Integrand size = 24, antiderivative size = 231 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{11 x^{11} (a+b x)}-\frac {a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^{10} (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^8 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)} \]
-1/11*a^5*((b*x+a)^2)^(1/2)/x^11/(b*x+a)-1/2*a^4*b*((b*x+a)^2)^(1/2)/x^10/ (b*x+a)-10/9*a^3*b^2*((b*x+a)^2)^(1/2)/x^9/(b*x+a)-5/4*a^2*b^3*((b*x+a)^2) ^(1/2)/x^8/(b*x+a)-5/7*a*b^4*((b*x+a)^2)^(1/2)/x^7/(b*x+a)-1/6*b^5*((b*x+a )^2)^(1/2)/x^6/(b*x+a)
Time = 0.66 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=-\frac {\sqrt {(a+b x)^2} \left (252 a^5+1386 a^4 b x+3080 a^3 b^2 x^2+3465 a^2 b^3 x^3+1980 a b^4 x^4+462 b^5 x^5\right )}{2772 x^{11} (a+b x)} \]
-1/2772*(Sqrt[(a + b*x)^2]*(252*a^5 + 1386*a^4*b*x + 3080*a^3*b^2*x^2 + 34 65*a^2*b^3*x^3 + 1980*a*b^4*x^4 + 462*b^5*x^5))/(x^11*(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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5}{x^{12}}dx}{b^5 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5}{x^{12}}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^5}{x^{12}}+\frac {5 b a^4}{x^{11}}+\frac {10 b^2 a^3}{x^{10}}+\frac {10 b^3 a^2}{x^9}+\frac {5 b^4 a}{x^8}+\frac {b^5}{x^7}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^5}{11 x^{11}}-\frac {a^4 b}{2 x^{10}}-\frac {10 a^3 b^2}{9 x^9}-\frac {5 a^2 b^3}{4 x^8}-\frac {5 a b^4}{7 x^7}-\frac {b^5}{6 x^6}\right ) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}\) |
((-1/11*a^5/x^11 - (a^4*b)/(2*x^10) - (10*a^3*b^2)/(9*x^9) - (5*a^2*b^3)/( 4*x^8) - (5*a*b^4)/(7*x^7) - b^5/(6*x^6))*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/( a + b*x)
3.2.80.3.1 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 = 2.59 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.32
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {1}{6} b^{5} x^{5}-\frac {5}{7} a \,b^{4} x^{4}-\frac {5}{4} a^{2} b^{3} x^{3}-\frac {10}{9} a^{3} b^{2} x^{2}-\frac {1}{2} a^{4} b x -\frac {1}{11} a^{5}\right )}{\left (b x +a \right ) x^{11}}\) | \(73\) |
gosper | \(-\frac {\left (462 b^{5} x^{5}+1980 a \,b^{4} x^{4}+3465 a^{2} b^{3} x^{3}+3080 a^{3} b^{2} x^{2}+1386 a^{4} b x +252 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{2772 x^{11} \left (b x +a \right )^{5}}\) | \(74\) |
default | \(-\frac {\left (462 b^{5} x^{5}+1980 a \,b^{4} x^{4}+3465 a^{2} b^{3} x^{3}+3080 a^{3} b^{2} x^{2}+1386 a^{4} b x +252 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{2772 x^{11} \left (b x +a \right )^{5}}\) | \(74\) |
((b*x+a)^2)^(1/2)/(b*x+a)/x^11*(-1/6*b^5*x^5-5/7*a*b^4*x^4-5/4*a^2*b^3*x^3 -10/9*a^3*b^2*x^2-1/2*a^4*b*x-1/11*a^5)
Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.25 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=-\frac {462 \, b^{5} x^{5} + 1980 \, a b^{4} x^{4} + 3465 \, a^{2} b^{3} x^{3} + 3080 \, a^{3} b^{2} x^{2} + 1386 \, a^{4} b x + 252 \, a^{5}}{2772 \, x^{11}} \]
-1/2772*(462*b^5*x^5 + 1980*a*b^4*x^4 + 3465*a^2*b^3*x^3 + 3080*a^3*b^2*x^ 2 + 1386*a^4*b*x + 252*a^5)/x^11
\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{12}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (153) = 306\).
Time = 0.21 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{11}}{6 \, a^{11}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{10}}{6 \, a^{10} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{9}}{6 \, a^{11} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{8}}{6 \, a^{10} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{7}}{6 \, a^{9} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{6}}{6 \, a^{8} x^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{5}}{6 \, a^{7} x^{6}} - \frac {461 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{4}}{2772 \, a^{6} x^{7}} + \frac {65 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{3}}{396 \, a^{5} x^{8}} - \frac {31 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{2}}{198 \, a^{4} x^{9}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{22 \, a^{3} x^{10}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{11 \, a^{2} x^{11}} \]
-1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^11/a^11 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^10/(a^10*x) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^9/(a^11*x ^2) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^8/(a^10*x^3) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^7/(a^9*x^4) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b ^6/(a^8*x^5) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^5/(a^7*x^6) - 461/277 2*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^4/(a^6*x^7) + 65/396*(b^2*x^2 + 2*a*b* x + a^2)^(7/2)*b^3/(a^5*x^8) - 31/198*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^2/ (a^4*x^9) + 3/22*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b/(a^3*x^10) - 1/11*(b^2* x^2 + 2*a*b*x + a^2)^(7/2)/(a^2*x^11)
Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.47 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=\frac {b^{11} \mathrm {sgn}\left (b x + a\right )}{2772 \, a^{6}} - \frac {462 \, b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + 1980 \, a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 3465 \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 3080 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 1386 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 252 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{2772 \, x^{11}} \]
1/2772*b^11*sgn(b*x + a)/a^6 - 1/2772*(462*b^5*x^5*sgn(b*x + a) + 1980*a*b ^4*x^4*sgn(b*x + a) + 3465*a^2*b^3*x^3*sgn(b*x + a) + 3080*a^3*b^2*x^2*sgn (b*x + a) + 1386*a^4*b*x*sgn(b*x + a) + 252*a^5*sgn(b*x + a))/x^11
Time = 10.14 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{11\,x^{11}\,\left (a+b\,x\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,x^6\,\left (a+b\,x\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^8\,\left (a+b\,x\right )}-\frac {10\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{9\,x^9\,\left (a+b\,x\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {a^4\,b\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,x^{10}\,\left (a+b\,x\right )} \]
- (a^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(11*x^11*(a + b*x)) - (b^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*x^6*(a + b*x)) - (5*a^2*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*x^8*(a + b*x)) - (10*a^3*b^2*(a^2 + b^2*x^2 + 2*a*b*x)^ (1/2))/(9*x^9*(a + b*x)) - (5*a*b^4*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*x^ 7*(a + b*x)) - (a^4*b*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(2*x^10*(a + b*x))