used = 11, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6741, 6742, 2100, 2081, 2079, 800, 634, 618, 206, 628, 2525} \begin {gather*} \frac {\log \left (-x^2-5 x+\frac {3}{x}+1\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + 3/x - 5*x - x^2]/x

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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + 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] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 2079

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> With[{r = Rt[-9*a*d^2 + S
qrt[3]*d*Sqrt[4*b^3*d + 27*a^2*d^2], 3]}, Dist[1/d^(2*p), Int[(e + f*x)^m*Simp[(18^(1/3)*b*d)/(3*r) - r/18^(1/
3) + d*x, x]^p*Simp[(b*d)/3 + (12^(1/3)*b^2*d^2)/(3*r^2) + r^2/(3*12^(1/3)) - d*((2^(1/3)*b*d)/(3^(1/3)*r) - r
/18^(1/3))*x + d^2*x^2, x]^p, x], x]] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[4*b^3 + 27*a^2*d, 0] && ILtQ[p, 0
]

Rule 2081

Int[(P3_)^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = C
oeff[P3, x, 2], d = Coeff[P3, x, 3]}, Subst[Int[((3*d*e - c*f)/(3*d) + f*x)^m*Simp[(2*c^3 - 9*b*c*d + 27*a*d^2
)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x, x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[{e, f, m, p}, x
] && PolyQ[P3, x, 3]

Rule 2100

Int[(Pm_)/(Qn_), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x]}, Simp[(Coeff[Pm, x, m]*Log[Qn])/(n*Coe
ff[Qn, x, n]), x] + Dist[1/(n*Coeff[Qn, x, n]), Int[ExpandToSum[n*Coeff[Qn, x, n]*Pm - Coeff[Pm, x, m]*D[Qn, x
], x]/Qn, x], x] /; EqQ[m, n - 1]] /; PolyQ[Pm, x] && PolyQ[Qn, x]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6741

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

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 \frac {-3-5 x^2-2 x^3-\left (3+x-5 x^2-x^3\right ) \log \left (\frac {3+x-5 x^2-x^3}{x}\right )}{x^2 \left (3+x-5 x^2-x^3\right )} \, dx\\ &=\int \left (\frac {3+5 x^2+2 x^3}{x^2 \left (-3-x+5 x^2+x^3\right )}-\frac {\log \left (1+\frac {3}{x}-5 x-x^2\right )}{x^2}\right ) \, dx\\ &=\int \frac {3+5 x^2+2 x^3}{x^2 \left (-3-x+5 x^2+x^3\right )} \, dx-\int \frac {\log \left (1+\frac {3}{x}-5 x-x^2\right )}{x^2} \, dx\\ &=\frac {\log \left (1+\frac {3}{x}-5 x-x^2\right )}{x}-\int \frac {-3-5 x^2-2 x^3}{x^2 \left (3+x-5 x^2-x^3\right )} \, dx+\int \left (-\frac {1}{x^2}+\frac {1}{3 x}+\frac {31+4 x-x^2}{3 \left (-3-x+5 x^2+x^3\right )}\right ) \, dx\\ &=\frac {1}{x}+\frac {\log (x)}{3}+\frac {\log \left (1+\frac {3}{x}-5 x-x^2\right )}{x}+\frac {1}{3} \int \frac {31+4 x-x^2}{-3-x+5 x^2+x^3} \, dx-\int \left (-\frac {1}{x^2}+\frac {1}{3 x}+\frac {31+4 x-x^2}{3 \left (-3-x+5 x^2+x^3\right )}\right ) \, dx\\ &=\frac {\log \left (1+\frac {3}{x}-5 x-x^2\right )}{x}-\frac {1}{9} \log \left (-3-x+5 x^2+x^3\right )+\frac {1}{9} \int \frac {92+22 x}{-3-x+5 x^2+x^3} \, dx-\frac {1}{3} \int \frac {31+4 x-x^2}{-3-x+5 x^2+x^3} \, dx\\ &=\frac {\log \left (1+\frac {3}{x}-5 x-x^2\right )}{x}-\frac {1}{9} \int \frac {92+22 x}{-3-x+5 x^2+x^3} \, dx+\frac {1}{9} \operatorname {Subst}\left (\int \frac {\frac {166}{3}+22 x}{\frac {214}{27}-\frac {28 x}{3}+x^3} \, dx,x,\frac {5}{3}+x\right )\\ &=\frac {\log \left (1+\frac {3}{x}-5 x-x^2\right )}{x}+\frac {1}{9} \operatorname {Subst}\left (\int \frac {\frac {166}{3}+22 x}{\left (\frac {28+\left (107-3 i \sqrt {1167}\right )^{2/3}}{3 \sqrt [3]{107-3 i \sqrt {1167}}}+x\right ) \left (\frac {1}{9} \left (-28+\frac {784}{\left (107-3 i \sqrt {1167}\right )^{2/3}}+\left (107-3 i \sqrt {1167}\right )^{2/3}\right )-\frac {\left (28+\left (107-3 i \sqrt {1167}\right )^{2/3}\right ) x}{3 \sqrt [3]{107-3 i \sqrt {1167}}}+x^2\right )} \, dx,x,\frac {5}{3}+x\right )-\frac {1}{9} \operatorname {Subst}\left (\int \frac {\frac {166}{3}+22 x}{\frac {214}{27}-\frac {28 x}{3}+x^3} \, dx,x,\frac {5}{3}+x\right )\\ &=\frac {\log \left (1+\frac {3}{x}-5 x-x^2\right )}{x}-\frac {1}{9} \operatorname {Subst}\left (\int \frac {\frac {166}{3}+22 x}{\left (\frac {28+\left (107-3 i \sqrt {1167}\right )^{2/3}}{3 \sqrt [3]{107-3 i \sqrt {1167}}}+x\right ) \left (\frac {1}{9} \left (-28+\frac {784}{\left (107-3 i \sqrt {1167}\right )^{2/3}}+\left (107-3 i \sqrt {1167}\right )^{2/3}\right )-\frac {\left (28+\left (107-3 i \sqrt {1167}\right )^{2/3}\right ) x}{3 \sqrt [3]{107-3 i \sqrt {1167}}}+x^2\right )} \, dx,x,\frac {5}{3}+x\right )+\frac {1}{9} \operatorname {Subst}\left (\int \left (\frac {6 \left (107-3 i \sqrt {1167}\right )^{2/3} \left (-308+83 \sqrt [3]{107-3 i \sqrt {1167}}-11 \left (107-3 i \sqrt {1167}\right )^{2/3}\right )}{\left (784+28 \left (107-3 i \sqrt {1167}\right )^{2/3}+\left (107-3 i \sqrt {1167}\right )^{4/3}\right ) \left (28+\left (107-3 i \sqrt {1167}\right )^{2/3}+3 \sqrt [3]{107-3 i \sqrt {1167}} x\right )}+\frac {6 \left (107-3 i \sqrt {1167}\right )^{2/3} \left (26386-498 i \sqrt {1167}-308 \left (107-3 i \sqrt {1167}\right )^{2/3}+\sqrt [3]{107-3 i \sqrt {1167}} \left (5825-33 i \sqrt {1167}\right )+3 \left (1177-33 i \sqrt {1167}+308 \sqrt [3]{107-3 i \sqrt {1167}}-83 \left (107-3 i \sqrt {1167}\right )^{2/3}\right ) x\right )}{\left (784+28 \left (107-3 i \sqrt {1167}\right )^{2/3}+\left (107-3 i \sqrt {1167}\right )^{4/3}\right ) \left (784-28 \left (107-3 i \sqrt {1167}\right )^{2/3}+\left (107-3 i \sqrt {1167}\right )^{4/3}-3 \left (107-3 i \sqrt {1167}+28 \sqrt [3]{107-3 i \sqrt {1167}}\right ) x+9 \left (107-3 i \sqrt {1167}\right )^{2/3} x^2\right )}\right ) \, dx,x,\frac {5}{3}+x\right )\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 20, normalized size = 0.74 \begin {gather*} \frac {\log \left (1+\frac {3}{x}-5 x-x^2\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + 3/x - 5*x - x^2]/x

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fricas [A]  time = 0.86, size = 23, normalized size = 0.85 \begin {gather*} \frac {\log \left (-\frac {x^{3} + 5 \, x^{2} - x - 3}{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-(x^3 + 5*x^2 - x - 3)/x)/x

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giac [A]  time = 0.21, size = 23, normalized size = 0.85 \begin {gather*} \frac {\log \left (-\frac {x^{3} + 5 \, x^{2} - x - 3}{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-(x^3 + 5*x^2 - x - 3)/x)/x

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maple [A]  time = 0.07, size = 23, normalized size = 0.85

method	result	size
default	\(\frac {\ln \left (\frac {-x^{3}-5 x^{2}+x +3}{x}\right )}{x}\)	\(23\)
norman	\(\frac {\ln \left (\frac {-x^{3}-5 x^{2}+x +3}{x}\right )}{x}\)	\(23\)
risch	\(\frac {\ln \left (\frac {-x^{3}-5 x^{2}+x +3}{x}\right )}{x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x*ln((-x^3-5*x^2+x+3)/x)

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maxima [A]  time = 0.35, size = 23, normalized size = 0.85 \begin {gather*} \frac {\log \left (-x^{3} - 5 \, x^{2} + x + 3\right ) - \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.40, size = 20, normalized size = 0.74 \begin {gather*} \frac {\ln \left (\frac {3}{x}-5\,x-x^2+1\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3/x - 5*x - x^2 + 1)/x

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sympy [A]  time = 0.18, size = 15, normalized size = 0.56 \begin {gather*} \frac {\log {\left (\frac {- x^{3} - 5 x^{2} + x + 3}{x} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((-x**3 - 5*x**2 + x + 3)/x)/x

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