\(\int \frac {1}{(a^2+2 a b x^2+b^2 x^4)^{9/4}} \, dx\) [7]

Optimal result

Integrand size = 22, antiderivative size = 135 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\frac {x \left (a+b x^2\right )}{7 a \left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}}+\frac {6 x}{35 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/4}}+\frac {8 x \left (a+b x^2\right )}{35 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/4}}+\frac {16 x}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \] Output:

1/7*x*(b*x^2+a)/a/(b^2*x^4+2*a*b*x^2+a^2)^(9/4)+6/35*x/a^2/(b^2*x^4+2*a*b* 
x^2+a^2)^(5/4)+8/35*x*(b*x^2+a)/a^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/4)+16/35*x/ 
a^4/(b^2*x^4+2*a*b*x^2+a^2)^(1/4)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\frac {\left (a+b x^2\right ) \left (35 a^3 x+70 a^2 b x^3+56 a b^2 x^5+16 b^3 x^7\right )}{35 a^4 \left (\left (a+b x^2\right )^2\right )^{9/4}} \] Input:

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-9/4),x]
 

Output:

((a + b*x^2)*(35*a^3*x + 70*a^2*b*x^3 + 56*a*b^2*x^5 + 16*b^3*x^7))/(35*a^ 
4*((a + b*x^2)^2)^(9/4))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1385, 209, 209, 209, 208}

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.

\(\displaystyle \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 1385

\(\displaystyle \frac {\sqrt {\frac {b x^2}{a}+1} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{9/2}}dx}{a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 209

\(\displaystyle \frac {\sqrt {\frac {b x^2}{a}+1} \left (\frac {6}{7} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{7/2}}dx+\frac {x}{7 \left (\frac {b x^2}{a}+1\right )^{7/2}}\right )}{a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 209

\(\displaystyle \frac {\sqrt {\frac {b x^2}{a}+1} \left (\frac {6}{7} \left (\frac {4}{5} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/2}}dx+\frac {x}{5 \left (\frac {b x^2}{a}+1\right )^{5/2}}\right )+\frac {x}{7 \left (\frac {b x^2}{a}+1\right )^{7/2}}\right )}{a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

\(\displaystyle \frac {\sqrt {\frac {b x^2}{a}+1} \left (\frac {x}{7 \left (\frac {b x^2}{a}+1\right )^{7/2}}+\frac {6}{7} \left (\frac {x}{5 \left (\frac {b x^2}{a}+1\right )^{5/2}}+\frac {4}{5} \left (\frac {2 x}{3 \sqrt {\frac {b x^2}{a}+1}}+\frac {x}{3 \left (\frac {b x^2}{a}+1\right )^{3/2}}\right )\right )\right )}{a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}\)

Input:

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-9/4),x]
 

Output:

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

Defintions of rubi rules used

rule 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 209

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
 

rule 1385

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.49

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right ) x \left (16 b^{3} x^{6}+56 b^{2} x^{4} a +70 a^{2} b \,x^{2}+35 a^{3}\right )}{35 a^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {9}{4}}}\)	\(66\)
orering	\(\frac {\left (b \,x^{2}+a \right ) x \left (16 b^{3} x^{6}+56 b^{2} x^{4} a +70 a^{2} b \,x^{2}+35 a^{3}\right )}{35 a^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {9}{4}}}\)	\(66\)

Input:

int(1/(b^2*x^4+2*a*b*x^2+a^2)^(9/4),x,method=_RETURNVERBOSE)
 

Output:

1/35*(b*x^2+a)*x*(16*b^3*x^6+56*a*b^2*x^4+70*a^2*b*x^2+35*a^3)/a^4/(b^2*x^ 
4+2*a*b*x^2+a^2)^(9/4)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\frac {{\left (16 \, b^{3} x^{7} + 56 \, a b^{2} x^{5} + 70 \, a^{2} b x^{3} + 35 \, a^{3} x\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}}}{35 \, {\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} \] Input:

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(9/4),x, algorithm="fricas")
 

Output:

1/35*(16*b^3*x^7 + 56*a*b^2*x^5 + 70*a^2*b*x^3 + 35*a^3*x)*(b^2*x^4 + 2*a* 
b*x^2 + a^2)^(1/4)/(a^4*b^4*x^8 + 4*a^5*b^3*x^6 + 6*a^6*b^2*x^4 + 4*a^7*b* 
x^2 + a^8)
 

Sympy [F]

\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\int \frac {1}{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {9}{4}}}\, dx \] Input:

integrate(1/(b**2*x**4+2*a*b*x**2+a**2)**(9/4),x)
 

Output:

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(-9/4), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(9/4),x, algorithm="maxima")
 

Output:

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(-9/4), x)
 

Giac [F]

\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(9/4),x, algorithm="giac")
 

Output:

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(-9/4), x)
 

Mupad [B] (verification not implemented)

Time = 18.92 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\frac {x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4}}{7\,a\,{\left (b\,x^2+a\right )}^5}+\frac {6\,x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4}}{35\,a^2\,{\left (b\,x^2+a\right )}^4}+\frac {8\,x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4}}{35\,a^3\,{\left (b\,x^2+a\right )}^3}+\frac {16\,x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4}}{35\,a^4\,{\left (b\,x^2+a\right )}^2} \] Input:

int(1/(a^2 + b^2*x^4 + 2*a*b*x^2)^(9/4),x)
 

Output:

(x*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/4))/(7*a*(a + b*x^2)^5) + (6*x*(a^2 + b^ 
2*x^4 + 2*a*b*x^2)^(3/4))/(35*a^2*(a + b*x^2)^4) + (8*x*(a^2 + b^2*x^4 + 2 
*a*b*x^2)^(3/4))/(35*a^3*(a + b*x^2)^3) + (16*x*(a^2 + b^2*x^4 + 2*a*b*x^2 
)^(3/4))/(35*a^4*(a + b*x^2)^2)
 

Reduce [F]

\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\int \frac {1}{\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{4}} a^{4}+4 \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{4}} a^{3} b \,x^{2}+6 \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{4}} a^{2} b^{2} x^{4}+4 \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{4}} a \,b^{3} x^{6}+\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{4}} b^{4} x^{8}}d x \] Input:

int(1/(b^2*x^4+2*a*b*x^2+a^2)^(9/4),x)
 

Output:

int(1/((a**2 + 2*a*b*x**2 + b**2*x**4)**(1/4)*a**4 + 4*(a**2 + 2*a*b*x**2 
+ b**2*x**4)**(1/4)*a**3*b*x**2 + 6*(a**2 + 2*a*b*x**2 + b**2*x**4)**(1/4) 
*a**2*b**2*x**4 + 4*(a**2 + 2*a*b*x**2 + b**2*x**4)**(1/4)*a*b**3*x**6 + ( 
a**2 + 2*a*b*x**2 + b**2*x**4)**(1/4)*b**4*x**8),x)