Integrand size = 26, antiderivative size = 113 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {a}{b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^2}{4 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
a/b^3/((b*x^2+a)^2)^(1/2)-1/4*a^2/b^3/(b*x^2+a)/((b*x^2+a)^2)^(1/2)+1/2*(b *x^2+a)*ln(b*x^2+a)/b^3/((b*x^2+a)^2)^(1/2)
Time = 0.47 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.70 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\frac {b x^2 \left (a \sqrt {\left (a+b x^2\right )^2} \left (-2 a^2-a b x^2+b^2 x^4\right )+\sqrt {a^2} \left (2 a^3+3 a^2 b x^2+b^3 x^6\right )\right )}{a^2 \left (a+b x^2\right ) \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )}+2 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-2 \log \left (b^3 \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{4 b^3} \] Input:
Integrate[x^5/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
Output:
((b*x^2*(a*Sqrt[(a + b*x^2)^2]*(-2*a^2 - a*b*x^2 + b^2*x^4) + Sqrt[a^2]*(2 *a^3 + 3*a^2*b*x^2 + b^3*x^6)))/(a^2*(a + b*x^2)*(a^2 + a*b*x^2 - Sqrt[a^2 ]*Sqrt[(a + b*x^2)^2])) + 2*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] - 2*Log[b^3*(Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2])])/(4*b^3)
Time = 0.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.71, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 27, 243, 49, 2009}
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.
\(\displaystyle \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \int \frac {x^5}{b^3 \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^5}{\left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^4}{\left (b x^2+a\right )^3}dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \left (\frac {a^2}{b^2 \left (b x^2+a\right )^3}-\frac {2 a}{b^2 \left (b x^2+a\right )^2}+\frac {1}{b^2 \left (b x^2+a\right )}\right )dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {a^2}{2 b^3 \left (a+b x^2\right )^2}+\frac {2 a}{b^3 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{b^3}\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[x^5/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
Output:
((a + b*x^2)*(-1/2*a^2/(b^3*(a + b*x^2)^2) + (2*a)/(b^3*(a + b*x^2)) + Log [a + b*x^2]/b^3))/(2*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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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)])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.48
method | result | size |
pseudoelliptic | \(\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\left (b \,x^{2}+a \right )^{2} \ln \left (b \,x^{2}+a \right )+2 a b \,x^{2}+\frac {3 a^{2}}{2}\right )}{2 \left (b \,x^{2}+a \right )^{2} b^{3}}\) | \(54\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {a \,x^{2}}{b^{2}}+\frac {3 a^{2}}{4 b^{3}}\right )}{\left (b \,x^{2}+a \right )^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b^{3}}\) | \(73\) |
default | \(\frac {\left (2 \ln \left (b \,x^{2}+a \right ) b^{2} x^{4}+4 \ln \left (b \,x^{2}+a \right ) a b \,x^{2}+4 a b \,x^{2}+2 a^{2} \ln \left (b \,x^{2}+a \right )+3 a^{2}\right ) \left (b \,x^{2}+a \right )}{4 b^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(81\) |
Input:
int(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2*csgn(b*x^2+a)*((b*x^2+a)^2*ln(b*x^2+a)+2*a*b*x^2+3/2*a^2)/(b*x^2+a)^2/ b^3
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.61 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {4 \, a b x^{2} + 3 \, a^{2} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} \] Input:
integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")
Output:
1/4*(4*a*b*x^2 + 3*a^2 + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)*log(b*x^2 + a))/(b^ 5*x^4 + 2*a*b^4*x^2 + a^2*b^3)
\[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^{5}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**5/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)
Output:
Integral(x**5/((a + b*x**2)**2)**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.49 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {4 \, a b x^{2} + 3 \, a^{2}}{4 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {\log \left (b x^{2} + a\right )}{2 \, b^{3}} \] Input:
integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")
Output:
1/4*(4*a*b*x^2 + 3*a^2)/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3) + 1/2*log(b*x^2 + a)/b^3
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.55 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {3 \, b x^{4} + 2 \, a x^{2}}{4 \, {\left (b x^{2} + a\right )}^{2} b^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:
integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")
Output:
1/2*log(abs(b*x^2 + a))/(b^3*sgn(b*x^2 + a)) - 1/4*(3*b*x^4 + 2*a*x^2)/((b *x^2 + a)^2*b^2*sgn(b*x^2 + a))
Timed out. \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^5}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \] Input:
int(x^5/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)
Output:
int(x^5/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)
Time = 0.17 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.72 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {2 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2}+4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a b \,x^{2}+2 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{2} x^{4}+a^{2}-2 b^{2} x^{4}}{4 b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)
Output:
(2*log(a + b*x**2)*a**2 + 4*log(a + b*x**2)*a*b*x**2 + 2*log(a + b*x**2)*b **2*x**4 + a**2 - 2*b**2*x**4)/(4*b**3*(a**2 + 2*a*b*x**2 + b**2*x**4))