3.5.44 \(\int \frac {1}{(15+\frac {2}{x^2}+\frac {13}{x}) x^2} \, dx\) [444]

Optimal. Leaf size=23 \[ \frac {1}{7} \log \left (5+\frac {1}{x}\right )-\frac {1}{7} \log \left (3+\frac {2}{x}\right ) \]

[Out]

1/7*ln(5+1/x)-1/7*ln(3+2/x)

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1366, 630, 31} \begin {gather*} \frac {1}{7} \log \left (\frac {1}{x}+5\right )-\frac {1}{7} \log \left (\frac {2}{x}+3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((15 + 2/x^2 + 13/x)*x^2),x]

[Out]

Log[5 + x^(-1)]/7 - Log[3 + 2/x]/7

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 1366

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (15+\frac {2}{x^2}+\frac {13}{x}\right ) x^2} \, dx &=-\text {Subst}\left (\int \frac {1}{15+13 x+2 x^2} \, dx,x,\frac {1}{x}\right )\\ &=-\left (\frac {2}{7} \text {Subst}\left (\int \frac {1}{3+2 x} \, dx,x,\frac {1}{x}\right )\right )+\frac {2}{7} \text {Subst}\left (\int \frac {1}{10+2 x} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{7} \log \left (5+\frac {1}{x}\right )-\frac {1}{7} \log \left (3+\frac {2}{x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.00, size = 21, normalized size = 0.91 \begin {gather*} -\frac {1}{7} \log (2+3 x)+\frac {1}{7} \log (1+5 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((15 + 2/x^2 + 13/x)*x^2),x]

[Out]

-1/7*Log[2 + 3*x] + Log[1 + 5*x]/7

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 18, normalized size = 0.78

method	result	size
default	\(\frac {\ln \left (1+5 x \right )}{7}-\frac {\ln \left (2+3 x \right )}{7}\)	\(18\)
norman	\(\frac {\ln \left (1+5 x \right )}{7}-\frac {\ln \left (2+3 x \right )}{7}\)	\(18\)
risch	\(\frac {\ln \left (1+5 x \right )}{7}-\frac {\ln \left (2+3 x \right )}{7}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(15+2/x^2+13/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/7*ln(1+5*x)-1/7*ln(2+3*x)

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 17, normalized size = 0.74 \begin {gather*} \frac {1}{7} \, \log \left (5 \, x + 1\right ) - \frac {1}{7} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(15+2/x^2+13/x)/x^2,x, algorithm="maxima")

[Out]

1/7*log(5*x + 1) - 1/7*log(3*x + 2)

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 17, normalized size = 0.74 \begin {gather*} \frac {1}{7} \, \log \left (5 \, x + 1\right ) - \frac {1}{7} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(15+2/x^2+13/x)/x^2,x, algorithm="fricas")

[Out]

1/7*log(5*x + 1) - 1/7*log(3*x + 2)

________________________________________________________________________________________

Sympy [A]
time = 0.03, size = 15, normalized size = 0.65 \begin {gather*} \frac {\log {\left (x + \frac {1}{5} \right )}}{7} - \frac {\log {\left (x + \frac {2}{3} \right )}}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(15+2/x**2+13/x)/x**2,x)

[Out]

log(x + 1/5)/7 - log(x + 2/3)/7

________________________________________________________________________________________

Giac [A]
time = 3.77, size = 19, normalized size = 0.83 \begin {gather*} \frac {1}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac {1}{7} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(15+2/x^2+13/x)/x^2,x, algorithm="giac")

[Out]

1/7*log(abs(5*x + 1)) - 1/7*log(abs(3*x + 2))

________________________________________________________________________________________

Mupad [B]
time = 1.37, size = 8, normalized size = 0.35 \begin {gather*} -\frac {2\,\mathrm {atanh}\left (\frac {30\,x}{7}+\frac {13}{7}\right )}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(13/x + 2/x^2 + 15)),x)

[Out]

-(2*atanh((30*x)/7 + 13/7))/7

________________________________________________________________________________________