3.78.36 \(\int \frac {-120+11 x-8 x^2}{125+5 x+8 x^2} \, dx\)

Optimal. Leaf size=19 \[ -x+\log \left (25+x+4 x \log \left (e^{2 x/5}\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1657, 628} \begin {gather*} \log \left (8 x^2+5 x+125\right )-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-120 + 11*x - 8*x^2)/(125 + 5*x + 8*x^2),x]

[Out]

-x + Log[125 + 5*x + 8*x^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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+\frac {5+16 x}{125+5 x+8 x^2}\right ) \, dx\\ &=-x+\int \frac {5+16 x}{125+5 x+8 x^2} \, dx\\ &=-x+\log \left (125+5 x+8 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.79 \begin {gather*} -x+\log \left (125+5 x+8 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-120 + 11*x - 8*x^2)/(125 + 5*x + 8*x^2),x]

[Out]

-x + Log[125 + 5*x + 8*x^2]

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fricas [A]  time = 0.54, size = 15, normalized size = 0.79 \begin {gather*} -x + \log \left (8 \, x^{2} + 5 \, x + 125\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^2+11*x-120)/(8*x^2+5*x+125),x, algorithm="fricas")

[Out]

-x + log(8*x^2 + 5*x + 125)

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giac [A]  time = 0.16, size = 15, normalized size = 0.79 \begin {gather*} -x + \log \left (8 \, x^{2} + 5 \, x + 125\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^2+11*x-120)/(8*x^2+5*x+125),x, algorithm="giac")

[Out]

-x + log(8*x^2 + 5*x + 125)

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maple [A]  time = 0.51, size = 16, normalized size = 0.84

method	result	size
default	\(-x +\ln \left (8 x^{2}+5 x +125\right )\)	\(16\)
norman	\(-x +\ln \left (8 x^{2}+5 x +125\right )\)	\(16\)
risch	\(-x +\ln \left (8 x^{2}+5 x +125\right )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*x^2+11*x-120)/(8*x^2+5*x+125),x,method=_RETURNVERBOSE)

[Out]

-x+ln(8*x^2+5*x+125)

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maxima [A]  time = 0.37, size = 15, normalized size = 0.79 \begin {gather*} -x + \log \left (8 \, x^{2} + 5 \, x + 125\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^2+11*x-120)/(8*x^2+5*x+125),x, algorithm="maxima")

[Out]

-x + log(8*x^2 + 5*x + 125)

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mupad [B]  time = 0.06, size = 15, normalized size = 0.79 \begin {gather*} \ln \left (8\,x^2+5\,x+125\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x^2 - 11*x + 120)/(5*x + 8*x^2 + 125),x)

[Out]

log(5*x + 8*x^2 + 125) - x

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sympy [A]  time = 0.10, size = 12, normalized size = 0.63 \begin {gather*} - x + \log {\left (8 x^{2} + 5 x + 125 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x**2+11*x-120)/(8*x**2+5*x+125),x)

[Out]

-x + log(8*x**2 + 5*x + 125)

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