Integrand size = 26, antiderivative size = 164 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^5} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^3 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \]
-1/4*a^3*((b*x^2+a)^2)^(1/2)/x^4/(b*x^2+a)-3/2*a^2*b*((b*x^2+a)^2)^(1/2)/x ^2/(b*x^2+a)+1/2*b^3*x^2*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+3*a*b^2*ln(x)*((b*x ^2+a)^2)^(1/2)/(b*x^2+a)
Leaf count is larger than twice the leaf count of optimal. \(612\) vs. \(2(164)=328\).
Time = 0.81 (sec) , antiderivative size = 612, normalized size of antiderivative = 3.73 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^5} \, dx=\frac {4 a^4 \sqrt {a^2}+28 a^3 \sqrt {a^2} b x^2+35 \left (a^2\right )^{3/2} b^2 x^4+3 a \sqrt {a^2} b^3 x^6-8 \sqrt {a^2} b^4 x^8-4 a^4 \sqrt {\left (a+b x^2\right )^2}-24 a^3 b x^2 \sqrt {\left (a+b x^2\right )^2}-11 a^2 b^2 x^4 \sqrt {\left (a+b x^2\right )^2}+8 a b^3 x^6 \sqrt {\left (a+b x^2\right )^2}-24 a b^2 x^4 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right ) \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-24 b^2 x^4 \left (\left (a^2\right )^{3/2}+a \sqrt {a^2} b x^2-a^2 \sqrt {\left (a+b x^2\right )^2}\right ) \log \left (x^2\right )+12 \left (a^2\right )^{3/2} b^2 x^4 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+12 a \sqrt {a^2} b^3 x^6 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-12 a^2 b^2 x^4 \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+12 \left (a^2\right )^{3/2} b^2 x^4 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+12 a \sqrt {a^2} b^3 x^6 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-12 a^2 b^2 x^4 \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )}{16 x^4 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]
(4*a^4*Sqrt[a^2] + 28*a^3*Sqrt[a^2]*b*x^2 + 35*(a^2)^(3/2)*b^2*x^4 + 3*a*S qrt[a^2]*b^3*x^6 - 8*Sqrt[a^2]*b^4*x^8 - 4*a^4*Sqrt[(a + b*x^2)^2] - 24*a^ 3*b*x^2*Sqrt[(a + b*x^2)^2] - 11*a^2*b^2*x^4*Sqrt[(a + b*x^2)^2] + 8*a*b^3 *x^6*Sqrt[(a + b*x^2)^2] - 24*a*b^2*x^4*(a^2 + a*b*x^2 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2])*ArcTanh[(b*x^2)/(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])] - 24*b^2* x^4*((a^2)^(3/2) + a*Sqrt[a^2]*b*x^2 - a^2*Sqrt[(a + b*x^2)^2])*Log[x^2] + 12*(a^2)^(3/2)*b^2*x^4*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] + 12* a*Sqrt[a^2]*b^3*x^6*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] - 12*a^2* b^2*x^4*Sqrt[(a + b*x^2)^2]*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] + 12*(a^2)^(3/2)*b^2*x^4*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]] + 12* a*Sqrt[a^2]*b^3*x^6*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]] - 12*a^2* b^2*x^4*Sqrt[(a + b*x^2)^2]*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]])/ (16*x^4*(a^2 + a*b*x^2 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.44, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^5} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^3 \left (b x^2+a\right )^3}{x^5}dx}{b^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{x^5}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{x^6}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^3}{x^6}+\frac {3 b a^2}{x^4}+\frac {3 b^2 a}{x^2}+b^3\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {a^3}{2 x^4}-\frac {3 a^2 b}{x^2}+3 a b^2 \log \left (x^2\right )+b^3 x^2\right )}{2 \left (a+b x^2\right )}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/2*a^3/x^4 - (3*a^2*b)/x^2 + b^3*x^2 + 3*a*b^2*Log[x^2]))/(2*(a + b*x^2))
3.6.65.3.1 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] :> 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 = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.29
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (-2 b^{3} x^{6}-6 b^{2} a \ln \left (x^{2}\right ) x^{4}+6 a^{2} b \,x^{2}+a^{3}\right )}{4 x^{4}}\) | \(48\) |
default | \(\frac {{\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}} \left (2 b^{3} x^{6}+12 \ln \left (x \right ) x^{4} a \,b^{2}-6 a^{2} b \,x^{2}-a^{3}\right )}{4 \left (b \,x^{2}+a \right )^{3} x^{4}}\) | \(60\) |
risch | \(\frac {b^{3} x^{2} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 b \,x^{2}+2 a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {3}{2} a^{2} b \,x^{2}-\frac {1}{4} a^{3}\right )}{\left (b \,x^{2}+a \right ) x^{4}}+\frac {3 a \,b^{2} \ln \left (x \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\) | \(97\) |
Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^5} \, dx=\frac {2 \, b^{3} x^{6} + 12 \, a b^{2} x^{4} \log \left (x\right ) - 6 \, a^{2} b x^{2} - a^{3}}{4 \, x^{4}} \]
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^5} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{5}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.21 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^5} \, dx=\frac {1}{2} \, b^{3} x^{2} + 3 \, a b^{2} \log \left (x\right ) - \frac {3 \, a^{2} b}{2 \, x^{2}} - \frac {a^{3}}{4 \, x^{4}} \]
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.53 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^5} \, dx=\frac {1}{2} \, b^{3} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{2} \, a b^{2} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {9 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 6 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{4 \, x^{4}} \]
1/2*b^3*x^2*sgn(b*x^2 + a) + 3/2*a*b^2*log(x^2)*sgn(b*x^2 + a) - 1/4*(9*a* b^2*x^4*sgn(b*x^2 + a) + 6*a^2*b*x^2*sgn(b*x^2 + a) + a^3*sgn(b*x^2 + a))/ x^4
Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^5} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}}{x^5} \,d x \]