Integrand size = 26, antiderivative size = 251 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^6 \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^4 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \]
-1/10*a^5*((b*x^2+a)^2)^(1/2)/x^10/(b*x^2+a)-5/8*a^4*b*((b*x^2+a)^2)^(1/2) /x^8/(b*x^2+a)-5/3*a^3*b^2*((b*x^2+a)^2)^(1/2)/x^6/(b*x^2+a)-5/2*a^2*b^3*( (b*x^2+a)^2)^(1/2)/x^4/(b*x^2+a)-5/2*a*b^4*((b*x^2+a)^2)^(1/2)/x^2/(b*x^2+ a)+b^5*ln(x)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)
Time = 0.49 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\frac {1}{240} \left (-\frac {\sqrt {\left (a+b x^2\right )^2} \left (12 a^4+63 a^3 b x^2+137 a^2 b^2 x^4+163 a b^3 x^6+137 b^4 x^8\right )}{x^{10}}+\frac {\sqrt {a^2} \left (12 a^4+75 a^3 b x^2+200 a^2 b^2 x^4+300 a b^3 x^6+300 b^4 x^8\right )}{x^{10}}-120 b^5 \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-\frac {120 \sqrt {a^2} b^5 \log \left (x^2\right )}{a}+\frac {60 \sqrt {a^2} b^5 \log \left (a \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{a}+\frac {60 \sqrt {a^2} b^5 \log \left (a \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{a}\right ) \]
(-((Sqrt[(a + b*x^2)^2]*(12*a^4 + 63*a^3*b*x^2 + 137*a^2*b^2*x^4 + 163*a*b ^3*x^6 + 137*b^4*x^8))/x^10) + (Sqrt[a^2]*(12*a^4 + 75*a^3*b*x^2 + 200*a^2 *b^2*x^4 + 300*a*b^3*x^6 + 300*b^4*x^8))/x^10 - 120*b^5*ArcTanh[(b*x^2)/(S qrt[a^2] - Sqrt[(a + b*x^2)^2])] - (120*Sqrt[a^2]*b^5*Log[x^2])/a + (60*Sq rt[a^2]*b^5*Log[a*(Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2])])/a + (60*Sqrt [a^2]*b^5*Log[a*(Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2])])/a)/240
Time = 0.24 (sec) , antiderivative size = 98, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^{11}}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{11}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{12}}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5}{x^{12}}+\frac {5 b a^4}{x^{10}}+\frac {10 b^2 a^3}{x^8}+\frac {10 b^3 a^2}{x^6}+\frac {5 b^4 a}{x^4}+\frac {b^5}{x^2}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {a^5}{5 x^{10}}-\frac {5 a^4 b}{4 x^8}-\frac {10 a^3 b^2}{3 x^6}-\frac {5 a^2 b^3}{x^4}-\frac {5 a b^4}{x^2}+b^5 \log \left (x^2\right )\right )}{2 \left (a+b x^2\right )}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/5*a^5/x^10 - (5*a^4*b)/(4*x^8) - (10* a^3*b^2)/(3*x^6) - (5*a^2*b^3)/x^4 - (5*a*b^4)/x^2 + b^5*Log[x^2]))/(2*(a + b*x^2))
3.6.97.3.1 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 = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.28
method | result | size |
pseudoelliptic | \(-\frac {\left (-5 b^{5} \ln \left (x^{2}\right ) x^{10}+a \left (25 b^{4} x^{8}+25 a \,b^{3} x^{6}+\frac {50}{3} a^{2} b^{2} x^{4}+\frac {25}{4} a^{3} b \,x^{2}+a^{4}\right )\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{10 x^{10}}\) | \(70\) |
default | \(\frac {{\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (120 \ln \left (x \right ) x^{10} b^{5}-300 a \,x^{8} b^{4}-300 a^{2} x^{6} b^{3}-200 a^{3} x^{4} b^{2}-75 x^{2} a^{4} b -12 a^{5}\right )}{120 \left (b \,x^{2}+a \right )^{5} x^{10}}\) | \(82\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {5}{2} a \,x^{8} b^{4}-\frac {5}{2} a^{2} x^{6} b^{3}-\frac {5}{3} a^{3} x^{4} b^{2}-\frac {5}{8} x^{2} a^{4} b -\frac {1}{10} a^{5}\right )}{\left (b \,x^{2}+a \right ) x^{10}}+\frac {b^{5} \ln \left (x \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\) | \(98\) |
-1/10*(-5*b^5*ln(x^2)*x^10+a*(25*b^4*x^8+25*a*b^3*x^6+50/3*a^2*b^2*x^4+25/ 4*a^3*b*x^2+a^4))*csgn(b*x^2+a)/x^10
Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\frac {120 \, b^{5} x^{10} \log \left (x\right ) - 300 \, a b^{4} x^{8} - 300 \, a^{2} b^{3} x^{6} - 200 \, a^{3} b^{2} x^{4} - 75 \, a^{4} b x^{2} - 12 \, a^{5}}{120 \, x^{10}} \]
1/120*(120*b^5*x^10*log(x) - 300*a*b^4*x^8 - 300*a^2*b^3*x^6 - 200*a^3*b^2 *x^4 - 75*a^4*b*x^2 - 12*a^5)/x^10
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{11}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=b^{5} \log \left (x\right ) - \frac {5 \, a b^{4}}{2 \, x^{2}} - \frac {5 \, a^{2} b^{3}}{2 \, x^{4}} - \frac {5 \, a^{3} b^{2}}{3 \, x^{6}} - \frac {5 \, a^{4} b}{8 \, x^{8}} - \frac {a^{5}}{10 \, x^{10}} \]
b^5*log(x) - 5/2*a*b^4/x^2 - 5/2*a^2*b^3/x^4 - 5/3*a^3*b^2/x^6 - 5/8*a^4*b /x^8 - 1/10*a^5/x^10
Time = 0.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\frac {1}{2} \, b^{5} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {137 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 300 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 300 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 200 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 75 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 12 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{120 \, x^{10}} \]
1/2*b^5*log(x^2)*sgn(b*x^2 + a) - 1/120*(137*b^5*x^10*sgn(b*x^2 + a) + 300 *a*b^4*x^8*sgn(b*x^2 + a) + 300*a^2*b^3*x^6*sgn(b*x^2 + a) + 200*a^3*b^2*x ^4*sgn(b*x^2 + a) + 75*a^4*b*x^2*sgn(b*x^2 + a) + 12*a^5*sgn(b*x^2 + a))/x ^10
Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^{11}} \,d x \]