\(\int \frac {-96+176 x-104 x^2}{16+4 x+13 x^2} \, dx\) [2245]

Optimal result

Integrand size = 23, antiderivative size = 17 \[ \int \frac {-96+176 x-104 x^2}{16+4 x+13 x^2} \, dx=8 \left (-x+\log \left (4+x+\frac {13 x^2}{4}\right )\right ) \]

[Out]

8*ln(4+13/4*x^2+x)-8*x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1671, 642} \[ \int \frac {-96+176 x-104 x^2}{16+4 x+13 x^2} \, dx=8 \log \left (13 x^2+4 x+16\right )-8 x \]

[In]

Int[(-96 + 176*x - 104*x^2)/(16 + 4*x + 13*x^2),x]

[Out]

-8*x + 8*Log[16 + 4*x + 13*x^2]

Rule 642

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 1671

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{align*} \text {integral}& = \int \left (-8+\frac {16 (2+13 x)}{16+4 x+13 x^2}\right ) \, dx \\ & = -8 x+16 \int \frac {2+13 x}{16+4 x+13 x^2} \, dx \\ & = -8 x+8 \log \left (16+4 x+13 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-96+176 x-104 x^2}{16+4 x+13 x^2} \, dx=-8 \left (x-\log \left (16+4 x+13 x^2\right )\right ) \]

[In]

Integrate[(-96 + 176*x - 104*x^2)/(16 + 4*x + 13*x^2),x]

[Out]

-8*(x - Log[16 + 4*x + 13*x^2])

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
parallelrisch	\(-8 x +8 \ln \left (x^{2}+\frac {4}{13} x +\frac {16}{13}\right )\)	\(16\)
default	\(-8 x +8 \ln \left (13 x^{2}+4 x +16\right )\)	\(18\)
norman	\(-8 x +8 \ln \left (13 x^{2}+4 x +16\right )\)	\(18\)
risch	\(-8 x +8 \ln \left (13 x^{2}+4 x +16\right )\)	\(18\)

[In]

int((-104*x^2+176*x-96)/(13*x^2+4*x+16),x,method=_RETURNVERBOSE)

[Out]

-8*x+8*ln(x^2+4/13*x+16/13)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-96+176 x-104 x^2}{16+4 x+13 x^2} \, dx=-8 \, x + 8 \, \log \left (13 \, x^{2} + 4 \, x + 16\right ) \]

[In]

integrate((-104*x^2+176*x-96)/(13*x^2+4*x+16),x, algorithm="fricas")

[Out]

-8*x + 8*log(13*x^2 + 4*x + 16)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-96+176 x-104 x^2}{16+4 x+13 x^2} \, dx=- 8 x + 8 \log {\left (13 x^{2} + 4 x + 16 \right )} \]

[In]

integrate((-104*x**2+176*x-96)/(13*x**2+4*x+16),x)

[Out]

-8*x + 8*log(13*x**2 + 4*x + 16)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-96+176 x-104 x^2}{16+4 x+13 x^2} \, dx=-8 \, x + 8 \, \log \left (13 \, x^{2} + 4 \, x + 16\right ) \]

[In]

integrate((-104*x^2+176*x-96)/(13*x^2+4*x+16),x, algorithm="maxima")

[Out]

-8*x + 8*log(13*x^2 + 4*x + 16)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-96+176 x-104 x^2}{16+4 x+13 x^2} \, dx=-8 \, x + 8 \, \log \left (13 \, x^{2} + 4 \, x + 16\right ) \]

[In]

integrate((-104*x^2+176*x-96)/(13*x^2+4*x+16),x, algorithm="giac")

[Out]

-8*x + 8*log(13*x^2 + 4*x + 16)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-96+176 x-104 x^2}{16+4 x+13 x^2} \, dx=8\,\ln \left (13\,x^2+4\,x+16\right )-8\,x \]

[In]

int(-(104*x^2 - 176*x + 96)/(4*x + 13*x^2 + 16),x)

[Out]

8*log(4*x + 13*x^2 + 16) - 8*x