3.1284 \(\int \frac {1}{x^{11} (a+b x^5)^2} \, dx\)

Optimal. Leaf size=69 \[ -\frac {3 b^2 \log \left (a+b x^5\right )}{5 a^4}+\frac {3 b^2 \log (x)}{a^4}+\frac {b^2}{5 a^3 \left (a+b x^5\right )}+\frac {2 b}{5 a^3 x^5}-\frac {1}{10 a^2 x^{10}} \]

[Out]

-1/10/a^2/x^10+2/5*b/a^3/x^5+1/5*b^2/a^3/(b*x^5+a)+3*b^2*ln(x)/a^4-3/5*b^2*ln(b*x^5+a)/a^4

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Rubi [A]  time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {266, 44} \[ \frac {b^2}{5 a^3 \left (a+b x^5\right )}-\frac {3 b^2 \log \left (a+b x^5\right )}{5 a^4}+\frac {3 b^2 \log (x)}{a^4}+\frac {2 b}{5 a^3 x^5}-\frac {1}{10 a^2 x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^11*(a + b*x^5)^2),x]

[Out]

-1/(10*a^2*x^10) + (2*b)/(5*a^3*x^5) + b^2/(5*a^3*(a + b*x^5)) + (3*b^2*Log[x])/a^4 - (3*b^2*Log[a + b*x^5])/(
5*a^4)

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 266

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

Rubi steps

\begin {align*} \int \frac {1}{x^{11} \left (a+b x^5\right )^2} \, dx &=\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{x^3 (a+b x)^2} \, dx,x,x^5\right )\\ &=\frac {1}{5} \operatorname {Subst}\left (\int \left (\frac {1}{a^2 x^3}-\frac {2 b}{a^3 x^2}+\frac {3 b^2}{a^4 x}-\frac {b^3}{a^3 (a+b x)^2}-\frac {3 b^3}{a^4 (a+b x)}\right ) \, dx,x,x^5\right )\\ &=-\frac {1}{10 a^2 x^{10}}+\frac {2 b}{5 a^3 x^5}+\frac {b^2}{5 a^3 \left (a+b x^5\right )}+\frac {3 b^2 \log (x)}{a^4}-\frac {3 b^2 \log \left (a+b x^5\right )}{5 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 57, normalized size = 0.83 \[ \frac {-6 b^2 \log \left (a+b x^5\right )+a \left (\frac {2 b^2}{a+b x^5}-\frac {a}{x^{10}}+\frac {4 b}{x^5}\right )+30 b^2 \log (x)}{10 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^11*(a + b*x^5)^2),x]

[Out]

(a*(-(a/x^10) + (4*b)/x^5 + (2*b^2)/(a + b*x^5)) + 30*b^2*Log[x] - 6*b^2*Log[a + b*x^5])/(10*a^4)

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fricas [A]  time = 0.91, size = 90, normalized size = 1.30 \[ \frac {6 \, a b^{2} x^{10} + 3 \, a^{2} b x^{5} - a^{3} - 6 \, {\left (b^{3} x^{15} + a b^{2} x^{10}\right )} \log \left (b x^{5} + a\right ) + 30 \, {\left (b^{3} x^{15} + a b^{2} x^{10}\right )} \log \relax (x)}{10 \, {\left (a^{4} b x^{15} + a^{5} x^{10}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^11/(b*x^5+a)^2,x, algorithm="fricas")

[Out]

1/10*(6*a*b^2*x^10 + 3*a^2*b*x^5 - a^3 - 6*(b^3*x^15 + a*b^2*x^10)*log(b*x^5 + a) + 30*(b^3*x^15 + a*b^2*x^10)
*log(x))/(a^4*b*x^15 + a^5*x^10)

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giac [A]  time = 0.18, size = 85, normalized size = 1.23 \[ -\frac {3 \, b^{2} \log \left ({\left | b x^{5} + a \right |}\right )}{5 \, a^{4}} + \frac {3 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{4}} + \frac {3 \, b^{3} x^{5} + 4 \, a b^{2}}{5 \, {\left (b x^{5} + a\right )} a^{4}} - \frac {9 \, b^{2} x^{10} - 4 \, a b x^{5} + a^{2}}{10 \, a^{4} x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^11/(b*x^5+a)^2,x, algorithm="giac")

[Out]

-3/5*b^2*log(abs(b*x^5 + a))/a^4 + 3*b^2*log(abs(x))/a^4 + 1/5*(3*b^3*x^5 + 4*a*b^2)/((b*x^5 + a)*a^4) - 1/10*
(9*b^2*x^10 - 4*a*b*x^5 + a^2)/(a^4*x^10)

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maple [A]  time = 0.02, size = 62, normalized size = 0.90 \[ \frac {b^{2}}{5 \left (b \,x^{5}+a \right ) a^{3}}+\frac {3 b^{2} \ln \relax (x )}{a^{4}}-\frac {3 b^{2} \ln \left (b \,x^{5}+a \right )}{5 a^{4}}+\frac {2 b}{5 a^{3} x^{5}}-\frac {1}{10 a^{2} x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^11/(b*x^5+a)^2,x)

[Out]

-1/10/a^2/x^10+2/5*b/a^3/x^5+1/5*b^2/a^3/(b*x^5+a)+3*b^2*ln(x)/a^4-3/5*b^2*ln(b*x^5+a)/a^4

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maxima [A]  time = 1.02, size = 70, normalized size = 1.01 \[ \frac {6 \, b^{2} x^{10} + 3 \, a b x^{5} - a^{2}}{10 \, {\left (a^{3} b x^{15} + a^{4} x^{10}\right )}} - \frac {3 \, b^{2} \log \left (b x^{5} + a\right )}{5 \, a^{4}} + \frac {3 \, b^{2} \log \left (x^{5}\right )}{5 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^11/(b*x^5+a)^2,x, algorithm="maxima")

[Out]

1/10*(6*b^2*x^10 + 3*a*b*x^5 - a^2)/(a^3*b*x^15 + a^4*x^10) - 3/5*b^2*log(b*x^5 + a)/a^4 + 3/5*b^2*log(x^5)/a^
4

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mupad [B]  time = 1.07, size = 67, normalized size = 0.97 \[ \frac {\frac {3\,b\,x^5}{10\,a^2}-\frac {1}{10\,a}+\frac {3\,b^2\,x^{10}}{5\,a^3}}{b\,x^{15}+a\,x^{10}}-\frac {3\,b^2\,\ln \left (b\,x^5+a\right )}{5\,a^4}+\frac {3\,b^2\,\ln \relax (x)}{a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^11*(a + b*x^5)^2),x)

[Out]

((3*b*x^5)/(10*a^2) - 1/(10*a) + (3*b^2*x^10)/(5*a^3))/(a*x^10 + b*x^15) - (3*b^2*log(a + b*x^5))/(5*a^4) + (3
*b^2*log(x))/a^4

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sympy [A]  time = 1.60, size = 68, normalized size = 0.99 \[ \frac {- a^{2} + 3 a b x^{5} + 6 b^{2} x^{10}}{10 a^{4} x^{10} + 10 a^{3} b x^{15}} + \frac {3 b^{2} \log {\relax (x )}}{a^{4}} - \frac {3 b^{2} \log {\left (\frac {a}{b} + x^{5} \right )}}{5 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**11/(b*x**5+a)**2,x)

[Out]

(-a**2 + 3*a*b*x**5 + 6*b**2*x**10)/(10*a**4*x**10 + 10*a**3*b*x**15) + 3*b**2*log(x)/a**4 - 3*b**2*log(a/b +
x**5)/(5*a**4)

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