Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 x^{22} \left (a+b x^2\right )}-\frac {a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{20} \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^{18} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^{16} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )} \] Output:
-1/22*a^5*((b*x^2+a)^2)^(1/2)/x^22/(b*x^2+a)-1/4*a^4*b*((b*x^2+a)^2)^(1/2) /x^20/(b*x^2+a)-5/9*a^3*b^2*((b*x^2+a)^2)^(1/2)/x^18/(b*x^2+a)-5/8*a^2*b^3 *((b*x^2+a)^2)^(1/2)/x^16/(b*x^2+a)-5/14*a*b^4*((b*x^2+a)^2)^(1/2)/x^14/(b *x^2+a)-1/12*b^5*((b*x^2+a)^2)^(1/2)/x^12/(b*x^2+a)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (252 a^5+1386 a^4 b x^2+3080 a^3 b^2 x^4+3465 a^2 b^3 x^6+1980 a b^4 x^8+462 b^5 x^{10}\right )}{5544 x^{22} \left (a+b x^2\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^23,x]
Output:
-1/5544*(Sqrt[(a + b*x^2)^2]*(252*a^5 + 1386*a^4*b*x^2 + 3080*a^3*b^2*x^4 + 3465*a^2*b^3*x^6 + 1980*a*b^4*x^8 + 462*b^5*x^10))/(x^22*(a + b*x^2))
Time = 0.41 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.41, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^{23}}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{23}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{24}}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5}{x^{24}}+\frac {5 b a^4}{x^{22}}+\frac {10 b^2 a^3}{x^{20}}+\frac {10 b^3 a^2}{x^{18}}+\frac {5 b^4 a}{x^{16}}+\frac {b^5}{x^{14}}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^5}{11 x^{22}}-\frac {a^4 b}{2 x^{20}}-\frac {10 a^3 b^2}{9 x^{18}}-\frac {5 a^2 b^3}{4 x^{16}}-\frac {5 a b^4}{7 x^{14}}-\frac {b^5}{6 x^{12}}\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^23,x]
Output:
((-1/11*a^5/x^22 - (a^4*b)/(2*x^20) - (10*a^3*b^2)/(9*x^18) - (5*a^2*b^3)/ (4*x^16) - (5*a*b^4)/(7*x^14) - b^5/(6*x^12))*Sqrt[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 = 2.71 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.26
method | result | size |
pseudoelliptic | \(-\frac {\left (\frac {11}{6} x^{10} b^{5}+\frac {55}{7} a \,x^{8} b^{4}+\frac {55}{4} a^{2} x^{6} b^{3}+\frac {110}{9} a^{3} x^{4} b^{2}+\frac {11}{2} x^{2} a^{4} b +a^{5}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{22 x^{22}}\) | \(66\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{22} a^{5}-\frac {1}{4} x^{2} a^{4} b -\frac {5}{9} a^{3} x^{4} b^{2}-\frac {5}{8} a^{2} x^{6} b^{3}-\frac {5}{14} a \,x^{8} b^{4}-\frac {1}{12} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{22}}\) | \(79\) |
gosper | \(-\frac {\left (462 x^{10} b^{5}+1980 a \,x^{8} b^{4}+3465 a^{2} x^{6} b^{3}+3080 a^{3} x^{4} b^{2}+1386 x^{2} a^{4} b +252 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{5544 x^{22} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (462 x^{10} b^{5}+1980 a \,x^{8} b^{4}+3465 a^{2} x^{6} b^{3}+3080 a^{3} x^{4} b^{2}+1386 x^{2} a^{4} b +252 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{5544 x^{22} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
orering | \(-\frac {\left (462 x^{10} b^{5}+1980 a \,x^{8} b^{4}+3465 a^{2} x^{6} b^{3}+3080 a^{3} x^{4} b^{2}+1386 x^{2} a^{4} b +252 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{5544 x^{22} \left (b \,x^{2}+a \right )^{5}}\) | \(89\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^23,x,method=_RETURNVERBOSE)
Output:
-1/22*(11/6*x^10*b^5+55/7*a*x^8*b^4+55/4*a^2*x^6*b^3+110/9*a^3*x^4*b^2+11/ 2*x^2*a^4*b+a^5)*csgn(b*x^2+a)/x^22
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {462 \, b^{5} x^{10} + 1980 \, a b^{4} x^{8} + 3465 \, a^{2} b^{3} x^{6} + 3080 \, a^{3} b^{2} x^{4} + 1386 \, a^{4} b x^{2} + 252 \, a^{5}}{5544 \, x^{22}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^23,x, algorithm="fricas")
Output:
-1/5544*(462*b^5*x^10 + 1980*a*b^4*x^8 + 3465*a^2*b^3*x^6 + 3080*a^3*b^2*x ^4 + 1386*a^4*b*x^2 + 252*a^5)/x^22
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{23}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**23,x)
Output:
Integral(((a + b*x**2)**2)**(5/2)/x**23, x)
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {b^{5}}{12 \, x^{12}} - \frac {5 \, a b^{4}}{14 \, x^{14}} - \frac {5 \, a^{2} b^{3}}{8 \, x^{16}} - \frac {5 \, a^{3} b^{2}}{9 \, x^{18}} - \frac {a^{4} b}{4 \, x^{20}} - \frac {a^{5}}{22 \, x^{22}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^23,x, algorithm="maxima")
Output:
-1/12*b^5/x^12 - 5/14*a*b^4/x^14 - 5/8*a^2*b^3/x^16 - 5/9*a^3*b^2/x^18 - 1 /4*a^4*b/x^20 - 1/22*a^5/x^22
Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {462 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 1980 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 3465 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 3080 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 1386 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 252 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{5544 \, x^{22}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^23,x, algorithm="giac")
Output:
-1/5544*(462*b^5*x^10*sgn(b*x^2 + a) + 1980*a*b^4*x^8*sgn(b*x^2 + a) + 346 5*a^2*b^3*x^6*sgn(b*x^2 + a) + 3080*a^3*b^2*x^4*sgn(b*x^2 + a) + 1386*a^4* b*x^2*sgn(b*x^2 + a) + 252*a^5*sgn(b*x^2 + a))/x^22
Time = 18.06 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{22\,x^{22}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{12\,x^{12}\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^{20}\,\left (b\,x^2+a\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^{16}\,\left (b\,x^2+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^{18}\,\left (b\,x^2+a\right )} \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^23,x)
Output:
- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(22*x^22*(a + b*x^2)) - (b^5*(a^ 2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(12*x^12*(a + b*x^2)) - (5*a*b^4*(a^2 + b^ 2*x^4 + 2*a*b*x^2)^(1/2))/(14*x^14*(a + b*x^2)) - (a^4*b*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(4*x^20*(a + b*x^2)) - (5*a^2*b^3*(a^2 + b^2*x^4 + 2*a*b *x^2)^(1/2))/(8*x^16*(a + b*x^2)) - (5*a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2) ^(1/2))/(9*x^18*(a + b*x^2))
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=\frac {-462 b^{5} x^{10}-1980 a \,b^{4} x^{8}-3465 a^{2} b^{3} x^{6}-3080 a^{3} b^{2} x^{4}-1386 a^{4} b \,x^{2}-252 a^{5}}{5544 x^{22}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^23,x)
Output:
( - 252*a**5 - 1386*a**4*b*x**2 - 3080*a**3*b**2*x**4 - 3465*a**2*b**3*x** 6 - 1980*a*b**4*x**8 - 462*b**5*x**10)/(5544*x**22)