Integrand size = 26, antiderivative size = 128 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{25}} \, dx=-\frac {\left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{24 a x^{24}}+\frac {b \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{84 a^2 x^{21}}-\frac {b^2 \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{504 a^3 x^{18}} \]
-1/24*(b*x^3+a)^5*((b*x^3+a)^2)^(1/2)/a/x^24+1/84*b*(b*x^3+a)^5*((b*x^3+a) ^2)^(1/2)/a^2/x^21-1/504*b^2*(b*x^3+a)^5*((b*x^3+a)^2)^(1/2)/a^3/x^18
Time = 1.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.65 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{25}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (21 a^5+120 a^4 b x^3+280 a^3 b^2 x^6+336 a^2 b^3 x^9+210 a b^4 x^{12}+56 b^5 x^{15}\right )}{504 x^{24} \left (a+b x^3\right )} \]
-1/504*(Sqrt[(a + b*x^3)^2]*(21*a^5 + 120*a^4*b*x^3 + 280*a^3*b^2*x^6 + 33 6*a^2*b^3*x^9 + 210*a*b^4*x^12 + 56*b^5*x^15))/(x^24*(a + b*x^3))
Time = 0.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.80, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1384, 27, 798, 55, 55, 48}
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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{25}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^5 \left (b x^3+a\right )^5}{x^{25}}dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^{25}}dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^{27}}dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {b \int \frac {\left (b x^3+a\right )^5}{x^{24}}dx^3}{4 a}-\frac {\left (a+b x^3\right )^6}{8 a x^{24}}\right )}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {b \left (-\frac {b \int \frac {\left (b x^3+a\right )^5}{x^{21}}dx^3}{7 a}-\frac {\left (a+b x^3\right )^6}{7 a x^{21}}\right )}{4 a}-\frac {\left (a+b x^3\right )^6}{8 a x^{24}}\right )}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {b \left (\frac {b \left (a+b x^3\right )^6}{42 a^2 x^{18}}-\frac {\left (a+b x^3\right )^6}{7 a x^{21}}\right )}{4 a}-\frac {\left (a+b x^3\right )^6}{8 a x^{24}}\right )}{3 \left (a+b x^3\right )}\) |
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-1/8*(a + b*x^3)^6/(a*x^24) - (b*(-1/7*( a + b*x^3)^6/(a*x^21) + (b*(a + b*x^3)^6)/(42*a^2*x^18)))/(4*a)))/(3*(a + b*x^3))
3.1.88.3.1 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] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
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 2.
Time = 4.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (\frac {8}{3} b^{5} x^{15}+10 a \,b^{4} x^{12}+16 a^{2} b^{3} x^{9}+\frac {40}{3} a^{3} b^{2} x^{6}+\frac {40}{7} a^{4} b \,x^{3}+a^{5}\right )}{24 x^{24}}\) | \(66\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{24} a^{5}-\frac {5}{9} a^{3} b^{2} x^{6}-\frac {2}{3} a^{2} b^{3} x^{9}-\frac {5}{12} a \,b^{4} x^{12}-\frac {1}{9} b^{5} x^{15}-\frac {5}{21} a^{4} b \,x^{3}\right )}{\left (b \,x^{3}+a \right ) x^{24}}\) | \(79\) |
gosper | \(-\frac {\left (56 b^{5} x^{15}+210 a \,b^{4} x^{12}+336 a^{2} b^{3} x^{9}+280 a^{3} b^{2} x^{6}+120 a^{4} b \,x^{3}+21 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{504 x^{24} \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (56 b^{5} x^{15}+210 a \,b^{4} x^{12}+336 a^{2} b^{3} x^{9}+280 a^{3} b^{2} x^{6}+120 a^{4} b \,x^{3}+21 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{504 x^{24} \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
-1/24*csgn(b*x^3+a)*(8/3*b^5*x^15+10*a*b^4*x^12+16*a^2*b^3*x^9+40/3*a^3*b^ 2*x^6+40/7*a^4*b*x^3+a^5)/x^24
Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.46 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{25}} \, dx=-\frac {56 \, b^{5} x^{15} + 210 \, a b^{4} x^{12} + 336 \, a^{2} b^{3} x^{9} + 280 \, a^{3} b^{2} x^{6} + 120 \, a^{4} b x^{3} + 21 \, a^{5}}{504 \, x^{24}} \]
-1/504*(56*b^5*x^15 + 210*a*b^4*x^12 + 336*a^2*b^3*x^9 + 280*a^3*b^2*x^6 + 120*a^4*b*x^3 + 21*a^5)/x^24
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{25}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{25}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (89) = 178\).
Time = 0.21 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.12 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{25}} \, dx=\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{8}}{18 \, a^{8}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{7}}{18 \, a^{7} x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{6}}{18 \, a^{8} x^{6}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{5}}{18 \, a^{7} x^{9}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{4}}{18 \, a^{6} x^{12}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{3}}{18 \, a^{5} x^{15}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{2}}{18 \, a^{4} x^{18}} + \frac {3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b}{56 \, a^{3} x^{21}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}}}{24 \, a^{2} x^{24}} \]
1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^8/a^8 + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^7/(a^7*x^3) - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^6/( a^8*x^6) + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^5/(a^7*x^9) - 1/18*(b^ 2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^4/(a^6*x^12) + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^3/(a^5*x^15) - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^2/( a^4*x^18) + 3/56*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b/(a^3*x^21) - 1/24*(b^ 2*x^6 + 2*a*b*x^3 + a^2)^(7/2)/(a^2*x^24)
Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{25}} \, dx=-\frac {56 \, b^{5} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + 210 \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + 336 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 280 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 120 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 21 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{504 \, x^{24}} \]
-1/504*(56*b^5*x^15*sgn(b*x^3 + a) + 210*a*b^4*x^12*sgn(b*x^3 + a) + 336*a ^2*b^3*x^9*sgn(b*x^3 + a) + 280*a^3*b^2*x^6*sgn(b*x^3 + a) + 120*a^4*b*x^3 *sgn(b*x^3 + a) + 21*a^5*sgn(b*x^3 + a))/x^24
Time = 8.35 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{25}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{24\,x^{24}\,\left (b\,x^3+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{9\,x^9\,\left (b\,x^3+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{12\,x^{12}\,\left (b\,x^3+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{21\,x^{21}\,\left (b\,x^3+a\right )}-\frac {2\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,x^{15}\,\left (b\,x^3+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{9\,x^{18}\,\left (b\,x^3+a\right )} \]
- (a^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(24*x^24*(a + b*x^3)) - (b^5*(a^ 2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(9*x^9*(a + b*x^3)) - (5*a*b^4*(a^2 + b^2* x^6 + 2*a*b*x^3)^(1/2))/(12*x^12*(a + b*x^3)) - (5*a^4*b*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(21*x^21*(a + b*x^3)) - (2*a^2*b^3*(a^2 + b^2*x^6 + 2*a* b*x^3)^(1/2))/(3*x^15*(a + b*x^3)) - (5*a^3*b^2*(a^2 + b^2*x^6 + 2*a*b*x^3 )^(1/2))/(9*x^18*(a + b*x^3))