\(\int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x^2+x \log (16)} \, dx\) [7767]

Optimal result

Integrand size = 28, antiderivative size = 22 \[ \int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x^2+x \log (16)} \, dx=4 \left (3 x+4 \left (x-\log \left (-\frac {9}{5} x (x+\log (16))\right )\right )\right ) \]

[Out]

28*x-16*ln(1/5*x*(-9*x-36*ln(2)))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1607, 1834} \[ \int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x^2+x \log (16)} \, dx=28 x-16 \log (x)-16 \log (x+\log (16)) \]

[In]

Int[(-32*x + 28*x^2 + (-16 + 28*x)*Log[16])/(x^2 + x*Log[16]),x]

[Out]

28*x - 16*Log[x] - 16*Log[x + Log[16]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1834

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x (x+\log (16))} \, dx \\ & = \int \left (28-\frac {16}{x}-\frac {16}{x+\log (16)}\right ) \, dx \\ & = 28 x-16 \log (x)-16 \log (x+\log (16)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x^2+x \log (16)} \, dx=4 (7 x-4 \log (x)-4 \log (x+\log (16))) \]

[In]

Integrate[(-32*x + 28*x^2 + (-16 + 28*x)*Log[16])/(x^2 + x*Log[16]),x]

[Out]

4*(7*x - 4*Log[x] - 4*Log[x + Log[16]])

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77

method	result	size
risch	\(28 x -16 \ln \left (4 x \ln \left (2\right )+x^{2}\right )\)	\(17\)
default	\(28 x -16 \ln \left (x \right )-16 \ln \left (x +4 \ln \left (2\right )\right )\)	\(18\)
norman	\(28 x -16 \ln \left (x \right )-16 \ln \left (x +4 \ln \left (2\right )\right )\)	\(18\)
parallelrisch	\(28 x -16 \ln \left (x \right )-16 \ln \left (x +4 \ln \left (2\right )\right )\)	\(18\)
meijerg	\(-16 \ln \left (x \right )+32 \ln \left (2\right )+16 \ln \left (\ln \left (2\right )\right )+16 \ln \left (1+\frac {x}{4 \ln \left (2\right )}\right )+112 \ln \left (2\right ) \left (\frac {x}{4 \ln \left (2\right )}-\ln \left (1+\frac {x}{4 \ln \left (2\right )}\right )\right )+4 \left (28 \ln \left (2\right )-8\right ) \ln \left (1+\frac {x}{4 \ln \left (2\right )}\right )\)	\(69\)

[In]

int((4*(28*x-16)*ln(2)+28*x^2-32*x)/(4*x*ln(2)+x^2),x,method=_RETURNVERBOSE)

[Out]

28*x-16*ln(4*x*ln(2)+x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x^2+x \log (16)} \, dx=28 \, x - 16 \, \log \left (x^{2} + 4 \, x \log \left (2\right )\right ) \]

[In]

integrate((4*(28*x-16)*log(2)+28*x^2-32*x)/(4*x*log(2)+x^2),x, algorithm="fricas")

[Out]

28*x - 16*log(x^2 + 4*x*log(2))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x^2+x \log (16)} \, dx=28 x - 16 \log {\left (x^{2} + 4 x \log {\left (2 \right )} \right )} \]

[In]

integrate((4*(28*x-16)*ln(2)+28*x**2-32*x)/(4*x*ln(2)+x**2),x)

[Out]

28*x - 16*log(x**2 + 4*x*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x^2+x \log (16)} \, dx=28 \, x - 16 \, \log \left (x + 4 \, \log \left (2\right )\right ) - 16 \, \log \left (x\right ) \]

[In]

integrate((4*(28*x-16)*log(2)+28*x^2-32*x)/(4*x*log(2)+x^2),x, algorithm="maxima")

[Out]

28*x - 16*log(x + 4*log(2)) - 16*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x^2+x \log (16)} \, dx=28 \, x - 16 \, \log \left ({\left | x + 4 \, \log \left (2\right ) \right |}\right ) - 16 \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((4*(28*x-16)*log(2)+28*x^2-32*x)/(4*x*log(2)+x^2),x, algorithm="giac")

[Out]

28*x - 16*log(abs(x + 4*log(2))) - 16*log(abs(x))

Mupad [B] (verification not implemented)

Time = 12.86 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x^2+x \log (16)} \, dx=28\,x-16\,\ln \left (x\,\left (x+\ln \left (16\right )\right )\right ) \]

[In]

int((4*log(2)*(28*x - 16) - 32*x + 28*x^2)/(4*x*log(2) + x^2),x)

[Out]

28*x - 16*log(x*(x + log(16)))