\(\int \frac {1}{(a+\frac {b}{x^2})^3 x^{11}} \, dx\) [1890]

Optimal result

Integrand size = 13, antiderivative size = 86 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{11}} \, dx=-\frac {1}{4 b^3 x^4}+\frac {3 a}{2 b^4 x^2}+\frac {a^2}{4 b^3 \left (b+a x^2\right )^2}+\frac {3 a^2}{2 b^4 \left (b+a x^2\right )}+\frac {6 a^2 \log (x)}{b^5}-\frac {3 a^2 \log \left (b+a x^2\right )}{b^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 86, 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} \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{11}} \, dx=-\frac {3 a^2 \log \left (a x^2+b\right )}{b^5}+\frac {6 a^2 \log (x)}{b^5}+\frac {3 a^2}{2 b^4 \left (a x^2+b\right )}+\frac {a^2}{4 b^3 \left (a x^2+b\right )^2}+\frac {3 a}{2 b^4 x^2}-\frac {1}{4 b^3 x^4} \]

[In]

Int[1/((a + b/x^2)^3*x^11),x]

[Out]

-1/4*1/(b^3*x^4) + (3*a)/(2*b^4*x^2) + a^2/(4*b^3*(b + a*x^2)^2) + (3*a^2)/(2*b^4*(b + a*x^2)) + (6*a^2*Log[x]
)/b^5 - (3*a^2*Log[b + a*x^2])/b^5

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^5 \left (b+a x^2\right )^3} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^3 (b+a x)^3} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{b^3 x^3}-\frac {3 a}{b^4 x^2}+\frac {6 a^2}{b^5 x}-\frac {a^3}{b^3 (b+a x)^3}-\frac {3 a^3}{b^4 (b+a x)^2}-\frac {6 a^3}{b^5 (b+a x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {1}{4 b^3 x^4}+\frac {3 a}{2 b^4 x^2}+\frac {a^2}{4 b^3 \left (b+a x^2\right )^2}+\frac {3 a^2}{2 b^4 \left (b+a x^2\right )}+\frac {6 a^2 \log (x)}{b^5}-\frac {3 a^2 \log \left (b+a x^2\right )}{b^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{11}} \, dx=\frac {\frac {b \left (-b^3+4 a b^2 x^2+18 a^2 b x^4+12 a^3 x^6\right )}{x^4 \left (b+a x^2\right )^2}+24 a^2 \log (x)-12 a^2 \log \left (b+a x^2\right )}{4 b^5} \]

[In]

Integrate[1/((a + b/x^2)^3*x^11),x]

[Out]

((b*(-b^3 + 4*a*b^2*x^2 + 18*a^2*b*x^4 + 12*a^3*x^6))/(x^4*(b + a*x^2)^2) + 24*a^2*Log[x] - 12*a^2*Log[b + a*x
^2])/(4*b^5)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {\frac {3 a^{3} x^{6}}{b^{4}}+\frac {9 a^{2} x^{4}}{2 b^{3}}+\frac {a \,x^{2}}{b^{2}}-\frac {1}{4 b}}{\left (a \,x^{2}+b \right )^{2} x^{4}}+\frac {6 a^{2} \ln \left (x \right )}{b^{5}}-\frac {3 a^{2} \ln \left (a \,x^{2}+b \right )}{b^{5}}\)	\(77\)
norman	\(\frac {\frac {a \,x^{8}}{b^{2}}-\frac {x^{6}}{4 b}-\frac {6 a^{3} x^{12}}{b^{4}}-\frac {9 a^{4} x^{14}}{2 b^{5}}}{\left (a \,x^{2}+b \right )^{2} x^{10}}+\frac {6 a^{2} \ln \left (x \right )}{b^{5}}-\frac {3 a^{2} \ln \left (a \,x^{2}+b \right )}{b^{5}}\)	\(80\)
default	\(-\frac {1}{4 b^{3} x^{4}}+\frac {3 a}{2 b^{4} x^{2}}+\frac {6 a^{2} \ln \left (x \right )}{b^{5}}-\frac {a^{3} \left (-\frac {3 b}{a \left (a \,x^{2}+b \right )}-\frac {b^{2}}{2 a \left (a \,x^{2}+b \right )^{2}}+\frac {6 \ln \left (a \,x^{2}+b \right )}{a}\right )}{2 b^{5}}\)	\(83\)
parallelrisch	\(\frac {24 \ln \left (x \right ) x^{8} a^{4}-12 \ln \left (a \,x^{2}+b \right ) x^{8} a^{4}-18 a^{4} x^{8}+48 \ln \left (x \right ) x^{6} a^{3} b -24 \ln \left (a \,x^{2}+b \right ) x^{6} a^{3} b -24 a^{3} b \,x^{6}+24 \ln \left (x \right ) x^{4} a^{2} b^{2}-12 \ln \left (a \,x^{2}+b \right ) x^{4} a^{2} b^{2}+4 a \,b^{3} x^{2}-b^{4}}{4 b^{5} x^{4} \left (a \,x^{2}+b \right )^{2}}\)	\(136\)

[In]

int(1/(a+b/x^2)^3/x^11,x,method=_RETURNVERBOSE)

[Out]

(3*a^3/b^4*x^6+9/2*a^2/b^3*x^4+a*x^2/b^2-1/4/b)/(a*x^2+b)^2/x^4+6*a^2*ln(x)/b^5-3*a^2*ln(a*x^2+b)/b^5

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.56 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{11}} \, dx=\frac {12 \, a^{3} b x^{6} + 18 \, a^{2} b^{2} x^{4} + 4 \, a b^{3} x^{2} - b^{4} - 12 \, {\left (a^{4} x^{8} + 2 \, a^{3} b x^{6} + a^{2} b^{2} x^{4}\right )} \log \left (a x^{2} + b\right ) + 24 \, {\left (a^{4} x^{8} + 2 \, a^{3} b x^{6} + a^{2} b^{2} x^{4}\right )} \log \left (x\right )}{4 \, {\left (a^{2} b^{5} x^{8} + 2 \, a b^{6} x^{6} + b^{7} x^{4}\right )}} \]

[In]

integrate(1/(a+b/x^2)^3/x^11,x, algorithm="fricas")

[Out]

1/4*(12*a^3*b*x^6 + 18*a^2*b^2*x^4 + 4*a*b^3*x^2 - b^4 - 12*(a^4*x^8 + 2*a^3*b*x^6 + a^2*b^2*x^4)*log(a*x^2 +
b) + 24*(a^4*x^8 + 2*a^3*b*x^6 + a^2*b^2*x^4)*log(x))/(a^2*b^5*x^8 + 2*a*b^6*x^6 + b^7*x^4)

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{11}} \, dx=\frac {6 a^{2} \log {\left (x \right )}}{b^{5}} - \frac {3 a^{2} \log {\left (x^{2} + \frac {b}{a} \right )}}{b^{5}} + \frac {12 a^{3} x^{6} + 18 a^{2} b x^{4} + 4 a b^{2} x^{2} - b^{3}}{4 a^{2} b^{4} x^{8} + 8 a b^{5} x^{6} + 4 b^{6} x^{4}} \]

[In]

integrate(1/(a+b/x**2)**3/x**11,x)

[Out]

6*a**2*log(x)/b**5 - 3*a**2*log(x**2 + b/a)/b**5 + (12*a**3*x**6 + 18*a**2*b*x**4 + 4*a*b**2*x**2 - b**3)/(4*a
**2*b**4*x**8 + 8*a*b**5*x**6 + 4*b**6*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{11}} \, dx=\frac {12 \, a^{3} x^{6} + 18 \, a^{2} b x^{4} + 4 \, a b^{2} x^{2} - b^{3}}{4 \, {\left (a^{2} b^{4} x^{8} + 2 \, a b^{5} x^{6} + b^{6} x^{4}\right )}} - \frac {3 \, a^{2} \log \left (a x^{2} + b\right )}{b^{5}} + \frac {3 \, a^{2} \log \left (x^{2}\right )}{b^{5}} \]

[In]

integrate(1/(a+b/x^2)^3/x^11,x, algorithm="maxima")

[Out]

1/4*(12*a^3*x^6 + 18*a^2*b*x^4 + 4*a*b^2*x^2 - b^3)/(a^2*b^4*x^8 + 2*a*b^5*x^6 + b^6*x^4) - 3*a^2*log(a*x^2 +
b)/b^5 + 3*a^2*log(x^2)/b^5

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{11}} \, dx=\frac {3 \, a^{2} \log \left (x^{2}\right )}{b^{5}} - \frac {3 \, a^{2} \log \left ({\left | a x^{2} + b \right |}\right )}{b^{5}} + \frac {12 \, a^{3} x^{6} + 18 \, a^{2} b x^{4} + 4 \, a b^{2} x^{2} - b^{3}}{4 \, {\left (a x^{4} + b x^{2}\right )}^{2} b^{4}} \]

[In]

integrate(1/(a+b/x^2)^3/x^11,x, algorithm="giac")

[Out]

3*a^2*log(x^2)/b^5 - 3*a^2*log(abs(a*x^2 + b))/b^5 + 1/4*(12*a^3*x^6 + 18*a^2*b*x^4 + 4*a*b^2*x^2 - b^3)/((a*x
^4 + b*x^2)^2*b^4)

Mupad [B] (verification not implemented)

Time = 5.88 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^{11}} \, dx=\frac {\frac {a\,x^2}{b^2}-\frac {1}{4\,b}+\frac {9\,a^2\,x^4}{2\,b^3}+\frac {3\,a^3\,x^6}{b^4}}{a^2\,x^8+2\,a\,b\,x^6+b^2\,x^4}-\frac {3\,a^2\,\ln \left (a\,x^2+b\right )}{b^5}+\frac {6\,a^2\,\ln \left (x\right )}{b^5} \]

[In]

int(1/(x^11*(a + b/x^2)^3),x)

[Out]

((a*x^2)/b^2 - 1/(4*b) + (9*a^2*x^4)/(2*b^3) + (3*a^3*x^6)/b^4)/(a^2*x^8 + b^2*x^4 + 2*a*b*x^6) - (3*a^2*log(b
+ a*x^2))/b^5 + (6*a^2*log(x))/b^5