Integrand size = 26, antiderivative size = 167 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{12} \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )} \] Output:
-1/14*a^3*((b*x^2+a)^2)^(1/2)/x^14/(b*x^2+a)-1/4*a^2*b*((b*x^2+a)^2)^(1/2) /x^12/(b*x^2+a)-3/10*a*b^2*((b*x^2+a)^2)^(1/2)/x^10/(b*x^2+a)-1/8*b^3*((b* x^2+a)^2)^(1/2)/x^8/(b*x^2+a)
Time = 1.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.37 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (20 a^3+70 a^2 b x^2+84 a b^2 x^4+35 b^3 x^6\right )}{280 x^{14} \left (a+b x^2\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^15,x]
Output:
-1/280*(Sqrt[(a + b*x^2)^2]*(20*a^3 + 70*a^2*b*x^2 + 84*a*b^2*x^4 + 35*b^3 *x^6))/(x^14*(a + b*x^2))
Time = 0.38 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.47, 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, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^3 \left (b x^2+a\right )^3}{x^{15}}dx}{b^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{x^{15}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{x^{16}}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^3}{x^{16}}+\frac {3 b a^2}{x^{14}}+\frac {3 b^2 a}{x^{12}}+\frac {b^3}{x^{10}}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^3}{7 x^{14}}-\frac {a^2 b}{2 x^{12}}-\frac {3 a b^2}{5 x^{10}}-\frac {b^3}{4 x^8}\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^15,x]
Output:
((-1/7*a^3/x^14 - (a^2*b)/(2*x^12) - (3*a*b^2)/(5*x^10) - b^3/(4*x^8))*Sqr t[a^2 + 2*a*b*x^2 + b^2*x^4])/(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] :> 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[(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 = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.28
method | result | size |
pseudoelliptic | \(-\frac {\left (35 b^{3} x^{6}+84 b^{2} x^{4} a +70 a^{2} b \,x^{2}+20 a^{3}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{280 x^{14}}\) | \(46\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{14} a^{3}-\frac {1}{4} a^{2} b \,x^{2}-\frac {3}{10} b^{2} x^{4} a -\frac {1}{8} b^{3} x^{6}\right )}{\left (b \,x^{2}+a \right ) x^{14}}\) | \(57\) |
gosper | \(-\frac {\left (35 b^{3} x^{6}+84 b^{2} x^{4} a +70 a^{2} b \,x^{2}+20 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{280 x^{14} \left (b \,x^{2}+a \right )^{3}}\) | \(58\) |
default | \(-\frac {\left (35 b^{3} x^{6}+84 b^{2} x^{4} a +70 a^{2} b \,x^{2}+20 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{280 x^{14} \left (b \,x^{2}+a \right )^{3}}\) | \(58\) |
orering | \(-\frac {\left (35 b^{3} x^{6}+84 b^{2} x^{4} a +70 a^{2} b \,x^{2}+20 a^{3}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {3}{2}}}{280 x^{14} \left (b \,x^{2}+a \right )^{3}}\) | \(67\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^15,x,method=_RETURNVERBOSE)
Output:
-1/280*(35*b^3*x^6+84*a*b^2*x^4+70*a^2*b*x^2+20*a^3)*csgn(b*x^2+a)/x^14
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {35 \, b^{3} x^{6} + 84 \, a b^{2} x^{4} + 70 \, a^{2} b x^{2} + 20 \, a^{3}}{280 \, x^{14}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^15,x, algorithm="fricas")
Output:
-1/280*(35*b^3*x^6 + 84*a*b^2*x^4 + 70*a^2*b*x^2 + 20*a^3)/x^14
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{15}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**15,x)
Output:
Integral(((a + b*x**2)**2)**(3/2)/x**15, x)
Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.21 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {b^{3}}{8 \, x^{8}} - \frac {3 \, a b^{2}}{10 \, x^{10}} - \frac {a^{2} b}{4 \, x^{12}} - \frac {a^{3}}{14 \, x^{14}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^15,x, algorithm="maxima")
Output:
-1/8*b^3/x^8 - 3/10*a*b^2/x^10 - 1/4*a^2*b/x^12 - 1/14*a^3/x^14
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.41 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {35 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 84 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 70 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{280 \, x^{14}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^15,x, algorithm="giac")
Output:
-1/280*(35*b^3*x^6*sgn(b*x^2 + a) + 84*a*b^2*x^4*sgn(b*x^2 + a) + 70*a^2*b *x^2*sgn(b*x^2 + a) + 20*a^3*sgn(b*x^2 + a))/x^14
Time = 17.79 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^8\,\left (b\,x^2+a\right )}-\frac {3\,a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{10\,x^{10}\,\left (b\,x^2+a\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^{12}\,\left (b\,x^2+a\right )} \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^15,x)
Output:
- (a^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(14*x^14*(a + b*x^2)) - (b^3*(a^ 2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(8*x^8*(a + b*x^2)) - (3*a*b^2*(a^2 + b^2* x^4 + 2*a*b*x^2)^(1/2))/(10*x^10*(a + b*x^2)) - (a^2*b*(a^2 + b^2*x^4 + 2* a*b*x^2)^(1/2))/(4*x^12*(a + b*x^2))
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=\frac {-35 b^{3} x^{6}-84 a \,b^{2} x^{4}-70 a^{2} b \,x^{2}-20 a^{3}}{280 x^{14}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^15,x)
Output:
( - 20*a**3 - 70*a**2*b*x**2 - 84*a*b**2*x**4 - 35*b**3*x**6)/(280*x**14)