\(\int \frac {x^{11}}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [645]

Optimal result

Integrand size = 26, antiderivative size = 238 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {5 a^2}{b^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^5}{8 b^6 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 a^4}{6 b^6 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 a^3}{2 b^6 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x^2 \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^6 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 660, 45} \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {5 a^2}{b^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x^2 \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^5}{8 b^6 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 a^4}{6 b^6 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 a^3}{2 b^6 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

Int[x^11/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

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

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 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 1125

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx,x,x^2\right ) \\ & = \frac {\left (b^4 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \frac {x^5}{\left (a b+b^2 x\right )^5} \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {\left (b^4 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \left (\frac {1}{b^{10}}-\frac {a^5}{b^{10} (a+b x)^5}+\frac {5 a^4}{b^{10} (a+b x)^4}-\frac {10 a^3}{b^{10} (a+b x)^3}+\frac {10 a^2}{b^{10} (a+b x)^2}-\frac {5 a}{b^{10} (a+b x)}\right ) \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {5 a^2}{b^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^5}{8 b^6 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 a^4}{6 b^6 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 a^3}{2 b^6 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x^2 \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^6 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.43 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {-77 a^5-248 a^4 b x^2-252 a^3 b^2 x^4-48 a^2 b^3 x^6+48 a b^4 x^8+12 b^5 x^{10}-60 a \left (a+b x^2\right )^4 \log \left (a+b x^2\right )}{24 b^6 \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \]

[In]

Integrate[x^11/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(-77*a^5 - 248*a^4*b*x^2 - 252*a^3*b^2*x^4 - 48*a^2*b^3*x^6 + 48*a*b^4*x^8 + 12*b^5*x^10 - 60*a*(a + b*x^2)^4*
Log[a + b*x^2])/(24*b^6*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.40

method	result	size
pseudoelliptic	\(-\frac {5 \left (a \left (b \,x^{2}+a \right )^{4} \ln \left (b \,x^{2}+a \right )-\frac {x^{10} b^{5}}{5}-\frac {4 a \,x^{8} b^{4}}{5}+\frac {4 a^{2} x^{6} b^{3}}{5}+\frac {21 a^{3} x^{4} b^{2}}{5}+\frac {62 x^{2} a^{4} b}{15}+\frac {77 a^{5}}{60}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right )^{4} b^{6}}\)	\(96\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{2}}{2 \left (b \,x^{2}+a \right ) b^{5}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-5 a^{2} b^{2} x^{6}-\frac {25 a^{3} b \,x^{4}}{2}-\frac {65 a^{4} x^{2}}{6}-\frac {77 a^{5}}{24 b}\right )}{\left (b \,x^{2}+a \right )^{5} b^{5}}-\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b^{6}}\)	\(125\)
default	\(-\frac {\left (-12 x^{10} b^{5}+60 \ln \left (b \,x^{2}+a \right ) x^{8} a \,b^{4}-48 a \,x^{8} b^{4}+240 \ln \left (b \,x^{2}+a \right ) x^{6} a^{2} b^{3}+48 a^{2} x^{6} b^{3}+360 \ln \left (b \,x^{2}+a \right ) x^{4} a^{3} b^{2}+252 a^{3} x^{4} b^{2}+240 \ln \left (b \,x^{2}+a \right ) x^{2} a^{4} b +248 x^{2} a^{4} b +60 \ln \left (b \,x^{2}+a \right ) a^{5}+77 a^{5}\right ) \left (b \,x^{2}+a \right )}{24 b^{6} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(163\)

[In]

int(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-5/2*(a*(b*x^2+a)^4*ln(b*x^2+a)-1/5*x^10*b^5-4/5*a*x^8*b^4+4/5*a^2*x^6*b^3+21/5*a^3*x^4*b^2+62/15*x^2*a^4*b+77
/60*a^5)*csgn(b*x^2+a)/(b*x^2+a)^4/b^6

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.66 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {12 \, b^{5} x^{10} + 48 \, a b^{4} x^{8} - 48 \, a^{2} b^{3} x^{6} - 252 \, a^{3} b^{2} x^{4} - 248 \, a^{4} b x^{2} - 77 \, a^{5} - 60 \, {\left (a b^{4} x^{8} + 4 \, a^{2} b^{3} x^{6} + 6 \, a^{3} b^{2} x^{4} + 4 \, a^{4} b x^{2} + a^{5}\right )} \log \left (b x^{2} + a\right )}{24 \, {\left (b^{10} x^{8} + 4 \, a b^{9} x^{6} + 6 \, a^{2} b^{8} x^{4} + 4 \, a^{3} b^{7} x^{2} + a^{4} b^{6}\right )}} \]

[In]

integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/24*(12*b^5*x^10 + 48*a*b^4*x^8 - 48*a^2*b^3*x^6 - 252*a^3*b^2*x^4 - 248*a^4*b*x^2 - 77*a^5 - 60*(a*b^4*x^8 +
4*a^2*b^3*x^6 + 6*a^3*b^2*x^4 + 4*a^4*b*x^2 + a^5)*log(b*x^2 + a))/(b^10*x^8 + 4*a*b^9*x^6 + 6*a^2*b^8*x^4 +
4*a^3*b^7*x^2 + a^4*b^6)

Sympy [F]

\[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{11}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x**11/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**11/((a + b*x**2)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.46 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {120 \, a^{2} b^{3} x^{6} + 300 \, a^{3} b^{2} x^{4} + 260 \, a^{4} b x^{2} + 77 \, a^{5}}{24 \, {\left (b^{10} x^{8} + 4 \, a b^{9} x^{6} + 6 \, a^{2} b^{8} x^{4} + 4 \, a^{3} b^{7} x^{2} + a^{4} b^{6}\right )}} + \frac {x^{2}}{2 \, b^{5}} - \frac {5 \, a \log \left (b x^{2} + a\right )}{2 \, b^{6}} \]

[In]

integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

-1/24*(120*a^2*b^3*x^6 + 300*a^3*b^2*x^4 + 260*a^4*b*x^2 + 77*a^5)/(b^10*x^8 + 4*a*b^9*x^6 + 6*a^2*b^8*x^4 + 4
*a^3*b^7*x^2 + a^4*b^6) + 1/2*x^2/b^5 - 5/2*a*log(b*x^2 + a)/b^6

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.48 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {x^{2}}{2 \, b^{5} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {5 \, a \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{6} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {125 \, a b^{4} x^{8} + 380 \, a^{2} b^{3} x^{6} + 450 \, a^{3} b^{2} x^{4} + 240 \, a^{4} b x^{2} + 48 \, a^{5}}{24 \, {\left (b x^{2} + a\right )}^{4} b^{6} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/2*x^2/(b^5*sgn(b*x^2 + a)) - 5/2*a*log(abs(b*x^2 + a))/(b^6*sgn(b*x^2 + a)) + 1/24*(125*a*b^4*x^8 + 380*a^2*
b^3*x^6 + 450*a^3*b^2*x^4 + 240*a^4*b*x^2 + 48*a^5)/((b*x^2 + a)^4*b^6*sgn(b*x^2 + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{11}}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \]

[In]

int(x^11/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int(x^11/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)