Integrand size = 26, antiderivative size = 69 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=-\frac {a \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 b^2}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{14 b^2} \] Output:
-1/12*a*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/b^2+1/14*(b^2*x^4+2*a*b*x^2+a^2)^( 7/2)/b^2
Time = 0.72 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.96 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^4 \left (21 a^5+70 a^4 b x^2+105 a^3 b^2 x^4+84 a^2 b^3 x^6+35 a b^4 x^8+6 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{84 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \] Input:
Integrate[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(x^4*(21*a^5 + 70*a^4*b*x^2 + 105*a^3*b^2*x^4 + 84*a^2*b^3*x^6 + 35*a*b^4* x^8 + 6*b^5*x^10)*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2]))) /(84*(-a^2 - a*b*x^2 + Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))
Time = 0.35 (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 = {1434, 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.
\(\displaystyle \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {1}{2} \int x^2 \left (b^2 x^4+2 a b x^2+a^2\right )^{5/2}dx^2\) |
\(\Big \downarrow \) 1100 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{7 b^2}-\frac {a \int \left (b^2 x^4+2 a b x^2+a^2\right )^{5/2}dx^2}{b}\right )\) |
\(\Big \downarrow \) 1079 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{7 b^2}-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^2 x^2+a b\right )^5dx^2}{b^6 \left (a+b x^2\right )}\right )\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{7 b^2}-\frac {a \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 b^2}\right )\) |
Input:
Int[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(-1/6*(a*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b^2 + (a^2 + 2*a*b *x^2 + b^2*x^4)^(7/2)/(7*b^2))/2
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_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp [1/2 Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96
method | result | size |
pseudoelliptic | \(\frac {x^{4} \left (\frac {2}{7} x^{10} b^{5}+\frac {5}{3} a \,x^{8} b^{4}+4 a^{2} x^{6} b^{3}+5 a^{3} x^{4} b^{2}+\frac {10}{3} x^{2} a^{4} b +a^{5}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{4}\) | \(66\) |
gosper | \(\frac {x^{4} \left (6 x^{10} b^{5}+35 a \,x^{8} b^{4}+84 a^{2} x^{6} b^{3}+105 a^{3} x^{4} b^{2}+70 x^{2} a^{4} b +21 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{84 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{4} \left (6 x^{10} b^{5}+35 a \,x^{8} b^{4}+84 a^{2} x^{6} b^{3}+105 a^{3} x^{4} b^{2}+70 x^{2} a^{4} b +21 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{84 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
orering | \(\frac {x^{4} \left (6 x^{10} b^{5}+35 a \,x^{8} b^{4}+84 a^{2} x^{6} b^{3}+105 a^{3} x^{4} b^{2}+70 x^{2} a^{4} b +21 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{84 \left (b \,x^{2}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{5} x^{4}}{4 b \,x^{2}+4 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \,a^{4} x^{6}}{6 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} b^{2} x^{8}}{4 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b^{3} x^{10}}{b \,x^{2}+a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{4} a \,x^{12}}{12 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{5} x^{14}}{14 b \,x^{2}+14 a}\) | \(177\) |
Input:
int(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/4*x^4*(2/7*x^10*b^5+5/3*a*x^8*b^4+4*a^2*x^6*b^3+5*a^3*x^4*b^2+10/3*x^2*a ^4*b+a^5)*csgn(b*x^2+a)
Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{14} \, b^{5} x^{14} + \frac {5}{12} \, a b^{4} x^{12} + a^{2} b^{3} x^{10} + \frac {5}{4} \, a^{3} b^{2} x^{8} + \frac {5}{6} \, a^{4} b x^{6} + \frac {1}{4} \, a^{5} x^{4} \] Input:
integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
1/14*b^5*x^14 + 5/12*a*b^4*x^12 + a^2*b^3*x^10 + 5/4*a^3*b^2*x^8 + 5/6*a^4 *b*x^6 + 1/4*a^5*x^4
\[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{3} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \] Input:
integrate(x**3*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral(x**3*((a + b*x**2)**2)**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{14} \, b^{5} x^{14} + \frac {5}{12} \, a b^{4} x^{12} + a^{2} b^{3} x^{10} + \frac {5}{4} \, a^{3} b^{2} x^{8} + \frac {5}{6} \, a^{4} b x^{6} + \frac {1}{4} \, a^{5} x^{4} \] Input:
integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
1/14*b^5*x^14 + 5/12*a*b^4*x^12 + a^2*b^3*x^10 + 5/4*a^3*b^2*x^8 + 5/6*a^4 *b*x^6 + 1/4*a^5*x^4
Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{84} \, {\left (6 \, b^{5} x^{14} + 35 \, a b^{4} x^{12} + 84 \, a^{2} b^{3} x^{10} + 105 \, a^{3} b^{2} x^{8} + 70 \, a^{4} b x^{6} + 21 \, a^{5} x^{4}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
1/84*(6*b^5*x^14 + 35*a*b^4*x^12 + 84*a^2*b^3*x^10 + 105*a^3*b^2*x^8 + 70* a^4*b*x^6 + 21*a^5*x^4)*sgn(b*x^2 + a)
Timed out. \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^3\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \] Input:
int(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)
Output:
int(x^3*(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 = 0.86 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^{4} \left (6 b^{5} x^{10}+35 a \,b^{4} x^{8}+84 a^{2} b^{3} x^{6}+105 a^{3} b^{2} x^{4}+70 a^{4} b \,x^{2}+21 a^{5}\right )}{84} \] Input:
int(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
(x**4*(21*a**5 + 70*a**4*b*x**2 + 105*a**3*b**2*x**4 + 84*a**2*b**3*x**6 + 35*a*b**4*x**8 + 6*b**5*x**10))/84