Integrand size = 15, antiderivative size = 92 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^{14}} \, dx=-\frac {\left (a+b x^2\right )^{7/2}}{13 a x^{13}}+\frac {6 b \left (a+b x^2\right )^{7/2}}{143 a^2 x^{11}}-\frac {8 b^2 \left (a+b x^2\right )^{7/2}}{429 a^3 x^9}+\frac {16 b^3 \left (a+b x^2\right )^{7/2}}{3003 a^4 x^7} \]
-1/13*(b*x^2+a)^(7/2)/a/x^13+6/143*b*(b*x^2+a)^(7/2)/a^2/x^11-8/429*b^2*(b *x^2+a)^(7/2)/a^3/x^9+16/3003*b^3*(b*x^2+a)^(7/2)/a^4/x^7
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^{14}} \, dx=\frac {\left (a+b x^2\right )^{7/2} \left (-231 a^3+126 a^2 b x^2-56 a b^2 x^4+16 b^3 x^6\right )}{3003 a^4 x^{13}} \]
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2}}{x^{14}} \, dx\) |
\(\Big \downarrow \) 245 |
\(\displaystyle -\frac {6 b \int \frac {\left (b x^2+a\right )^{5/2}}{x^{12}}dx}{13 a}-\frac {\left (a+b x^2\right )^{7/2}}{13 a x^{13}}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \int \frac {\left (b x^2+a\right )^{5/2}}{x^{10}}dx}{11 a}-\frac {\left (a+b x^2\right )^{7/2}}{11 a x^{11}}\right )}{13 a}-\frac {\left (a+b x^2\right )^{7/2}}{13 a x^{13}}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \left (-\frac {2 b \int \frac {\left (b x^2+a\right )^{5/2}}{x^8}dx}{9 a}-\frac {\left (a+b x^2\right )^{7/2}}{9 a x^9}\right )}{11 a}-\frac {\left (a+b x^2\right )^{7/2}}{11 a x^{11}}\right )}{13 a}-\frac {\left (a+b x^2\right )^{7/2}}{13 a x^{13}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \left (\frac {2 b \left (a+b x^2\right )^{7/2}}{63 a^2 x^7}-\frac {\left (a+b x^2\right )^{7/2}}{9 a x^9}\right )}{11 a}-\frac {\left (a+b x^2\right )^{7/2}}{11 a x^{11}}\right )}{13 a}-\frac {\left (a+b x^2\right )^{7/2}}{13 a x^{13}}\) |
-1/13*(a + b*x^2)^(7/2)/(a*x^13) - (6*b*(-1/11*(a + b*x^2)^(7/2)/(a*x^11) - (4*b*(-1/9*(a + b*x^2)^(7/2)/(a*x^9) + (2*b*(a + b*x^2)^(7/2))/(63*a^2*x ^7)))/(11*a)))/(13*a)
3.5.6.3.1 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 = 2.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} \left (-16 b^{3} x^{6}+56 a \,b^{2} x^{4}-126 a^{2} b \,x^{2}+231 a^{3}\right )}{3003 x^{13} a^{4}}\) | \(50\) |
pseudoelliptic | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} \left (-16 b^{3} x^{6}+56 a \,b^{2} x^{4}-126 a^{2} b \,x^{2}+231 a^{3}\right )}{3003 x^{13} a^{4}}\) | \(50\) |
trager | \(-\frac {\left (-16 b^{6} x^{12}+8 a \,b^{5} x^{10}-6 a^{2} b^{4} x^{8}+5 a^{3} b^{3} x^{6}+371 a^{4} b^{2} x^{4}+567 a^{5} b \,x^{2}+231 a^{6}\right ) \sqrt {b \,x^{2}+a}}{3003 x^{13} a^{4}}\) | \(83\) |
risch | \(-\frac {\left (-16 b^{6} x^{12}+8 a \,b^{5} x^{10}-6 a^{2} b^{4} x^{8}+5 a^{3} b^{3} x^{6}+371 a^{4} b^{2} x^{4}+567 a^{5} b \,x^{2}+231 a^{6}\right ) \sqrt {b \,x^{2}+a}}{3003 x^{13} a^{4}}\) | \(83\) |
default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{13 a \,x^{13}}-\frac {6 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{11 a \,x^{11}}-\frac {4 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{9 a \,x^{9}}+\frac {2 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{63 a^{2} x^{7}}\right )}{11 a}\right )}{13 a}\) | \(85\) |
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^{14}} \, dx=\frac {{\left (16 \, b^{6} x^{12} - 8 \, a b^{5} x^{10} + 6 \, a^{2} b^{4} x^{8} - 5 \, a^{3} b^{3} x^{6} - 371 \, a^{4} b^{2} x^{4} - 567 \, a^{5} b x^{2} - 231 \, a^{6}\right )} \sqrt {b x^{2} + a}}{3003 \, a^{4} x^{13}} \]
1/3003*(16*b^6*x^12 - 8*a*b^5*x^10 + 6*a^2*b^4*x^8 - 5*a^3*b^3*x^6 - 371*a ^4*b^2*x^4 - 567*a^5*b*x^2 - 231*a^6)*sqrt(b*x^2 + a)/(a^4*x^13)
Leaf count of result is larger than twice the leaf count of optimal. 721 vs. \(2 (85) = 170\).
Time = 1.42 (sec) , antiderivative size = 721, normalized size of antiderivative = 7.84 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^{14}} \, dx=- \frac {231 a^{9} b^{\frac {19}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3003 a^{7} b^{9} x^{12} + 9009 a^{6} b^{10} x^{14} + 9009 a^{5} b^{11} x^{16} + 3003 a^{4} b^{12} x^{18}} - \frac {1260 a^{8} b^{\frac {21}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{3003 a^{7} b^{9} x^{12} + 9009 a^{6} b^{10} x^{14} + 9009 a^{5} b^{11} x^{16} + 3003 a^{4} b^{12} x^{18}} - \frac {2765 a^{7} b^{\frac {23}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{3003 a^{7} b^{9} x^{12} + 9009 a^{6} b^{10} x^{14} + 9009 a^{5} b^{11} x^{16} + 3003 a^{4} b^{12} x^{18}} - \frac {3050 a^{6} b^{\frac {25}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{3003 a^{7} b^{9} x^{12} + 9009 a^{6} b^{10} x^{14} + 9009 a^{5} b^{11} x^{16} + 3003 a^{4} b^{12} x^{18}} - \frac {1689 a^{5} b^{\frac {27}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{3003 a^{7} b^{9} x^{12} + 9009 a^{6} b^{10} x^{14} + 9009 a^{5} b^{11} x^{16} + 3003 a^{4} b^{12} x^{18}} - \frac {376 a^{4} b^{\frac {29}{2}} x^{10} \sqrt {\frac {a}{b x^{2}} + 1}}{3003 a^{7} b^{9} x^{12} + 9009 a^{6} b^{10} x^{14} + 9009 a^{5} b^{11} x^{16} + 3003 a^{4} b^{12} x^{18}} + \frac {5 a^{3} b^{\frac {31}{2}} x^{12} \sqrt {\frac {a}{b x^{2}} + 1}}{3003 a^{7} b^{9} x^{12} + 9009 a^{6} b^{10} x^{14} + 9009 a^{5} b^{11} x^{16} + 3003 a^{4} b^{12} x^{18}} + \frac {30 a^{2} b^{\frac {33}{2}} x^{14} \sqrt {\frac {a}{b x^{2}} + 1}}{3003 a^{7} b^{9} x^{12} + 9009 a^{6} b^{10} x^{14} + 9009 a^{5} b^{11} x^{16} + 3003 a^{4} b^{12} x^{18}} + \frac {40 a b^{\frac {35}{2}} x^{16} \sqrt {\frac {a}{b x^{2}} + 1}}{3003 a^{7} b^{9} x^{12} + 9009 a^{6} b^{10} x^{14} + 9009 a^{5} b^{11} x^{16} + 3003 a^{4} b^{12} x^{18}} + \frac {16 b^{\frac {37}{2}} x^{18} \sqrt {\frac {a}{b x^{2}} + 1}}{3003 a^{7} b^{9} x^{12} + 9009 a^{6} b^{10} x^{14} + 9009 a^{5} b^{11} x^{16} + 3003 a^{4} b^{12} x^{18}} \]
-231*a**9*b**(19/2)*sqrt(a/(b*x**2) + 1)/(3003*a**7*b**9*x**12 + 9009*a**6 *b**10*x**14 + 9009*a**5*b**11*x**16 + 3003*a**4*b**12*x**18) - 1260*a**8* b**(21/2)*x**2*sqrt(a/(b*x**2) + 1)/(3003*a**7*b**9*x**12 + 9009*a**6*b**1 0*x**14 + 9009*a**5*b**11*x**16 + 3003*a**4*b**12*x**18) - 2765*a**7*b**(2 3/2)*x**4*sqrt(a/(b*x**2) + 1)/(3003*a**7*b**9*x**12 + 9009*a**6*b**10*x** 14 + 9009*a**5*b**11*x**16 + 3003*a**4*b**12*x**18) - 3050*a**6*b**(25/2)* x**6*sqrt(a/(b*x**2) + 1)/(3003*a**7*b**9*x**12 + 9009*a**6*b**10*x**14 + 9009*a**5*b**11*x**16 + 3003*a**4*b**12*x**18) - 1689*a**5*b**(27/2)*x**8* sqrt(a/(b*x**2) + 1)/(3003*a**7*b**9*x**12 + 9009*a**6*b**10*x**14 + 9009* a**5*b**11*x**16 + 3003*a**4*b**12*x**18) - 376*a**4*b**(29/2)*x**10*sqrt( a/(b*x**2) + 1)/(3003*a**7*b**9*x**12 + 9009*a**6*b**10*x**14 + 9009*a**5* b**11*x**16 + 3003*a**4*b**12*x**18) + 5*a**3*b**(31/2)*x**12*sqrt(a/(b*x* *2) + 1)/(3003*a**7*b**9*x**12 + 9009*a**6*b**10*x**14 + 9009*a**5*b**11*x **16 + 3003*a**4*b**12*x**18) + 30*a**2*b**(33/2)*x**14*sqrt(a/(b*x**2) + 1)/(3003*a**7*b**9*x**12 + 9009*a**6*b**10*x**14 + 9009*a**5*b**11*x**16 + 3003*a**4*b**12*x**18) + 40*a*b**(35/2)*x**16*sqrt(a/(b*x**2) + 1)/(3003* a**7*b**9*x**12 + 9009*a**6*b**10*x**14 + 9009*a**5*b**11*x**16 + 3003*a** 4*b**12*x**18) + 16*b**(37/2)*x**18*sqrt(a/(b*x**2) + 1)/(3003*a**7*b**9*x **12 + 9009*a**6*b**10*x**14 + 9009*a**5*b**11*x**16 + 3003*a**4*b**12*x** 18)
Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^{14}} \, dx=\frac {16 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}}{3003 \, a^{4} x^{7}} - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}}{429 \, a^{3} x^{9}} + \frac {6 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b}{143 \, a^{2} x^{11}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{13 \, a x^{13}} \]
16/3003*(b*x^2 + a)^(7/2)*b^3/(a^4*x^7) - 8/429*(b*x^2 + a)^(7/2)*b^2/(a^3 *x^9) + 6/143*(b*x^2 + a)^(7/2)*b/(a^2*x^11) - 1/13*(b*x^2 + a)^(7/2)/(a*x ^13)
Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (76) = 152\).
Time = 0.33 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.98 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^{14}} \, dx=\frac {32 \, {\left (3003 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{18} b^{\frac {13}{2}} + 9009 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{16} a b^{\frac {13}{2}} + 18018 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{14} a^{2} b^{\frac {13}{2}} + 16302 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{12} a^{3} b^{\frac {13}{2}} + 10296 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} a^{4} b^{\frac {13}{2}} + 2288 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a^{5} b^{\frac {13}{2}} + 286 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{6} b^{\frac {13}{2}} - 78 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{7} b^{\frac {13}{2}} + 13 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{8} b^{\frac {13}{2}} - a^{9} b^{\frac {13}{2}}\right )}}{3003 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{13}} \]
32/3003*(3003*(sqrt(b)*x - sqrt(b*x^2 + a))^18*b^(13/2) + 9009*(sqrt(b)*x - sqrt(b*x^2 + a))^16*a*b^(13/2) + 18018*(sqrt(b)*x - sqrt(b*x^2 + a))^14* a^2*b^(13/2) + 16302*(sqrt(b)*x - sqrt(b*x^2 + a))^12*a^3*b^(13/2) + 10296 *(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^4*b^(13/2) + 2288*(sqrt(b)*x - sqrt(b* x^2 + a))^8*a^5*b^(13/2) + 286*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^6*b^(13/2 ) - 78*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^7*b^(13/2) + 13*(sqrt(b)*x - sqrt (b*x^2 + a))^2*a^8*b^(13/2) - a^9*b^(13/2))/((sqrt(b)*x - sqrt(b*x^2 + a)) ^2 - a)^13
Time = 6.06 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.42 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^{14}} \, dx=\frac {2\,b^4\,\sqrt {b\,x^2+a}}{1001\,a^2\,x^5}-\frac {53\,b^2\,\sqrt {b\,x^2+a}}{429\,x^9}-\frac {5\,b^3\,\sqrt {b\,x^2+a}}{3003\,a\,x^7}-\frac {a^2\,\sqrt {b\,x^2+a}}{13\,x^{13}}-\frac {8\,b^5\,\sqrt {b\,x^2+a}}{3003\,a^3\,x^3}+\frac {16\,b^6\,\sqrt {b\,x^2+a}}{3003\,a^4\,x}-\frac {27\,a\,b\,\sqrt {b\,x^2+a}}{143\,x^{11}} \]