Integrand size = 26, antiderivative size = 128 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {\left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 a x^{16}}+\frac {b \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{56 a^2 x^{14}}-\frac {b^2 \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{336 a^3 x^{12}} \]
-1/16*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/a/x^16+1/56*b*(b*x^2+a)^5*((b*x^2+a) ^2)^(1/2)/a^2/x^14-1/336*b^2*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/a^3/x^12
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.65 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (21 a^5+120 a^4 b x^2+280 a^3 b^2 x^4+336 a^2 b^3 x^6+210 a b^4 x^8+56 b^5 x^{10}\right )}{336 x^{16} \left (a+b x^2\right )} \]
-1/336*(Sqrt[(a + b*x^2)^2]*(21*a^5 + 120*a^4*b*x^2 + 280*a^3*b^2*x^4 + 33 6*a^2*b^3*x^6 + 210*a*b^4*x^8 + 56*b^5*x^10))/(x^16*(a + b*x^2))
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, 243, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^{17}}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{17}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{18}}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {b \int \frac {\left (b x^2+a\right )^5}{x^{16}}dx^2}{4 a}-\frac {\left (a+b x^2\right )^6}{8 a x^{16}}\right )}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {b \left (-\frac {b \int \frac {\left (b x^2+a\right )^5}{x^{14}}dx^2}{7 a}-\frac {\left (a+b x^2\right )^6}{7 a x^{14}}\right )}{4 a}-\frac {\left (a+b x^2\right )^6}{8 a x^{16}}\right )}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {b \left (\frac {b \left (a+b x^2\right )^6}{42 a^2 x^{12}}-\frac {\left (a+b x^2\right )^6}{7 a x^{14}}\right )}{4 a}-\frac {\left (a+b x^2\right )^6}{8 a x^{16}}\right )}{2 \left (a+b x^2\right )}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/8*(a + b*x^2)^6/(a*x^16) - (b*(-1/7*( a + b*x^2)^6/(a*x^14) + (b*(a + b*x^2)^6)/(42*a^2*x^12)))/(4*a)))/(2*(a + b*x^2))
3.6.100.3.1 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_)^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 2.
Time = 0.74 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52
method | result | size |
pseudoelliptic | \(-\frac {\left (\frac {8}{3} x^{10} b^{5}+10 a \,x^{8} b^{4}+16 a^{2} x^{6} b^{3}+\frac {40}{3} a^{3} x^{4} b^{2}+\frac {40}{7} x^{2} a^{4} b +a^{5}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{16 x^{16}}\) | \(66\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{16} a^{5}-\frac {5}{14} x^{2} a^{4} b -\frac {5}{6} a^{3} x^{4} b^{2}-a^{2} x^{6} b^{3}-\frac {5}{8} a \,x^{8} b^{4}-\frac {1}{6} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{16}}\) | \(79\) |
gosper | \(-\frac {\left (56 x^{10} b^{5}+210 a \,x^{8} b^{4}+336 a^{2} x^{6} b^{3}+280 a^{3} x^{4} b^{2}+120 x^{2} a^{4} b +21 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{336 x^{16} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (56 x^{10} b^{5}+210 a \,x^{8} b^{4}+336 a^{2} x^{6} b^{3}+280 a^{3} x^{4} b^{2}+120 x^{2} a^{4} b +21 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{336 x^{16} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
-1/16*(8/3*x^10*b^5+10*a*x^8*b^4+16*a^2*x^6*b^3+40/3*a^3*x^4*b^2+40/7*x^2* a^4*b+a^5)*csgn(b*x^2+a)/x^16
Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.46 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {56 \, b^{5} x^{10} + 210 \, a b^{4} x^{8} + 336 \, a^{2} b^{3} x^{6} + 280 \, a^{3} b^{2} x^{4} + 120 \, a^{4} b x^{2} + 21 \, a^{5}}{336 \, x^{16}} \]
-1/336*(56*b^5*x^10 + 210*a*b^4*x^8 + 336*a^2*b^3*x^6 + 280*a^3*b^2*x^4 + 120*a^4*b*x^2 + 21*a^5)/x^16
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{17}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.45 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {b^{5}}{6 \, x^{6}} - \frac {5 \, a b^{4}}{8 \, x^{8}} - \frac {a^{2} b^{3}}{x^{10}} - \frac {5 \, a^{3} b^{2}}{6 \, x^{12}} - \frac {5 \, a^{4} b}{14 \, x^{14}} - \frac {a^{5}}{16 \, x^{16}} \]
-1/6*b^5/x^6 - 5/8*a*b^4/x^8 - a^2*b^3/x^10 - 5/6*a^3*b^2/x^12 - 5/14*a^4* b/x^14 - 1/16*a^5/x^16
Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {56 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 210 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 336 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 280 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 120 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 21 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{336 \, x^{16}} \]
-1/336*(56*b^5*x^10*sgn(b*x^2 + a) + 210*a*b^4*x^8*sgn(b*x^2 + a) + 336*a^ 2*b^3*x^6*sgn(b*x^2 + a) + 280*a^3*b^2*x^4*sgn(b*x^2 + a) + 120*a^4*b*x^2* sgn(b*x^2 + a) + 21*a^5*sgn(b*x^2 + a))/x^16
Time = 13.60 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{16\,x^{16}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{6\,x^6\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^8\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x^{10}\,\left (b\,x^2+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{6\,x^{12}\,\left (b\,x^2+a\right )} \]
- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(16*x^16*(a + b*x^2)) - (b^5*(a^ 2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(6*x^6*(a + b*x^2)) - (5*a*b^4*(a^2 + b^2* x^4 + 2*a*b*x^2)^(1/2))/(8*x^8*(a + b*x^2)) - (5*a^4*b*(a^2 + b^2*x^4 + 2* a*b*x^2)^(1/2))/(14*x^14*(a + b*x^2)) - (a^2*b^3*(a^2 + b^2*x^4 + 2*a*b*x^ 2)^(1/2))/(x^10*(a + b*x^2)) - (5*a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2 ))/(6*x^12*(a + b*x^2))