Integrand size = 22, antiderivative size = 150 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^{10} \, dx=-\frac {16 d^3 (11 b c-8 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^3}{3465 c^5}+\frac {8 d^2 (11 b c-8 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^5}{1155 c^4}-\frac {2 d (11 b c-8 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^7}{231 c^3}+\frac {(11 b c-8 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^9}{99 c^2}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^{11}}{11 c} \] Output:
-16/3465*d^3*(-8*a*d+11*b*c)*(c+d/x^2)^(3/2)*x^3/c^5+8/1155*d^2*(-8*a*d+11 *b*c)*(c+d/x^2)^(3/2)*x^5/c^4-2/231*d*(-8*a*d+11*b*c)*(c+d/x^2)^(3/2)*x^7/ c^3+1/99*(-8*a*d+11*b*c)*(c+d/x^2)^(3/2)*x^9/c^2+1/11*a*(c+d/x^2)^(3/2)*x^ 11/c
Time = 0.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.72 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^{10} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} x \left (d+c x^2\right ) \left (11 b c \left (-16 d^3+24 c d^2 x^2-30 c^2 d x^4+35 c^3 x^6\right )+a \left (128 d^4-192 c d^3 x^2+240 c^2 d^2 x^4-280 c^3 d x^6+315 c^4 x^8\right )\right )}{3465 c^5} \] Input:
Integrate[(a + b/x^2)*Sqrt[c + d/x^2]*x^10,x]
Output:
(Sqrt[c + d/x^2]*x*(d + c*x^2)*(11*b*c*(-16*d^3 + 24*c*d^2*x^2 - 30*c^2*d* x^4 + 35*c^3*x^6) + a*(128*d^4 - 192*c*d^3*x^2 + 240*c^2*d^2*x^4 - 280*c^3 *d*x^6 + 315*c^4*x^8)))/(3465*c^5)
Time = 0.45 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {955, 803, 803, 803, 796}
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.
\(\displaystyle \int x^{10} \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle \frac {(11 b c-8 a d) \int \sqrt {c+\frac {d}{x^2}} x^8dx}{11 c}+\frac {a x^{11} \left (c+\frac {d}{x^2}\right )^{3/2}}{11 c}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {(11 b c-8 a d) \left (\frac {x^9 \left (c+\frac {d}{x^2}\right )^{3/2}}{9 c}-\frac {2 d \int \sqrt {c+\frac {d}{x^2}} x^6dx}{3 c}\right )}{11 c}+\frac {a x^{11} \left (c+\frac {d}{x^2}\right )^{3/2}}{11 c}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {(11 b c-8 a d) \left (\frac {x^9 \left (c+\frac {d}{x^2}\right )^{3/2}}{9 c}-\frac {2 d \left (\frac {x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c}-\frac {4 d \int \sqrt {c+\frac {d}{x^2}} x^4dx}{7 c}\right )}{3 c}\right )}{11 c}+\frac {a x^{11} \left (c+\frac {d}{x^2}\right )^{3/2}}{11 c}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {(11 b c-8 a d) \left (\frac {x^9 \left (c+\frac {d}{x^2}\right )^{3/2}}{9 c}-\frac {2 d \left (\frac {x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c}-\frac {4 d \left (\frac {x^5 \left (c+\frac {d}{x^2}\right )^{3/2}}{5 c}-\frac {2 d \int \sqrt {c+\frac {d}{x^2}} x^2dx}{5 c}\right )}{7 c}\right )}{3 c}\right )}{11 c}+\frac {a x^{11} \left (c+\frac {d}{x^2}\right )^{3/2}}{11 c}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {\left (\frac {x^9 \left (c+\frac {d}{x^2}\right )^{3/2}}{9 c}-\frac {2 d \left (\frac {x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c}-\frac {4 d \left (\frac {x^5 \left (c+\frac {d}{x^2}\right )^{3/2}}{5 c}-\frac {2 d x^3 \left (c+\frac {d}{x^2}\right )^{3/2}}{15 c^2}\right )}{7 c}\right )}{3 c}\right ) (11 b c-8 a d)}{11 c}+\frac {a x^{11} \left (c+\frac {d}{x^2}\right )^{3/2}}{11 c}\) |
Input:
Int[(a + b/x^2)*Sqrt[c + d/x^2]*x^10,x]
Output:
(a*(c + d/x^2)^(3/2)*x^11)/(11*c) + ((11*b*c - 8*a*d)*(((c + d/x^2)^(3/2)* x^9)/(9*c) - (2*d*(((c + d/x^2)^(3/2)*x^7)/(7*c) - (4*d*((-2*d*(c + d/x^2) ^(3/2)*x^3)/(15*c^2) + ((c + d/x^2)^(3/2)*x^5)/(5*c)))/(7*c)))/(3*c)))/(11 *c)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 ))) Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Time = 0.10 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.75
method | result | size |
gosper | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \left (315 a \,x^{8} c^{4}-280 a \,c^{3} d \,x^{6}+385 b \,c^{4} x^{6}+240 a \,c^{2} d^{2} x^{4}-330 b \,c^{3} d \,x^{4}-192 a c \,d^{3} x^{2}+264 b \,c^{2} d^{2} x^{2}+128 a \,d^{4}-176 b c \,d^{3}\right ) \left (c \,x^{2}+d \right )}{3465 c^{5}}\) | \(113\) |
default | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \left (315 a \,x^{8} c^{4}-280 a \,c^{3} d \,x^{6}+385 b \,c^{4} x^{6}+240 a \,c^{2} d^{2} x^{4}-330 b \,c^{3} d \,x^{4}-192 a c \,d^{3} x^{2}+264 b \,c^{2} d^{2} x^{2}+128 a \,d^{4}-176 b c \,d^{3}\right ) \left (c \,x^{2}+d \right )}{3465 c^{5}}\) | \(113\) |
orering | \(\frac {\left (315 a \,x^{8} c^{4}-280 a \,c^{3} d \,x^{6}+385 b \,c^{4} x^{6}+240 a \,c^{2} d^{2} x^{4}-330 b \,c^{3} d \,x^{4}-192 a c \,d^{3} x^{2}+264 b \,c^{2} d^{2} x^{2}+128 a \,d^{4}-176 b c \,d^{3}\right ) \left (c \,x^{2}+d \right ) x^{3} \left (a +\frac {b}{x^{2}}\right ) \sqrt {c +\frac {d}{x^{2}}}}{3465 c^{5} \left (a \,x^{2}+b \right )}\) | \(127\) |
risch | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \left (315 a \,c^{5} x^{10}+35 a \,c^{4} d \,x^{8}+385 b \,c^{5} x^{8}-40 a \,c^{3} d^{2} x^{6}+55 b \,c^{4} d \,x^{6}+48 a \,c^{2} d^{3} x^{4}-66 b \,c^{3} d^{2} x^{4}-64 a c \,d^{4} x^{2}+88 b \,c^{2} d^{3} x^{2}+128 a \,d^{5}-176 b c \,d^{4}\right )}{3465 c^{5}}\) | \(130\) |
trager | \(\frac {\left (315 a \,c^{5} x^{10}+35 a \,c^{4} d \,x^{8}+385 b \,c^{5} x^{8}-40 a \,c^{3} d^{2} x^{6}+55 b \,c^{4} d \,x^{6}+48 a \,c^{2} d^{3} x^{4}-66 b \,c^{3} d^{2} x^{4}-64 a c \,d^{4} x^{2}+88 b \,c^{2} d^{3} x^{2}+128 a \,d^{5}-176 b c \,d^{4}\right ) x \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{3465 c^{5}}\) | \(134\) |
Input:
int((a+b/x^2)*(c+d/x^2)^(1/2)*x^10,x,method=_RETURNVERBOSE)
Output:
1/3465*((c*x^2+d)/x^2)^(1/2)*x*(315*a*c^4*x^8-280*a*c^3*d*x^6+385*b*c^4*x^ 6+240*a*c^2*d^2*x^4-330*b*c^3*d*x^4-192*a*c*d^3*x^2+264*b*c^2*d^2*x^2+128* a*d^4-176*b*c*d^3)*(c*x^2+d)/c^5
Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^{10} \, dx=\frac {{\left (315 \, a c^{5} x^{11} + 35 \, {\left (11 \, b c^{5} + a c^{4} d\right )} x^{9} + 5 \, {\left (11 \, b c^{4} d - 8 \, a c^{3} d^{2}\right )} x^{7} - 6 \, {\left (11 \, b c^{3} d^{2} - 8 \, a c^{2} d^{3}\right )} x^{5} + 8 \, {\left (11 \, b c^{2} d^{3} - 8 \, a c d^{4}\right )} x^{3} - 16 \, {\left (11 \, b c d^{4} - 8 \, a d^{5}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{3465 \, c^{5}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)*x^10,x, algorithm="fricas")
Output:
1/3465*(315*a*c^5*x^11 + 35*(11*b*c^5 + a*c^4*d)*x^9 + 5*(11*b*c^4*d - 8*a *c^3*d^2)*x^7 - 6*(11*b*c^3*d^2 - 8*a*c^2*d^3)*x^5 + 8*(11*b*c^2*d^3 - 8*a *c*d^4)*x^3 - 16*(11*b*c*d^4 - 8*a*d^5)*x)*sqrt((c*x^2 + d)/x^2)/c^5
Leaf count of result is larger than twice the leaf count of optimal. 1386 vs. \(2 (146) = 292\).
Time = 4.79 (sec) , antiderivative size = 1386, normalized size of antiderivative = 9.24 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^{10} \, dx=\text {Too large to display} \] Input:
integrate((a+b/x**2)*(c+d/x**2)**(1/2)*x**10,x)
Output:
315*a*c**9*d**(33/2)*x**18*sqrt(c*x**2/d + 1)/(3465*c**9*d**16*x**8 + 1386 0*c**8*d**17*x**6 + 20790*c**7*d**18*x**4 + 13860*c**6*d**19*x**2 + 3465*c **5*d**20) + 1295*a*c**8*d**(35/2)*x**16*sqrt(c*x**2/d + 1)/(3465*c**9*d** 16*x**8 + 13860*c**8*d**17*x**6 + 20790*c**7*d**18*x**4 + 13860*c**6*d**19 *x**2 + 3465*c**5*d**20) + 1990*a*c**7*d**(37/2)*x**14*sqrt(c*x**2/d + 1)/ (3465*c**9*d**16*x**8 + 13860*c**8*d**17*x**6 + 20790*c**7*d**18*x**4 + 13 860*c**6*d**19*x**2 + 3465*c**5*d**20) + 1358*a*c**6*d**(39/2)*x**12*sqrt( c*x**2/d + 1)/(3465*c**9*d**16*x**8 + 13860*c**8*d**17*x**6 + 20790*c**7*d **18*x**4 + 13860*c**6*d**19*x**2 + 3465*c**5*d**20) + 343*a*c**5*d**(41/2 )*x**10*sqrt(c*x**2/d + 1)/(3465*c**9*d**16*x**8 + 13860*c**8*d**17*x**6 + 20790*c**7*d**18*x**4 + 13860*c**6*d**19*x**2 + 3465*c**5*d**20) + 35*a*c **4*d**(43/2)*x**8*sqrt(c*x**2/d + 1)/(3465*c**9*d**16*x**8 + 13860*c**8*d **17*x**6 + 20790*c**7*d**18*x**4 + 13860*c**6*d**19*x**2 + 3465*c**5*d**2 0) + 280*a*c**3*d**(45/2)*x**6*sqrt(c*x**2/d + 1)/(3465*c**9*d**16*x**8 + 13860*c**8*d**17*x**6 + 20790*c**7*d**18*x**4 + 13860*c**6*d**19*x**2 + 34 65*c**5*d**20) + 560*a*c**2*d**(47/2)*x**4*sqrt(c*x**2/d + 1)/(3465*c**9*d **16*x**8 + 13860*c**8*d**17*x**6 + 20790*c**7*d**18*x**4 + 13860*c**6*d** 19*x**2 + 3465*c**5*d**20) + 448*a*c*d**(49/2)*x**2*sqrt(c*x**2/d + 1)/(34 65*c**9*d**16*x**8 + 13860*c**8*d**17*x**6 + 20790*c**7*d**18*x**4 + 13860 *c**6*d**19*x**2 + 3465*c**5*d**20) + 128*a*d**(51/2)*sqrt(c*x**2/d + 1...
Time = 0.03 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.05 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^{10} \, dx=\frac {{\left (35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {9}{2}} x^{9} - 135 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} d x^{7} + 189 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d^{2} x^{5} - 105 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{3} x^{3}\right )} b}{315 \, c^{4}} + \frac {{\left (315 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {11}{2}} x^{11} - 1540 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {9}{2}} d x^{9} + 2970 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} d^{2} x^{7} - 2772 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d^{3} x^{5} + 1155 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{4} x^{3}\right )} a}{3465 \, c^{5}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)*x^10,x, algorithm="maxima")
Output:
1/315*(35*(c + d/x^2)^(9/2)*x^9 - 135*(c + d/x^2)^(7/2)*d*x^7 + 189*(c + d /x^2)^(5/2)*d^2*x^5 - 105*(c + d/x^2)^(3/2)*d^3*x^3)*b/c^4 + 1/3465*(315*( c + d/x^2)^(11/2)*x^11 - 1540*(c + d/x^2)^(9/2)*d*x^9 + 2970*(c + d/x^2)^( 7/2)*d^2*x^7 - 2772*(c + d/x^2)^(5/2)*d^3*x^5 + 1155*(c + d/x^2)^(3/2)*d^4 *x^3)*a/c^5
Time = 0.13 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.17 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^{10} \, dx=\frac {16 \, {\left (11 \, b c d^{\frac {9}{2}} - 8 \, a d^{\frac {11}{2}}\right )} \mathrm {sgn}\left (x\right )}{3465 \, c^{5}} + \frac {315 \, {\left (c x^{2} + d\right )}^{\frac {11}{2}} a \mathrm {sgn}\left (x\right ) + 385 \, {\left (c x^{2} + d\right )}^{\frac {9}{2}} b c \mathrm {sgn}\left (x\right ) - 1540 \, {\left (c x^{2} + d\right )}^{\frac {9}{2}} a d \mathrm {sgn}\left (x\right ) - 1485 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} b c d \mathrm {sgn}\left (x\right ) + 2970 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} a d^{2} \mathrm {sgn}\left (x\right ) + 2079 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} b c d^{2} \mathrm {sgn}\left (x\right ) - 2772 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a d^{3} \mathrm {sgn}\left (x\right ) - 1155 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c d^{3} \mathrm {sgn}\left (x\right ) + 1155 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a d^{4} \mathrm {sgn}\left (x\right )}{3465 \, c^{5}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)*x^10,x, algorithm="giac")
Output:
16/3465*(11*b*c*d^(9/2) - 8*a*d^(11/2))*sgn(x)/c^5 + 1/3465*(315*(c*x^2 + d)^(11/2)*a*sgn(x) + 385*(c*x^2 + d)^(9/2)*b*c*sgn(x) - 1540*(c*x^2 + d)^( 9/2)*a*d*sgn(x) - 1485*(c*x^2 + d)^(7/2)*b*c*d*sgn(x) + 2970*(c*x^2 + d)^( 7/2)*a*d^2*sgn(x) + 2079*(c*x^2 + d)^(5/2)*b*c*d^2*sgn(x) - 2772*(c*x^2 + d)^(5/2)*a*d^3*sgn(x) - 1155*(c*x^2 + d)^(3/2)*b*c*d^3*sgn(x) + 1155*(c*x^ 2 + d)^(3/2)*a*d^4*sgn(x))/c^5
Time = 3.84 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.78 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^{10} \, dx=\sqrt {c+\frac {d}{x^2}}\,\left (\frac {a\,x^{11}}{11}+\frac {x\,\left (128\,a\,d^5-176\,b\,c\,d^4\right )}{3465\,c^5}+\frac {x^9\,\left (385\,b\,c^5+35\,a\,d\,c^4\right )}{3465\,c^5}-\frac {d\,x^7\,\left (8\,a\,d-11\,b\,c\right )}{693\,c^2}+\frac {2\,d^2\,x^5\,\left (8\,a\,d-11\,b\,c\right )}{1155\,c^3}-\frac {8\,d^3\,x^3\,\left (8\,a\,d-11\,b\,c\right )}{3465\,c^4}\right ) \] Input:
int(x^10*(a + b/x^2)*(c + d/x^2)^(1/2),x)
Output:
(c + d/x^2)^(1/2)*((a*x^11)/11 + (x*(128*a*d^5 - 176*b*c*d^4))/(3465*c^5) + (x^9*(385*b*c^5 + 35*a*c^4*d))/(3465*c^5) - (d*x^7*(8*a*d - 11*b*c))/(69 3*c^2) + (2*d^2*x^5*(8*a*d - 11*b*c))/(1155*c^3) - (8*d^3*x^3*(8*a*d - 11* b*c))/(3465*c^4))
Time = 0.19 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.82 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^{10} \, dx=\frac {\sqrt {c \,x^{2}+d}\, \left (315 a \,c^{5} x^{10}+35 a \,c^{4} d \,x^{8}+385 b \,c^{5} x^{8}-40 a \,c^{3} d^{2} x^{6}+55 b \,c^{4} d \,x^{6}+48 a \,c^{2} d^{3} x^{4}-66 b \,c^{3} d^{2} x^{4}-64 a c \,d^{4} x^{2}+88 b \,c^{2} d^{3} x^{2}+128 a \,d^{5}-176 b c \,d^{4}\right )}{3465 c^{5}} \] Input:
int((a+b/x^2)*(c+d/x^2)^(1/2)*x^10,x)
Output:
(sqrt(c*x**2 + d)*(315*a*c**5*x**10 + 35*a*c**4*d*x**8 - 40*a*c**3*d**2*x* *6 + 48*a*c**2*d**3*x**4 - 64*a*c*d**4*x**2 + 128*a*d**5 + 385*b*c**5*x**8 + 55*b*c**4*d*x**6 - 66*b*c**3*d**2*x**4 + 88*b*c**2*d**3*x**2 - 176*b*c* d**4))/(3465*c**5)