3.56.25 \(\int \frac {50-27 x-23 x^2+x^3+(24 x-x^2) \log (x^2)}{25 x^2-x^3+(-25 x+x^2) \log (x^2)} \, dx\)

Optimal. Leaf size=25 \[ \log \left (\frac {12 e^{-x}}{(25-x) \left (-x+\log \left (x^2\right )\right )}\right ) \]

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Rubi [A]  time = 0.45, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 4, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {6741, 6742, 43, 6684} \begin {gather*} -\log \left (x-\log \left (x^2\right )\right )-x-\log (25-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(50 - 27*x - 23*x^2 + x^3 + (24*x - x^2)*Log[x^2])/(25*x^2 - x^3 + (-25*x + x^2)*Log[x^2]),x]

[Out]

-x - Log[25 - x] - Log[x - Log[x^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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 {50-27 x-23 x^2+x^3+\left (24 x-x^2\right ) \log \left (x^2\right )}{(25-x) x \left (x-\log \left (x^2\right )\right )} \, dx\\ &=\int \left (\frac {24-x}{-25+x}+\frac {2-x}{x \left (x-\log \left (x^2\right )\right )}\right ) \, dx\\ &=\int \frac {24-x}{-25+x} \, dx+\int \frac {2-x}{x \left (x-\log \left (x^2\right )\right )} \, dx\\ &=-\log \left (x-\log \left (x^2\right )\right )+\int \left (-1+\frac {1}{25-x}\right ) \, dx\\ &=-x-\log (25-x)-\log \left (x-\log \left (x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 23, normalized size = 0.92 \begin {gather*} -x-\log (25-x)-\log \left (x-\log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50 - 27*x - 23*x^2 + x^3 + (24*x - x^2)*Log[x^2])/(25*x^2 - x^3 + (-25*x + x^2)*Log[x^2]),x]

[Out]

-x - Log[25 - x] - Log[x - Log[x^2]]

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fricas [A]  time = 0.99, size = 21, normalized size = 0.84 \begin {gather*} -x - \log \left (x - 25\right ) - \log \left (-x + \log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+24*x)*log(x^2)+x^3-23*x^2-27*x+50)/((x^2-25*x)*log(x^2)-x^3+25*x^2),x, algorithm="fricas")

[Out]

-x - log(x - 25) - log(-x + log(x^2))

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giac [A]  time = 0.20, size = 21, normalized size = 0.84 \begin {gather*} -x - \log \left (x - 25\right ) - \log \left (-x + \log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+24*x)*log(x^2)+x^3-23*x^2-27*x+50)/((x^2-25*x)*log(x^2)-x^3+25*x^2),x, algorithm="giac")

[Out]

-x - log(x - 25) - log(-x + log(x^2))

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maple [A]  time = 0.04, size = 22, normalized size = 0.88

method	result	size
norman	\(-x -\ln \left (x -25\right )-\ln \left (-\ln \left (x^{2}\right )+x \right )\)	\(22\)
risch	\(-x -\ln \left (x -25\right )-\ln \left (\ln \left (x^{2}\right )-x \right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+24*x)*ln(x^2)+x^3-23*x^2-27*x+50)/((x^2-25*x)*ln(x^2)-x^3+25*x^2),x,method=_RETURNVERBOSE)

[Out]

-x-ln(x-25)-ln(-ln(x^2)+x)

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maxima [A]  time = 0.40, size = 19, normalized size = 0.76 \begin {gather*} -x - \log \left (x - 25\right ) - \log \left (-\frac {1}{2} \, x + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+24*x)*log(x^2)+x^3-23*x^2-27*x+50)/((x^2-25*x)*log(x^2)-x^3+25*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.58, size = 21, normalized size = 0.84 \begin {gather*} -x-\ln \left (x-25\right )-\ln \left (\ln \left (x^2\right )-x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^2)*(24*x - x^2) - 27*x - 23*x^2 + x^3 + 50)/(log(x^2)*(25*x - x^2) - 25*x^2 + x^3),x)

[Out]

- x - log(x - 25) - log(log(x^2) - x)

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sympy [A]  time = 0.14, size = 15, normalized size = 0.60 \begin {gather*} - x - \log {\left (- x + \log {\left (x^{2} \right )} \right )} - \log {\left (x - 25 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+24*x)*ln(x**2)+x**3-23*x**2-27*x+50)/((x**2-25*x)*ln(x**2)-x**3+25*x**2),x)

[Out]

-x - log(-x + log(x**2)) - log(x - 25)

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