3.93.99 \(\int \frac {-480+280 x-40 x^2+15 x^3-5 x^4+(-80 x+40 x^2-48 x^3-5 x^4) \log (\frac {80-40 x+48 x^2+5 x^3}{5 x^2})}{80 x-40 x^2+48 x^3+5 x^4} \, dx\)

Optimal. Leaf size=23 \[ (3-x) \log \left (\frac {43}{5}+\frac {(4-x)^2}{x^2}+x\right ) \]

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Rubi [A]  time = 7.77, antiderivative size = 42, normalized size of antiderivative = 1.83, number of steps used = 26, number of rules used = 12, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.141, Rules used = {6688, 6742, 2100, 2081, 2079, 800, 634, 618, 204, 628, 2523, 12} \begin {gather*} -x \log \left (\frac {16}{x^2}+x-\frac {8}{x}+\frac {48}{5}\right )+3 \log \left (5 x^3+48 x^2-40 x+80\right )-6 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-480 + 280*x - 40*x^2 + 15*x^3 - 5*x^4 + (-80*x + 40*x^2 - 48*x^3 - 5*x^4)*Log[(80 - 40*x + 48*x^2 + 5*x^
3)/(5*x^2)])/(80*x - 40*x^2 + 48*x^3 + 5*x^4),x]

[Out]

-6*Log[x] - x*Log[48/5 + 16/x^2 - 8/x + x] + 3*Log[80 - 40*x + 48*x^2 + 5*x^3]

Rule 12

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

Rule 204

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

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 (-\frac {5 \left (96-56 x+8 x^2-3 x^3+x^4\right )}{x \left (80-40 x+48 x^2+5 x^3\right )}-\log \left (\frac {48}{5}+\frac {16}{x^2}-\frac {8}{x}+x\right )\right ) \, dx\\ &=-\left (5 \int \frac {96-56 x+8 x^2-3 x^3+x^4}{x \left (80-40 x+48 x^2+5 x^3\right )} \, dx\right )-\int \log \left (\frac {48}{5}+\frac {16}{x^2}-\frac {8}{x}+x\right ) \, dx\\ &=-x \log \left (\frac {48}{5}+\frac {16}{x^2}-\frac {8}{x}+x\right )-5 \int \left (\frac {1}{5}+\frac {6}{5 x}+\frac {-120-208 x-93 x^2}{5 \left (80-40 x+48 x^2+5 x^3\right )}\right ) \, dx+\int \frac {5 \left (-32+8 x+x^3\right )}{80-40 x+48 x^2+5 x^3} \, dx\\ &=-x-6 \log (x)-x \log \left (\frac {48}{5}+\frac {16}{x^2}-\frac {8}{x}+x\right )+5 \int \frac {-32+8 x+x^3}{80-40 x+48 x^2+5 x^3} \, dx-\int \frac {-120-208 x-93 x^2}{80-40 x+48 x^2+5 x^3} \, dx\\ &=-x-6 \log (x)-x \log \left (\frac {48}{5}+\frac {16}{x^2}-\frac {8}{x}+x\right )+\frac {31}{5} \log \left (80-40 x+48 x^2+5 x^3\right )-\frac {1}{15} \int \frac {-5520+5808 x}{80-40 x+48 x^2+5 x^3} \, dx+5 \int \left (\frac {1}{5}-\frac {16 \left (15-5 x+3 x^2\right )}{5 \left (80-40 x+48 x^2+5 x^3\right )}\right ) \, dx\\ &=-6 \log (x)-x \log \left (\frac {48}{5}+\frac {16}{x^2}-\frac {8}{x}+x\right )+\frac {31}{5} \log \left (80-40 x+48 x^2+5 x^3\right )-\frac {1}{15} \operatorname {Subst}\left (\int \frac {-\frac {120528}{5}+5808 x}{\frac {13392}{25}-\frac {968 x}{5}+5 x^3} \, dx,x,\frac {16}{5}+x\right )-16 \int \frac {15-5 x+3 x^2}{80-40 x+48 x^2+5 x^3} \, dx\\ &=-6 \log (x)-x \log \left (\frac {48}{5}+\frac {16}{x^2}-\frac {8}{x}+x\right )+3 \log \left (80-40 x+48 x^2+5 x^3\right )-\frac {16}{15} \int \frac {345-363 x}{80-40 x+48 x^2+5 x^3} \, dx-\frac {5}{3} \operatorname {Subst}\left (\int \frac {-\frac {120528}{5}+5808 x}{\left (\frac {2}{3} \left (\frac {242\ 3^{2/3}}{\sqrt [3]{7533-5 \sqrt {569145}}}+\sqrt [3]{3 \left (7533-5 \sqrt {569145}\right )}\right )+5 x\right ) \left (-\frac {4}{9} \left (726-\frac {175692 \sqrt [3]{3}}{\left (7533-5 \sqrt {569145}\right )^{2/3}}-\left (3 \left (7533-5 \sqrt {569145}\right )\right )^{2/3}\right )-\frac {10 \left (242 \sqrt [3]{\frac {3}{7533-5 \sqrt {569145}}}+\sqrt [3]{7533-5 \sqrt {569145}}\right ) x}{3^{2/3}}+25 x^2\right )} \, dx,x,\frac {16}{5}+x\right )\\ &=-6 \log (x)-x \log \left (\frac {48}{5}+\frac {16}{x^2}-\frac {8}{x}+x\right )+3 \log \left (80-40 x+48 x^2+5 x^3\right )-\frac {16}{15} \operatorname {Subst}\left (\int \frac {\frac {7533}{5}-363 x}{\frac {13392}{25}-\frac {968 x}{5}+5 x^3} \, dx,x,\frac {16}{5}+x\right )-\frac {5}{3} \operatorname {Subst}\left (\int \left (\frac {12 \left (7533-5 \sqrt {569145}\right )^{2/3} \left (-58564 3^{2/3}-7533 \sqrt [3]{7533-5 \sqrt {569145}}-242 \sqrt [3]{3} \left (7533-5 \sqrt {569145}\right )^{2/3}\right )}{5 \left (58564 \sqrt [3]{3}+\left (2511\ 3^{2/3}-5 \sqrt [6]{3} \sqrt {189715}\right ) \sqrt [3]{7533-5 \sqrt {569145}}+242 \left (7533-5 \sqrt {569145}\right )^{2/3}\right ) \left (484\ 3^{2/3}+2 \sqrt [3]{3} \left (7533-5 \sqrt {569145}\right )^{2/3}+15 \sqrt [3]{7533-5 \sqrt {569145}} x\right )}+\frac {12 \left (7533-5 \sqrt {569145}\right )^{2/3} \left (-4 \left (\sqrt [3]{3} \left (11829119-12555 \sqrt {569145}\right )+121 \left (2511\ 3^{2/3}+5 \sqrt [6]{3} \sqrt {189715}\right ) \sqrt [3]{7533-5 \sqrt {569145}}+29282 \left (7533-5 \sqrt {569145}\right )^{2/3}\right )+5 \left (58564\ 3^{2/3} \sqrt [3]{7533-5 \sqrt {569145}}+7533 \left (7533-5 \sqrt {569145}\right )^{2/3}+242 \sqrt [3]{3} \left (7533-5 \sqrt {569145}\right )\right ) x\right )}{5 \left (58564 \sqrt [3]{3}+\left (2511\ 3^{2/3}-5 \sqrt [6]{3} \sqrt {189715}\right ) \sqrt [3]{7533-5 \sqrt {569145}}+242 \left (7533-5 \sqrt {569145}\right )^{2/3}\right ) \left (4 \left (58564 \sqrt [3]{3}+\left (2511\ 3^{2/3}-5 \sqrt [6]{3} \sqrt {189715}\right ) \sqrt [3]{7533-5 \sqrt {569145}}-242 \left (7533-5 \sqrt {569145}\right )^{2/3}\right )-10\ 3^{2/3} \left (2511\ 3^{2/3}-5 \sqrt [6]{3} \sqrt {189715}+242 \sqrt [3]{7533-5 \sqrt {569145}}\right ) x+75 \left (7533-5 \sqrt {569145}\right )^{2/3} x^2\right )}\right ) \, dx,x,\frac {16}{5}+x\right )\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.10, size = 166, normalized size = 7.22 \begin {gather*} -6 \log (x)-x \log \left (\frac {48}{5}+\frac {16}{x^2}-\frac {8}{x}+x\right )-16 \text {RootSum}\left [80-40 \text {$\#$1}+48 \text {$\#$1}^2+5 \text {$\#$1}^3\&,\frac {15 \log (x-\text {$\#$1})-5 \log (x-\text {$\#$1}) \text {$\#$1}+3 \log (x-\text {$\#$1}) \text {$\#$1}^2}{-40+96 \text {$\#$1}+15 \text {$\#$1}^2}\&\right ]+\text {RootSum}\left [80-40 \text {$\#$1}+48 \text {$\#$1}^2+5 \text {$\#$1}^3\&,\frac {120 \log (x-\text {$\#$1})+208 \log (x-\text {$\#$1}) \text {$\#$1}+93 \log (x-\text {$\#$1}) \text {$\#$1}^2}{-40+96 \text {$\#$1}+15 \text {$\#$1}^2}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-480 + 280*x - 40*x^2 + 15*x^3 - 5*x^4 + (-80*x + 40*x^2 - 48*x^3 - 5*x^4)*Log[(80 - 40*x + 48*x^2
+ 5*x^3)/(5*x^2)])/(80*x - 40*x^2 + 48*x^3 + 5*x^4),x]

[Out]

-6*Log[x] - x*Log[48/5 + 16/x^2 - 8/x + x] - 16*RootSum[80 - 40*#1 + 48*#1^2 + 5*#1^3 & , (15*Log[x - #1] - 5*
Log[x - #1]*#1 + 3*Log[x - #1]*#1^2)/(-40 + 96*#1 + 15*#1^2) & ] + RootSum[80 - 40*#1 + 48*#1^2 + 5*#1^3 & , (
120*Log[x - #1] + 208*Log[x - #1]*#1 + 93*Log[x - #1]*#1^2)/(-40 + 96*#1 + 15*#1^2) & ]

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fricas [A]  time = 0.56, size = 26, normalized size = 1.13 \begin {gather*} -{\left (x - 3\right )} \log \left (\frac {5 \, x^{3} + 48 \, x^{2} - 40 \, x + 80}{5 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^4-48*x^3+40*x^2-80*x)*log(1/5*(5*x^3+48*x^2-40*x+80)/x^2)-5*x^4+15*x^3-40*x^2+280*x-480)/(5*x
^4+48*x^3-40*x^2+80*x),x, algorithm="fricas")

[Out]

-(x - 3)*log(1/5*(5*x^3 + 48*x^2 - 40*x + 80)/x^2)

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giac [B]  time = 0.17, size = 47, normalized size = 2.04 \begin {gather*} -x \log \left (\frac {5 \, x^{3} + 48 \, x^{2} - 40 \, x + 80}{5 \, x^{2}}\right ) + 3 \, \log \left (5 \, x^{3} + 48 \, x^{2} - 40 \, x + 80\right ) - 6 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^4-48*x^3+40*x^2-80*x)*log(1/5*(5*x^3+48*x^2-40*x+80)/x^2)-5*x^4+15*x^3-40*x^2+280*x-480)/(5*x
^4+48*x^3-40*x^2+80*x),x, algorithm="giac")

[Out]

-x*log(1/5*(5*x^3 + 48*x^2 - 40*x + 80)/x^2) + 3*log(5*x^3 + 48*x^2 - 40*x + 80) - 6*log(x)

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maple [B]  time = 0.07, size = 48, normalized size = 2.09

method	result	size
risch	\(-\ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right ) x -6 \ln \relax (x )+3 \ln \left (5 x^{3}+48 x^{2}-40 x +80\right )\)	\(48\)
norman	\(3 \ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right )-\ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right ) x\)	\(49\)
default	\(-6 \ln \relax (x )-\left (\munderset {\textit {\_R} =\RootOf \left (5 \textit {\_Z}^{3}+48 \textit {\_Z}^{2}-40 \textit {\_Z} +80\right )}{\sum }\frac {\left (-93 \textit {\_R}^{2}-208 \textit {\_R} -120\right ) \ln \left (x -\textit {\_R} \right )}{15 \textit {\_R}^{2}+96 \textit {\_R} -40}\right )-x \ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{x^{2}}\right )+16 \left (\munderset {\textit {\_R} =\RootOf \left (5 \textit {\_Z}^{3}+48 \textit {\_Z}^{2}-40 \textit {\_Z} +80\right )}{\sum }\frac {\left (-3 \textit {\_R}^{2}+5 \textit {\_R} -15\right ) \ln \left (x -\textit {\_R} \right )}{15 \textit {\_R}^{2}+96 \textit {\_R} -40}\right )+x \ln \relax (5)\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x^4-48*x^3+40*x^2-80*x)*ln(1/5*(5*x^3+48*x^2-40*x+80)/x^2)-5*x^4+15*x^3-40*x^2+280*x-480)/(5*x^4+48*x
^3-40*x^2+80*x),x,method=_RETURNVERBOSE)

[Out]

-ln(1/5*(5*x^3+48*x^2-40*x+80)/x^2)*x-6*ln(x)+3*ln(5*x^3+48*x^2-40*x+80)

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maxima [A]  time = 0.43, size = 33, normalized size = 1.43 \begin {gather*} x \log \relax (5) - {\left (x - 3\right )} \log \left (5 \, x^{3} + 48 \, x^{2} - 40 \, x + 80\right ) + 2 \, {\left (x - 3\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^4-48*x^3+40*x^2-80*x)*log(1/5*(5*x^3+48*x^2-40*x+80)/x^2)-5*x^4+15*x^3-40*x^2+280*x-480)/(5*x
^4+48*x^3-40*x^2+80*x),x, algorithm="maxima")

[Out]

x*log(5) - (x - 3)*log(5*x^3 + 48*x^2 - 40*x + 80) + 2*(x - 3)*log(x)

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mupad [B]  time = 9.17, size = 42, normalized size = 1.83 \begin {gather*} 3\,\ln \left (x^3+\frac {48\,x^2}{5}-8\,x+16\right )-6\,\ln \relax (x)-x\,\ln \left (\frac {x^3+\frac {48\,x^2}{5}-8\,x+16}{x^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(((48*x^2)/5 - 8*x + x^3 + 16)/x^2)*(80*x - 40*x^2 + 48*x^3 + 5*x^4) - 280*x + 40*x^2 - 15*x^3 + 5*x^
4 + 480)/(80*x - 40*x^2 + 48*x^3 + 5*x^4),x)

[Out]

3*log((48*x^2)/5 - 8*x + x^3 + 16) - 6*log(x) - x*log(((48*x^2)/5 - 8*x + x^3 + 16)/x^2)

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sympy [B]  time = 0.24, size = 44, normalized size = 1.91 \begin {gather*} - x \log {\left (\frac {x^{3} + \frac {48 x^{2}}{5} - 8 x + 16}{x^{2}} \right )} - 6 \log {\relax (x )} + 3 \log {\left (5 x^{3} + 48 x^{2} - 40 x + 80 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x**4-48*x**3+40*x**2-80*x)*ln(1/5*(5*x**3+48*x**2-40*x+80)/x**2)-5*x**4+15*x**3-40*x**2+280*x-4
80)/(5*x**4+48*x**3-40*x**2+80*x),x)

[Out]

-x*log((x**3 + 48*x**2/5 - 8*x + 16)/x**2) - 6*log(x) + 3*log(5*x**3 + 48*x**2 - 40*x + 80)

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