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

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Rubi [A]  time = 0.03, antiderivative size = 41, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.00, size = 41, normalized size = 1.00 \[ -\frac {1}{4 x^2}+\frac {13}{4 x}+\frac {139 \log (x)}{8}+\frac {27}{56} \log (3 x+2)-\frac {125}{7} \log (5 x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 39, normalized size = 0.95 \[ -\frac {1000 \, x^{2} \log \left (5 \, x + 1\right ) - 27 \, x^{2} \log \left (3 \, x + 2\right ) - 973 \, x^{2} \log \relax (x) - 182 \, x + 14}{56 \, x^{2}} \]

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

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giac [A]  time = 0.28, size = 34, normalized size = 0.83 \[ \frac {13 \, x - 1}{4 \, x^{2}} - \frac {125}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) + \frac {27}{56} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {139}{8} \, \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.01, size = 32, normalized size = 0.78 \[ \frac {139 \ln \relax (x )}{8}-\frac {125 \ln \left (5 x +1\right )}{7}+\frac {27 \ln \left (3 x +2\right )}{56}+\frac {13}{4 x}-\frac {1}{4 x^{2}} \]

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(3*x+2)-125/7*ln(5*x+1)

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maxima [A]  time = 1.02, size = 31, normalized size = 0.76 \[ \frac {13 \, x - 1}{4 \, x^{2}} - \frac {125}{7} \, \log \left (5 \, x + 1\right ) + \frac {27}{56} \, \log \left (3 \, x + 2\right ) + \frac {139}{8} \, \log \relax (x) \]

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)

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mupad [B]  time = 0.04, size = 26, normalized size = 0.63 \[ \frac {27\,\ln \left (x+\frac {2}{3}\right )}{56}-\frac {125\,\ln \left (x+\frac {1}{5}\right )}{7}+\frac {139\,\ln \relax (x)}{8}+\frac {\frac {13\,x}{4}-\frac {1}{4}}{x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(15*x + 2/x + 13)),x)

[Out]

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

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sympy [A]  time = 0.17, size = 36, normalized size = 0.88 \[ \frac {139 \log {\relax (x )}}{8} - \frac {125 \log {\left (x + \frac {1}{5} \right )}}{7} + \frac {27 \log {\left (x + \frac {2}{3} \right )}}{56} + \frac {13 x - 1}{4 x^{2}} \]

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)

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