Integrand size = 24, antiderivative size = 252 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {a^5 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {5 a^3 b^2 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a b^4 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {b^5 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )} \] Output:
a^5*x^2*((b*x^3+a)^2)^(1/2)/(2*b*x^3+2*a)+a^4*b*x^5*((b*x^3+a)^2)^(1/2)/(b *x^3+a)+5*a^3*b^2*x^8*((b*x^3+a)^2)^(1/2)/(4*b*x^3+4*a)+10*a^2*b^3*x^11*(( b*x^3+a)^2)^(1/2)/(11*b*x^3+11*a)+5*a*b^4*x^14*((b*x^3+a)^2)^(1/2)/(14*b*x ^3+14*a)+b^5*x^17*((b*x^3+a)^2)^(1/2)/(17*b*x^3+17*a)
Time = 1.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^2 \sqrt {\left (a+b x^3\right )^2} \left (2618 a^5+5236 a^4 b x^3+6545 a^3 b^2 x^6+4760 a^2 b^3 x^9+1870 a b^4 x^{12}+308 b^5 x^{15}\right )}{5236 \left (a+b x^3\right )} \] Input:
Integrate[x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]
Output:
(x^2*Sqrt[(a + b*x^3)^2]*(2618*a^5 + 5236*a^4*b*x^3 + 6545*a^3*b^2*x^6 + 4 760*a^2*b^3*x^9 + 1870*a*b^4*x^12 + 308*b^5*x^15))/(5236*(a + b*x^3))
Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.39, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1384, 27, 802, 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.
\(\displaystyle \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int b^5 x \left (b x^3+a\right )^5dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x \left (b x^3+a\right )^5dx}{a+b x^3}\) |
\(\Big \downarrow \) 802 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (b^5 x^{16}+5 a b^4 x^{13}+10 a^2 b^3 x^{10}+10 a^3 b^2 x^7+5 a^4 b x^4+a^5 x\right )dx}{a+b x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (\frac {a^5 x^2}{2}+a^4 b x^5+\frac {5}{4} a^3 b^2 x^8+\frac {10}{11} a^2 b^3 x^{11}+\frac {5}{14} a b^4 x^{14}+\frac {b^5 x^{17}}{17}\right )}{a+b x^3}\) |
Input:
Int[x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*((a^5*x^2)/2 + a^4*b*x^5 + (5*a^3*b^2*x^8 )/4 + (10*a^2*b^3*x^11)/11 + (5*a*b^4*x^14)/14 + (b^5*x^17)/17))/(a + b*x^ 3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[Exp andIntegrand[(c*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
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)])
Time = 1.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.32
method | result | size |
gosper | \(\frac {x^{2} \left (308 b^{5} x^{15}+1870 a \,b^{4} x^{12}+4760 a^{2} b^{3} x^{9}+6545 a^{3} b^{2} x^{6}+5236 a^{4} b \,x^{3}+2618 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{5236 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{2} \left (308 b^{5} x^{15}+1870 a \,b^{4} x^{12}+4760 a^{2} b^{3} x^{9}+6545 a^{3} b^{2} x^{6}+5236 a^{4} b \,x^{3}+2618 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{5236 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
orering | \(\frac {x^{2} \left (308 b^{5} x^{15}+1870 a \,b^{4} x^{12}+4760 a^{2} b^{3} x^{9}+6545 a^{3} b^{2} x^{6}+5236 a^{4} b \,x^{3}+2618 a^{5}\right ) \left (b^{2} x^{6}+2 a \,x^{3} b +a^{2}\right )^{\frac {5}{2}}}{5236 \left (b \,x^{3}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {a^{5} x^{2} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{2 b \,x^{3}+2 a}+\frac {a^{4} b \,x^{5} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} b^{2} x^{8}}{4 \left (b \,x^{3}+a \right )}+\frac {10 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} b^{3} x^{11}}{11 \left (b \,x^{3}+a \right )}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a \,b^{4} x^{14}}{14 \left (b \,x^{3}+a \right )}+\frac {b^{5} x^{17} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{17 b \,x^{3}+17 a}\) | \(177\) |
Input:
int(x*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/5236*x^2*(308*b^5*x^15+1870*a*b^4*x^12+4760*a^2*b^3*x^9+6545*a^3*b^2*x^6 +5236*a^4*b*x^3+2618*a^5)*((b*x^3+a)^2)^(5/2)/(b*x^3+a)^5
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{17} \, b^{5} x^{17} + \frac {5}{14} \, a b^{4} x^{14} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {5}{4} \, a^{3} b^{2} x^{8} + a^{4} b x^{5} + \frac {1}{2} \, a^{5} x^{2} \] Input:
integrate(x*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")
Output:
1/17*b^5*x^17 + 5/14*a*b^4*x^14 + 10/11*a^2*b^3*x^11 + 5/4*a^3*b^2*x^8 + a ^4*b*x^5 + 1/2*a^5*x^2
\[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \] Input:
integrate(x*(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)
Output:
Integral(x*((a + b*x**3)**2)**(5/2), x)
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{17} \, b^{5} x^{17} + \frac {5}{14} \, a b^{4} x^{14} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {5}{4} \, a^{3} b^{2} x^{8} + a^{4} b x^{5} + \frac {1}{2} \, a^{5} x^{2} \] Input:
integrate(x*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")
Output:
1/17*b^5*x^17 + 5/14*a*b^4*x^14 + 10/11*a^2*b^3*x^11 + 5/4*a^3*b^2*x^8 + a ^4*b*x^5 + 1/2*a^5*x^2
Time = 0.12 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.41 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{17} \, b^{5} x^{17} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{14} \, a b^{4} x^{14} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{11} \, a^{2} b^{3} x^{11} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{4} \, a^{3} b^{2} x^{8} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{4} b x^{5} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{2} \, a^{5} x^{2} \mathrm {sgn}\left (b x^{3} + a\right ) \] Input:
integrate(x*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")
Output:
1/17*b^5*x^17*sgn(b*x^3 + a) + 5/14*a*b^4*x^14*sgn(b*x^3 + a) + 10/11*a^2* b^3*x^11*sgn(b*x^3 + a) + 5/4*a^3*b^2*x^8*sgn(b*x^3 + a) + a^4*b*x^5*sgn(b *x^3 + a) + 1/2*a^5*x^2*sgn(b*x^3 + a)
Timed out. \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \] Input:
int(x*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)
Output:
int(x*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2), x)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^{2} \left (308 b^{5} x^{15}+1870 a \,b^{4} x^{12}+4760 a^{2} b^{3} x^{9}+6545 a^{3} b^{2} x^{6}+5236 a^{4} b \,x^{3}+2618 a^{5}\right )}{5236} \] Input:
int(x*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)
Output:
(x**2*(2618*a**5 + 5236*a**4*b*x**3 + 6545*a**3*b**2*x**6 + 4760*a**2*b**3 *x**9 + 1870*a*b**4*x**12 + 308*b**5*x**15))/5236