\(\int \frac {1}{(a+\frac {b}{x^3})^2 x^7} \, dx\) [1990]

Optimal result

Integrand size = 13, antiderivative size = 38 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^7} \, dx=\frac {1}{3 b \left (b+a x^3\right )}+\frac {\log (x)}{b^2}-\frac {\log \left (b+a x^3\right )}{3 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {269, 272, 46} \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^7} \, dx=-\frac {\log \left (a x^3+b\right )}{3 b^2}+\frac {1}{3 b \left (a x^3+b\right )}+\frac {\log (x)}{b^2} \]

[In]

Int[1/((a + b/x^3)^2*x^7),x]

[Out]

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

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 269

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^7} \, dx=\frac {\frac {b}{b+a x^3}+3 \log (x)-\log \left (b+a x^3\right )}{3 b^2} \]

[In]

Integrate[1/((a + b/x^3)^2*x^7),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {1}{3 b \left (a \,x^{3}+b \right )}+\frac {\ln \left (x \right )}{b^{2}}-\frac {\ln \left (a \,x^{3}+b \right )}{3 b^{2}}\)	\(35\)
norman	\(-\frac {a \,x^{3}}{3 b^{2} \left (a \,x^{3}+b \right )}+\frac {\ln \left (x \right )}{b^{2}}-\frac {\ln \left (a \,x^{3}+b \right )}{3 b^{2}}\)	\(39\)
default	\(-\frac {a \left (-\frac {b}{a \left (a \,x^{3}+b \right )}+\frac {\ln \left (a \,x^{3}+b \right )}{a}\right )}{3 b^{2}}+\frac {\ln \left (x \right )}{b^{2}}\)	\(42\)
parallelrisch	\(\frac {3 a \ln \left (x \right ) x^{3}-a \ln \left (a \,x^{3}+b \right ) x^{3}-a \,x^{3}+3 b \ln \left (x \right )-b \ln \left (a \,x^{3}+b \right )}{3 b^{2} \left (a \,x^{3}+b \right )}\)	\(60\)

[In]

int(1/(a+b/x^3)^2/x^7,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^7} \, dx=-\frac {{\left (a x^{3} + b\right )} \log \left (a x^{3} + b\right ) - 3 \, {\left (a x^{3} + b\right )} \log \left (x\right ) - b}{3 \, {\left (a b^{2} x^{3} + b^{3}\right )}} \]

[In]

integrate(1/(a+b/x^3)^2/x^7,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^7} \, dx=\frac {1}{3 a b x^{3} + 3 b^{2}} + \frac {\log {\left (x \right )}}{b^{2}} - \frac {\log {\left (x^{3} + \frac {b}{a} \right )}}{3 b^{2}} \]

[In]

integrate(1/(a+b/x**3)**2/x**7,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^7} \, dx=\frac {1}{3 \, {\left (a b x^{3} + b^{2}\right )}} - \frac {\log \left (a x^{3} + b\right )}{3 \, b^{2}} + \frac {\log \left (x^{3}\right )}{3 \, b^{2}} \]

[In]

integrate(1/(a+b/x^3)^2/x^7,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^7} \, dx=-\frac {\log \left ({\left | a x^{3} + b \right |}\right )}{3 \, b^{2}} + \frac {\log \left ({\left | x \right |}\right )}{b^{2}} + \frac {a x^{3} + 2 \, b}{3 \, {\left (a x^{3} + b\right )} b^{2}} \]

[In]

integrate(1/(a+b/x^3)^2/x^7,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^7} \, dx=\frac {\ln \left (x\right )}{b^2}+\frac {1}{3\,b\,\left (a\,x^3+b\right )}-\frac {\ln \left (a\,x^3+b\right )}{3\,b^2} \]

[In]

int(1/(x^7*(a + b/x^3)^2),x)

[Out]

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