Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{21}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{20 x^{20} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 x^{17} \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^{14} \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 x^{11} \left (a+b x^3\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^8 \left (a+b x^3\right )}-\frac {b^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 x^5 \left (a+b x^3\right )} \]
-1/20*a^5*((b*x^3+a)^2)^(1/2)/x^20/(b*x^3+a)-5/17*a^4*b*((b*x^3+a)^2)^(1/2 )/x^17/(b*x^3+a)-5/7*a^3*b^2*((b*x^3+a)^2)^(1/2)/x^14/(b*x^3+a)-10/11*a^2* b^3*((b*x^3+a)^2)^(1/2)/x^11/(b*x^3+a)-5/8*a*b^4*((b*x^3+a)^2)^(1/2)/x^8/( b*x^3+a)-1/5*b^5*((b*x^3+a)^2)^(1/2)/x^5/(b*x^3+a)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{21}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (2618 a^5+15400 a^4 b x^3+37400 a^3 b^2 x^6+47600 a^2 b^3 x^9+32725 a b^4 x^{12}+10472 b^5 x^{15}\right )}{52360 x^{20} \left (a+b x^3\right )} \]
-1/52360*(Sqrt[(a + b*x^3)^2]*(2618*a^5 + 15400*a^4*b*x^3 + 37400*a^3*b^2* x^6 + 47600*a^2*b^3*x^9 + 32725*a*b^4*x^12 + 10472*b^5*x^15))/(x^20*(a + b *x^3))
Time = 0.24 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.40, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1384, 27, 802, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{21}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^5 \left (b x^3+a\right )^5}{x^{21}}dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^{21}}dx}{a+b x^3}\) |
\(\Big \downarrow \) 802 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a^5}{x^{21}}+\frac {5 b a^4}{x^{18}}+\frac {10 b^2 a^3}{x^{15}}+\frac {10 b^3 a^2}{x^{12}}+\frac {5 b^4 a}{x^9}+\frac {b^5}{x^6}\right )dx}{a+b x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^5}{20 x^{20}}-\frac {5 a^4 b}{17 x^{17}}-\frac {5 a^3 b^2}{7 x^{14}}-\frac {10 a^2 b^3}{11 x^{11}}-\frac {5 a b^4}{8 x^8}-\frac {b^5}{5 x^5}\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}\) |
((-1/20*a^5/x^20 - (5*a^4*b)/(17*x^17) - (5*a^3*b^2)/(7*x^14) - (10*a^2*b^ 3)/(11*x^11) - (5*a*b^4)/(8*x^8) - b^5/(5*x^5))*Sqrt[a^2 + 2*a*b*x^3 + b^2 *x^6])/(a + b*x^3)
3.1.84.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[Exp andIntegrand[(c*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
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)])
Time = 38.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.31
method | result | size |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{20} a^{5}-\frac {5}{17} a^{4} b \,x^{3}-\frac {5}{7} a^{3} b^{2} x^{6}-\frac {10}{11} a^{2} b^{3} x^{9}-\frac {5}{8} a \,b^{4} x^{12}-\frac {1}{5} b^{5} x^{15}\right )}{\left (b \,x^{3}+a \right ) x^{20}}\) | \(79\) |
gosper | \(-\frac {\left (10472 b^{5} x^{15}+32725 a \,b^{4} x^{12}+47600 a^{2} b^{3} x^{9}+37400 a^{3} b^{2} x^{6}+15400 a^{4} b \,x^{3}+2618 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{52360 x^{20} \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (10472 b^{5} x^{15}+32725 a \,b^{4} x^{12}+47600 a^{2} b^{3} x^{9}+37400 a^{3} b^{2} x^{6}+15400 a^{4} b \,x^{3}+2618 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{52360 x^{20} \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
((b*x^3+a)^2)^(1/2)/(b*x^3+a)*(-1/20*a^5-5/17*a^4*b*x^3-5/7*a^3*b^2*x^6-10 /11*a^2*b^3*x^9-5/8*a*b^4*x^12-1/5*b^5*x^15)/x^20
Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{21}} \, dx=-\frac {10472 \, b^{5} x^{15} + 32725 \, a b^{4} x^{12} + 47600 \, a^{2} b^{3} x^{9} + 37400 \, a^{3} b^{2} x^{6} + 15400 \, a^{4} b x^{3} + 2618 \, a^{5}}{52360 \, x^{20}} \]
-1/52360*(10472*b^5*x^15 + 32725*a*b^4*x^12 + 47600*a^2*b^3*x^9 + 37400*a^ 3*b^2*x^6 + 15400*a^4*b*x^3 + 2618*a^5)/x^20
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{21}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{21}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{21}} \, dx=-\frac {10472 \, b^{5} x^{15} + 32725 \, a b^{4} x^{12} + 47600 \, a^{2} b^{3} x^{9} + 37400 \, a^{3} b^{2} x^{6} + 15400 \, a^{4} b x^{3} + 2618 \, a^{5}}{52360 \, x^{20}} \]
-1/52360*(10472*b^5*x^15 + 32725*a*b^4*x^12 + 47600*a^2*b^3*x^9 + 37400*a^ 3*b^2*x^6 + 15400*a^4*b*x^3 + 2618*a^5)/x^20
Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{21}} \, dx=-\frac {10472 \, b^{5} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + 32725 \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + 47600 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 37400 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 15400 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 2618 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{52360 \, x^{20}} \]
-1/52360*(10472*b^5*x^15*sgn(b*x^3 + a) + 32725*a*b^4*x^12*sgn(b*x^3 + a) + 47600*a^2*b^3*x^9*sgn(b*x^3 + a) + 37400*a^3*b^2*x^6*sgn(b*x^3 + a) + 15 400*a^4*b*x^3*sgn(b*x^3 + a) + 2618*a^5*sgn(b*x^3 + a))/x^20
Time = 8.45 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{21}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{20\,x^{20}\,\left (b\,x^3+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{5\,x^5\,\left (b\,x^3+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{8\,x^8\,\left (b\,x^3+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{17\,x^{17}\,\left (b\,x^3+a\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{11\,x^{11}\,\left (b\,x^3+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{7\,x^{14}\,\left (b\,x^3+a\right )} \]
- (a^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(20*x^20*(a + b*x^3)) - (b^5*(a^ 2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(5*x^5*(a + b*x^3)) - (5*a*b^4*(a^2 + b^2* x^6 + 2*a*b*x^3)^(1/2))/(8*x^8*(a + b*x^3)) - (5*a^4*b*(a^2 + b^2*x^6 + 2* a*b*x^3)^(1/2))/(17*x^17*(a + b*x^3)) - (10*a^2*b^3*(a^2 + b^2*x^6 + 2*a*b *x^3)^(1/2))/(11*x^11*(a + b*x^3)) - (5*a^3*b^2*(a^2 + b^2*x^6 + 2*a*b*x^3 )^(1/2))/(7*x^14*(a + b*x^3))