Integrand size = 26, antiderivative size = 128 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {3 x}{8 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x^3}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {3 \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
-3/8*x/b^2/((b*x^2+a)^2)^(1/2)-1/4*x^3/b/(b*x^2+a)/((b*x^2+a)^2)^(1/2)+3/8 *(b*x^2+a)*arctan(b^(1/2)*x/a^(1/2))/a^(1/2)/b^(5/2)/((b*x^2+a)^2)^(1/2)
Time = 1.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.66 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {-\sqrt {a} \sqrt {b} x \left (3 a+5 b x^2\right )+3 \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{5/2} \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \] Input:
Integrate[x^4/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
Output:
(-(Sqrt[a]*Sqrt[b]*x*(3*a + 5*b*x^2)) + 3*(a + b*x^2)^2*ArcTan[(Sqrt[b]*x) /Sqrt[a]])/(8*Sqrt[a]*b^(5/2)*(a + b*x^2)*Sqrt[(a + b*x^2)^2])
Time = 0.36 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 27, 252, 252, 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.
\(\displaystyle \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \int \frac {x^4}{b^3 \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^4}{\left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {3 \int \frac {x^2}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^3}{4 b \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {3 \left (\frac {\int \frac {1}{b x^2+a}dx}{2 b}-\frac {x}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^3}{4 b \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{3/2}}-\frac {x}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^3}{4 b \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[x^4/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
Output:
((a + b*x^2)*(-1/4*x^3/(b*(a + b*x^2)^2) + (3*(-1/2*x/(b*(a + b*x^2)) + Ar cTan[(Sqrt[b]*x)/Sqrt[a]]/(2*Sqrt[a]*b^(3/2))))/(4*b)))/Sqrt[a^2 + 2*a*b*x ^2 + b^2*x^4]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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)])
Time = 1.45 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.76
method | result | size |
default | \(-\frac {\left (-3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{2} x^{4}+5 \sqrt {a b}\, b \,x^{3}-6 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a b \,x^{2}+3 \sqrt {a b}\, a x -3 a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )\right ) \left (b \,x^{2}+a \right )}{8 \sqrt {a b}\, b^{2} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(97\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {5 x^{3}}{8 b}-\frac {3 a x}{8 b^{2}}\right )}{\left (b \,x^{2}+a \right )^{3}}-\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b x +\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, b^{2}}+\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (-b x +\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, b^{2}}\) | \(124\) |
Input:
int(x^4/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/8*(-3*arctan(b/(a*b)^(1/2)*x)*b^2*x^4+5*(a*b)^(1/2)*b*x^3-6*arctan(b/(a *b)^(1/2)*x)*a*b*x^2+3*(a*b)^(1/2)*a*x-3*a^2*arctan(b/(a*b)^(1/2)*x))*(b*x ^2+a)/(a*b)^(1/2)/b^2/((b*x^2+a)^2)^(3/2)
Time = 0.08 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.47 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\left [-\frac {10 \, a b^{2} x^{3} + 6 \, a^{2} b x + 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}}, -\frac {5 \, a b^{2} x^{3} + 3 \, a^{2} b x - 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}}\right ] \] Input:
integrate(x^4/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")
Output:
[-1/16*(10*a*b^2*x^3 + 6*a^2*b*x + 3*(b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(-a*b )*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a*b^5*x^4 + 2*a^2*b^4*x^ 2 + a^3*b^3), -1/8*(5*a*b^2*x^3 + 3*a^2*b*x - 3*(b^2*x^4 + 2*a*b*x^2 + a^2 )*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a*b^5*x^4 + 2*a^2*b^4*x^2 + a^3*b^3)]
\[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^{4}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**4/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)
Output:
Integral(x**4/((a + b*x**2)**2)**(3/2), x)
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.46 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {5 \, b x^{3} + 3 \, a x}{8 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} + \frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{2}} \] Input:
integrate(x^4/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")
Output:
-1/8*(5*b*x^3 + 3*a*x)/(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2) + 3/8*arctan(b*x/ sqrt(a*b))/(sqrt(a*b)*b^2)
Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.51 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{2} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {5 \, b x^{3} + 3 \, a x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:
integrate(x^4/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")
Output:
3/8*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2*sgn(b*x^2 + a)) - 1/8*(5*b*x^3 + 3*a*x)/((b*x^2 + a)^2*b^2*sgn(b*x^2 + a))
Timed out. \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^4}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \] Input:
int(x^4/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)
Output:
int(x^4/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{4}-3 a^{2} b x -5 a \,b^{2} x^{3}}{8 a \,b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^4/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)
Output:
(3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2 + 6*sqrt(b)*sqrt(a)* atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*x**2 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqr t(b)*sqrt(a)))*b**2*x**4 - 3*a**2*b*x - 5*a*b**2*x**3)/(8*a*b**3*(a**2 + 2 *a*b*x**2 + b**2*x**4))