Integrand size = 22, antiderivative size = 105 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \, dx=\frac {x \left (a+b x^2\right )}{5 a \left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}}+\frac {4 x}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac {8 x \left (a+b x^2\right )}{15 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}} \] Output:
1/5*x*(b*x^2+a)/a/(b^2*x^4+2*a*b*x^2+a^2)^(7/4)+4/15*x/a^2/(b^2*x^4+2*a*b* x^2+a^2)^(3/4)+8/15*x*(b*x^2+a)/a^3/(b^2*x^4+2*a*b*x^2+a^2)^(3/4)
Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.47 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \, dx=\frac {\left (a+b x^2\right ) \left (15 a^2 x+20 a b x^3+8 b^2 x^5\right )}{15 a^3 \left (\left (a+b x^2\right )^2\right )^{7/4}} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-7/4),x]
Output:
((a + b*x^2)*(15*a^2*x + 20*a*b*x^3 + 8*b^2*x^5))/(15*a^3*((a + b*x^2)^2)^ (7/4))
Time = 0.32 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1385, 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.
\(\displaystyle \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 1385 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/2} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{7/2}}dx}{a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/2} \left (\frac {4}{5} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/2}}dx+\frac {x}{5 \left (\frac {b x^2}{a}+1\right )^{5/2}}\right )}{a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/2} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/2}}dx+\frac {x}{3 \left (\frac {b x^2}{a}+1\right )^{3/2}}\right )+\frac {x}{5 \left (\frac {b x^2}{a}+1\right )^{5/2}}\right )}{a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/2} \left (\frac {x}{5 \left (\frac {b x^2}{a}+1\right )^{5/2}}+\frac {4}{5} \left (\frac {2 x}{3 \sqrt {\frac {b x^2}{a}+1}}+\frac {x}{3 \left (\frac {b x^2}{a}+1\right )^{3/2}}\right )\right )}{a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-7/4),x]
Output:
((1 + (b*x^2)/a)^(3/2)*(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))/(a^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4))
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
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]
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)]
Time = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.52
method | result | size |
gosper | \(\frac {\left (b \,x^{2}+a \right ) x \left (8 b^{2} x^{4}+20 a b \,x^{2}+15 a^{2}\right )}{15 a^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {7}{4}}}\) | \(55\) |
orering | \(\frac {\left (b \,x^{2}+a \right ) x \left (8 b^{2} x^{4}+20 a b \,x^{2}+15 a^{2}\right )}{15 a^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {7}{4}}}\) | \(55\) |
Input:
int(1/(b^2*x^4+2*a*b*x^2+a^2)^(7/4),x,method=_RETURNVERBOSE)
Output:
1/15*(b*x^2+a)*x*(8*b^2*x^4+20*a*b*x^2+15*a^2)/a^3/(b^2*x^4+2*a*b*x^2+a^2) ^(7/4)
Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \, dx=\frac {{\left (8 \, b^{2} x^{5} + 20 \, a b x^{3} + 15 \, a^{2} x\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}}}{15 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} \] Input:
integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(7/4),x, algorithm="fricas")
Output:
1/15*(8*b^2*x^5 + 20*a*b*x^3 + 15*a^2*x)*(b^2*x^4 + 2*a*b*x^2 + a^2)^(1/4) /(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^5*b*x^2 + a^6)
\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \, dx=\int \frac {1}{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {7}{4}}}\, dx \] Input:
integrate(1/(b**2*x**4+2*a*b*x**2+a**2)**(7/4),x)
Output:
Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(-7/4), x)
\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(7/4),x, algorithm="maxima")
Output:
integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(-7/4), x)
\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(7/4),x, algorithm="giac")
Output:
integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(-7/4), x)
Time = 18.60 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \, dx=\frac {x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/4}\,\left (15\,a^2+20\,a\,b\,x^2+8\,b^2\,x^4\right )}{15\,a^3\,{\left (b\,x^2+a\right )}^3} \] Input:
int(1/(a^2 + b^2*x^4 + 2*a*b*x^2)^(7/4),x)
Output:
(x*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/4)*(15*a^2 + 8*b^2*x^4 + 20*a*b*x^2))/(1 5*a^3*(a + b*x^2)^3)
\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \, dx=\int \frac {1}{\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {3}{4}} a^{2}+2 \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {3}{4}} a b \,x^{2}+\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {3}{4}} b^{2} x^{4}}d x \] Input:
int(1/(b^2*x^4+2*a*b*x^2+a^2)^(7/4),x)
Output:
int(1/((a**2 + 2*a*b*x**2 + b**2*x**4)**(3/4)*a**2 + 2*(a**2 + 2*a*b*x**2 + b**2*x**4)**(3/4)*a*b*x**2 + (a**2 + 2*a*b*x**2 + b**2*x**4)**(3/4)*b**2 *x**4),x)