Integrand size = 26, antiderivative size = 160 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=-\frac {a^3 \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 b^4}+\frac {3 a^2 \left (a+b x^2\right )^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 b^4}-\frac {3 a \left (a+b x^2\right )^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 b^4}+\frac {\left (a+b x^2\right )^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{18 b^4} \]
-1/12*a^3*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/b^4+3/14*a^2*(b*x^2+a)^6*((b*x^2 +a)^2)^(1/2)/b^4-3/16*a*(b*x^2+a)^7*((b*x^2+a)^2)^(1/2)/b^4+1/18*(b*x^2+a) ^8*((b*x^2+a)^2)^(1/2)/b^4
Time = 0.76 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.84 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^8 \left (126 a^5+504 a^4 b x^2+840 a^3 b^2 x^4+720 a^2 b^3 x^6+315 a b^4 x^8+56 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{1008 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]
(x^8*(126*a^5 + 504*a^4*b*x^2 + 840*a^3*b^2*x^4 + 720*a^2*b^3*x^6 + 315*a* b^4*x^8 + 56*b^5*x^10)*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^ 2])))/(1008*(-a^2 - a*b*x^2 + Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))
Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.67, 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.
\(\displaystyle \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^5 x^7 \left (b x^2+a\right )^5dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^7 \left (b x^2+a\right )^5dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^6 \left (b x^2+a\right )^5dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {\left (b x^2+a\right )^8}{b^3}-\frac {3 a \left (b x^2+a\right )^7}{b^3}+\frac {3 a^2 \left (b x^2+a\right )^6}{b^3}-\frac {a^3 \left (b x^2+a\right )^5}{b^3}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {a^3 \left (a+b x^2\right )^6}{6 b^4}+\frac {3 a^2 \left (a+b x^2\right )^7}{7 b^4}+\frac {\left (a+b x^2\right )^9}{9 b^4}-\frac {3 a \left (a+b x^2\right )^8}{8 b^4}\right )}{2 \left (a+b x^2\right )}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/6*(a^3*(a + b*x^2)^6)/b^4 + (3*a^2*(a + b*x^2)^7)/(7*b^4) - (3*a*(a + b*x^2)^8)/(8*b^4) + (a + b*x^2)^9/(9*b^4) ))/(2*(a + b*x^2))
3.6.88.3.1 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 2.
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.41
method | result | size |
pseudoelliptic | \(\frac {\left (\frac {4}{9} x^{10} b^{5}+\frac {5}{2} a \,x^{8} b^{4}+\frac {40}{7} a^{2} x^{6} b^{3}+\frac {20}{3} a^{3} x^{4} b^{2}+4 x^{2} a^{4} b +a^{5}\right ) x^{8} \operatorname {csgn}\left (b \,x^{2}+a \right )}{8}\) | \(66\) |
gosper | \(\frac {x^{8} \left (56 x^{10} b^{5}+315 a \,x^{8} b^{4}+720 a^{2} x^{6} b^{3}+840 a^{3} x^{4} b^{2}+504 x^{2} a^{4} b +126 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{1008 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{8} \left (56 x^{10} b^{5}+315 a \,x^{8} b^{4}+720 a^{2} x^{6} b^{3}+840 a^{3} x^{4} b^{2}+504 x^{2} a^{4} b +126 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{1008 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{5} x^{8}}{8 b \,x^{2}+8 a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \,a^{4} x^{10}}{2 b \,x^{2}+2 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} b^{2} x^{12}}{6 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b^{3} x^{14}}{7 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{4} a \,x^{16}}{16 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{5} x^{18}}{18 b \,x^{2}+18 a}\) | \(178\) |
1/8*(4/9*x^10*b^5+5/2*a*x^8*b^4+40/7*a^2*x^6*b^3+20/3*a^3*x^4*b^2+4*x^2*a^ 4*b+a^5)*x^8*csgn(b*x^2+a)
Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{18} \, b^{5} x^{18} + \frac {5}{16} \, a b^{4} x^{16} + \frac {5}{7} \, a^{2} b^{3} x^{14} + \frac {5}{6} \, a^{3} b^{2} x^{12} + \frac {1}{2} \, a^{4} b x^{10} + \frac {1}{8} \, a^{5} x^{8} \]
1/18*b^5*x^18 + 5/16*a*b^4*x^16 + 5/7*a^2*b^3*x^14 + 5/6*a^3*b^2*x^12 + 1/ 2*a^4*b*x^10 + 1/8*a^5*x^8
\[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{7} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{18} \, b^{5} x^{18} + \frac {5}{16} \, a b^{4} x^{16} + \frac {5}{7} \, a^{2} b^{3} x^{14} + \frac {5}{6} \, a^{3} b^{2} x^{12} + \frac {1}{2} \, a^{4} b x^{10} + \frac {1}{8} \, a^{5} x^{8} \]
1/18*b^5*x^18 + 5/16*a*b^4*x^16 + 5/7*a^2*b^3*x^14 + 5/6*a^3*b^2*x^12 + 1/ 2*a^4*b*x^10 + 1/8*a^5*x^8
Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.66 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{18} \, b^{5} x^{18} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{16} \, a b^{4} x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{7} \, a^{2} b^{3} x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{6} \, a^{3} b^{2} x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{2} \, a^{4} b x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{8} \, a^{5} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) \]
1/18*b^5*x^18*sgn(b*x^2 + a) + 5/16*a*b^4*x^16*sgn(b*x^2 + a) + 5/7*a^2*b^ 3*x^14*sgn(b*x^2 + a) + 5/6*a^3*b^2*x^12*sgn(b*x^2 + a) + 1/2*a^4*b*x^10*s gn(b*x^2 + a) + 1/8*a^5*x^8*sgn(b*x^2 + a)
Timed out. \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^7\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]