3.22.37 \(\int \frac {-25-50 x-25 x^2+(124+100 x+25 x^2) \log (2)+\log ^2(2)}{(125+100 x+25 x^2) \log (2)+\log ^2(2)} \, dx\)

Optimal. Leaf size=26 \[ x+\frac {-x+\log \left (5+x+x (3+x)+\frac {\log (2)}{25}\right )}{\log (2)} \]

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Rubi [A]  time = 0.08, antiderivative size = 32, normalized size of antiderivative = 1.23, number of steps used = 4, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1984, 1657, 628} \begin {gather*} \frac {\log \left (25 x^2+100 x+125+\log (2)\right )}{\log (2)}-\frac {x (1-\log (2))}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 - 50*x - 25*x^2 + (124 + 100*x + 25*x^2)*Log[2] + Log[2]^2)/((125 + 100*x + 25*x^2)*Log[2] + Log[2]^2
),x]

[Out]

-((x*(1 - Log[2]))/Log[2]) + Log[125 + 100*x + 25*x^2 + Log[2]]/Log[2]

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 1657

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

Rule 1984

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
QuadraticQ[{u, v}, x] &&  !QuadraticMatchQ[{u, v}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-25-25 x^2 (1-\log (2))+124 \log (2)+\log ^2(2)-50 x (1-\log (4))}{100 x \log (2)+25 x^2 \log (2)+\log (2) (125+\log (2))} \, dx\\ &=\int \left (1-\frac {1}{\log (2)}+\frac {50 (2+x)}{100 x \log (2)+25 x^2 \log (2)+\log (2) (125+\log (2))}\right ) \, dx\\ &=-\frac {x (1-\log (2))}{\log (2)}+50 \int \frac {2+x}{100 x \log (2)+25 x^2 \log (2)+\log (2) (125+\log (2))} \, dx\\ &=-\frac {x (1-\log (2))}{\log (2)}+\frac {\log \left (125+100 x+25 x^2+\log (2)\right )}{\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 25, normalized size = 0.96 \begin {gather*} \frac {x (-1+\log (2))+\log \left (125+100 x+25 x^2+\log (2)\right )}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 - 50*x - 25*x^2 + (124 + 100*x + 25*x^2)*Log[2] + Log[2]^2)/((125 + 100*x + 25*x^2)*Log[2] + Lo
g[2]^2),x]

[Out]

(x*(-1 + Log[2]) + Log[125 + 100*x + 25*x^2 + Log[2]])/Log[2]

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fricas [A]  time = 0.79, size = 26, normalized size = 1.00 \begin {gather*} \frac {x \log \relax (2) - x + \log \left (25 \, x^{2} + 100 \, x + \log \relax (2) + 125\right )}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)^2+(25*x^2+100*x+124)*log(2)-25*x^2-50*x-25)/(log(2)^2+(25*x^2+100*x+125)*log(2)),x, algorith
m="fricas")

[Out]

(x*log(2) - x + log(25*x^2 + 100*x + log(2) + 125))/log(2)

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giac [A]  time = 0.24, size = 32, normalized size = 1.23 \begin {gather*} \frac {x \log \relax (2) - x}{\log \relax (2)} + \frac {\log \left (25 \, x^{2} + 100 \, x + \log \relax (2) + 125\right )}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)^2+(25*x^2+100*x+124)*log(2)-25*x^2-50*x-25)/(log(2)^2+(25*x^2+100*x+125)*log(2)),x, algorith
m="giac")

[Out]

(x*log(2) - x)/log(2) + log(25*x^2 + 100*x + log(2) + 125)/log(2)

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maple [A]  time = 0.76, size = 27, normalized size = 1.04

method	result	size
default	\(\frac {x \ln \relax (2)-x +\ln \left (25 x^{2}+\ln \relax (2)+100 x +125\right )}{\ln \relax (2)}\)	\(27\)
risch	\(x -\frac {x}{\ln \relax (2)}+\frac {\ln \left (25 x^{2}+\ln \relax (2)+100 x +125\right )}{\ln \relax (2)}\)	\(28\)
norman	\(\frac {\left (\ln \relax (2)-1\right ) x}{\ln \relax (2)}+\frac {\ln \left (25 x^{2}+\ln \relax (2)+100 x +125\right )}{\ln \relax (2)}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(2)^2+(25*x^2+100*x+124)*ln(2)-25*x^2-50*x-25)/(ln(2)^2+(25*x^2+100*x+125)*ln(2)),x,method=_RETURNVERBO
SE)

[Out]

1/ln(2)*(x*ln(2)-x+ln(25*x^2+ln(2)+100*x+125))

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maxima [A]  time = 0.59, size = 29, normalized size = 1.12 \begin {gather*} \frac {x {\left (\log \relax (2) - 1\right )}}{\log \relax (2)} + \frac {\log \left (25 \, x^{2} + 100 \, x + \log \relax (2) + 125\right )}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)^2+(25*x^2+100*x+124)*log(2)-25*x^2-50*x-25)/(log(2)^2+(25*x^2+100*x+125)*log(2)),x, algorith
m="maxima")

[Out]

x*(log(2) - 1)/log(2) + log(25*x^2 + 100*x + log(2) + 125)/log(2)

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mupad [B]  time = 0.16, size = 32, normalized size = 1.23 \begin {gather*} \frac {\ln \left (x^2+4\,x+\frac {\ln \relax (2)}{25}+5\right )}{\ln \relax (2)}+\frac {x\,\left (25\,\ln \relax (2)-25\right )}{25\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(50*x - log(2)*(100*x + 25*x^2 + 124) - log(2)^2 + 25*x^2 + 25)/(log(2)*(100*x + 25*x^2 + 125) + log(2)^2
),x)

[Out]

log(4*x + log(2)/25 + x^2 + 5)/log(2) + (x*(25*log(2) - 25))/(25*log(2))

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sympy [A]  time = 0.23, size = 26, normalized size = 1.00 \begin {gather*} - x \left (-1 + \frac {1}{\log {\relax (2 )}}\right ) + \frac {\log {\left (25 x^{2} + 100 x + \log {\relax (2 )} + 125 \right )}}{\log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(2)**2+(25*x**2+100*x+124)*ln(2)-25*x**2-50*x-25)/(ln(2)**2+(25*x**2+100*x+125)*ln(2)),x)

[Out]

-x*(-1 + 1/log(2)) + log(25*x**2 + 100*x + log(2) + 125)/log(2)

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