\(\int \frac {x^{10}}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [601]

Optimal result

Integrand size = 26, antiderivative size = 248 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {325 a x}{128 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^4 x}{8 b^5 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {11 a^3 x}{16 b^5 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {105 a^2 x}{64 b^5 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x \left (a+b x^2\right )}{b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {315 \sqrt {a} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 b^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:

325/128*a*x/b^5/((b*x^2+a)^2)^(1/2)-1/8*a^4*x/b^5/(b*x^2+a)^3/((b*x^2+a)^2 
)^(1/2)+11/16*a^3*x/b^5/(b*x^2+a)^2/((b*x^2+a)^2)^(1/2)-105/64*a^2*x/b^5/( 
b*x^2+a)/((b*x^2+a)^2)^(1/2)+x*(b*x^2+a)/b^5/((b*x^2+a)^2)^(1/2)-315/128*a 
^(1/2)*(b*x^2+a)*arctan(b^(1/2)*x/a^(1/2))/b^(11/2)/((b*x^2+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.45 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {\sqrt {b} x \left (315 a^4+1155 a^3 b x^2+1533 a^2 b^2 x^4+837 a b^3 x^6+128 b^4 x^8\right )-315 \sqrt {a} \left (a+b x^2\right )^4 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 b^{11/2} \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \] Input:

Integrate[x^10/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
 

Output:

(Sqrt[b]*x*(315*a^4 + 1155*a^3*b*x^2 + 1533*a^2*b^2*x^4 + 837*a*b^3*x^6 + 
128*b^4*x^8) - 315*Sqrt[a]*(a + b*x^2)^4*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(128 
*b^(11/2)*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.68, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1384, 27, 252, 252, 252, 252, 262, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1384

\(\displaystyle \frac {b^5 \left (a+b x^2\right ) \int \frac {x^{10}}{b^5 \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^{10}}{\left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \int \frac {x^8}{\left (b x^2+a\right )^4}dx}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \left (\frac {7 \int \frac {x^6}{\left (b x^2+a\right )^3}dx}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \left (\frac {7 \left (\frac {5 \int \frac {x^4}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \int \frac {x^2}{b x^2+a}dx}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {x}{b}-\frac {a \int \frac {1}{b x^2+a}dx}{b}\right )}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {x}{b}-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}\right )}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^7}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^9}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

Input:

Int[x^10/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
 

Output:

((a + b*x^2)*(-1/8*x^9/(b*(a + b*x^2)^4) + (9*(-1/6*x^7/(b*(a + b*x^2)^3) 
+ (7*(-1/4*x^5/(b*(a + b*x^2)^2) + (5*(-1/2*x^3/(b*(a + b*x^2)) + (3*(x/b 
- (Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(3/2)))/(2*b)))/(4*b)))/(6*b)))/ 
(8*b)))/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 1384

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

Maple [A] (verified)

Time = 4.20 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.71

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, x}{\left (b \,x^{2}+a \right ) b^{5}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {325}{128} a \,b^{3} x^{7}+\frac {765}{128} a^{2} b^{2} x^{5}+\frac {643}{128} a^{3} b \,x^{3}+\frac {187}{128} a^{4} x \right )}{\left (b \,x^{2}+a \right )^{5} b^{5}}+\frac {315 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right )}{256 \left (b \,x^{2}+a \right ) b^{6}}-\frac {315 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right )}{256 \left (b \,x^{2}+a \right ) b^{6}}\)	\(176\)
default	\(\frac {\left (128 \sqrt {a b}\, b^{4} x^{9}-315 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{4} x^{8}+837 \sqrt {a b}\, a \,b^{3} x^{7}-1260 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b^{3} x^{6}+1533 \sqrt {a b}\, a^{2} b^{2} x^{5}-1890 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{3} b^{2} x^{4}+1155 \sqrt {a b}\, a^{3} b \,x^{3}-1260 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{4} b \,x^{2}+315 \sqrt {a b}\, a^{4} x -315 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{5}\right ) \left (b \,x^{2}+a \right )}{128 \sqrt {a b}\, b^{5} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(188\)

Input:

int(x^10/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

((b*x^2+a)^2)^(1/2)/(b*x^2+a)*x/b^5+((b*x^2+a)^2)^(1/2)/(b*x^2+a)^5*(325/1 
28*a*b^3*x^7+765/128*a^2*b^2*x^5+643/128*a^3*b*x^3+187/128*a^4*x)/b^5+315/ 
256*((b*x^2+a)^2)^(1/2)/(b*x^2+a)/b^6*(-a*b)^(1/2)*ln(-(-a*b)^(1/2)*x-a)-3 
15/256*((b*x^2+a)^2)^(1/2)/(b*x^2+a)/b^6*(-a*b)^(1/2)*ln((-a*b)^(1/2)*x-a)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.35 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\left [\frac {256 \, b^{4} x^{9} + 1674 \, a b^{3} x^{7} + 3066 \, a^{2} b^{2} x^{5} + 2310 \, a^{3} b x^{3} + 630 \, a^{4} x + 315 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{256 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}, \frac {128 \, b^{4} x^{9} + 837 \, a b^{3} x^{7} + 1533 \, a^{2} b^{2} x^{5} + 1155 \, a^{3} b x^{3} + 315 \, a^{4} x - 315 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{128 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}\right ] \] Input:

integrate(x^10/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
 

Output:

[1/256*(256*b^4*x^9 + 1674*a*b^3*x^7 + 3066*a^2*b^2*x^5 + 2310*a^3*b*x^3 + 
 630*a^4*x + 315*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^ 
4)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)))/(b^9*x^8 + 
4*a*b^8*x^6 + 6*a^2*b^7*x^4 + 4*a^3*b^6*x^2 + a^4*b^5), 1/128*(128*b^4*x^9 
 + 837*a*b^3*x^7 + 1533*a^2*b^2*x^5 + 1155*a^3*b*x^3 + 315*a^4*x - 315*(b^ 
4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(a/b)*arctan( 
b*x*sqrt(a/b)/a))/(b^9*x^8 + 4*a*b^8*x^6 + 6*a^2*b^7*x^4 + 4*a^3*b^6*x^2 + 
 a^4*b^5)]
 

Sympy [F]

\[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{10}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**10/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
 

Output:

Integral(x**10/((a + b*x**2)**2)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.45 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {325 \, a b^{3} x^{7} + 765 \, a^{2} b^{2} x^{5} + 643 \, a^{3} b x^{3} + 187 \, a^{4} x}{128 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}} - \frac {315 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} b^{5}} + \frac {x}{b^{5}} \] Input:

integrate(x^10/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
 

Output:

1/128*(325*a*b^3*x^7 + 765*a^2*b^2*x^5 + 643*a^3*b*x^3 + 187*a^4*x)/(b^9*x 
^8 + 4*a*b^8*x^6 + 6*a^2*b^7*x^4 + 4*a^3*b^6*x^2 + a^4*b^5) - 315/128*a*ar 
ctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) + x/b^5
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.43 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {315 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} b^{5} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {x}{b^{5} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {325 \, a b^{3} x^{7} + 765 \, a^{2} b^{2} x^{5} + 643 \, a^{3} b x^{3} + 187 \, a^{4} x}{128 \, {\left (b x^{2} + a\right )}^{4} b^{5} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:

integrate(x^10/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
 

Output:

-315/128*a*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5*sgn(b*x^2 + a)) + x/(b^5*s 
gn(b*x^2 + a)) + 1/128*(325*a*b^3*x^7 + 765*a^2*b^2*x^5 + 643*a^3*b*x^3 + 
187*a^4*x)/((b*x^2 + a)^4*b^5*sgn(b*x^2 + a))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{10}}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \] Input:

int(x^10/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)
 

Output:

int(x^10/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.87 \[ \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {-315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4}-1260 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b \,x^{2}-1890 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{4}-1260 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x^{6}-315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{8}+315 a^{4} b x +1155 a^{3} b^{2} x^{3}+1533 a^{2} b^{3} x^{5}+837 a \,b^{4} x^{7}+128 b^{5} x^{9}}{128 b^{6} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:

int(x^10/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
 

Output:

( - 315*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4 - 1260*sqrt(b)* 
sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*x**2 - 1890*sqrt(b)*sqrt(a)*a 
tan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*x**4 - 1260*sqrt(b)*sqrt(a)*atan((b 
*x)/(sqrt(b)*sqrt(a)))*a*b**3*x**6 - 315*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt( 
b)*sqrt(a)))*b**4*x**8 + 315*a**4*b*x + 1155*a**3*b**2*x**3 + 1533*a**2*b* 
*3*x**5 + 837*a*b**4*x**7 + 128*b**5*x**9)/(128*b**6*(a**4 + 4*a**3*b*x**2 
 + 6*a**2*b**2*x**4 + 4*a*b**3*x**6 + b**4*x**8))