\(\int \frac {(a+\frac {b}{x^2})^3}{x} \, dx\) [1837]

Optimal result

Integrand size = 13, antiderivative size = 39 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^3}{x} \, dx=-\frac {b^3}{6 x^6}-\frac {3 a b^2}{4 x^4}-\frac {3 a^2 b}{2 x^2}+a^3 \log (x) \]

[Out]

-1/6*b^3/x^6-3/4*a*b^2/x^4-3/2*a^2*b/x^2+a^3*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, 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, 45} \[ \int \frac {\left (a+\frac {b}{x^2}\right )^3}{x} \, dx=a^3 \log (x)-\frac {3 a^2 b}{2 x^2}-\frac {3 a b^2}{4 x^4}-\frac {b^3}{6 x^6} \]

[In]

Int[(a + b/x^2)^3/x,x]

[Out]

-1/6*b^3/x^6 - (3*a*b^2)/(4*x^4) - (3*a^2*b)/(2*x^2) + a^3*Log[x]

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 {\left (b+a x^2\right )^3}{x^7} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {(b+a x)^3}{x^4} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {b^3}{x^4}+\frac {3 a b^2}{x^3}+\frac {3 a^2 b}{x^2}+\frac {a^3}{x}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b^3}{6 x^6}-\frac {3 a b^2}{4 x^4}-\frac {3 a^2 b}{2 x^2}+a^3 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^3}{x} \, dx=-\frac {b^3}{6 x^6}-\frac {3 a b^2}{4 x^4}-\frac {3 a^2 b}{2 x^2}+a^3 \log (x) \]

[In]

Integrate[(a + b/x^2)^3/x,x]

[Out]

-1/6*b^3/x^6 - (3*a*b^2)/(4*x^4) - (3*a^2*b)/(2*x^2) + a^3*Log[x]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {b^{3}}{6 x^{6}}-\frac {3 a \,b^{2}}{4 x^{4}}-\frac {3 a^{2} b}{2 x^{2}}+a^{3} \ln \left (x \right )\)	\(34\)
norman	\(\frac {-\frac {1}{6} b^{3}-\frac {3}{4} a \,b^{2} x^{2}-\frac {3}{2} a^{2} b \,x^{4}}{x^{6}}+a^{3} \ln \left (x \right )\)	\(36\)
risch	\(\frac {-\frac {1}{6} b^{3}-\frac {3}{4} a \,b^{2} x^{2}-\frac {3}{2} a^{2} b \,x^{4}}{x^{6}}+a^{3} \ln \left (x \right )\)	\(36\)
parallelrisch	\(\frac {12 a^{3} \ln \left (x \right ) x^{6}-18 a^{2} b \,x^{4}-9 a \,b^{2} x^{2}-2 b^{3}}{12 x^{6}}\)	\(40\)

[In]

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

[Out]

-1/6*b^3/x^6-3/4*a*b^2/x^4-3/2*a^2*b/x^2+a^3*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^3}{x} \, dx=\frac {12 \, a^{3} x^{6} \log \left (x\right ) - 18 \, a^{2} b x^{4} - 9 \, a b^{2} x^{2} - 2 \, b^{3}}{12 \, x^{6}} \]

[In]

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

[Out]

1/12*(12*a^3*x^6*log(x) - 18*a^2*b*x^4 - 9*a*b^2*x^2 - 2*b^3)/x^6

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^3}{x} \, dx=a^{3} \log {\left (x \right )} + \frac {- 18 a^{2} b x^{4} - 9 a b^{2} x^{2} - 2 b^{3}}{12 x^{6}} \]

[In]

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

[Out]

a**3*log(x) + (-18*a**2*b*x**4 - 9*a*b**2*x**2 - 2*b**3)/(12*x**6)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^3}{x} \, dx=\frac {1}{2} \, a^{3} \log \left (x^{2}\right ) - \frac {18 \, a^{2} b x^{4} + 9 \, a b^{2} x^{2} + 2 \, b^{3}}{12 \, x^{6}} \]

[In]

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

[Out]

1/2*a^3*log(x^2) - 1/12*(18*a^2*b*x^4 + 9*a*b^2*x^2 + 2*b^3)/x^6

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^3}{x} \, dx=\frac {1}{2} \, a^{3} \log \left (x^{2}\right ) - \frac {11 \, a^{3} x^{6} + 18 \, a^{2} b x^{4} + 9 \, a b^{2} x^{2} + 2 \, b^{3}}{12 \, x^{6}} \]

[In]

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

[Out]

1/2*a^3*log(x^2) - 1/12*(11*a^3*x^6 + 18*a^2*b*x^4 + 9*a*b^2*x^2 + 2*b^3)/x^6

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^3}{x} \, dx=a^3\,\ln \left (x\right )-\frac {\frac {3\,a^2\,b\,x^4}{2}+\frac {3\,a\,b^2\,x^2}{4}+\frac {b^3}{6}}{x^6} \]

[In]

int((a + b/x^2)^3/x,x)

[Out]

a^3*log(x) - (b^3/6 + (3*a*b^2*x^2)/4 + (3*a^2*b*x^4)/2)/x^6