Integrand size = 26, antiderivative size = 84 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {\left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 a x^{10}}+\frac {b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{40 a^2 x^8} \] Output:
-1/10*(b*x^2+a)^3*((b*x^2+a)^2)^(1/2)/a/x^10+1/40*b*(b*x^2+a)^3*((b*x^2+a) ^2)^(1/2)/a^2/x^8
Time = 1.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (4 a^3+15 a^2 b x^2+20 a b^2 x^4+10 b^3 x^6\right )}{40 x^{10} \left (a+b x^2\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^11,x]
Output:
-1/40*(Sqrt[(a + b*x^2)^2]*(4*a^3 + 15*a^2*b*x^2 + 20*a*b^2*x^4 + 10*b^3*x ^6))/(x^10*(a + b*x^2))
Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 27, 243, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^3 \left (b x^2+a\right )^3}{x^{11}}dx}{b^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{x^{11}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{x^{12}}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {b \int \frac {\left (b x^2+a\right )^3}{x^{10}}dx^2}{5 a}-\frac {\left (a+b x^2\right )^4}{5 a x^{10}}\right )}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b \left (a+b x^2\right )^4}{20 a^2 x^8}-\frac {\left (a+b x^2\right )^4}{5 a x^{10}}\right )}{2 \left (a+b x^2\right )}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^11,x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/5*(a + b*x^2)^4/(a*x^10) + (b*(a + b* x^2)^4)/(20*a^2*x^8)))/(2*(a + b*x^2))
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 = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.52
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\frac {5}{2} b^{3} x^{6}+5 b^{2} x^{4} a +\frac {15}{4} a^{2} b \,x^{2}+a^{3}\right )}{10 x^{10}}\) | \(44\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{4} b^{3} x^{6}-\frac {1}{2} b^{2} x^{4} a -\frac {3}{8} a^{2} b \,x^{2}-\frac {1}{10} a^{3}\right )}{\left (b \,x^{2}+a \right ) x^{10}}\) | \(57\) |
gosper | \(-\frac {\left (10 b^{3} x^{6}+20 b^{2} x^{4} a +15 a^{2} b \,x^{2}+4 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{40 x^{10} \left (b \,x^{2}+a \right )^{3}}\) | \(58\) |
default | \(-\frac {\left (10 b^{3} x^{6}+20 b^{2} x^{4} a +15 a^{2} b \,x^{2}+4 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{40 x^{10} \left (b \,x^{2}+a \right )^{3}}\) | \(58\) |
orering | \(-\frac {\left (10 b^{3} x^{6}+20 b^{2} x^{4} a +15 a^{2} b \,x^{2}+4 a^{3}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {3}{2}}}{40 x^{10} \left (b \,x^{2}+a \right )^{3}}\) | \(67\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^11,x,method=_RETURNVERBOSE)
Output:
-1/10*csgn(b*x^2+a)*(5/2*b^3*x^6+5*b^2*x^4*a+15/4*a^2*b*x^2+a^3)/x^10
Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.44 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {10 \, b^{3} x^{6} + 20 \, a b^{2} x^{4} + 15 \, a^{2} b x^{2} + 4 \, a^{3}}{40 \, x^{10}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^11,x, algorithm="fricas")
Output:
-1/40*(10*b^3*x^6 + 20*a*b^2*x^4 + 15*a^2*b*x^2 + 4*a^3)/x^10
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{11}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**11,x)
Output:
Integral(((a + b*x**2)**2)**(3/2)/x**11, x)
Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {b^{3}}{4 \, x^{4}} - \frac {a b^{2}}{2 \, x^{6}} - \frac {3 \, a^{2} b}{8 \, x^{8}} - \frac {a^{3}}{10 \, x^{10}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^11,x, algorithm="maxima")
Output:
-1/4*b^3/x^4 - 1/2*a*b^2/x^6 - 3/8*a^2*b/x^8 - 1/10*a^3/x^10
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {10 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{40 \, x^{10}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^11,x, algorithm="giac")
Output:
-1/40*(10*b^3*x^6*sgn(b*x^2 + a) + 20*a*b^2*x^4*sgn(b*x^2 + a) + 15*a^2*b* x^2*sgn(b*x^2 + a) + 4*a^3*sgn(b*x^2 + a))/x^10
Time = 17.81 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{10\,x^{10}\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^4\,\left (b\,x^2+a\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^6\,\left (b\,x^2+a\right )}-\frac {3\,a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^8\,\left (b\,x^2+a\right )} \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^11,x)
Output:
- (a^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(10*x^10*(a + b*x^2)) - (b^3*(a^ 2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(4*x^4*(a + b*x^2)) - (a*b^2*(a^2 + b^2*x^ 4 + 2*a*b*x^2)^(1/2))/(2*x^6*(a + b*x^2)) - (3*a^2*b*(a^2 + b^2*x^4 + 2*a* b*x^2)^(1/2))/(8*x^8*(a + b*x^2))
Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.44 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=\frac {-10 b^{3} x^{6}-20 a \,b^{2} x^{4}-15 a^{2} b \,x^{2}-4 a^{3}}{40 x^{10}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^11,x)
Output:
( - 4*a**3 - 15*a**2*b*x**2 - 20*a*b**2*x**4 - 10*b**3*x**6)/(40*x**10)