\(\int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4 \log ^2(16)} \, dx\) [6484]

Optimal result

Integrand size = 25, antiderivative size = 19 \[ \int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4 \log ^2(16)} \, dx=\frac {\left (-x+\frac {4}{x \log (16)}\right )^2}{x} \]

[Out]

(1/x/ln(2)-x)^2/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 14} \[ \int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4 \log ^2(16)} \, dx=\frac {16}{x^3 \log ^2(16)}+x-\frac {8}{x \log (16)} \]

[In]

Int[(-48 + 8*x^2*Log[16] + x^4*Log[16]^2)/(x^4*Log[16]^2),x]

[Out]

x + 16/(x^3*Log[16]^2) - 8/(x*Log[16])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4} \, dx}{\log ^2(16)} \\ & = \frac {\int \left (-\frac {48}{x^4}+\frac {8 \log (16)}{x^2}+\log ^2(16)\right ) \, dx}{\log ^2(16)} \\ & = x+\frac {16}{x^3 \log ^2(16)}-\frac {8}{x \log (16)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4 \log ^2(16)} \, dx=x+\frac {16}{x^3 \log ^2(16)}-\frac {8}{x \log (16)} \]

[In]

Integrate[(-48 + 8*x^2*Log[16] + x^4*Log[16]^2)/(x^4*Log[16]^2),x]

[Out]

x + 16/(x^3*Log[16]^2) - 8/(x*Log[16])

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11

method	result	size
risch	\(x +\frac {-32 x^{2} \ln \left (2\right )+16}{16 \ln \left (2\right )^{2} x^{3}}\)	\(21\)
default	\(\frac {x \ln \left (2\right )^{2}+\frac {1}{x^{3}}-\frac {2 \ln \left (2\right )}{x}}{\ln \left (2\right )^{2}}\)	\(23\)
norman	\(\frac {\frac {1}{\ln \left (2\right )}+x^{4} \ln \left (2\right )-2 x^{2}}{x^{3} \ln \left (2\right )}\)	\(25\)
gosper	\(\frac {x^{4} \ln \left (2\right )^{2}+1-2 x^{2} \ln \left (2\right )}{x^{3} \ln \left (2\right )^{2}}\)	\(26\)
parallelrisch	\(\frac {16+16 x^{4} \ln \left (2\right )^{2}-32 x^{2} \ln \left (2\right )}{16 \ln \left (2\right )^{2} x^{3}}\)	\(28\)

[In]

int(1/16*(16*x^4*ln(2)^2+32*x^2*ln(2)-48)/x^4/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

x+1/16/ln(2)^2*(-32*x^2*ln(2)+16)/x^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4 \log ^2(16)} \, dx=\frac {x^{4} \log \left (2\right )^{2} - 2 \, x^{2} \log \left (2\right ) + 1}{x^{3} \log \left (2\right )^{2}} \]

[In]

integrate(1/16*(16*x^4*log(2)^2+32*x^2*log(2)-48)/x^4/log(2)^2,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 24 vs. \(2 (10) = 20\).

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4 \log ^2(16)} \, dx=\frac {x \log {\left (2 \right )}^{2} + \frac {- 2 x^{2} \log {\left (2 \right )} + 1}{x^{3}}}{\log {\left (2 \right )}^{2}} \]

[In]

integrate(1/16*(16*x**4*ln(2)**2+32*x**2*ln(2)-48)/x**4/ln(2)**2,x)

[Out]

(x*log(2)**2 + (-2*x**2*log(2) + 1)/x**3)/log(2)**2

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4 \log ^2(16)} \, dx=\frac {x \log \left (2\right )^{2} - \frac {2 \, x^{2} \log \left (2\right ) - 1}{x^{3}}}{\log \left (2\right )^{2}} \]

[In]

integrate(1/16*(16*x^4*log(2)^2+32*x^2*log(2)-48)/x^4/log(2)^2,x, algorithm="maxima")

[Out]

(x*log(2)^2 - (2*x^2*log(2) - 1)/x^3)/log(2)^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4 \log ^2(16)} \, dx=\frac {x \log \left (2\right )^{2} - \frac {2 \, x^{2} \log \left (2\right ) - 1}{x^{3}}}{\log \left (2\right )^{2}} \]

[In]

integrate(1/16*(16*x^4*log(2)^2+32*x^2*log(2)-48)/x^4/log(2)^2,x, algorithm="giac")

[Out]

(x*log(2)^2 - (2*x^2*log(2) - 1)/x^3)/log(2)^2

Mupad [B] (verification not implemented)

Time = 12.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4 \log ^2(16)} \, dx=\frac {{\left (x^2\,\ln \left (2\right )-1\right )}^2}{x^3\,{\ln \left (2\right )}^2} \]

[In]

int((x^4*log(2)^2 + 2*x^2*log(2) - 3)/(x^4*log(2)^2),x)

[Out]

(x^2*log(2) - 1)^2/(x^3*log(2)^2)