Integrand size = 15, antiderivative size = 92 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^{12}} \, dx=-\frac {\left (a+b x^2\right )^{5/2}}{11 a x^{11}}+\frac {2 b \left (a+b x^2\right )^{5/2}}{33 a^2 x^9}-\frac {8 b^2 \left (a+b x^2\right )^{5/2}}{231 a^3 x^7}+\frac {16 b^3 \left (a+b x^2\right )^{5/2}}{1155 a^4 x^5} \]
-1/11*(b*x^2+a)^(5/2)/a/x^11+2/33*b*(b*x^2+a)^(5/2)/a^2/x^9-8/231*b^2*(b*x ^2+a)^(5/2)/a^3/x^7+16/1155*b^3*(b*x^2+a)^(5/2)/a^4/x^5
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^{12}} \, dx=\frac {\left (a+b x^2\right )^{5/2} \left (-105 a^3+70 a^2 b x^2-40 a b^2 x^4+16 b^3 x^6\right )}{1155 a^4 x^{11}} \]
Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {245, 245, 245, 242}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2}}{x^{12}} \, dx\) |
\(\Big \downarrow \) 245 |
\(\displaystyle -\frac {6 b \int \frac {\left (b x^2+a\right )^{3/2}}{x^{10}}dx}{11 a}-\frac {\left (a+b x^2\right )^{5/2}}{11 a x^{11}}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \int \frac {\left (b x^2+a\right )^{3/2}}{x^8}dx}{9 a}-\frac {\left (a+b x^2\right )^{5/2}}{9 a x^9}\right )}{11 a}-\frac {\left (a+b x^2\right )^{5/2}}{11 a x^{11}}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \left (-\frac {2 b \int \frac {\left (b x^2+a\right )^{3/2}}{x^6}dx}{7 a}-\frac {\left (a+b x^2\right )^{5/2}}{7 a x^7}\right )}{9 a}-\frac {\left (a+b x^2\right )^{5/2}}{9 a x^9}\right )}{11 a}-\frac {\left (a+b x^2\right )^{5/2}}{11 a x^{11}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \left (\frac {2 b \left (a+b x^2\right )^{5/2}}{35 a^2 x^5}-\frac {\left (a+b x^2\right )^{5/2}}{7 a x^7}\right )}{9 a}-\frac {\left (a+b x^2\right )^{5/2}}{9 a x^9}\right )}{11 a}-\frac {\left (a+b x^2\right )^{5/2}}{11 a x^{11}}\) |
-1/11*(a + b*x^2)^(5/2)/(a*x^11) - (6*b*(-1/9*(a + b*x^2)^(5/2)/(a*x^9) - (4*b*(-1/7*(a + b*x^2)^(5/2)/(a*x^7) + (2*b*(a + b*x^2)^(5/2))/(35*a^2*x^5 )))/(9*a)))/(11*a)
3.4.86.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Time = 1.89 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (-16 b^{3} x^{6}+40 a \,b^{2} x^{4}-70 a^{2} b \,x^{2}+105 a^{3}\right )}{1155 x^{11} a^{4}}\) | \(50\) |
pseudoelliptic | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (-16 b^{3} x^{6}+40 a \,b^{2} x^{4}-70 a^{2} b \,x^{2}+105 a^{3}\right )}{1155 x^{11} a^{4}}\) | \(50\) |
trager | \(-\frac {\left (-16 b^{5} x^{10}+8 a \,b^{4} x^{8}-6 a^{2} b^{3} x^{6}+5 a^{3} b^{2} x^{4}+140 a^{4} b \,x^{2}+105 a^{5}\right ) \sqrt {b \,x^{2}+a}}{1155 x^{11} a^{4}}\) | \(72\) |
risch | \(-\frac {\left (-16 b^{5} x^{10}+8 a \,b^{4} x^{8}-6 a^{2} b^{3} x^{6}+5 a^{3} b^{2} x^{4}+140 a^{4} b \,x^{2}+105 a^{5}\right ) \sqrt {b \,x^{2}+a}}{1155 x^{11} a^{4}}\) | \(72\) |
default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{11 a \,x^{11}}-\frac {6 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{9 a \,x^{9}}-\frac {4 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 a \,x^{7}}+\frac {2 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 a^{2} x^{5}}\right )}{9 a}\right )}{11 a}\) | \(85\) |
Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^{12}} \, dx=\frac {{\left (16 \, b^{5} x^{10} - 8 \, a b^{4} x^{8} + 6 \, a^{2} b^{3} x^{6} - 5 \, a^{3} b^{2} x^{4} - 140 \, a^{4} b x^{2} - 105 \, a^{5}\right )} \sqrt {b x^{2} + a}}{1155 \, a^{4} x^{11}} \]
1/1155*(16*b^5*x^10 - 8*a*b^4*x^8 + 6*a^2*b^3*x^6 - 5*a^3*b^2*x^4 - 140*a^ 4*b*x^2 - 105*a^5)*sqrt(b*x^2 + a)/(a^4*x^11)
Leaf count of result is larger than twice the leaf count of optimal. 648 vs. \(2 (85) = 170\).
Time = 1.14 (sec) , antiderivative size = 648, normalized size of antiderivative = 7.04 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^{12}} \, dx=- \frac {105 a^{8} b^{\frac {19}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{1155 a^{7} b^{9} x^{10} + 3465 a^{6} b^{10} x^{12} + 3465 a^{5} b^{11} x^{14} + 1155 a^{4} b^{12} x^{16}} - \frac {455 a^{7} b^{\frac {21}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{1155 a^{7} b^{9} x^{10} + 3465 a^{6} b^{10} x^{12} + 3465 a^{5} b^{11} x^{14} + 1155 a^{4} b^{12} x^{16}} - \frac {740 a^{6} b^{\frac {23}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{1155 a^{7} b^{9} x^{10} + 3465 a^{6} b^{10} x^{12} + 3465 a^{5} b^{11} x^{14} + 1155 a^{4} b^{12} x^{16}} - \frac {534 a^{5} b^{\frac {25}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{1155 a^{7} b^{9} x^{10} + 3465 a^{6} b^{10} x^{12} + 3465 a^{5} b^{11} x^{14} + 1155 a^{4} b^{12} x^{16}} - \frac {145 a^{4} b^{\frac {27}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{1155 a^{7} b^{9} x^{10} + 3465 a^{6} b^{10} x^{12} + 3465 a^{5} b^{11} x^{14} + 1155 a^{4} b^{12} x^{16}} + \frac {5 a^{3} b^{\frac {29}{2}} x^{10} \sqrt {\frac {a}{b x^{2}} + 1}}{1155 a^{7} b^{9} x^{10} + 3465 a^{6} b^{10} x^{12} + 3465 a^{5} b^{11} x^{14} + 1155 a^{4} b^{12} x^{16}} + \frac {30 a^{2} b^{\frac {31}{2}} x^{12} \sqrt {\frac {a}{b x^{2}} + 1}}{1155 a^{7} b^{9} x^{10} + 3465 a^{6} b^{10} x^{12} + 3465 a^{5} b^{11} x^{14} + 1155 a^{4} b^{12} x^{16}} + \frac {40 a b^{\frac {33}{2}} x^{14} \sqrt {\frac {a}{b x^{2}} + 1}}{1155 a^{7} b^{9} x^{10} + 3465 a^{6} b^{10} x^{12} + 3465 a^{5} b^{11} x^{14} + 1155 a^{4} b^{12} x^{16}} + \frac {16 b^{\frac {35}{2}} x^{16} \sqrt {\frac {a}{b x^{2}} + 1}}{1155 a^{7} b^{9} x^{10} + 3465 a^{6} b^{10} x^{12} + 3465 a^{5} b^{11} x^{14} + 1155 a^{4} b^{12} x^{16}} \]
-105*a**8*b**(19/2)*sqrt(a/(b*x**2) + 1)/(1155*a**7*b**9*x**10 + 3465*a**6 *b**10*x**12 + 3465*a**5*b**11*x**14 + 1155*a**4*b**12*x**16) - 455*a**7*b **(21/2)*x**2*sqrt(a/(b*x**2) + 1)/(1155*a**7*b**9*x**10 + 3465*a**6*b**10 *x**12 + 3465*a**5*b**11*x**14 + 1155*a**4*b**12*x**16) - 740*a**6*b**(23/ 2)*x**4*sqrt(a/(b*x**2) + 1)/(1155*a**7*b**9*x**10 + 3465*a**6*b**10*x**12 + 3465*a**5*b**11*x**14 + 1155*a**4*b**12*x**16) - 534*a**5*b**(25/2)*x** 6*sqrt(a/(b*x**2) + 1)/(1155*a**7*b**9*x**10 + 3465*a**6*b**10*x**12 + 346 5*a**5*b**11*x**14 + 1155*a**4*b**12*x**16) - 145*a**4*b**(27/2)*x**8*sqrt (a/(b*x**2) + 1)/(1155*a**7*b**9*x**10 + 3465*a**6*b**10*x**12 + 3465*a**5 *b**11*x**14 + 1155*a**4*b**12*x**16) + 5*a**3*b**(29/2)*x**10*sqrt(a/(b*x **2) + 1)/(1155*a**7*b**9*x**10 + 3465*a**6*b**10*x**12 + 3465*a**5*b**11* x**14 + 1155*a**4*b**12*x**16) + 30*a**2*b**(31/2)*x**12*sqrt(a/(b*x**2) + 1)/(1155*a**7*b**9*x**10 + 3465*a**6*b**10*x**12 + 3465*a**5*b**11*x**14 + 1155*a**4*b**12*x**16) + 40*a*b**(33/2)*x**14*sqrt(a/(b*x**2) + 1)/(1155 *a**7*b**9*x**10 + 3465*a**6*b**10*x**12 + 3465*a**5*b**11*x**14 + 1155*a* *4*b**12*x**16) + 16*b**(35/2)*x**16*sqrt(a/(b*x**2) + 1)/(1155*a**7*b**9* x**10 + 3465*a**6*b**10*x**12 + 3465*a**5*b**11*x**14 + 1155*a**4*b**12*x* *16)
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^{12}} \, dx=\frac {16 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}}{1155 \, a^{4} x^{5}} - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}}{231 \, a^{3} x^{7}} + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{33 \, a^{2} x^{9}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{11 \, a x^{11}} \]
16/1155*(b*x^2 + a)^(5/2)*b^3/(a^4*x^5) - 8/231*(b*x^2 + a)^(5/2)*b^2/(a^3 *x^7) + 2/33*(b*x^2 + a)^(5/2)*b/(a^2*x^9) - 1/11*(b*x^2 + a)^(5/2)/(a*x^1 1)
Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (76) = 152\).
Time = 0.31 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.39 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^{12}} \, dx=\frac {32 \, {\left (1155 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{14} b^{\frac {11}{2}} + 2079 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{12} a b^{\frac {11}{2}} + 2541 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} a^{2} b^{\frac {11}{2}} + 825 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a^{3} b^{\frac {11}{2}} + 165 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{4} b^{\frac {11}{2}} - 55 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{5} b^{\frac {11}{2}} + 11 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{6} b^{\frac {11}{2}} - a^{7} b^{\frac {11}{2}}\right )}}{1155 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{11}} \]
32/1155*(1155*(sqrt(b)*x - sqrt(b*x^2 + a))^14*b^(11/2) + 2079*(sqrt(b)*x - sqrt(b*x^2 + a))^12*a*b^(11/2) + 2541*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a ^2*b^(11/2) + 825*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^3*b^(11/2) + 165*(sqrt (b)*x - sqrt(b*x^2 + a))^6*a^4*b^(11/2) - 55*(sqrt(b)*x - sqrt(b*x^2 + a)) ^4*a^5*b^(11/2) + 11*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^6*b^(11/2) - a^7*b^ (11/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^11
Time = 5.40 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^{12}} \, dx=\frac {2\,b^3\,\sqrt {b\,x^2+a}}{385\,a^2\,x^5}-\frac {4\,b\,\sqrt {b\,x^2+a}}{33\,x^9}-\frac {b^2\,\sqrt {b\,x^2+a}}{231\,a\,x^7}-\frac {a\,\sqrt {b\,x^2+a}}{11\,x^{11}}-\frac {8\,b^4\,\sqrt {b\,x^2+a}}{1155\,a^3\,x^3}+\frac {16\,b^5\,\sqrt {b\,x^2+a}}{1155\,a^4\,x} \]