\(\int \frac {x^8}{(a+b x^2)^{9/2}} \, dx\) [529]

Optimal result

Integrand size = 15, antiderivative size = 106 \[ \int \frac {x^8}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {x^7}{7 b \left (a+b x^2\right )^{7/2}}-\frac {x^5}{5 b^2 \left (a+b x^2\right )^{5/2}}-\frac {x^3}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac {x}{b^4 \sqrt {a+b x^2}}+\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{9/2}} \] Output:

-1/7*x^7/b/(b*x^2+a)^(7/2)-1/5*x^5/b^2/(b*x^2+a)^(5/2)-1/3*x^3/b^3/(b*x^2+ 
a)^(3/2)-x/b^4/(b*x^2+a)^(1/2)+arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(9/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.83 \[ \int \frac {x^8}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {x \left (105 a^3+350 a^2 b x^2+406 a b^2 x^4+176 b^3 x^6\right )}{105 b^4 \left (a+b x^2\right )^{7/2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{b^{9/2}} \] Input:

Integrate[x^8/(a + b*x^2)^(9/2),x]
 

Output:

-1/105*(x*(105*a^3 + 350*a^2*b*x^2 + 406*a*b^2*x^4 + 176*b^3*x^6))/(b^4*(a 
 + b*x^2)^(7/2)) + (2*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/b 
^(9/2)
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {252, 252, 252, 252, 224, 219}

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.

\(\displaystyle \int \frac {x^8}{\left (a+b x^2\right )^{9/2}} \, dx\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {\int \frac {x^6}{\left (b x^2+a\right )^{7/2}}dx}{b}-\frac {x^7}{7 b \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {\frac {\int \frac {x^4}{\left (b x^2+a\right )^{5/2}}dx}{b}-\frac {x^5}{5 b \left (a+b x^2\right )^{5/2}}}{b}-\frac {x^7}{7 b \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {\frac {\frac {\int \frac {x^2}{\left (b x^2+a\right )^{3/2}}dx}{b}-\frac {x^3}{3 b \left (a+b x^2\right )^{3/2}}}{b}-\frac {x^5}{5 b \left (a+b x^2\right )^{5/2}}}{b}-\frac {x^7}{7 b \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{b}-\frac {x}{b \sqrt {a+b x^2}}}{b}-\frac {x^3}{3 b \left (a+b x^2\right )^{3/2}}}{b}-\frac {x^5}{5 b \left (a+b x^2\right )^{5/2}}}{b}-\frac {x^7}{7 b \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}}-\frac {x}{b \sqrt {a+b x^2}}}{b}-\frac {x^3}{3 b \left (a+b x^2\right )^{3/2}}}{b}-\frac {x^5}{5 b \left (a+b x^2\right )^{5/2}}}{b}-\frac {x^7}{7 b \left (a+b x^2\right )^{7/2}}\)

Input:

Int[x^8/(a + b*x^2)^(9/2),x]
 

Output:

-1/7*x^7/(b*(a + b*x^2)^(7/2)) + (-1/5*x^5/(b*(a + b*x^2)^(5/2)) + (-1/3*x 
^3/(b*(a + b*x^2)^(3/2)) + (-(x/(b*Sqrt[a + b*x^2])) + ArcTanh[(Sqrt[b]*x) 
/Sqrt[a + b*x^2]]/b^(3/2))/b)/b)/b
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

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]
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-\frac {176 x^{7} b^{\frac {7}{2}}}{105}-\frac {58 b^{\frac {5}{2}} a \,x^{5}}{15}-\frac {10 b^{\frac {3}{2}} a^{2} x^{3}}{3}-\sqrt {b}\, a^{3} x}{b^{\frac {9}{2}} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}\)	\(79\)
default	\(-\frac {x^{7}}{7 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {-\frac {x^{5}}{5 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}}{b}}{b}\)	\(103\)

Input:

int(x^8/(b*x^2+a)^(9/2),x,method=_RETURNVERBOSE)
 

Output:

1/b^(9/2)*((b*x^2+a)^(7/2)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))-176/105*x^7* 
b^(7/2)-58/15*b^(5/2)*a*x^5-10/3*b^(3/2)*a^2*x^3-b^(1/2)*a^3*x)/(b*x^2+a)^ 
(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 331, normalized size of antiderivative = 3.12 \[ \int \frac {x^8}{\left (a+b x^2\right )^{9/2}} \, dx=\left [\frac {105 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (176 \, b^{4} x^{7} + 406 \, a b^{3} x^{5} + 350 \, a^{2} b^{2} x^{3} + 105 \, a^{3} b x\right )} \sqrt {b x^{2} + a}}{210 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}, -\frac {105 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (176 \, b^{4} x^{7} + 406 \, a b^{3} x^{5} + 350 \, a^{2} b^{2} x^{3} + 105 \, a^{3} b x\right )} \sqrt {b x^{2} + a}}{105 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}\right ] \] Input:

integrate(x^8/(b*x^2+a)^(9/2),x, algorithm="fricas")
 

Output:

[1/210*(105*(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)*sq 
rt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(176*b^4*x^7 + 4 
06*a*b^3*x^5 + 350*a^2*b^2*x^3 + 105*a^3*b*x)*sqrt(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), -1/105*(105*(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(-b)*arctan(sqrt 
(-b)*x/sqrt(b*x^2 + a)) + (176*b^4*x^7 + 406*a*b^3*x^5 + 350*a^2*b^2*x^3 + 
 105*a^3*b*x)*sqrt(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)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2980 vs. \(2 (92) = 184\).

Time = 4.78 (sec) , antiderivative size = 2980, normalized size of antiderivative = 28.11 \[ \int \frac {x^8}{\left (a+b x^2\right )^{9/2}} \, dx=\text {Too large to display} \] Input:

integrate(x**8/(b*x**2+a)**(9/2),x)
 

Output:

105*a**(205/2)*b**45*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(105*a**( 
205/2)*b**(99/2)*sqrt(1 + b*x**2/a) + 630*a**(203/2)*b**(101/2)*x**2*sqrt( 
1 + b*x**2/a) + 1575*a**(201/2)*b**(103/2)*x**4*sqrt(1 + b*x**2/a) + 2100* 
a**(199/2)*b**(105/2)*x**6*sqrt(1 + b*x**2/a) + 1575*a**(197/2)*b**(107/2) 
*x**8*sqrt(1 + b*x**2/a) + 630*a**(195/2)*b**(109/2)*x**10*sqrt(1 + b*x**2 
/a) + 105*a**(193/2)*b**(111/2)*x**12*sqrt(1 + b*x**2/a)) + 630*a**(203/2) 
*b**46*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(105*a**(205/2)*b* 
*(99/2)*sqrt(1 + b*x**2/a) + 630*a**(203/2)*b**(101/2)*x**2*sqrt(1 + b*x** 
2/a) + 1575*a**(201/2)*b**(103/2)*x**4*sqrt(1 + b*x**2/a) + 2100*a**(199/2 
)*b**(105/2)*x**6*sqrt(1 + b*x**2/a) + 1575*a**(197/2)*b**(107/2)*x**8*sqr 
t(1 + b*x**2/a) + 630*a**(195/2)*b**(109/2)*x**10*sqrt(1 + b*x**2/a) + 105 
*a**(193/2)*b**(111/2)*x**12*sqrt(1 + b*x**2/a)) + 1575*a**(201/2)*b**47*x 
**4*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(105*a**(205/2)*b**(99/2)* 
sqrt(1 + b*x**2/a) + 630*a**(203/2)*b**(101/2)*x**2*sqrt(1 + b*x**2/a) + 1 
575*a**(201/2)*b**(103/2)*x**4*sqrt(1 + b*x**2/a) + 2100*a**(199/2)*b**(10 
5/2)*x**6*sqrt(1 + b*x**2/a) + 1575*a**(197/2)*b**(107/2)*x**8*sqrt(1 + b* 
x**2/a) + 630*a**(195/2)*b**(109/2)*x**10*sqrt(1 + b*x**2/a) + 105*a**(193 
/2)*b**(111/2)*x**12*sqrt(1 + b*x**2/a)) + 2100*a**(199/2)*b**48*x**6*sqrt 
(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(105*a**(205/2)*b**(99/2)*sqrt(1 + 
 b*x**2/a) + 630*a**(203/2)*b**(101/2)*x**2*sqrt(1 + b*x**2/a) + 1575*a...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (86) = 172\).

Time = 0.06 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.41 \[ \int \frac {x^8}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {1}{35} \, {\left (\frac {35 \, x^{6}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {70 \, a x^{4}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} + \frac {56 \, a^{2} x^{2}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}} + \frac {16 \, a^{3}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{4}}\right )} x - \frac {x {\left (\frac {15 \, x^{4}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b} + \frac {20 \, a x^{2}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}}\right )}}{15 \, b} - \frac {x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{3 \, b^{2}} - \frac {a x^{3}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}} + \frac {139 \, x}{105 \, \sqrt {b x^{2} + a} b^{4}} + \frac {17 \, a x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}} - \frac {29 \, a^{2} x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{4}} + \frac {\operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {9}{2}}} \] Input:

integrate(x^8/(b*x^2+a)^(9/2),x, algorithm="maxima")
 

Output:

-1/35*(35*x^6/((b*x^2 + a)^(7/2)*b) + 70*a*x^4/((b*x^2 + a)^(7/2)*b^2) + 5 
6*a^2*x^2/((b*x^2 + a)^(7/2)*b^3) + 16*a^3/((b*x^2 + a)^(7/2)*b^4))*x - 1/ 
15*x*(15*x^4/((b*x^2 + a)^(5/2)*b) + 20*a*x^2/((b*x^2 + a)^(5/2)*b^2) + 8* 
a^2/((b*x^2 + a)^(5/2)*b^3))/b - 1/3*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/ 
((b*x^2 + a)^(3/2)*b^2))/b^2 - a*x^3/((b*x^2 + a)^(5/2)*b^3) + 139/105*x/( 
sqrt(b*x^2 + a)*b^4) + 17/105*a*x/((b*x^2 + a)^(3/2)*b^4) - 29/35*a^2*x/(( 
b*x^2 + a)^(5/2)*b^4) + arcsinh(b*x/sqrt(a*b))/b^(9/2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.74 \[ \int \frac {x^8}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {{\left (2 \, {\left (x^{2} {\left (\frac {88 \, x^{2}}{b} + \frac {203 \, a}{b^{2}}\right )} + \frac {175 \, a^{2}}{b^{3}}\right )} x^{2} + \frac {105 \, a^{3}}{b^{4}}\right )} x}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} - \frac {\log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {9}{2}}} \] Input:

integrate(x^8/(b*x^2+a)^(9/2),x, algorithm="giac")
 

Output:

-1/105*(2*(x^2*(88*x^2/b + 203*a/b^2) + 175*a^2/b^3)*x^2 + 105*a^3/b^4)*x/ 
(b*x^2 + a)^(7/2) - log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(9/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8}{\left (a+b x^2\right )^{9/2}} \, dx=\int \frac {x^8}{{\left (b\,x^2+a\right )}^{9/2}} \,d x \] Input:

int(x^8/(a + b*x^2)^(9/2),x)
 

Output:

int(x^8/(a + b*x^2)^(9/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.96 \[ \int \frac {x^8}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-105 \sqrt {b \,x^{2}+a}\, a^{3} b x -350 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{3}-406 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{5}-176 \sqrt {b \,x^{2}+a}\, b^{4} x^{7}+105 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4}+420 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b \,x^{2}+630 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} x^{4}+420 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3} x^{6}+105 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} x^{8}+56 \sqrt {b}\, a^{4}+224 \sqrt {b}\, a^{3} b \,x^{2}+336 \sqrt {b}\, a^{2} b^{2} x^{4}+224 \sqrt {b}\, a \,b^{3} x^{6}+56 \sqrt {b}\, b^{4} x^{8}}{105 b^{5} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:

int(x^8/(b*x^2+a)^(9/2),x)
 

Output:

( - 105*sqrt(a + b*x**2)*a**3*b*x - 350*sqrt(a + b*x**2)*a**2*b**2*x**3 - 
406*sqrt(a + b*x**2)*a*b**3*x**5 - 176*sqrt(a + b*x**2)*b**4*x**7 + 105*sq 
rt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4 + 420*sqrt(b)*log(( 
sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b*x**2 + 630*sqrt(b)*log((sqrt 
(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b**2*x**4 + 420*sqrt(b)*log((sqrt( 
a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**3*x**6 + 105*sqrt(b)*log((sqrt(a + 
b*x**2) + sqrt(b)*x)/sqrt(a))*b**4*x**8 + 56*sqrt(b)*a**4 + 224*sqrt(b)*a* 
*3*b*x**2 + 336*sqrt(b)*a**2*b**2*x**4 + 224*sqrt(b)*a*b**3*x**6 + 56*sqrt 
(b)*b**4*x**8)/(105*b**5*(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))