\(\int \frac {x^{13}}{(a^2+2 a b x^2+b^2 x^4)^3} \, dx\) [513]

Optimal result

Integrand size = 24, antiderivative size = 118 \[ \int \frac {x^{13}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {x^2}{2 b^6}-\frac {a^6}{10 b^7 \left (a+b x^2\right )^5}+\frac {3 a^5}{4 b^7 \left (a+b x^2\right )^4}-\frac {5 a^4}{2 b^7 \left (a+b x^2\right )^3}+\frac {5 a^3}{b^7 \left (a+b x^2\right )^2}-\frac {15 a^2}{2 b^7 \left (a+b x^2\right )}-\frac {3 a \log \left (a+b x^2\right )}{b^7} \]

[Out]

1/2*x^2/b^6-1/10*a^6/b^7/(b*x^2+a)^5+3/4*a^5/b^7/(b*x^2+a)^4-5/2*a^4/b^7/(b*x^2+a)^3+5*a^3/b^7/(b*x^2+a)^2-15/
2*a^2/b^7/(b*x^2+a)-3*a*ln(b*x^2+a)/b^7

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 118, 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} \[ \int \frac {x^{13}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=-\frac {a^6}{10 b^7 \left (a+b x^2\right )^5}+\frac {3 a^5}{4 b^7 \left (a+b x^2\right )^4}-\frac {5 a^4}{2 b^7 \left (a+b x^2\right )^3}+\frac {5 a^3}{b^7 \left (a+b x^2\right )^2}-\frac {15 a^2}{2 b^7 \left (a+b x^2\right )}-\frac {3 a \log \left (a+b x^2\right )}{b^7}+\frac {x^2}{2 b^6} \]

[In]

Int[x^13/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

x^2/(2*b^6) - a^6/(10*b^7*(a + b*x^2)^5) + (3*a^5)/(4*b^7*(a + b*x^2)^4) - (5*a^4)/(2*b^7*(a + b*x^2)^3) + (5*
a^3)/(b^7*(a + b*x^2)^2) - (15*a^2)/(2*b^7*(a + b*x^2)) - (3*a*Log[a + b*x^2])/b^7

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^6 \int \frac {x^{13}}{\left (a b+b^2 x^2\right )^6} \, dx \\ & = \frac {1}{2} b^6 \text {Subst}\left (\int \frac {x^6}{\left (a b+b^2 x\right )^6} \, dx,x,x^2\right ) \\ & = \frac {1}{2} b^6 \text {Subst}\left (\int \left (\frac {1}{b^{12}}+\frac {a^6}{b^{12} (a+b x)^6}-\frac {6 a^5}{b^{12} (a+b x)^5}+\frac {15 a^4}{b^{12} (a+b x)^4}-\frac {20 a^3}{b^{12} (a+b x)^3}+\frac {15 a^2}{b^{12} (a+b x)^2}-\frac {6 a}{b^{12} (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = \frac {x^2}{2 b^6}-\frac {a^6}{10 b^7 \left (a+b x^2\right )^5}+\frac {3 a^5}{4 b^7 \left (a+b x^2\right )^4}-\frac {5 a^4}{2 b^7 \left (a+b x^2\right )^3}+\frac {5 a^3}{b^7 \left (a+b x^2\right )^2}-\frac {15 a^2}{2 b^7 \left (a+b x^2\right )}-\frac {3 a \log \left (a+b x^2\right )}{b^7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.86 \[ \int \frac {x^{13}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=-\frac {87 a^6+375 a^5 b x^2+600 a^4 b^2 x^4+400 a^3 b^3 x^6+50 a^2 b^4 x^8-50 a b^5 x^{10}-10 b^6 x^{12}+60 a \left (a+b x^2\right )^5 \log \left (a+b x^2\right )}{20 b^7 \left (a+b x^2\right )^5} \]

[In]

Integrate[x^13/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

-1/20*(87*a^6 + 375*a^5*b*x^2 + 600*a^4*b^2*x^4 + 400*a^3*b^3*x^6 + 50*a^2*b^4*x^8 - 50*a*b^5*x^10 - 10*b^6*x^
12 + 60*a*(a + b*x^2)^5*Log[a + b*x^2])/(b^7*(a + b*x^2)^5)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.74

method	result	size
norman	\(\frac {\frac {x^{12}}{2 b}-\frac {137 a^{6}}{20 b^{7}}-\frac {15 a^{2} x^{8}}{b^{3}}-\frac {45 a^{3} x^{6}}{b^{4}}-\frac {55 a^{4} x^{4}}{b^{5}}-\frac {125 a^{5} x^{2}}{4 b^{6}}}{\left (b \,x^{2}+a \right )^{5}}-\frac {3 a \ln \left (b \,x^{2}+a \right )}{b^{7}}\)	\(87\)
risch	\(\frac {x^{2}}{2 b^{6}}+\frac {-\frac {15 a^{2} b^{3} x^{8}}{2}-25 a^{3} b^{2} x^{6}-\frac {65 b \,a^{4} x^{4}}{2}-\frac {77 a^{5} x^{2}}{4}-\frac {87 a^{6}}{20 b}}{b^{6} \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{2}}-\frac {3 a \ln \left (b \,x^{2}+a \right )}{b^{7}}\)	\(105\)
default	\(\frac {x^{2}}{2 b^{6}}-\frac {a \left (\frac {a^{5}}{5 b \left (b \,x^{2}+a \right )^{5}}+\frac {6 \ln \left (b \,x^{2}+a \right )}{b}-\frac {3 a^{4}}{2 b \left (b \,x^{2}+a \right )^{4}}+\frac {5 a^{3}}{b \left (b \,x^{2}+a \right )^{3}}+\frac {15 a}{b \left (b \,x^{2}+a \right )}-\frac {10 a^{2}}{b \left (b \,x^{2}+a \right )^{2}}\right )}{2 b^{6}}\)	\(113\)
parallelrisch	\(-\frac {-10 b^{6} x^{12}+60 \ln \left (b \,x^{2}+a \right ) x^{10} a \,b^{5}+300 \ln \left (b \,x^{2}+a \right ) x^{8} a^{2} b^{4}+300 a^{2} b^{4} x^{8}+600 \ln \left (b \,x^{2}+a \right ) x^{6} a^{3} b^{3}+900 a^{3} b^{3} x^{6}+600 \ln \left (b \,x^{2}+a \right ) x^{4} a^{4} b^{2}+1100 a^{4} b^{2} x^{4}+300 \ln \left (b \,x^{2}+a \right ) x^{2} a^{5} b +625 a^{5} b \,x^{2}+60 \ln \left (b \,x^{2}+a \right ) a^{6}+137 a^{6}}{20 b^{7} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{2} \left (b \,x^{2}+a \right )}\)	\(195\)

[In]

int(x^13/(b^2*x^4+2*a*b*x^2+a^2)^3,x,method=_RETURNVERBOSE)

[Out]

(1/2/b*x^12-137/20*a^6/b^7-15*a^2/b^3*x^8-45*a^3/b^4*x^6-55*a^4/b^5*x^4-125/4*a^5/b^6*x^2)/(b*x^2+a)^5-3*a*ln(
b*x^2+a)/b^7

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.61 \[ \int \frac {x^{13}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {10 \, b^{6} x^{12} + 50 \, a b^{5} x^{10} - 50 \, a^{2} b^{4} x^{8} - 400 \, a^{3} b^{3} x^{6} - 600 \, a^{4} b^{2} x^{4} - 375 \, a^{5} b x^{2} - 87 \, a^{6} - 60 \, {\left (a b^{5} x^{10} + 5 \, a^{2} b^{4} x^{8} + 10 \, a^{3} b^{3} x^{6} + 10 \, a^{4} b^{2} x^{4} + 5 \, a^{5} b x^{2} + a^{6}\right )} \log \left (b x^{2} + a\right )}{20 \, {\left (b^{12} x^{10} + 5 \, a b^{11} x^{8} + 10 \, a^{2} b^{10} x^{6} + 10 \, a^{3} b^{9} x^{4} + 5 \, a^{4} b^{8} x^{2} + a^{5} b^{7}\right )}} \]

[In]

integrate(x^13/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="fricas")

[Out]

1/20*(10*b^6*x^12 + 50*a*b^5*x^10 - 50*a^2*b^4*x^8 - 400*a^3*b^3*x^6 - 600*a^4*b^2*x^4 - 375*a^5*b*x^2 - 87*a^
6 - 60*(a*b^5*x^10 + 5*a^2*b^4*x^8 + 10*a^3*b^3*x^6 + 10*a^4*b^2*x^4 + 5*a^5*b*x^2 + a^6)*log(b*x^2 + a))/(b^1
2*x^10 + 5*a*b^11*x^8 + 10*a^2*b^10*x^6 + 10*a^3*b^9*x^4 + 5*a^4*b^8*x^2 + a^5*b^7)

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.17 \[ \int \frac {x^{13}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=- \frac {3 a \log {\left (a + b x^{2} \right )}}{b^{7}} + \frac {- 87 a^{6} - 385 a^{5} b x^{2} - 650 a^{4} b^{2} x^{4} - 500 a^{3} b^{3} x^{6} - 150 a^{2} b^{4} x^{8}}{20 a^{5} b^{7} + 100 a^{4} b^{8} x^{2} + 200 a^{3} b^{9} x^{4} + 200 a^{2} b^{10} x^{6} + 100 a b^{11} x^{8} + 20 b^{12} x^{10}} + \frac {x^{2}}{2 b^{6}} \]

[In]

integrate(x**13/(b**2*x**4+2*a*b*x**2+a**2)**3,x)

[Out]

-3*a*log(a + b*x**2)/b**7 + (-87*a**6 - 385*a**5*b*x**2 - 650*a**4*b**2*x**4 - 500*a**3*b**3*x**6 - 150*a**2*b
**4*x**8)/(20*a**5*b**7 + 100*a**4*b**8*x**2 + 200*a**3*b**9*x**4 + 200*a**2*b**10*x**6 + 100*a*b**11*x**8 + 2
0*b**12*x**10) + x**2/(2*b**6)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.12 \[ \int \frac {x^{13}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=-\frac {150 \, a^{2} b^{4} x^{8} + 500 \, a^{3} b^{3} x^{6} + 650 \, a^{4} b^{2} x^{4} + 385 \, a^{5} b x^{2} + 87 \, a^{6}}{20 \, {\left (b^{12} x^{10} + 5 \, a b^{11} x^{8} + 10 \, a^{2} b^{10} x^{6} + 10 \, a^{3} b^{9} x^{4} + 5 \, a^{4} b^{8} x^{2} + a^{5} b^{7}\right )}} + \frac {x^{2}}{2 \, b^{6}} - \frac {3 \, a \log \left (b x^{2} + a\right )}{b^{7}} \]

[In]

integrate(x^13/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="maxima")

[Out]

-1/20*(150*a^2*b^4*x^8 + 500*a^3*b^3*x^6 + 650*a^4*b^2*x^4 + 385*a^5*b*x^2 + 87*a^6)/(b^12*x^10 + 5*a*b^11*x^8
+ 10*a^2*b^10*x^6 + 10*a^3*b^9*x^4 + 5*a^4*b^8*x^2 + a^5*b^7) + 1/2*x^2/b^6 - 3*a*log(b*x^2 + a)/b^7

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.81 \[ \int \frac {x^{13}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {x^{2}}{2 \, b^{6}} - \frac {3 \, a \log \left ({\left | b x^{2} + a \right |}\right )}{b^{7}} + \frac {137 \, a b^{5} x^{10} + 535 \, a^{2} b^{4} x^{8} + 870 \, a^{3} b^{3} x^{6} + 720 \, a^{4} b^{2} x^{4} + 300 \, a^{5} b x^{2} + 50 \, a^{6}}{20 \, {\left (b x^{2} + a\right )}^{5} b^{7}} \]

[In]

integrate(x^13/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="giac")

[Out]

1/2*x^2/b^6 - 3*a*log(abs(b*x^2 + a))/b^7 + 1/20*(137*a*b^5*x^10 + 535*a^2*b^4*x^8 + 870*a^3*b^3*x^6 + 720*a^4
*b^2*x^4 + 300*a^5*b*x^2 + 50*a^6)/((b*x^2 + a)^5*b^7)

Mupad [B] (verification not implemented)

Time = 13.61 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.12 \[ \int \frac {x^{13}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {x^2}{2\,b^6}-\frac {\frac {87\,a^6}{20\,b}+\frac {77\,a^5\,x^2}{4}+\frac {65\,a^4\,b\,x^4}{2}+25\,a^3\,b^2\,x^6+\frac {15\,a^2\,b^3\,x^8}{2}}{a^5\,b^6+5\,a^4\,b^7\,x^2+10\,a^3\,b^8\,x^4+10\,a^2\,b^9\,x^6+5\,a\,b^{10}\,x^8+b^{11}\,x^{10}}-\frac {3\,a\,\ln \left (b\,x^2+a\right )}{b^7} \]

[In]

int(x^13/(a^2 + b^2*x^4 + 2*a*b*x^2)^3,x)

[Out]

x^2/(2*b^6) - ((87*a^6)/(20*b) + (77*a^5*x^2)/4 + (65*a^4*b*x^4)/2 + 25*a^3*b^2*x^6 + (15*a^2*b^3*x^8)/2)/(a^5
*b^6 + b^11*x^10 + 5*a*b^10*x^8 + 5*a^4*b^7*x^2 + 10*a^3*b^8*x^4 + 10*a^2*b^9*x^6) - (3*a*log(a + b*x^2))/b^7