Integrand size = 26, antiderivative size = 255 \[ \int x^{13} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {a^5 x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {5 a^4 b x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^{18} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {a^2 b^3 x^{20} \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a b^4 x^{22} \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )}+\frac {b^5 x^{24} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 \left (a+b x^2\right )} \] Output:
a^5*x^14*((b*x^2+a)^2)^(1/2)/(14*b*x^2+14*a)+5*a^4*b*x^16*((b*x^2+a)^2)^(1 /2)/(16*b*x^2+16*a)+5*a^3*b^2*x^18*((b*x^2+a)^2)^(1/2)/(9*b*x^2+9*a)+a^2*b ^3*x^20*((b*x^2+a)^2)^(1/2)/(2*b*x^2+2*a)+5*a*b^4*x^22*((b*x^2+a)^2)^(1/2) /(22*b*x^2+22*a)+b^5*x^24*((b*x^2+a)^2)^(1/2)/(24*b*x^2+24*a)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x^{13} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^{14} \sqrt {\left (a+b x^2\right )^2} \left (792 a^5+3465 a^4 b x^2+6160 a^3 b^2 x^4+5544 a^2 b^3 x^6+2520 a b^4 x^8+462 b^5 x^{10}\right )}{11088 \left (a+b x^2\right )} \] Input:
Integrate[x^13*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(x^14*Sqrt[(a + b*x^2)^2]*(792*a^5 + 3465*a^4*b*x^2 + 6160*a^3*b^2*x^4 + 5 544*a^2*b^3*x^6 + 2520*a*b^4*x^8 + 462*b^5*x^10))/(11088*(a + b*x^2))
Time = 0.43 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.40, 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.
\(\displaystyle \int x^{13} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^5 x^{13} \left (b x^2+a\right )^5dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^{13} \left (b x^2+a\right )^5dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^{12} \left (b x^2+a\right )^5dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^5 x^{22}+5 a b^4 x^{20}+10 a^2 b^3 x^{18}+10 a^3 b^2 x^{16}+5 a^4 b x^{14}+a^5 x^{12}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {a^5 x^{14}}{7}+\frac {5}{8} a^4 b x^{16}+\frac {10}{9} a^3 b^2 x^{18}+a^2 b^3 x^{20}+\frac {5}{11} a b^4 x^{22}+\frac {b^5 x^{24}}{12}\right )}{2 \left (a+b x^2\right )}\) |
Input:
Int[x^13*(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^5*x^14)/7 + (5*a^4*b*x^16)/8 + (10*a^ 3*b^2*x^18)/9 + a^2*b^3*x^20 + (5*a*b^4*x^22)/11 + (b^5*x^24)/12))/(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 = 0.43 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.26
method | result | size |
pseudoelliptic | \(\frac {x^{14} \left (\frac {7}{12} x^{10} b^{5}+\frac {35}{11} a \,x^{8} b^{4}+7 a^{2} x^{6} b^{3}+\frac {70}{9} a^{3} x^{4} b^{2}+\frac {35}{8} x^{2} a^{4} b +a^{5}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{14}\) | \(66\) |
gosper | \(\frac {x^{14} \left (462 x^{10} b^{5}+2520 a \,x^{8} b^{4}+5544 a^{2} x^{6} b^{3}+6160 a^{3} x^{4} b^{2}+3465 x^{2} a^{4} b +792 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{11088 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{14} \left (462 x^{10} b^{5}+2520 a \,x^{8} b^{4}+5544 a^{2} x^{6} b^{3}+6160 a^{3} x^{4} b^{2}+3465 x^{2} a^{4} b +792 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{11088 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
orering | \(\frac {x^{14} \left (462 x^{10} b^{5}+2520 a \,x^{8} b^{4}+5544 a^{2} x^{6} b^{3}+6160 a^{3} x^{4} b^{2}+3465 x^{2} a^{4} b +792 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{11088 \left (b \,x^{2}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {b^{5} x^{24} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{24 b \,x^{2}+24 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} b^{2} x^{18}}{9 \left (b \,x^{2}+a \right )}+\frac {a^{5} x^{14} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{14 b \,x^{2}+14 a}+\frac {a^{2} b^{3} x^{20} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 b \,x^{2}+2 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \,a^{4} x^{16}}{16 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{4} a \,x^{22}}{22 \left (b \,x^{2}+a \right )}\) | \(178\) |
Input:
int(x^13*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/14*x^14*(7/12*x^10*b^5+35/11*a*x^8*b^4+7*a^2*x^6*b^3+70/9*a^3*x^4*b^2+35 /8*x^2*a^4*b+a^5)*csgn(b*x^2+a)
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{13} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{24} \, b^{5} x^{24} + \frac {5}{22} \, a b^{4} x^{22} + \frac {1}{2} \, a^{2} b^{3} x^{20} + \frac {5}{9} \, a^{3} b^{2} x^{18} + \frac {5}{16} \, a^{4} b x^{16} + \frac {1}{14} \, a^{5} x^{14} \] Input:
integrate(x^13*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
1/24*b^5*x^24 + 5/22*a*b^4*x^22 + 1/2*a^2*b^3*x^20 + 5/9*a^3*b^2*x^18 + 5/ 16*a^4*b*x^16 + 1/14*a^5*x^14
\[ \int x^{13} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{13} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \] Input:
integrate(x**13*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral(x**13*((a + b*x**2)**2)**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{13} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{24} \, b^{5} x^{24} + \frac {5}{22} \, a b^{4} x^{22} + \frac {1}{2} \, a^{2} b^{3} x^{20} + \frac {5}{9} \, a^{3} b^{2} x^{18} + \frac {5}{16} \, a^{4} b x^{16} + \frac {1}{14} \, a^{5} x^{14} \] Input:
integrate(x^13*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
1/24*b^5*x^24 + 5/22*a*b^4*x^22 + 1/2*a^2*b^3*x^20 + 5/9*a^3*b^2*x^18 + 5/ 16*a^4*b*x^16 + 1/14*a^5*x^14
Time = 0.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.41 \[ \int x^{13} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{24} \, b^{5} x^{24} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{22} \, a b^{4} x^{22} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{2} \, a^{2} b^{3} x^{20} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{9} \, a^{3} b^{2} x^{18} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{16} \, a^{4} b x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{14} \, a^{5} x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate(x^13*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
1/24*b^5*x^24*sgn(b*x^2 + a) + 5/22*a*b^4*x^22*sgn(b*x^2 + a) + 1/2*a^2*b^ 3*x^20*sgn(b*x^2 + a) + 5/9*a^3*b^2*x^18*sgn(b*x^2 + a) + 5/16*a^4*b*x^16* sgn(b*x^2 + a) + 1/14*a^5*x^14*sgn(b*x^2 + a)
Timed out. \[ \int x^{13} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{13}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \] Input:
int(x^13*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)
Output:
int(x^13*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)
Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int x^{13} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^{14} \left (462 b^{5} x^{10}+2520 a \,b^{4} x^{8}+5544 a^{2} b^{3} x^{6}+6160 a^{3} b^{2} x^{4}+3465 a^{4} b \,x^{2}+792 a^{5}\right )}{11088} \] Input:
int(x^13*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
(x**14*(792*a**5 + 3465*a**4*b*x**2 + 6160*a**3*b**2*x**4 + 5544*a**2*b**3 *x**6 + 2520*a*b**4*x**8 + 462*b**5*x**10))/11088