Integrand size = 26, antiderivative size = 119 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {a^2 \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 b^3}-\frac {a \left (a+b x^2\right )^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 b^3}+\frac {\left (a+b x^2\right )^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 b^3} \] Output:
1/12*a^2*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/b^3-1/7*a*(b*x^2+a)^6*((b*x^2+a)^ 2)^(1/2)/b^3+1/16*(b*x^2+a)^7*((b*x^2+a)^2)^(1/2)/b^3
Time = 0.78 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.13 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^6 \left (56 a^5+210 a^4 b x^2+336 a^3 b^2 x^4+280 a^2 b^3 x^6+120 a b^4 x^8+21 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{336 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \] Input:
Integrate[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(x^6*(56*a^5 + 210*a^4*b*x^2 + 336*a^3*b^2*x^4 + 280*a^2*b^3*x^6 + 120*a*b ^4*x^8 + 21*b^5*x^10)*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2 ])))/(336*(-a^2 - a*b*x^2 + Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))
Time = 0.40 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.74, 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.
\(\displaystyle \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^5 x^5 \left (b x^2+a\right )^5dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^5 \left (b x^2+a\right )^5dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^4 \left (b x^2+a\right )^5dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {\left (b x^2+a\right )^7}{b^2}-\frac {2 a \left (b x^2+a\right )^6}{b^2}+\frac {a^2 \left (b x^2+a\right )^5}{b^2}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {a^2 \left (a+b x^2\right )^6}{6 b^3}+\frac {\left (a+b x^2\right )^8}{8 b^3}-\frac {2 a \left (a+b x^2\right )^7}{7 b^3}\right )}{2 \left (a+b x^2\right )}\) |
Input:
Int[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((a^2*(a + b*x^2)^6)/(6*b^3) - (2*a*(a + b*x^2)^7)/(7*b^3) + (a + b*x^2)^8/(8*b^3)))/(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 2.
Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.55
method | result | size |
pseudoelliptic | \(\frac {x^{6} \left (\frac {3}{8} x^{10} b^{5}+\frac {15}{7} a \,x^{8} b^{4}+5 a^{2} x^{6} b^{3}+6 a^{3} x^{4} b^{2}+\frac {15}{4} x^{2} a^{4} b +a^{5}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{6}\) | \(66\) |
gosper | \(\frac {x^{6} \left (21 x^{10} b^{5}+120 a \,x^{8} b^{4}+280 a^{2} x^{6} b^{3}+336 a^{3} x^{4} b^{2}+210 x^{2} a^{4} b +56 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{336 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{6} \left (21 x^{10} b^{5}+120 a \,x^{8} b^{4}+280 a^{2} x^{6} b^{3}+336 a^{3} x^{4} b^{2}+210 x^{2} a^{4} b +56 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{336 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
orering | \(\frac {x^{6} \left (21 x^{10} b^{5}+120 a \,x^{8} b^{4}+280 a^{2} x^{6} b^{3}+336 a^{3} x^{4} b^{2}+210 x^{2} a^{4} b +56 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{336 \left (b \,x^{2}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{5} x^{6}}{6 b \,x^{2}+6 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \,a^{4} x^{8}}{8 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} b^{2} x^{10}}{b \,x^{2}+a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b^{3} x^{12}}{6 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{4} a \,x^{14}}{14 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{5} x^{16}}{16 b \,x^{2}+16 a}\) | \(177\) |
Input:
int(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/6*x^6*(3/8*x^10*b^5+15/7*a*x^8*b^4+5*a^2*x^6*b^3+6*a^3*x^4*b^2+15/4*x^2* a^4*b+a^5)*csgn(b*x^2+a)
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.47 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} + \frac {5}{14} \, a b^{4} x^{14} + \frac {5}{6} \, a^{2} b^{3} x^{12} + a^{3} b^{2} x^{10} + \frac {5}{8} \, a^{4} b x^{8} + \frac {1}{6} \, a^{5} x^{6} \] Input:
integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
1/16*b^5*x^16 + 5/14*a*b^4*x^14 + 5/6*a^2*b^3*x^12 + a^3*b^2*x^10 + 5/8*a^ 4*b*x^8 + 1/6*a^5*x^6
\[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{5} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \] Input:
integrate(x**5*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral(x**5*((a + b*x**2)**2)**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.47 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} + \frac {5}{14} \, a b^{4} x^{14} + \frac {5}{6} \, a^{2} b^{3} x^{12} + a^{3} b^{2} x^{10} + \frac {5}{8} \, a^{4} b x^{8} + \frac {1}{6} \, a^{5} x^{6} \] Input:
integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
1/16*b^5*x^16 + 5/14*a*b^4*x^14 + 5/6*a^2*b^3*x^12 + a^3*b^2*x^10 + 5/8*a^ 4*b*x^8 + 1/6*a^5*x^6
Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{14} \, a b^{4} x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{6} \, a^{2} b^{3} x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{3} b^{2} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{8} \, a^{4} b x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{6} \, a^{5} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
1/16*b^5*x^16*sgn(b*x^2 + a) + 5/14*a*b^4*x^14*sgn(b*x^2 + a) + 5/6*a^2*b^ 3*x^12*sgn(b*x^2 + a) + a^3*b^2*x^10*sgn(b*x^2 + a) + 5/8*a^4*b*x^8*sgn(b* x^2 + a) + 1/6*a^5*x^6*sgn(b*x^2 + a)
Timed out. \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^5\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \] Input:
int(x^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)
Output:
int(x^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.50 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^{6} \left (21 b^{5} x^{10}+120 a \,b^{4} x^{8}+280 a^{2} b^{3} x^{6}+336 a^{3} b^{2} x^{4}+210 a^{4} b \,x^{2}+56 a^{5}\right )}{336} \] Input:
int(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
(x**6*(56*a**5 + 210*a**4*b*x**2 + 336*a**3*b**2*x**4 + 280*a**2*b**3*x**6 + 120*a*b**4*x**8 + 21*b**5*x**10))/336