Integrand size = 26, antiderivative size = 69 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=-\frac {a \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 b^2}+\frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{7/2}}{21 b^2} \] Output:
-1/18*a*(b*x^3+a)^5*((b*x^3+a)^2)^(1/2)/b^2+1/21*(b^2*x^6+2*a*b*x^3+a^2)^( 7/2)/b^2
Time = 1.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.20 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^6 \sqrt {\left (a+b x^3\right )^2} \left (21 a^5+70 a^4 b x^3+105 a^3 b^2 x^6+84 a^2 b^3 x^9+35 a b^4 x^{12}+6 b^5 x^{15}\right )}{126 \left (a+b x^3\right )} \] Input:
Integrate[x^5*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]
Output:
(x^6*Sqrt[(a + b*x^3)^2]*(21*a^5 + 70*a^4*b*x^3 + 105*a^3*b^2*x^6 + 84*a^2 *b^3*x^9 + 35*a*b^4*x^12 + 6*b^5*x^15))/(126*(a + b*x^3))
Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1693, 1100, 1079, 17}
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.
\(\displaystyle \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1693 |
\(\displaystyle \frac {1}{3} \int x^3 \left (b^2 x^6+2 a b x^3+a^2\right )^{5/2}dx^3\) |
\(\Big \downarrow \) 1100 |
\(\displaystyle \frac {1}{3} \left (\frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{7/2}}{7 b^2}-\frac {a \int \left (b^2 x^6+2 a b x^3+a^2\right )^{5/2}dx^3}{b}\right )\) |
\(\Big \downarrow \) 1079 |
\(\displaystyle \frac {1}{3} \left (\frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{7/2}}{7 b^2}-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (b^2 x^3+a b\right )^5dx^3}{b^6 \left (a+b x^3\right )}\right )\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {1}{3} \left (\frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{7/2}}{7 b^2}-\frac {a \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 b^2}\right )\) |
Input:
Int[x^5*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]
Output:
(-1/6*(a*(a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/b^2 + (a^2 + 2*a*b *x^3 + b^2*x^6)^(7/2)/(7*b^2))/3
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c *x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])) Int[(b/2 + c *x)^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0]
Int[((d_.) + (e_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b* e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c, 0]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && IntegerQ [Simplify[(m + 1)/n]]
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.45
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{6} \left (-6 b \,x^{3}+a \right )}{126 b^{2}}\) | \(31\) |
gosper | \(\frac {x^{6} \left (6 b^{5} x^{15}+35 a \,b^{4} x^{12}+84 a^{2} b^{3} x^{9}+105 a^{3} b^{2} x^{6}+70 a^{4} b \,x^{3}+21 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{126 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{6} \left (6 b^{5} x^{15}+35 a \,b^{4} x^{12}+84 a^{2} b^{3} x^{9}+105 a^{3} b^{2} x^{6}+70 a^{4} b \,x^{3}+21 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{126 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
orering | \(\frac {x^{6} \left (6 b^{5} x^{15}+35 a \,b^{4} x^{12}+84 a^{2} b^{3} x^{9}+105 a^{3} b^{2} x^{6}+70 a^{4} b \,x^{3}+21 a^{5}\right ) \left (b^{2} x^{6}+2 a \,x^{3} b +a^{2}\right )^{\frac {5}{2}}}{126 \left (b \,x^{3}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{5} x^{6}}{6 b \,x^{3}+6 a}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{4} b \,x^{9}}{9 \left (b \,x^{3}+a \right )}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} b^{2} x^{12}}{6 \left (b \,x^{3}+a \right )}+\frac {2 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} b^{3} x^{15}}{3 \left (b \,x^{3}+a \right )}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a \,b^{4} x^{18}}{18 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{5} x^{21}}{21 b \,x^{3}+21 a}\) | \(178\) |
Input:
int(x^5*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/126*csgn(b*x^3+a)*(b*x^3+a)^6*(-6*b*x^3+a)/b^2
Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{21} \, b^{5} x^{21} + \frac {5}{18} \, a b^{4} x^{18} + \frac {2}{3} \, a^{2} b^{3} x^{15} + \frac {5}{6} \, a^{3} b^{2} x^{12} + \frac {5}{9} \, a^{4} b x^{9} + \frac {1}{6} \, a^{5} x^{6} \] Input:
integrate(x^5*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")
Output:
1/21*b^5*x^21 + 5/18*a*b^4*x^18 + 2/3*a^2*b^3*x^15 + 5/6*a^3*b^2*x^12 + 5/ 9*a^4*b*x^9 + 1/6*a^5*x^6
\[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{5} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \] Input:
integrate(x**5*(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)
Output:
Integral(x**5*((a + b*x**3)**2)**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.20 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=-\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a x^{3}}{18 \, b} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a^{2}}{18 \, b^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}}}{21 \, b^{2}} \] Input:
integrate(x^5*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")
Output:
-1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a*x^3/b - 1/18*(b^2*x^6 + 2*a*b*x^ 3 + a^2)^(5/2)*a^2/b^2 + 1/21*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)/b^2
Time = 0.13 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{126} \, {\left (6 \, b^{5} x^{21} + 35 \, a b^{4} x^{18} + 84 \, a^{2} b^{3} x^{15} + 105 \, a^{3} b^{2} x^{12} + 70 \, a^{4} b x^{9} + 21 \, a^{5} x^{6}\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \] Input:
integrate(x^5*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")
Output:
1/126*(6*b^5*x^21 + 35*a*b^4*x^18 + 84*a^2*b^3*x^15 + 105*a^3*b^2*x^12 + 7 0*a^4*b*x^9 + 21*a^5*x^6)*sgn(b*x^3 + a)
Timed out. \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^5\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \] Input:
int(x^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)
Output:
int(x^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2), x)
Time = 0.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^{6} \left (6 b^{5} x^{15}+35 a \,b^{4} x^{12}+84 a^{2} b^{3} x^{9}+105 a^{3} b^{2} x^{6}+70 a^{4} b \,x^{3}+21 a^{5}\right )}{126} \] Input:
int(x^5*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)
Output:
(x**6*(21*a**5 + 70*a**4*b*x**3 + 105*a**3*b**2*x**6 + 84*a**2*b**3*x**9 + 35*a*b**4*x**12 + 6*b**5*x**15))/126