Integrand size = 26, antiderivative size = 250 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^4 \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2 \left (a+b x^2\right )}+\frac {b^5 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \] Output:
-1/8*a^5*((b*x^2+a)^2)^(1/2)/x^8/(b*x^2+a)-5/6*a^4*b*((b*x^2+a)^2)^(1/2)/x ^6/(b*x^2+a)-5/2*a^3*b^2*((b*x^2+a)^2)^(1/2)/x^4/(b*x^2+a)-5*a^2*b^3*((b*x ^2+a)^2)^(1/2)/x^2/(b*x^2+a)+b^5*x^2*((b*x^2+a)^2)^(1/2)/(2*b*x^2+2*a)+5*a *b^4*((b*x^2+a)^2)^(1/2)*ln(x)/(b*x^2+a)
Leaf count is larger than twice the leaf count of optimal. \(696\) vs. \(2(250)=500\).
Time = 1.02 (sec) , antiderivative size = 696, normalized size of antiderivative = 2.78 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\frac {48 a^6 \sqrt {a^2}+368 a^5 \sqrt {a^2} b x^2+1280 a^4 \sqrt {a^2} b^2 x^4+2880 a^3 \sqrt {a^2} b^3 x^6+2677 \left (a^2\right )^{3/2} b^4 x^8+565 a \sqrt {a^2} b^5 x^{10}-192 \sqrt {a^2} b^6 x^{12}-48 a^6 \sqrt {\left (a+b x^2\right )^2}-320 a^5 b x^2 \sqrt {\left (a+b x^2\right )^2}-960 a^4 b^2 x^4 \sqrt {\left (a+b x^2\right )^2}-1920 a^3 b^3 x^6 \sqrt {\left (a+b x^2\right )^2}-757 a^2 b^4 x^8 \sqrt {\left (a+b x^2\right )^2}+192 a b^5 x^{10} \sqrt {\left (a+b x^2\right )^2}-960 a b^4 x^8 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right ) \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-960 b^4 x^8 \left (\left (a^2\right )^{3/2}+a \sqrt {a^2} b x^2-a^2 \sqrt {\left (a+b x^2\right )^2}\right ) \log \left (x^2\right )+480 \left (a^2\right )^{3/2} b^4 x^8 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+480 a \sqrt {a^2} b^5 x^{10} \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-480 a^2 b^4 x^8 \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+480 \left (a^2\right )^{3/2} b^4 x^8 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+480 a \sqrt {a^2} b^5 x^{10} \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-480 a^2 b^4 x^8 \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )}{384 x^8 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^9,x]
Output:
(48*a^6*Sqrt[a^2] + 368*a^5*Sqrt[a^2]*b*x^2 + 1280*a^4*Sqrt[a^2]*b^2*x^4 + 2880*a^3*Sqrt[a^2]*b^3*x^6 + 2677*(a^2)^(3/2)*b^4*x^8 + 565*a*Sqrt[a^2]*b ^5*x^10 - 192*Sqrt[a^2]*b^6*x^12 - 48*a^6*Sqrt[(a + b*x^2)^2] - 320*a^5*b* x^2*Sqrt[(a + b*x^2)^2] - 960*a^4*b^2*x^4*Sqrt[(a + b*x^2)^2] - 1920*a^3*b ^3*x^6*Sqrt[(a + b*x^2)^2] - 757*a^2*b^4*x^8*Sqrt[(a + b*x^2)^2] + 192*a*b ^5*x^10*Sqrt[(a + b*x^2)^2] - 960*a*b^4*x^8*(a^2 + a*b*x^2 - Sqrt[a^2]*Sqr t[(a + b*x^2)^2])*ArcTanh[(b*x^2)/(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])] - 960 *b^4*x^8*((a^2)^(3/2) + a*Sqrt[a^2]*b*x^2 - a^2*Sqrt[(a + b*x^2)^2])*Log[x ^2] + 480*(a^2)^(3/2)*b^4*x^8*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] + 480*a*Sqrt[a^2]*b^5*x^10*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] - 480*a^2*b^4*x^8*Sqrt[(a + b*x^2)^2]*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x ^2)^2]] + 480*(a^2)^(3/2)*b^4*x^8*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2) ^2]] + 480*a*Sqrt[a^2]*b^5*x^10*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2 ]] - 480*a^2*b^4*x^8*Sqrt[(a + b*x^2)^2]*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]])/(384*x^8*(a^2 + a*b*x^2 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))
Time = 0.40 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.38, 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, 49, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^9}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^9}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{10}}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5}{x^{10}}+\frac {5 b a^4}{x^8}+\frac {10 b^2 a^3}{x^6}+\frac {10 b^3 a^2}{x^4}+\frac {5 b^4 a}{x^2}+b^5\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {a^5}{4 x^8}-\frac {5 a^4 b}{3 x^6}-\frac {5 a^3 b^2}{x^4}-\frac {10 a^2 b^3}{x^2}+5 a b^4 \log \left (x^2\right )+b^5 x^2\right )}{2 \left (a+b x^2\right )}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^9,x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/4*a^5/x^8 - (5*a^4*b)/(3*x^6) - (5*a^ 3*b^2)/x^4 - (10*a^2*b^3)/x^2 + b^5*x^2 + 5*a*b^4*Log[x^2]))/(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}, x] && IGtQ[m, 0] && IGtQ[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 3.
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.28
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (-4 x^{10} b^{5}-20 b^{4} a \ln \left (x^{2}\right ) x^{8}+40 a^{2} x^{6} b^{3}+20 a^{3} x^{4} b^{2}+\frac {20 x^{2} a^{4} b}{3}+a^{5}\right )}{8 x^{8}}\) | \(70\) |
default | \(\frac {{\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (12 x^{10} b^{5}+120 \ln \left (x \right ) x^{8} a \,b^{4}-120 a^{2} x^{6} b^{3}-60 a^{3} x^{4} b^{2}-20 x^{2} a^{4} b -3 a^{5}\right )}{24 x^{8} \left (b \,x^{2}+a \right )^{5}}\) | \(82\) |
risch | \(\frac {b^{5} x^{2} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 b \,x^{2}+2 a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-5 a^{2} x^{6} b^{3}-\frac {5}{2} a^{3} x^{4} b^{2}-\frac {5}{6} x^{2} a^{4} b -\frac {1}{8} a^{5}\right )}{\left (b \,x^{2}+a \right ) x^{8}}+\frac {5 a \,b^{4} \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (x \right )}{b \,x^{2}+a}\) | \(119\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^9,x,method=_RETURNVERBOSE)
Output:
-1/8*csgn(b*x^2+a)*(-4*x^10*b^5-20*b^4*a*ln(x^2)*x^8+40*a^2*x^6*b^3+20*a^3 *x^4*b^2+20/3*x^2*a^4*b+a^5)/x^8
Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\frac {12 \, b^{5} x^{10} + 120 \, a b^{4} x^{8} \log \left (x\right ) - 120 \, a^{2} b^{3} x^{6} - 60 \, a^{3} b^{2} x^{4} - 20 \, a^{4} b x^{2} - 3 \, a^{5}}{24 \, x^{8}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^9,x, algorithm="fricas")
Output:
1/24*(12*b^5*x^10 + 120*a*b^4*x^8*log(x) - 120*a^2*b^3*x^6 - 60*a^3*b^2*x^ 4 - 20*a^4*b*x^2 - 3*a^5)/x^8
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{9}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**9,x)
Output:
Integral(((a + b*x**2)**2)**(5/2)/x**9, x)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\frac {1}{2} \, b^{5} x^{2} + 5 \, a b^{4} \log \left (x\right ) - \frac {5 \, a^{2} b^{3}}{x^{2}} - \frac {5 \, a^{3} b^{2}}{2 \, x^{4}} - \frac {5 \, a^{4} b}{6 \, x^{6}} - \frac {a^{5}}{8 \, x^{8}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^9,x, algorithm="maxima")
Output:
1/2*b^5*x^2 + 5*a*b^4*log(x) - 5*a^2*b^3/x^2 - 5/2*a^3*b^2/x^4 - 5/6*a^4*b /x^6 - 1/8*a^5/x^8
Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\frac {1}{2} \, b^{5} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{2} \, a b^{4} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {125 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 120 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 60 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{24 \, x^{8}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^9,x, algorithm="giac")
Output:
1/2*b^5*x^2*sgn(b*x^2 + a) + 5/2*a*b^4*log(x^2)*sgn(b*x^2 + a) - 1/24*(125 *a*b^4*x^8*sgn(b*x^2 + a) + 120*a^2*b^3*x^6*sgn(b*x^2 + a) + 60*a^3*b^2*x^ 4*sgn(b*x^2 + a) + 20*a^4*b*x^2*sgn(b*x^2 + a) + 3*a^5*sgn(b*x^2 + a))/x^8
Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^9} \,d x \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^9,x)
Output:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^9, x)
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\frac {120 \,\mathrm {log}\left (x \right ) a \,b^{4} x^{8}-3 a^{5}-20 a^{4} b \,x^{2}-60 a^{3} b^{2} x^{4}-120 a^{2} b^{3} x^{6}+12 b^{5} x^{10}}{24 x^{8}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^9,x)
Output:
(120*log(x)*a*b**4*x**8 - 3*a**5 - 20*a**4*b*x**2 - 60*a**3*b**2*x**4 - 12 0*a**2*b**3*x**6 + 12*b**5*x**10)/(24*x**8)