\(\int \frac {-1+4 x^3+8 x^2 \log (16)+4 x \log ^2(16)}{x^2+2 x \log (16)+\log ^2(16)} \, dx\) [5372]

Optimal result

Integrand size = 37, antiderivative size = 12 \[ \int \frac {-1+4 x^3+8 x^2 \log (16)+4 x \log ^2(16)}{x^2+2 x \log (16)+\log ^2(16)} \, dx=2 x^2+\frac {1}{x+\log (16)} \]

[Out]

1/(x+4*ln(2))+2*x^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {27, 1864} \[ \int \frac {-1+4 x^3+8 x^2 \log (16)+4 x \log ^2(16)}{x^2+2 x \log (16)+\log ^2(16)} \, dx=2 x^2+\frac {1}{x+\log (16)} \]

[In]

Int[(-1 + 4*x^3 + 8*x^2*Log[16] + 4*x*Log[16]^2)/(x^2 + 2*x*Log[16] + Log[16]^2),x]

[Out]

2*x^2 + (x + Log[16])^(-1)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1864

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+4 x^3+8 x^2 \log (16)+4 x \log ^2(16)}{(x+\log (16))^2} \, dx \\ & = \int \left (4 x-\frac {1}{(x+\log (16))^2}\right ) \, dx \\ & = 2 x^2+\frac {1}{x+\log (16)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.50 \[ \int \frac {-1+4 x^3+8 x^2 \log (16)+4 x \log ^2(16)}{x^2+2 x \log (16)+\log ^2(16)} \, dx=2 x^2-2 \log ^2(16)+\frac {1}{x+\log (16)} \]

[In]

Integrate[(-1 + 4*x^3 + 8*x^2*Log[16] + 4*x*Log[16]^2)/(x^2 + 2*x*Log[16] + Log[16]^2),x]

[Out]

2*x^2 - 2*Log[16]^2 + (x + Log[16])^(-1)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {1}{x +4 \ln \left (2\right )}+2 x^{2}\)	\(15\)
risch	\(\frac {1}{x +4 \ln \left (2\right )}+2 x^{2}\)	\(17\)
gosper	\(\frac {2 x^{3}+8 x^{2} \ln \left (2\right )+1}{x +4 \ln \left (2\right )}\)	\(24\)
norman	\(\frac {2 x^{3}+8 x^{2} \ln \left (2\right )+1}{x +4 \ln \left (2\right )}\)	\(24\)
parallelrisch	\(\frac {2 x^{3}+8 x^{2} \ln \left (2\right )+1}{x +4 \ln \left (2\right )}\)	\(24\)
meijerg	\(-\frac {x}{16 \ln \left (2\right )^{2} \left (1+\frac {x}{4 \ln \left (2\right )}\right )}+64 \ln \left (2\right )^{2} \left (-\frac {x}{4 \ln \left (2\right ) \left (1+\frac {x}{4 \ln \left (2\right )}\right )}+\ln \left (1+\frac {x}{4 \ln \left (2\right )}\right )\right )+128 \ln \left (2\right )^{2} \left (\frac {x \left (\frac {3 x}{4 \ln \left (2\right )}+6\right )}{12 \ln \left (2\right ) \left (1+\frac {x}{4 \ln \left (2\right )}\right )}-2 \ln \left (1+\frac {x}{4 \ln \left (2\right )}\right )\right )+64 \ln \left (2\right )^{2} \left (-\frac {x \left (-\frac {x^{2}}{8 \ln \left (2\right )^{2}}+\frac {3 x}{2 \ln \left (2\right )}+12\right )}{16 \ln \left (2\right ) \left (1+\frac {x}{4 \ln \left (2\right )}\right )}+3 \ln \left (1+\frac {x}{4 \ln \left (2\right )}\right )\right )\)	\(156\)

[In]

int((64*x*ln(2)^2+32*x^2*ln(2)+4*x^3-1)/(16*ln(2)^2+8*x*ln(2)+x^2),x,method=_RETURNVERBOSE)

[Out]

1/(x+4*ln(2))+2*x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.92 \[ \int \frac {-1+4 x^3+8 x^2 \log (16)+4 x \log ^2(16)}{x^2+2 x \log (16)+\log ^2(16)} \, dx=\frac {2 \, x^{3} + 8 \, x^{2} \log \left (2\right ) + 1}{x + 4 \, \log \left (2\right )} \]

[In]

integrate((64*x*log(2)^2+32*x^2*log(2)+4*x^3-1)/(16*log(2)^2+8*x*log(2)+x^2),x, algorithm="fricas")

[Out]

(2*x^3 + 8*x^2*log(2) + 1)/(x + 4*log(2))

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {-1+4 x^3+8 x^2 \log (16)+4 x \log ^2(16)}{x^2+2 x \log (16)+\log ^2(16)} \, dx=2 x^{2} + \frac {1}{x + 4 \log {\left (2 \right )}} \]

[In]

integrate((64*x*ln(2)**2+32*x**2*ln(2)+4*x**3-1)/(16*ln(2)**2+8*x*ln(2)+x**2),x)

[Out]

2*x**2 + 1/(x + 4*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {-1+4 x^3+8 x^2 \log (16)+4 x \log ^2(16)}{x^2+2 x \log (16)+\log ^2(16)} \, dx=2 \, x^{2} + \frac {1}{x + 4 \, \log \left (2\right )} \]

[In]

integrate((64*x*log(2)^2+32*x^2*log(2)+4*x^3-1)/(16*log(2)^2+8*x*log(2)+x^2),x, algorithm="maxima")

[Out]

2*x^2 + 1/(x + 4*log(2))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {-1+4 x^3+8 x^2 \log (16)+4 x \log ^2(16)}{x^2+2 x \log (16)+\log ^2(16)} \, dx=2 \, x^{2} + \frac {1}{x + 4 \, \log \left (2\right )} \]

[In]

integrate((64*x*log(2)^2+32*x^2*log(2)+4*x^3-1)/(16*log(2)^2+8*x*log(2)+x^2),x, algorithm="giac")

[Out]

2*x^2 + 1/(x + 4*log(2))

Mupad [B] (verification not implemented)

Time = 12.89 (sec) , antiderivative size = 58, normalized size of antiderivative = 4.83 \[ \int \frac {-1+4 x^3+8 x^2 \log (16)+4 x \log ^2(16)}{x^2+2 x \log (16)+\log ^2(16)} \, dx=2\,x^2+\frac {\mathrm {atanh}\left (\frac {2\,x+8\,\ln \left (2\right )}{2\,\sqrt {4\,\ln \left (2\right )+\ln \left (16\right )}\,\sqrt {4\,\ln \left (2\right )-\ln \left (16\right )}}\right )}{\sqrt {4\,\ln \left (2\right )+\ln \left (16\right )}\,\sqrt {4\,\ln \left (2\right )-\ln \left (16\right )}} \]

[In]

int((64*x*log(2)^2 + 32*x^2*log(2) + 4*x^3 - 1)/(8*x*log(2) + 16*log(2)^2 + x^2),x)

[Out]

2*x^2 + atanh((2*x + 8*log(2))/(2*(4*log(2) + log(16))^(1/2)*(4*log(2) - log(16))^(1/2)))/((4*log(2) + log(16)
)^(1/2)*(4*log(2) - log(16))^(1/2))