Integrand size = 26, antiderivative size = 250 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^3} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a^2 b^3 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a b^4 x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac {b^5 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \] Output:
-1/2*a^5*((b*x^2+a)^2)^(1/2)/x^2/(b*x^2+a)+5*a^3*b^2*x^2*((b*x^2+a)^2)^(1/ 2)/(b*x^2+a)+5*a^2*b^3*x^4*((b*x^2+a)^2)^(1/2)/(2*b*x^2+2*a)+5*a*b^4*x^6*( (b*x^2+a)^2)^(1/2)/(6*b*x^2+6*a)+b^5*x^8*((b*x^2+a)^2)^(1/2)/(8*b*x^2+8*a) +5*a^4*b*((b*x^2+a)^2)^(1/2)*ln(x)/(b*x^2+a)
Time = 1.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.34 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^3} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (-12 a^5+120 a^3 b^2 x^4+60 a^2 b^3 x^6+20 a b^4 x^8+3 b^5 x^{10}+120 a^4 b x^2 \log (x)\right )}{24 x^2 \left (a+b x^2\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^3,x]
Output:
(Sqrt[(a + b*x^2)^2]*(-12*a^5 + 120*a^3*b^2*x^4 + 60*a^2*b^3*x^6 + 20*a*b^ 4*x^8 + 3*b^5*x^10 + 120*a^4*b*x^2*Log[x]))/(24*x^2*(a + b*x^2))
Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.39, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^3}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^3}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^4}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^5 x^6+5 a b^4 x^4+10 a^2 b^3 x^2+10 a^3 b^2+\frac {5 a^4 b}{x^2}+\frac {a^5}{x^4}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {a^5}{x^2}+5 a^4 b \log \left (x^2\right )+10 a^3 b^2 x^2+5 a^2 b^3 x^4+\frac {5}{3} a b^4 x^6+\frac {b^5 x^8}{4}\right )}{2 \left (a+b x^2\right )}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^3,x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-(a^5/x^2) + 10*a^3*b^2*x^2 + 5*a^2*b^3* x^4 + (5*a*b^4*x^6)/3 + (b^5*x^8)/4 + 5*a^4*b*Log[x^2]))/(2*(a + b*x^2))
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 = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.28
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (-\frac {x^{10} b^{5}}{4}-\frac {5 a \,x^{8} b^{4}}{3}-5 a^{2} x^{6} b^{3}-10 a^{3} x^{4} b^{2}-5 b \,a^{4} \ln \left (x^{2}\right ) x^{2}+a^{5}\right )}{2 x^{2}}\) | \(70\) |
default | \(\frac {{\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (3 x^{10} b^{5}+20 a \,x^{8} b^{4}+60 a^{2} x^{6} b^{3}+120 a^{3} x^{4} b^{2}+120 b \,a^{4} \ln \left (x \right ) x^{2}-12 a^{5}\right )}{24 x^{2} \left (b \,x^{2}+a \right )^{5}}\) | \(82\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{2} \left (\frac {1}{8} b^{3} x^{8}+\frac {5}{6} b^{2} x^{6} a +\frac {5}{2} a^{2} b \,x^{4}+5 a^{3} x^{2}\right )}{b \,x^{2}+a}-\frac {a^{5} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 x^{2} \left (b \,x^{2}+a \right )}+\frac {5 a^{4} b \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (x \right )}{b \,x^{2}+a}\) | \(117\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^3,x,method=_RETURNVERBOSE)
Output:
-1/2*csgn(b*x^2+a)*(-1/4*x^10*b^5-5/3*a*x^8*b^4-5*a^2*x^6*b^3-10*a^3*x^4*b ^2-5*b*a^4*ln(x^2)*x^2+a^5)/x^2
Time = 0.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^3} \, dx=\frac {3 \, b^{5} x^{10} + 20 \, a b^{4} x^{8} + 60 \, a^{2} b^{3} x^{6} + 120 \, a^{3} b^{2} x^{4} + 120 \, a^{4} b x^{2} \log \left (x\right ) - 12 \, a^{5}}{24 \, x^{2}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^3,x, algorithm="fricas")
Output:
1/24*(3*b^5*x^10 + 20*a*b^4*x^8 + 60*a^2*b^3*x^6 + 120*a^3*b^2*x^4 + 120*a ^4*b*x^2*log(x) - 12*a^5)/x^2
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^3} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{3}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**3,x)
Output:
Integral(((a + b*x**2)**2)**(5/2)/x**3, x)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^3} \, dx=\frac {1}{8} \, b^{5} x^{8} + \frac {5}{6} \, a b^{4} x^{6} + \frac {5}{2} \, a^{2} b^{3} x^{4} + 5 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b \log \left (x\right ) - \frac {a^{5}}{2 \, x^{2}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^3,x, algorithm="maxima")
Output:
1/8*b^5*x^8 + 5/6*a*b^4*x^6 + 5/2*a^2*b^3*x^4 + 5*a^3*b^2*x^2 + 5*a^4*b*lo g(x) - 1/2*a^5/x^2
Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^3} \, dx=\frac {1}{8} \, b^{5} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{6} \, a b^{4} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{2} \, a^{2} b^{3} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{2} \, a^{4} b \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {5 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, x^{2}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^3,x, algorithm="giac")
Output:
1/8*b^5*x^8*sgn(b*x^2 + a) + 5/6*a*b^4*x^6*sgn(b*x^2 + a) + 5/2*a^2*b^3*x^ 4*sgn(b*x^2 + a) + 5*a^3*b^2*x^2*sgn(b*x^2 + a) + 5/2*a^4*b*log(x^2)*sgn(b *x^2 + a) - 1/2*(5*a^4*b*x^2*sgn(b*x^2 + a) + a^5*sgn(b*x^2 + a))/x^2
Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^3} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^3} \,d x \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^3,x)
Output:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^3, x)
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^3} \, dx=\frac {120 \,\mathrm {log}\left (x \right ) a^{4} b \,x^{2}-12 a^{5}+120 a^{3} b^{2} x^{4}+60 a^{2} b^{3} x^{6}+20 a \,b^{4} x^{8}+3 b^{5} x^{10}}{24 x^{2}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^3,x)
Output:
(120*log(x)*a**4*b*x**2 - 12*a**5 + 120*a**3*b**2*x**4 + 60*a**2*b**3*x**6 + 20*a*b**4*x**8 + 3*b**5*x**10)/(24*x**2)