\(\int \frac {x^{11}}{a^2+2 a b x^2+b^2 x^4} \, dx\) [475]

Optimal result

Integrand size = 24, antiderivative size = 83 \[ \int \frac {x^{11}}{a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {2 a^3 x^2}{b^5}+\frac {3 a^2 x^4}{4 b^4}-\frac {a x^6}{3 b^3}+\frac {x^8}{8 b^2}+\frac {a^5}{2 b^6 \left (a+b x^2\right )}+\frac {5 a^4 \log \left (a+b x^2\right )}{2 b^6} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 272, 45} \[ \int \frac {x^{11}}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {a^5}{2 b^6 \left (a+b x^2\right )}+\frac {5 a^4 \log \left (a+b x^2\right )}{2 b^6}-\frac {2 a^3 x^2}{b^5}+\frac {3 a^2 x^4}{4 b^4}-\frac {a x^6}{3 b^3}+\frac {x^8}{8 b^2} \]

[In]

Int[x^11/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

(-2*a^3*x^2)/b^5 + (3*a^2*x^4)/(4*b^4) - (a*x^6)/(3*b^3) + x^8/(8*b^2) + a^5/(2*b^6*(a + b*x^2)) + (5*a^4*Log[
a + b*x^2])/(2*b^6)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 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}& = b^2 \int \frac {x^{11}}{\left (a b+b^2 x^2\right )^2} \, dx \\ & = \frac {1}{2} b^2 \text {Subst}\left (\int \frac {x^5}{\left (a b+b^2 x\right )^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} b^2 \text {Subst}\left (\int \left (-\frac {4 a^3}{b^7}+\frac {3 a^2 x}{b^6}-\frac {2 a x^2}{b^5}+\frac {x^3}{b^4}-\frac {a^5}{b^7 (a+b x)^2}+\frac {5 a^4}{b^7 (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {2 a^3 x^2}{b^5}+\frac {3 a^2 x^4}{4 b^4}-\frac {a x^6}{3 b^3}+\frac {x^8}{8 b^2}+\frac {a^5}{2 b^6 \left (a+b x^2\right )}+\frac {5 a^4 \log \left (a+b x^2\right )}{2 b^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.87 \[ \int \frac {x^{11}}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {-48 a^3 b x^2+18 a^2 b^2 x^4-8 a b^3 x^6+3 b^4 x^8+\frac {12 a^5}{a+b x^2}+60 a^4 \log \left (a+b x^2\right )}{24 b^6} \]

[In]

Integrate[x^11/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

(-48*a^3*b*x^2 + 18*a^2*b^2*x^4 - 8*a*b^3*x^6 + 3*b^4*x^8 + (12*a^5)/(a + b*x^2) + 60*a^4*Log[a + b*x^2])/(24*
b^6)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {2 a^{3} x^{2}}{b^{5}}+\frac {3 a^{2} x^{4}}{4 b^{4}}-\frac {a \,x^{6}}{3 b^{3}}+\frac {x^{8}}{8 b^{2}}+\frac {a^{5}}{2 b^{6} \left (b \,x^{2}+a \right )}+\frac {5 a^{4} \ln \left (b \,x^{2}+a \right )}{2 b^{6}}\)	\(74\)
default	\(-\frac {-\frac {1}{4} b^{3} x^{8}+\frac {2}{3} b^{2} x^{6} a -\frac {3}{2} a^{2} b \,x^{4}+4 a^{3} x^{2}}{2 b^{5}}+\frac {5 a^{4} \ln \left (b \,x^{2}+a \right )}{2 b^{6}}+\frac {a^{5}}{2 b^{6} \left (b \,x^{2}+a \right )}\)	\(75\)
norman	\(\frac {\frac {x^{10}}{8 b}-\frac {5 a \,x^{8}}{24 b^{2}}+\frac {5 a^{5}}{2 b^{6}}+\frac {5 a^{2} x^{6}}{12 b^{3}}-\frac {5 a^{3} x^{4}}{4 b^{4}}}{b \,x^{2}+a}+\frac {5 a^{4} \ln \left (b \,x^{2}+a \right )}{2 b^{6}}\)	\(76\)
parallelrisch	\(\frac {3 x^{10} b^{5}-5 a \,x^{8} b^{4}+10 a^{2} x^{6} b^{3}-30 a^{3} x^{4} b^{2}+60 \ln \left (b \,x^{2}+a \right ) x^{2} a^{4} b +60 \ln \left (b \,x^{2}+a \right ) a^{5}+60 a^{5}}{24 b^{6} \left (b \,x^{2}+a \right )}\)	\(90\)

[In]

int(x^11/(b^2*x^4+2*a*b*x^2+a^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int \frac {x^{11}}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {3 \, b^{5} x^{10} - 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} - 30 \, a^{3} b^{2} x^{4} - 48 \, a^{4} b x^{2} + 12 \, a^{5} + 60 \, {\left (a^{4} b x^{2} + a^{5}\right )} \log \left (b x^{2} + a\right )}{24 \, {\left (b^{7} x^{2} + a b^{6}\right )}} \]

[In]

integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

1/24*(3*b^5*x^10 - 5*a*b^4*x^8 + 10*a^2*b^3*x^6 - 30*a^3*b^2*x^4 - 48*a^4*b*x^2 + 12*a^5 + 60*(a^4*b*x^2 + a^5
)*log(b*x^2 + a))/(b^7*x^2 + a*b^6)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {x^{11}}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {a^{5}}{2 a b^{6} + 2 b^{7} x^{2}} + \frac {5 a^{4} \log {\left (a + b x^{2} \right )}}{2 b^{6}} - \frac {2 a^{3} x^{2}}{b^{5}} + \frac {3 a^{2} x^{4}}{4 b^{4}} - \frac {a x^{6}}{3 b^{3}} + \frac {x^{8}}{8 b^{2}} \]

[In]

integrate(x**11/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

a**5/(2*a*b**6 + 2*b**7*x**2) + 5*a**4*log(a + b*x**2)/(2*b**6) - 2*a**3*x**2/b**5 + 3*a**2*x**4/(4*b**4) - a*
x**6/(3*b**3) + x**8/(8*b**2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93 \[ \int \frac {x^{11}}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {a^{5}}{2 \, {\left (b^{7} x^{2} + a b^{6}\right )}} + \frac {5 \, a^{4} \log \left (b x^{2} + a\right )}{2 \, b^{6}} + \frac {3 \, b^{3} x^{8} - 8 \, a b^{2} x^{6} + 18 \, a^{2} b x^{4} - 48 \, a^{3} x^{2}}{24 \, b^{5}} \]

[In]

integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

1/2*a^5/(b^7*x^2 + a*b^6) + 5/2*a^4*log(b*x^2 + a)/b^6 + 1/24*(3*b^3*x^8 - 8*a*b^2*x^6 + 18*a^2*b*x^4 - 48*a^3
*x^2)/b^5

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.11 \[ \int \frac {x^{11}}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {5 \, a^{4} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{6}} - \frac {5 \, a^{4} b x^{2} + 4 \, a^{5}}{2 \, {\left (b x^{2} + a\right )} b^{6}} + \frac {3 \, b^{6} x^{8} - 8 \, a b^{5} x^{6} + 18 \, a^{2} b^{4} x^{4} - 48 \, a^{3} b^{3} x^{2}}{24 \, b^{8}} \]

[In]

integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

5/2*a^4*log(abs(b*x^2 + a))/b^6 - 1/2*(5*a^4*b*x^2 + 4*a^5)/((b*x^2 + a)*b^6) + 1/24*(3*b^6*x^8 - 8*a*b^5*x^6
+ 18*a^2*b^4*x^4 - 48*a^3*b^3*x^2)/b^8

Mupad [B] (verification not implemented)

Time = 13.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {x^{11}}{a^2+2 a b x^2+b^2 x^4} \, dx=\frac {x^8}{8\,b^2}+\frac {a^5}{2\,b\,\left (b^6\,x^2+a\,b^5\right )}-\frac {a\,x^6}{3\,b^3}+\frac {5\,a^4\,\ln \left (b\,x^2+a\right )}{2\,b^6}+\frac {3\,a^2\,x^4}{4\,b^4}-\frac {2\,a^3\,x^2}{b^5} \]

[In]

int(x^11/(a^2 + b^2*x^4 + 2*a*b*x^2),x)

[Out]

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