Integrand size = 16, antiderivative size = 93 \[ \int \frac {1}{x^{11} \left (a-b x^4\right )^{3/2}} \, dx=\frac {1}{2 a x^{10} \sqrt {a-b x^4}}-\frac {3 \sqrt {a-b x^4}}{5 a^2 x^{10}}-\frac {4 b \sqrt {a-b x^4}}{5 a^3 x^6}-\frac {8 b^2 \sqrt {a-b x^4}}{5 a^4 x^2} \] Output:
1/2/a/x^10/(-b*x^4+a)^(1/2)-3/5*(-b*x^4+a)^(1/2)/a^2/x^10-4/5*b*(-b*x^4+a) ^(1/2)/a^3/x^6-8/5*b^2*(-b*x^4+a)^(1/2)/a^4/x^2
Time = 0.42 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^{11} \left (a-b x^4\right )^{3/2}} \, dx=-\frac {a^3+2 a^2 b x^4+8 a b^2 x^8-16 b^3 x^{12}}{10 a^4 x^{10} \sqrt {a-b x^4}} \] Input:
Integrate[1/(x^11*(a - b*x^4)^(3/2)),x]
Output:
-1/10*(a^3 + 2*a^2*b*x^4 + 8*a*b^2*x^8 - 16*b^3*x^12)/(a^4*x^10*Sqrt[a - b *x^4])
Time = 0.36 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.
\(\displaystyle \int \frac {1}{x^{11} \left (a-b x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {6 b \int \frac {1}{x^7 \left (a-b x^4\right )^{3/2}}dx}{5 a}-\frac {1}{10 a x^{10} \sqrt {a-b x^4}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {6 b \left (\frac {4 b \int \frac {1}{x^3 \left (a-b x^4\right )^{3/2}}dx}{3 a}-\frac {1}{6 a x^6 \sqrt {a-b x^4}}\right )}{5 a}-\frac {1}{10 a x^{10} \sqrt {a-b x^4}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {6 b \left (\frac {4 b \left (\frac {2 b \int \frac {x}{\left (a-b x^4\right )^{3/2}}dx}{a}-\frac {1}{2 a x^2 \sqrt {a-b x^4}}\right )}{3 a}-\frac {1}{6 a x^6 \sqrt {a-b x^4}}\right )}{5 a}-\frac {1}{10 a x^{10} \sqrt {a-b x^4}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {6 b \left (\frac {4 b \left (\frac {b x^2}{a^2 \sqrt {a-b x^4}}-\frac {1}{2 a x^2 \sqrt {a-b x^4}}\right )}{3 a}-\frac {1}{6 a x^6 \sqrt {a-b x^4}}\right )}{5 a}-\frac {1}{10 a x^{10} \sqrt {a-b x^4}}\) |
Input:
Int[1/(x^11*(a - b*x^4)^(3/2)),x]
Output:
-1/10*1/(a*x^10*Sqrt[a - b*x^4]) + (6*b*(-1/6*1/(a*x^6*Sqrt[a - b*x^4]) + (4*b*(-1/2*1/(a*x^2*Sqrt[a - b*x^4]) + (b*x^2)/(a^2*Sqrt[a - b*x^4])))/(3* a)))/(5*a)
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]
Time = 0.98 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.53
method | result | size |
gosper | \(-\frac {-16 b^{3} x^{12}+8 a \,b^{2} x^{8}+2 a^{2} b \,x^{4}+a^{3}}{10 x^{10} \sqrt {-b \,x^{4}+a}\, a^{4}}\) | \(49\) |
default | \(-\frac {-16 b^{3} x^{12}+8 a \,b^{2} x^{8}+2 a^{2} b \,x^{4}+a^{3}}{10 x^{10} \sqrt {-b \,x^{4}+a}\, a^{4}}\) | \(49\) |
trager | \(-\frac {-16 b^{3} x^{12}+8 a \,b^{2} x^{8}+2 a^{2} b \,x^{4}+a^{3}}{10 x^{10} \sqrt {-b \,x^{4}+a}\, a^{4}}\) | \(49\) |
elliptic | \(-\frac {-16 b^{3} x^{12}+8 a \,b^{2} x^{8}+2 a^{2} b \,x^{4}+a^{3}}{10 x^{10} \sqrt {-b \,x^{4}+a}\, a^{4}}\) | \(49\) |
pseudoelliptic | \(-\frac {-16 b^{3} x^{12}+8 a \,b^{2} x^{8}+2 a^{2} b \,x^{4}+a^{3}}{10 x^{10} \sqrt {-b \,x^{4}+a}\, a^{4}}\) | \(49\) |
orering | \(-\frac {-16 b^{3} x^{12}+8 a \,b^{2} x^{8}+2 a^{2} b \,x^{4}+a^{3}}{10 x^{10} \sqrt {-b \,x^{4}+a}\, a^{4}}\) | \(49\) |
risch | \(-\frac {\sqrt {-b \,x^{4}+a}\, \left (11 b^{2} x^{8}+3 a b \,x^{4}+a^{2}\right )}{10 a^{4} x^{10}}+\frac {x^{2} b^{3}}{2 \sqrt {-b \,x^{4}+a}\, a^{4}}\) | \(60\) |
Input:
int(1/x^11/(-b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/10*(-16*b^3*x^12+8*a*b^2*x^8+2*a^2*b*x^4+a^3)/x^10/(-b*x^4+a)^(1/2)/a^4
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x^{11} \left (a-b x^4\right )^{3/2}} \, dx=-\frac {{\left (16 \, b^{3} x^{12} - 8 \, a b^{2} x^{8} - 2 \, a^{2} b x^{4} - a^{3}\right )} \sqrt {-b x^{4} + a}}{10 \, {\left (a^{4} b x^{14} - a^{5} x^{10}\right )}} \] Input:
integrate(1/x^11/(-b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
-1/10*(16*b^3*x^12 - 8*a*b^2*x^8 - 2*a^2*b*x^4 - a^3)*sqrt(-b*x^4 + a)/(a^ 4*b*x^14 - a^5*x^10)
Result contains complex when optimal does not.
Time = 6.21 (sec) , antiderivative size = 724, normalized size of antiderivative = 7.78 \[ \int \frac {1}{x^{11} \left (a-b x^4\right )^{3/2}} \, dx=\begin {cases} \frac {a^{5} b^{\frac {19}{2}} \sqrt {\frac {a}{b x^{4}} - 1}}{- 10 a^{7} b^{9} x^{8} + 30 a^{6} b^{10} x^{12} - 30 a^{5} b^{11} x^{16} + 10 a^{4} b^{12} x^{20}} + \frac {5 a^{3} b^{\frac {23}{2}} x^{8} \sqrt {\frac {a}{b x^{4}} - 1}}{- 10 a^{7} b^{9} x^{8} + 30 a^{6} b^{10} x^{12} - 30 a^{5} b^{11} x^{16} + 10 a^{4} b^{12} x^{20}} - \frac {30 a^{2} b^{\frac {25}{2}} x^{12} \sqrt {\frac {a}{b x^{4}} - 1}}{- 10 a^{7} b^{9} x^{8} + 30 a^{6} b^{10} x^{12} - 30 a^{5} b^{11} x^{16} + 10 a^{4} b^{12} x^{20}} + \frac {40 a b^{\frac {27}{2}} x^{16} \sqrt {\frac {a}{b x^{4}} - 1}}{- 10 a^{7} b^{9} x^{8} + 30 a^{6} b^{10} x^{12} - 30 a^{5} b^{11} x^{16} + 10 a^{4} b^{12} x^{20}} - \frac {16 b^{\frac {29}{2}} x^{20} \sqrt {\frac {a}{b x^{4}} - 1}}{- 10 a^{7} b^{9} x^{8} + 30 a^{6} b^{10} x^{12} - 30 a^{5} b^{11} x^{16} + 10 a^{4} b^{12} x^{20}} & \text {for}\: \left |{\frac {a}{b x^{4}}}\right | > 1 \\\frac {i a^{5} b^{\frac {19}{2}} \sqrt {- \frac {a}{b x^{4}} + 1}}{- 10 a^{7} b^{9} x^{8} + 30 a^{6} b^{10} x^{12} - 30 a^{5} b^{11} x^{16} + 10 a^{4} b^{12} x^{20}} + \frac {5 i a^{3} b^{\frac {23}{2}} x^{8} \sqrt {- \frac {a}{b x^{4}} + 1}}{- 10 a^{7} b^{9} x^{8} + 30 a^{6} b^{10} x^{12} - 30 a^{5} b^{11} x^{16} + 10 a^{4} b^{12} x^{20}} - \frac {30 i a^{2} b^{\frac {25}{2}} x^{12} \sqrt {- \frac {a}{b x^{4}} + 1}}{- 10 a^{7} b^{9} x^{8} + 30 a^{6} b^{10} x^{12} - 30 a^{5} b^{11} x^{16} + 10 a^{4} b^{12} x^{20}} + \frac {40 i a b^{\frac {27}{2}} x^{16} \sqrt {- \frac {a}{b x^{4}} + 1}}{- 10 a^{7} b^{9} x^{8} + 30 a^{6} b^{10} x^{12} - 30 a^{5} b^{11} x^{16} + 10 a^{4} b^{12} x^{20}} - \frac {16 i b^{\frac {29}{2}} x^{20} \sqrt {- \frac {a}{b x^{4}} + 1}}{- 10 a^{7} b^{9} x^{8} + 30 a^{6} b^{10} x^{12} - 30 a^{5} b^{11} x^{16} + 10 a^{4} b^{12} x^{20}} & \text {otherwise} \end {cases} \] Input:
integrate(1/x**11/(-b*x**4+a)**(3/2),x)
Output:
Piecewise((a**5*b**(19/2)*sqrt(a/(b*x**4) - 1)/(-10*a**7*b**9*x**8 + 30*a* *6*b**10*x**12 - 30*a**5*b**11*x**16 + 10*a**4*b**12*x**20) + 5*a**3*b**(2 3/2)*x**8*sqrt(a/(b*x**4) - 1)/(-10*a**7*b**9*x**8 + 30*a**6*b**10*x**12 - 30*a**5*b**11*x**16 + 10*a**4*b**12*x**20) - 30*a**2*b**(25/2)*x**12*sqrt (a/(b*x**4) - 1)/(-10*a**7*b**9*x**8 + 30*a**6*b**10*x**12 - 30*a**5*b**11 *x**16 + 10*a**4*b**12*x**20) + 40*a*b**(27/2)*x**16*sqrt(a/(b*x**4) - 1)/ (-10*a**7*b**9*x**8 + 30*a**6*b**10*x**12 - 30*a**5*b**11*x**16 + 10*a**4* b**12*x**20) - 16*b**(29/2)*x**20*sqrt(a/(b*x**4) - 1)/(-10*a**7*b**9*x**8 + 30*a**6*b**10*x**12 - 30*a**5*b**11*x**16 + 10*a**4*b**12*x**20), Abs(a /(b*x**4)) > 1), (I*a**5*b**(19/2)*sqrt(-a/(b*x**4) + 1)/(-10*a**7*b**9*x* *8 + 30*a**6*b**10*x**12 - 30*a**5*b**11*x**16 + 10*a**4*b**12*x**20) + 5* I*a**3*b**(23/2)*x**8*sqrt(-a/(b*x**4) + 1)/(-10*a**7*b**9*x**8 + 30*a**6* b**10*x**12 - 30*a**5*b**11*x**16 + 10*a**4*b**12*x**20) - 30*I*a**2*b**(2 5/2)*x**12*sqrt(-a/(b*x**4) + 1)/(-10*a**7*b**9*x**8 + 30*a**6*b**10*x**12 - 30*a**5*b**11*x**16 + 10*a**4*b**12*x**20) + 40*I*a*b**(27/2)*x**16*sqr t(-a/(b*x**4) + 1)/(-10*a**7*b**9*x**8 + 30*a**6*b**10*x**12 - 30*a**5*b** 11*x**16 + 10*a**4*b**12*x**20) - 16*I*b**(29/2)*x**20*sqrt(-a/(b*x**4) + 1)/(-10*a**7*b**9*x**8 + 30*a**6*b**10*x**12 - 30*a**5*b**11*x**16 + 10*a* *4*b**12*x**20), True))
Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^{11} \left (a-b x^4\right )^{3/2}} \, dx=\frac {b^{3} x^{2}}{2 \, \sqrt {-b x^{4} + a} a^{4}} - \frac {\frac {15 \, \sqrt {-b x^{4} + a} b^{2}}{x^{2}} + \frac {5 \, {\left (-b x^{4} + a\right )}^{\frac {3}{2}} b}{x^{6}} + \frac {{\left (-b x^{4} + a\right )}^{\frac {5}{2}}}{x^{10}}}{10 \, a^{4}} \] Input:
integrate(1/x^11/(-b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
1/2*b^3*x^2/(sqrt(-b*x^4 + a)*a^4) - 1/10*(15*sqrt(-b*x^4 + a)*b^2/x^2 + 5 *(-b*x^4 + a)^(3/2)*b/x^6 + (-b*x^4 + a)^(5/2)/x^10)/a^4
Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (77) = 154\).
Time = 0.14 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.42 \[ \int \frac {1}{x^{11} \left (a-b x^4\right )^{3/2}} \, dx=-\frac {\sqrt {-b x^{4} + a} b^{3} x^{2}}{2 \, {\left (b x^{4} - a\right )} a^{4}} + \frac {5 \, {\left (\sqrt {-b} x^{2} - \sqrt {-b x^{4} + a}\right )}^{8} \sqrt {-b} b^{2} - 30 \, {\left (\sqrt {-b} x^{2} - \sqrt {-b x^{4} + a}\right )}^{6} a \sqrt {-b} b^{2} + 80 \, {\left (\sqrt {-b} x^{2} - \sqrt {-b x^{4} + a}\right )}^{4} a^{2} \sqrt {-b} b^{2} - 50 \, {\left (\sqrt {-b} x^{2} - \sqrt {-b x^{4} + a}\right )}^{2} a^{3} \sqrt {-b} b^{2} + 11 \, a^{4} \sqrt {-b} b^{2}}{5 \, {\left ({\left (\sqrt {-b} x^{2} - \sqrt {-b x^{4} + a}\right )}^{2} - a\right )}^{5} a^{3}} \] Input:
integrate(1/x^11/(-b*x^4+a)^(3/2),x, algorithm="giac")
Output:
-1/2*sqrt(-b*x^4 + a)*b^3*x^2/((b*x^4 - a)*a^4) + 1/5*(5*(sqrt(-b)*x^2 - s qrt(-b*x^4 + a))^8*sqrt(-b)*b^2 - 30*(sqrt(-b)*x^2 - sqrt(-b*x^4 + a))^6*a *sqrt(-b)*b^2 + 80*(sqrt(-b)*x^2 - sqrt(-b*x^4 + a))^4*a^2*sqrt(-b)*b^2 - 50*(sqrt(-b)*x^2 - sqrt(-b*x^4 + a))^2*a^3*sqrt(-b)*b^2 + 11*a^4*sqrt(-b)* b^2)/(((sqrt(-b)*x^2 - sqrt(-b*x^4 + a))^2 - a)^5*a^3)
Time = 0.71 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^{11} \left (a-b x^4\right )^{3/2}} \, dx=\frac {b^3\,x^2}{2\,a^4\,\sqrt {a-b\,x^4}}-\frac {3\,b\,\sqrt {a-b\,x^4}}{10\,a^3\,x^6}-\frac {11\,b^2\,\sqrt {a-b\,x^4}}{10\,a^4\,x^2}-\frac {\sqrt {a-b\,x^4}}{10\,a^2\,x^{10}} \] Input:
int(1/(x^11*(a - b*x^4)^(3/2)),x)
Output:
(b^3*x^2)/(2*a^4*(a - b*x^4)^(1/2)) - (3*b*(a - b*x^4)^(1/2))/(10*a^3*x^6) - (11*b^2*(a - b*x^4)^(1/2))/(10*a^4*x^2) - (a - b*x^4)^(1/2)/(10*a^2*x^1 0)
Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x^{11} \left (a-b x^4\right )^{3/2}} \, dx=\frac {16 b^{3} x^{12}-8 a \,b^{2} x^{8}-2 a^{2} b \,x^{4}-a^{3}}{10 \sqrt {-b \,x^{4}+a}\, a^{4} x^{10}} \] Input:
int(1/x^11/(-b*x^4+a)^(3/2),x)
Output:
( - a**3 - 2*a**2*b*x**4 - 8*a*b**2*x**8 + 16*b**3*x**12)/(10*sqrt(a - b*x **4)*a**4*x**10)