3.385 \(\int \frac{1}{x^4 (13+\frac{2}{x}+15 x)} \, dx\)

Optimal. Leaf size=41 \[ -\frac{1}{4 x^2}+\frac{13}{4 x}+\frac{139 \log (x)}{8}+\frac{27}{56} \log (3 x+2)-\frac{125}{7} \log (5 x+1) \]

[Out]

-1/(4*x^2) + 13/(4*x) + (139*Log[x])/8 + (27*Log[2 + 3*x])/56 - (125*Log[1 + 5*x])/7

________________________________________________________________________________________

Rubi [A]  time = 0.0338796, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1386, 709, 800} \[ -\frac{1}{4 x^2}+\frac{13}{4 x}+\frac{139 \log (x)}{8}+\frac{27}{56} \log (3 x+2)-\frac{125}{7} \log (5 x+1) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(13 + 2/x + 15*x)),x]

[Out]

-1/(4*x^2) + 13/(4*x) + (139*Log[x])/8 + (27*Log[2 + 3*x])/56 - (125*Log[1 + 5*x])/7

Rule 1386

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n_.) + (b_.)*(x_)^(mn_))^(p_.), x_Symbol] :> Int[x^(m - n*p)*(b + a*x^n + c
*x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n}, x] && EqQ[mn, -n] && IntegerQ[p] && PosQ[n]

Rule 709

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d + e*x)^(m + 1)*Simp[c*d - b*e - c
*e*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[m, -1]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (13+\frac{2}{x}+15 x\right )} \, dx &=\int \frac{1}{x^3 \left (2+13 x+15 x^2\right )} \, dx\\ &=-\frac{1}{4 x^2}+\frac{1}{2} \int \frac{-13-15 x}{x^2 \left (2+13 x+15 x^2\right )} \, dx\\ &=-\frac{1}{4 x^2}+\frac{1}{2} \int \left (-\frac{13}{2 x^2}+\frac{139}{4 x}+\frac{81}{28 (2+3 x)}-\frac{1250}{7 (1+5 x)}\right ) \, dx\\ &=-\frac{1}{4 x^2}+\frac{13}{4 x}+\frac{139 \log (x)}{8}+\frac{27}{56} \log (2+3 x)-\frac{125}{7} \log (1+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0045968, size = 41, normalized size = 1. \[ -\frac{1}{4 x^2}+\frac{13}{4 x}+\frac{139 \log (x)}{8}+\frac{27}{56} \log (3 x+2)-\frac{125}{7} \log (5 x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(13 + 2/x + 15*x)),x]

[Out]

-1/(4*x^2) + 13/(4*x) + (139*Log[x])/8 + (27*Log[2 + 3*x])/56 - (125*Log[1 + 5*x])/7

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 32, normalized size = 0.8 \begin{align*} -{\frac{1}{4\,{x}^{2}}}+{\frac{13}{4\,x}}+{\frac{139\,\ln \left ( x \right ) }{8}}+{\frac{27\,\ln \left ( 2+3\,x \right ) }{56}}-{\frac{125\,\ln \left ( 1+5\,x \right ) }{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(13+2/x+15*x),x)

[Out]

-1/4/x^2+13/4/x+139/8*ln(x)+27/56*ln(2+3*x)-125/7*ln(1+5*x)

________________________________________________________________________________________

Maxima [A]  time = 0.9811, size = 42, normalized size = 1.02 \begin{align*} \frac{13 \, x - 1}{4 \, x^{2}} - \frac{125}{7} \, \log \left (5 \, x + 1\right ) + \frac{27}{56} \, \log \left (3 \, x + 2\right ) + \frac{139}{8} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(13+2/x+15*x),x, algorithm="maxima")

[Out]

1/4*(13*x - 1)/x^2 - 125/7*log(5*x + 1) + 27/56*log(3*x + 2) + 139/8*log(x)

________________________________________________________________________________________

Fricas [A]  time = 1.25822, size = 117, normalized size = 2.85 \begin{align*} -\frac{1000 \, x^{2} \log \left (5 \, x + 1\right ) - 27 \, x^{2} \log \left (3 \, x + 2\right ) - 973 \, x^{2} \log \left (x\right ) - 182 \, x + 14}{56 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(13+2/x+15*x),x, algorithm="fricas")

[Out]

-1/56*(1000*x^2*log(5*x + 1) - 27*x^2*log(3*x + 2) - 973*x^2*log(x) - 182*x + 14)/x^2

________________________________________________________________________________________

Sympy [A]  time = 0.153319, size = 36, normalized size = 0.88 \begin{align*} \frac{139 \log{\left (x \right )}}{8} - \frac{125 \log{\left (x + \frac{1}{5} \right )}}{7} + \frac{27 \log{\left (x + \frac{2}{3} \right )}}{56} + \frac{13 x - 1}{4 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(13+2/x+15*x),x)

[Out]

139*log(x)/8 - 125*log(x + 1/5)/7 + 27*log(x + 2/3)/56 + (13*x - 1)/(4*x**2)

________________________________________________________________________________________

Giac [A]  time = 1.14518, size = 46, normalized size = 1.12 \begin{align*} \frac{13 \, x - 1}{4 \, x^{2}} - \frac{125}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) + \frac{27}{56} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac{139}{8} \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(13+2/x+15*x),x, algorithm="giac")

[Out]

1/4*(13*x - 1)/x^2 - 125/7*log(abs(5*x + 1)) + 27/56*log(abs(3*x + 2)) + 139/8*log(abs(x))