\(\int \frac {4+5 x^2+8 x^3+x^2 \log (\frac {5}{3})}{64 x^2} \, dx\) [3816]

Optimal result

Integrand size = 27, antiderivative size = 22 \[ \int \frac {4+5 x^2+8 x^3+x^2 \log \left (\frac {5}{3}\right )}{64 x^2} \, dx=\frac {1}{16} x \left (-\frac {1}{x^2}+x+\frac {1}{4} \left (5+\log \left (\frac {5}{3}\right )\right )\right ) \]

[Out]

1/16*x*(x-1/x^2-1/4*ln(3/5)+5/4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6, 12, 14} \[ \int \frac {4+5 x^2+8 x^3+x^2 \log \left (\frac {5}{3}\right )}{64 x^2} \, dx=\frac {x^2}{16}-\frac {1}{16 x}+\frac {1}{64} x \left (5+\log \left (\frac {5}{3}\right )\right ) \]

[In]

Int[(4 + 5*x^2 + 8*x^3 + x^2*Log[5/3])/(64*x^2),x]

[Out]

-1/16*1/x + x^2/16 + (x*(5 + Log[5/3]))/64

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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}& = \int \frac {4+8 x^3+x^2 \left (5+\log \left (\frac {5}{3}\right )\right )}{64 x^2} \, dx \\ & = \frac {1}{64} \int \frac {4+8 x^3+x^2 \left (5+\log \left (\frac {5}{3}\right )\right )}{x^2} \, dx \\ & = \frac {1}{64} \int \left (\frac {4}{x^2}+8 x+5 \left (1+\frac {1}{5} \log \left (\frac {5}{3}\right )\right )\right ) \, dx \\ & = -\frac {1}{16 x}+\frac {x^2}{16}+\frac {1}{64} x \left (5+\log \left (\frac {5}{3}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {4+5 x^2+8 x^3+x^2 \log \left (\frac {5}{3}\right )}{64 x^2} \, dx=\frac {1}{64} \left (-\frac {4}{x}+4 x^2+x \left (5+\log \left (\frac {5}{3}\right )\right )\right ) \]

[In]

Integrate[(4 + 5*x^2 + 8*x^3 + x^2*Log[5/3])/(64*x^2),x]

[Out]

(-4/x + 4*x^2 + x*(5 + Log[5/3]))/64

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {x^{2}}{16}-\frac {x \ln \left (\frac {3}{5}\right )}{64}+\frac {5 x}{64}-\frac {1}{16 x}\)	\(20\)
gosper	\(-\frac {x^{2} \ln \left (\frac {3}{5}\right )-4 x^{3}-5 x^{2}+4}{64 x}\)	\(24\)
parallelrisch	\(-\frac {x^{2} \ln \left (\frac {3}{5}\right )-4 x^{3}-5 x^{2}+4}{64 x}\)	\(24\)
risch	\(\frac {x \ln \left (5\right )}{64}-\frac {x \ln \left (3\right )}{64}+\frac {x^{2}}{16}+\frac {5 x}{64}-\frac {1}{16 x}\)	\(25\)
norman	\(\frac {-\frac {1}{16}+\left (-\frac {\ln \left (3\right )}{64}+\frac {\ln \left (5\right )}{64}+\frac {5}{64}\right ) x^{2}+\frac {x^{3}}{16}}{x}\)	\(26\)

[In]

int(1/64*(-x^2*ln(3/5)+8*x^3+5*x^2+4)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/16*x^2-1/64*x*ln(3/5)+5/64*x-1/16/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {4+5 x^2+8 x^3+x^2 \log \left (\frac {5}{3}\right )}{64 x^2} \, dx=\frac {4 \, x^{3} - x^{2} \log \left (\frac {3}{5}\right ) + 5 \, x^{2} - 4}{64 \, x} \]

[In]

integrate(1/64*(-x^2*log(3/5)+8*x^3+5*x^2+4)/x^2,x, algorithm="fricas")

[Out]

1/64*(4*x^3 - x^2*log(3/5) + 5*x^2 - 4)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {4+5 x^2+8 x^3+x^2 \log \left (\frac {5}{3}\right )}{64 x^2} \, dx=\frac {x^{2}}{16} + \frac {x \left (- \log {\left (3 \right )} + \log {\left (5 \right )} + 5\right )}{64} - \frac {1}{16 x} \]

[In]

integrate(1/64*(-x**2*ln(3/5)+8*x**3+5*x**2+4)/x**2,x)

[Out]

x**2/16 + x*(-log(3) + log(5) + 5)/64 - 1/(16*x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {4+5 x^2+8 x^3+x^2 \log \left (\frac {5}{3}\right )}{64 x^2} \, dx=\frac {1}{16} \, x^{2} - \frac {1}{64} \, x {\left (\log \left (\frac {3}{5}\right ) - 5\right )} - \frac {1}{16 \, x} \]

[In]

integrate(1/64*(-x^2*log(3/5)+8*x^3+5*x^2+4)/x^2,x, algorithm="maxima")

[Out]

1/16*x^2 - 1/64*x*(log(3/5) - 5) - 1/16/x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {4+5 x^2+8 x^3+x^2 \log \left (\frac {5}{3}\right )}{64 x^2} \, dx=\frac {1}{16} \, x^{2} - \frac {1}{64} \, x \log \left (\frac {3}{5}\right ) + \frac {5}{64} \, x - \frac {1}{16 \, x} \]

[In]

integrate(1/64*(-x^2*log(3/5)+8*x^3+5*x^2+4)/x^2,x, algorithm="giac")

[Out]

1/16*x^2 - 1/64*x*log(3/5) + 5/64*x - 1/16/x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {4+5 x^2+8 x^3+x^2 \log \left (\frac {5}{3}\right )}{64 x^2} \, dx=x\,\left (\frac {\ln \left (\frac {5}{3}\right )}{64}+\frac {5}{64}\right )-\frac {1}{16\,x}+\frac {x^2}{16} \]

[In]

int(((5*x^2)/64 - (x^2*log(3/5))/64 + x^3/8 + 1/16)/x^2,x)

[Out]

x*(log(5/3)/64 + 5/64) - 1/(16*x) + x^2/16