3.65.80 \(\int \frac {100-60 x-140 x^2-70 x^3-14 x^4-x^5+(-25+15 x+9 x^2+x^3) \log (2)}{-100 x-120 x^2-60 x^3-13 x^4-x^5+(25 x+10 x^2+x^3) \log (2)} \, dx\)

Optimal. Leaf size=26 \[ x+\log \left (\frac {x-(2+x)^2-\frac {x}{5+x}+\log (2)}{x}\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 38, normalized size of antiderivative = 1.46, number of steps used = 3, number of rules used = 2, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2074, 1587} \begin {gather*} \log \left (x^3+8 x^2+x (20-\log (2))+5 (4-\log (2))\right )+x-\log (x)-\log (x+5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(100 - 60*x - 140*x^2 - 70*x^3 - 14*x^4 - x^5 + (-25 + 15*x + 9*x^2 + x^3)*Log[2])/(-100*x - 120*x^2 - 60*
x^3 - 13*x^4 - x^5 + (25*x + 10*x^2 + x^3)*Log[2]),x]

[Out]

x - Log[x] - Log[5 + x] + Log[8*x^2 + x^3 + 5*(4 - Log[2]) + x*(20 - Log[2])]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 35, normalized size = 1.35 \begin {gather*} x-\log (x)-\log (5+x)+\log \left (20+20 x+8 x^2+x^3-5 \log (2)-x \log (2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100 - 60*x - 140*x^2 - 70*x^3 - 14*x^4 - x^5 + (-25 + 15*x + 9*x^2 + x^3)*Log[2])/(-100*x - 120*x^2
- 60*x^3 - 13*x^4 - x^5 + (25*x + 10*x^2 + x^3)*Log[2]),x]

[Out]

x - Log[x] - Log[5 + x] + Log[20 + 20*x + 8*x^2 + x^3 - 5*Log[2] - x*Log[2]]

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fricas [A]  time = 0.70, size = 33, normalized size = 1.27 \begin {gather*} x + \log \left (x^{3} + 8 \, x^{2} - {\left (x + 5\right )} \log \relax (2) + 20 \, x + 20\right ) - \log \left (x^{2} + 5 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+9*x^2+15*x-25)*log(2)-x^5-14*x^4-70*x^3-140*x^2-60*x+100)/((x^3+10*x^2+25*x)*log(2)-x^5-13*x^4
-60*x^3-120*x^2-100*x),x, algorithm="fricas")

[Out]

x + log(x^3 + 8*x^2 - (x + 5)*log(2) + 20*x + 20) - log(x^2 + 5*x)

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giac [A]  time = 0.19, size = 38, normalized size = 1.46 \begin {gather*} x + \log \left ({\left | x^{3} + 8 \, x^{2} - x \log \relax (2) + 20 \, x - 5 \, \log \relax (2) + 20 \right |}\right ) - \log \left ({\left | x + 5 \right |}\right ) - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+9*x^2+15*x-25)*log(2)-x^5-14*x^4-70*x^3-140*x^2-60*x+100)/((x^3+10*x^2+25*x)*log(2)-x^5-13*x^4
-60*x^3-120*x^2-100*x),x, algorithm="giac")

[Out]

x + log(abs(x^3 + 8*x^2 - x*log(2) + 20*x - 5*log(2) + 20)) - log(abs(x + 5)) - log(abs(x))

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maple [A]  time = 0.14, size = 36, normalized size = 1.38

method	result	size
default	\(x +\ln \left (x^{3}-x \ln \relax (2)+8 x^{2}-5 \ln \relax (2)+20 x +20\right )-\ln \left (5+x \right )-\ln \relax (x )\)	\(36\)
norman	\(x -\ln \relax (x )-\ln \left (5+x \right )+\ln \left (-x^{3}+x \ln \relax (2)-8 x^{2}+5 \ln \relax (2)-20 x -20\right )\)	\(37\)
risch	\(x -\ln \left (-x^{2}-5 x \right )+\ln \left (x^{3}+8 x^{2}+\left (-\ln \relax (2)+20\right ) x -5 \ln \relax (2)+20\right )\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+9*x^2+15*x-25)*ln(2)-x^5-14*x^4-70*x^3-140*x^2-60*x+100)/((x^3+10*x^2+25*x)*ln(2)-x^5-13*x^4-60*x^3-
120*x^2-100*x),x,method=_RETURNVERBOSE)

[Out]

x+ln(x^3-x*ln(2)+8*x^2-5*ln(2)+20*x+20)-ln(5+x)-ln(x)

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maxima [A]  time = 0.39, size = 34, normalized size = 1.31 \begin {gather*} x + \log \left (x^{3} + 8 \, x^{2} - x {\left (\log \relax (2) - 20\right )} - 5 \, \log \relax (2) + 20\right ) - \log \left (x + 5\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+9*x^2+15*x-25)*log(2)-x^5-14*x^4-70*x^3-140*x^2-60*x+100)/((x^3+10*x^2+25*x)*log(2)-x^5-13*x^4
-60*x^3-120*x^2-100*x),x, algorithm="maxima")

[Out]

x + log(x^3 + 8*x^2 - x*(log(2) - 20) - 5*log(2) + 20) - log(x + 5) - log(x)

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mupad [B]  time = 4.16, size = 33, normalized size = 1.27 \begin {gather*} x-\ln \left (x\,\left (x+5\right )\right )+\ln \left (20\,x-\ln \left (32\right )-x\,\ln \relax (2)+8\,x^2+x^3+20\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((60*x - log(2)*(15*x + 9*x^2 + x^3 - 25) + 140*x^2 + 70*x^3 + 14*x^4 + x^5 - 100)/(100*x - log(2)*(25*x +
10*x^2 + x^3) + 120*x^2 + 60*x^3 + 13*x^4 + x^5),x)

[Out]

x - log(x*(x + 5)) + log(20*x - log(32) - x*log(2) + 8*x^2 + x^3 + 20)

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sympy [A]  time = 1.67, size = 32, normalized size = 1.23 \begin {gather*} x - \log {\left (x^{2} + 5 x \right )} + \log {\left (x^{3} + 8 x^{2} + x \left (20 - \log {\relax (2 )}\right ) - 5 \log {\relax (2 )} + 20 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+9*x**2+15*x-25)*ln(2)-x**5-14*x**4-70*x**3-140*x**2-60*x+100)/((x**3+10*x**2+25*x)*ln(2)-x**5
-13*x**4-60*x**3-120*x**2-100*x),x)

[Out]

x - log(x**2 + 5*x) + log(x**3 + 8*x**2 + x*(20 - log(2)) - 5*log(2) + 20)

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