Integrand size = 24, antiderivative size = 231 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^8 (a+b x)}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)} \]
-1/10*a^5*((b*x+a)^2)^(1/2)/x^10/(b*x+a)-5/9*a^4*b*((b*x+a)^2)^(1/2)/x^9/( b*x+a)-5/4*a^3*b^2*((b*x+a)^2)^(1/2)/x^8/(b*x+a)-10/7*a^2*b^3*((b*x+a)^2)^ (1/2)/x^7/(b*x+a)-5/6*a*b^4*((b*x+a)^2)^(1/2)/x^6/(b*x+a)-1/5*b^5*((b*x+a) ^2)^(1/2)/x^5/(b*x+a)
Time = 0.67 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx=-\frac {\sqrt {(a+b x)^2} \left (126 a^5+700 a^4 b x+1575 a^3 b^2 x^2+1800 a^2 b^3 x^3+1050 a b^4 x^4+252 b^5 x^5\right )}{1260 x^{10} (a+b x)} \]
-1/1260*(Sqrt[(a + b*x)^2]*(126*a^5 + 700*a^4*b*x + 1575*a^3*b^2*x^2 + 180 0*a^2*b^3*x^3 + 1050*a*b^4*x^4 + 252*b^5*x^5))/(x^10*(a + b*x))
Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.42, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1102, 27, 53, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5}{x^{11}}dx}{b^5 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5}{x^{11}}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^5}{x^{11}}+\frac {5 b a^4}{x^{10}}+\frac {10 b^2 a^3}{x^9}+\frac {10 b^3 a^2}{x^8}+\frac {5 b^4 a}{x^7}+\frac {b^5}{x^6}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^5}{10 x^{10}}-\frac {5 a^4 b}{9 x^9}-\frac {5 a^3 b^2}{4 x^8}-\frac {10 a^2 b^3}{7 x^7}-\frac {5 a b^4}{6 x^6}-\frac {b^5}{5 x^5}\right ) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}\) |
((-1/10*a^5/x^10 - (5*a^4*b)/(9*x^9) - (5*a^3*b^2)/(4*x^8) - (10*a^2*b^3)/ (7*x^7) - (5*a*b^4)/(6*x^6) - b^5/(5*x^5))*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/ (a + b*x)
3.2.79.3.1 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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.52 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.32
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {1}{5} b^{5} x^{5}-\frac {5}{6} a \,b^{4} x^{4}-\frac {10}{7} a^{2} b^{3} x^{3}-\frac {5}{4} a^{3} b^{2} x^{2}-\frac {5}{9} a^{4} b x -\frac {1}{10} a^{5}\right )}{\left (b x +a \right ) x^{10}}\) | \(73\) |
gosper | \(-\frac {\left (252 b^{5} x^{5}+1050 a \,b^{4} x^{4}+1800 a^{2} b^{3} x^{3}+1575 a^{3} b^{2} x^{2}+700 a^{4} b x +126 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{1260 x^{10} \left (b x +a \right )^{5}}\) | \(74\) |
default | \(-\frac {\left (252 b^{5} x^{5}+1050 a \,b^{4} x^{4}+1800 a^{2} b^{3} x^{3}+1575 a^{3} b^{2} x^{2}+700 a^{4} b x +126 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{1260 x^{10} \left (b x +a \right )^{5}}\) | \(74\) |
((b*x+a)^2)^(1/2)/(b*x+a)/x^10*(-1/5*b^5*x^5-5/6*a*b^4*x^4-10/7*a^2*b^3*x^ 3-5/4*a^3*b^2*x^2-5/9*a^4*b*x-1/10*a^5)
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.25 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx=-\frac {252 \, b^{5} x^{5} + 1050 \, a b^{4} x^{4} + 1800 \, a^{2} b^{3} x^{3} + 1575 \, a^{3} b^{2} x^{2} + 700 \, a^{4} b x + 126 \, a^{5}}{1260 \, x^{10}} \]
-1/1260*(252*b^5*x^5 + 1050*a*b^4*x^4 + 1800*a^2*b^3*x^3 + 1575*a^3*b^2*x^ 2 + 700*a^4*b*x + 126*a^5)/x^10
\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{11}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (153) = 306\).
Time = 0.21 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{10}}{6 \, a^{10}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{9}}{6 \, a^{9} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{8}}{6 \, a^{10} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{7}}{6 \, a^{9} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{6}}{6 \, a^{8} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{5}}{6 \, a^{7} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{4}}{6 \, a^{6} x^{6}} + \frac {209 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{3}}{1260 \, a^{5} x^{7}} - \frac {29 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{2}}{180 \, a^{4} x^{8}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{90 \, a^{3} x^{9}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{10 \, a^{2} x^{10}} \]
1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^10/a^10 + 1/6*(b^2*x^2 + 2*a*b*x + a ^2)^(5/2)*b^9/(a^9*x) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^8/(a^10*x^2) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^7/(a^9*x^3) - 1/6*(b^2*x^2 + 2*a* b*x + a^2)^(7/2)*b^6/(a^8*x^4) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^5/( a^7*x^5) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^4/(a^6*x^6) + 209/1260*(b ^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^3/(a^5*x^7) - 29/180*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^2/(a^4*x^8) + 13/90*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b/(a^3*x^ 9) - 1/10*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)/(a^2*x^10)
Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.47 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx=-\frac {b^{10} \mathrm {sgn}\left (b x + a\right )}{1260 \, a^{5}} - \frac {252 \, b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + 1050 \, a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 1800 \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 1575 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 700 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 126 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{1260 \, x^{10}} \]
-1/1260*b^10*sgn(b*x + a)/a^5 - 1/1260*(252*b^5*x^5*sgn(b*x + a) + 1050*a* b^4*x^4*sgn(b*x + a) + 1800*a^2*b^3*x^3*sgn(b*x + a) + 1575*a^3*b^2*x^2*sg n(b*x + a) + 700*a^4*b*x*sgn(b*x + a) + 126*a^5*sgn(b*x + a))/x^10
Time = 10.24 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{10\,x^{10}\,\left (a+b\,x\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^8\,\left (a+b\,x\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,x^6\,\left (a+b\,x\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{9\,x^9\,\left (a+b\,x\right )} \]
- (a^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(10*x^10*(a + b*x)) - (b^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(5*x^5*(a + b*x)) - (10*a^2*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*x^7*(a + b*x)) - (5*a^3*b^2*(a^2 + b^2*x^2 + 2*a*b*x)^ (1/2))/(4*x^8*(a + b*x)) - (5*a*b^4*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*x^ 6*(a + b*x)) - (5*a^4*b*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(9*x^9*(a + b*x))