3.384 \(\int \frac{1}{x^3 (13+\frac{2}{x}+15 x)} \, dx\)

Optimal. Leaf size=34 \[ -\frac{1}{2 x}-\frac{13 \log (x)}{4}-\frac{9}{28} \log (3 x+2)+\frac{25}{7} \log (5 x+1) \]

[Out]

-1/(2*x) - (13*Log[x])/4 - (9*Log[2 + 3*x])/28 + (25*Log[1 + 5*x])/7

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Rubi [A]  time = 0.0297196, antiderivative size = 34, 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} \[ -\frac{1}{2 x}-\frac{13 \log (x)}{4}-\frac{9}{28} \log (3 x+2)+\frac{25}{7} \log (5 x+1) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(13 + 2/x + 15*x)),x]

[Out]

-1/(2*x) - (13*Log[x])/4 - (9*Log[2 + 3*x])/28 + (25*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^3 \left (13+\frac{2}{x}+15 x\right )} \, dx &=\int \frac{1}{x^2 \left (2+13 x+15 x^2\right )} \, dx\\ &=-\frac{1}{2 x}+\frac{1}{2} \int \frac{-13-15 x}{x \left (2+13 x+15 x^2\right )} \, dx\\ &=-\frac{1}{2 x}+\frac{1}{2} \int \left (-\frac{13}{2 x}-\frac{27}{14 (2+3 x)}+\frac{250}{7 (1+5 x)}\right ) \, dx\\ &=-\frac{1}{2 x}-\frac{13 \log (x)}{4}-\frac{9}{28} \log (2+3 x)+\frac{25}{7} \log (1+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0040524, size = 34, normalized size = 1. \[ -\frac{1}{2 x}-\frac{13 \log (x)}{4}-\frac{9}{28} \log (3 x+2)+\frac{25}{7} \log (5 x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(13 + 2/x + 15*x)),x]

[Out]

-1/(2*x) - (13*Log[x])/4 - (9*Log[2 + 3*x])/28 + (25*Log[1 + 5*x])/7

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Maple [A]  time = 0.006, size = 27, normalized size = 0.8 \begin{align*} -{\frac{1}{2\,x}}-{\frac{13\,\ln \left ( x \right ) }{4}}-{\frac{9\,\ln \left ( 2+3\,x \right ) }{28}}+{\frac{25\,\ln \left ( 1+5\,x \right ) }{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(13+2/x+15*x),x)

[Out]

-1/2/x-13/4*ln(x)-9/28*ln(2+3*x)+25/7*ln(1+5*x)

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Maxima [A]  time = 0.99963, size = 35, normalized size = 1.03 \begin{align*} -\frac{1}{2 \, x} + \frac{25}{7} \, \log \left (5 \, x + 1\right ) - \frac{9}{28} \, \log \left (3 \, x + 2\right ) - \frac{13}{4} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(13+2/x+15*x),x, algorithm="maxima")

[Out]

-1/2/x + 25/7*log(5*x + 1) - 9/28*log(3*x + 2) - 13/4*log(x)

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Fricas [A]  time = 1.2629, size = 90, normalized size = 2.65 \begin{align*} \frac{100 \, x \log \left (5 \, x + 1\right ) - 9 \, x \log \left (3 \, x + 2\right ) - 91 \, x \log \left (x\right ) - 14}{28 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(13+2/x+15*x),x, algorithm="fricas")

[Out]

1/28*(100*x*log(5*x + 1) - 9*x*log(3*x + 2) - 91*x*log(x) - 14)/x

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Sympy [A]  time = 0.14262, size = 31, normalized size = 0.91 \begin{align*} - \frac{13 \log{\left (x \right )}}{4} + \frac{25 \log{\left (x + \frac{1}{5} \right )}}{7} - \frac{9 \log{\left (x + \frac{2}{3} \right )}}{28} - \frac{1}{2 x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(13+2/x+15*x),x)

[Out]

-13*log(x)/4 + 25*log(x + 1/5)/7 - 9*log(x + 2/3)/28 - 1/(2*x)

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Giac [A]  time = 1.12182, size = 39, normalized size = 1.15 \begin{align*} -\frac{1}{2 \, x} + \frac{25}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac{9}{28} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac{13}{4} \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(13+2/x+15*x),x, algorithm="giac")

[Out]

-1/2/x + 25/7*log(abs(5*x + 1)) - 9/28*log(abs(3*x + 2)) - 13/4*log(abs(x))