Integrand size = 26, antiderivative size = 250 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^7} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2 \left (a+b x^2\right )}+\frac {5 a b^4 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {b^5 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \]
-1/6*a^5*((b*x^2+a)^2)^(1/2)/x^6/(b*x^2+a)-5/4*a^4*b*((b*x^2+a)^2)^(1/2)/x ^4/(b*x^2+a)-5*a^3*b^2*((b*x^2+a)^2)^(1/2)/x^2/(b*x^2+a)+5/2*a*b^4*x^2*((b *x^2+a)^2)^(1/2)/(b*x^2+a)+1/4*b^5*x^4*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+10*a^ 2*b^3*ln(x)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)
Time = 0.68 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^7} \, dx=\frac {\left (4 a^5+30 a^4 b x^2+120 a^3 b^2 x^4+53 a^2 b^3 x^6-60 a b^4 x^8-6 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{24 x^6 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )}-5 a^2 b^3 \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-5 a \sqrt {a^2} b^3 \log \left (x^2\right )+\frac {5}{2} a \sqrt {a^2} b^3 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+\frac {5}{2} a \sqrt {a^2} b^3 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right ) \]
((4*a^5 + 30*a^4*b*x^2 + 120*a^3*b^2*x^4 + 53*a^2*b^3*x^6 - 60*a*b^4*x^8 - 6*b^5*x^10)*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])))/(24* x^6*(a^2 + a*b*x^2 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2])) - 5*a^2*b^3*ArcTanh[( b*x^2)/(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])] - 5*a*Sqrt[a^2]*b^3*Log[x^2] + ( 5*a*Sqrt[a^2]*b^3*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]])/2 + (5*a*S qrt[a^2]*b^3*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]])/2
Time = 0.24 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.40, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^7} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^7}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^7}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^8}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5}{x^8}+\frac {5 b a^4}{x^6}+\frac {10 b^2 a^3}{x^4}+\frac {10 b^3 a^2}{x^2}+5 b^4 a+b^5 x^2\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {a^5}{3 x^6}-\frac {5 a^4 b}{2 x^4}-\frac {10 a^3 b^2}{x^2}+10 a^2 b^3 \log \left (x^2\right )+5 a b^4 x^2+\frac {b^5 x^4}{2}\right )}{2 \left (a+b x^2\right )}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/3*a^5/x^6 - (5*a^4*b)/(2*x^4) - (10*a ^3*b^2)/x^2 + 5*a*b^4*x^2 + (b^5*x^4)/2 + 10*a^2*b^3*Log[x^2]))/(2*(a + b* x^2))
3.6.95.3.1 Defintions of rubi rules used
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 = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.28
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (-\frac {3 x^{10} b^{5}}{2}-15 a \,x^{8} b^{4}-30 a^{2} b^{3} \ln \left (x^{2}\right ) x^{6}+30 a^{3} x^{4} b^{2}+\frac {15 x^{2} a^{4} b}{2}+a^{5}\right )}{6 x^{6}}\) | \(70\) |
default | \(\frac {{\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (3 x^{10} b^{5}+30 a \,x^{8} b^{4}+120 \ln \left (x \right ) x^{6} a^{2} b^{3}-60 a^{3} x^{4} b^{2}-15 x^{2} a^{4} b -2 a^{5}\right )}{12 \left (b \,x^{2}+a \right )^{5} x^{6}}\) | \(82\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{3} \left (b \,x^{2}+5 a \right )^{2}}{4 b \,x^{2}+4 a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-5 a^{3} x^{4} b^{2}-\frac {5}{4} x^{2} a^{4} b -\frac {1}{6} a^{5}\right )}{\left (b \,x^{2}+a \right ) x^{6}}+\frac {10 a^{2} b^{3} \ln \left (x \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\) | \(118\) |
-1/6*csgn(b*x^2+a)*(-3/2*x^10*b^5-15*a*x^8*b^4-30*a^2*b^3*ln(x^2)*x^6+30*a ^3*x^4*b^2+15/2*x^2*a^4*b+a^5)/x^6
Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^7} \, dx=\frac {3 \, b^{5} x^{10} + 30 \, a b^{4} x^{8} + 120 \, a^{2} b^{3} x^{6} \log \left (x\right ) - 60 \, a^{3} b^{2} x^{4} - 15 \, a^{4} b x^{2} - 2 \, a^{5}}{12 \, x^{6}} \]
1/12*(3*b^5*x^10 + 30*a*b^4*x^8 + 120*a^2*b^3*x^6*log(x) - 60*a^3*b^2*x^4 - 15*a^4*b*x^2 - 2*a^5)/x^6
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^7} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{7}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^7} \, dx=\frac {1}{4} \, b^{5} x^{4} + \frac {5}{2} \, a b^{4} x^{2} + 10 \, a^{2} b^{3} \log \left (x\right ) - \frac {5 \, a^{3} b^{2}}{x^{2}} - \frac {5 \, a^{4} b}{4 \, x^{4}} - \frac {a^{5}}{6 \, x^{6}} \]
1/4*b^5*x^4 + 5/2*a*b^4*x^2 + 10*a^2*b^3*log(x) - 5*a^3*b^2/x^2 - 5/4*a^4* b/x^4 - 1/6*a^5/x^6
Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^7} \, dx=\frac {1}{4} \, b^{5} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{2} \, a b^{4} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, a^{2} b^{3} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {110 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 60 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{12 \, x^{6}} \]
1/4*b^5*x^4*sgn(b*x^2 + a) + 5/2*a*b^4*x^2*sgn(b*x^2 + a) + 5*a^2*b^3*log( x^2)*sgn(b*x^2 + a) - 1/12*(110*a^2*b^3*x^6*sgn(b*x^2 + a) + 60*a^3*b^2*x^ 4*sgn(b*x^2 + a) + 15*a^4*b*x^2*sgn(b*x^2 + a) + 2*a^5*sgn(b*x^2 + a))/x^6
Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^7} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^7} \,d x \]