3.13.83 \(\int \frac {1}{x^6 (a+b x^5)^2} \, dx\) [1283]

Optimal. Leaf size=52 \[ -\frac {1}{5 a^2 x^5}-\frac {b}{5 a^2 \left (a+b x^5\right )}-\frac {2 b \log (x)}{a^3}+\frac {2 b \log \left (a+b x^5\right )}{5 a^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 52, 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 = {272, 46} \begin {gather*} \frac {2 b \log \left (a+b x^5\right )}{5 a^3}-\frac {2 b \log (x)}{a^3}-\frac {b}{5 a^2 \left (a+b x^5\right )}-\frac {1}{5 a^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a + b*x^5)^2),x]

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 272

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^6 \left (a+b x^5\right )^2} \, dx &=\frac {1}{5} \text {Subst}\left (\int \frac {1}{x^2 (a+b x)^2} \, dx,x,x^5\right )\\ &=\frac {1}{5} \text {Subst}\left (\int \left (\frac {1}{a^2 x^2}-\frac {2 b}{a^3 x}+\frac {b^2}{a^2 (a+b x)^2}+\frac {2 b^2}{a^3 (a+b x)}\right ) \, dx,x,x^5\right )\\ &=-\frac {1}{5 a^2 x^5}-\frac {b}{5 a^2 \left (a+b x^5\right )}-\frac {2 b \log (x)}{a^3}+\frac {2 b \log \left (a+b x^5\right )}{5 a^3}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 41, normalized size = 0.79 \begin {gather*} -\frac {a \left (\frac {1}{x^5}+\frac {b}{a+b x^5}\right )+10 b \log (x)-2 b \log \left (a+b x^5\right )}{5 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(a + b*x^5)^2),x]

[Out]

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

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Maple [A]
time = 0.17, size = 55, normalized size = 1.06

method	result	size
norman	\(\frac {-\frac {1}{5 a}+\frac {2 b^{2} x^{10}}{5 a^{3}}}{x^{5} \left (b \,x^{5}+a \right )}-\frac {2 b \ln \left (x \right )}{a^{3}}+\frac {2 b \ln \left (b \,x^{5}+a \right )}{5 a^{3}}\)	\(54\)
default	\(\frac {b^{2} \left (-\frac {a}{b \left (b \,x^{5}+a \right )}+\frac {2 \ln \left (b \,x^{5}+a \right )}{b}\right )}{5 a^{3}}-\frac {1}{5 a^{2} x^{5}}-\frac {2 b \ln \left (x \right )}{a^{3}}\)	\(55\)
risch	\(\frac {-\frac {2 b \,x^{5}}{5 a^{2}}-\frac {1}{5 a}}{x^{5} \left (b \,x^{5}+a \right )}-\frac {2 b \ln \left (x \right )}{a^{3}}+\frac {2 b \ln \left (-b \,x^{5}-a \right )}{5 a^{3}}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(b*x^5+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 53, normalized size = 1.02 \begin {gather*} -\frac {2 \, b x^{5} + a}{5 \, {\left (a^{2} b x^{10} + a^{3} x^{5}\right )}} + \frac {2 \, b \log \left (b x^{5} + a\right )}{5 \, a^{3}} - \frac {2 \, b \log \left (x^{5}\right )}{5 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^5+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 73, normalized size = 1.40 \begin {gather*} -\frac {2 \, a b x^{5} + a^{2} - 2 \, {\left (b^{2} x^{10} + a b x^{5}\right )} \log \left (b x^{5} + a\right ) + 10 \, {\left (b^{2} x^{10} + a b x^{5}\right )} \log \left (x\right )}{5 \, {\left (a^{3} b x^{10} + a^{4} x^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^5+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.32, size = 54, normalized size = 1.04 \begin {gather*} \frac {- a - 2 b x^{5}}{5 a^{3} x^{5} + 5 a^{2} b x^{10}} - \frac {2 b \log {\left (x \right )}}{a^{3}} + \frac {2 b \log {\left (\frac {a}{b} + x^{5} \right )}}{5 a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(b*x**5+a)**2,x)

[Out]

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

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Giac [A]
time = 1.56, size = 51, normalized size = 0.98 \begin {gather*} \frac {2 \, b \log \left ({\left | b x^{5} + a \right |}\right )}{5 \, a^{3}} - \frac {2 \, b \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {2 \, b x^{5} + a}{5 \, {\left (b x^{10} + a x^{5}\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^5+a)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.07, size = 53, normalized size = 1.02 \begin {gather*} \frac {2\,b\,\ln \left (b\,x^5+a\right )}{5\,a^3}-\frac {\frac {1}{5\,a}+\frac {2\,b\,x^5}{5\,a^2}}{b\,x^{10}+a\,x^5}-\frac {2\,b\,\ln \left (x\right )}{a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a + b*x^5)^2),x)

[Out]

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

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