Integrand size = 13, antiderivative size = 100 \[ \int \frac {x^{13}}{\left (a+b x^2\right )^3} \, dx=-\frac {5 a^3 x^2}{b^6}+\frac {3 a^2 x^4}{2 b^5}-\frac {a x^6}{2 b^4}+\frac {x^8}{8 b^3}-\frac {a^6}{4 b^7 \left (a+b x^2\right )^2}+\frac {3 a^5}{b^7 \left (a+b x^2\right )}+\frac {15 a^4 \log \left (a+b x^2\right )}{2 b^7} \] Output:
-5*a^3*x^2/b^6+3/2*a^2*x^4/b^5-1/2*a*x^6/b^4+1/8*x^8/b^3-1/4*a^6/b^7/(b*x^ 2+a)^2+3*a^5/b^7/(b*x^2+a)+15/2*a^4*ln(b*x^2+a)/b^7
Time = 0.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.85 \[ \int \frac {x^{13}}{\left (a+b x^2\right )^3} \, dx=\frac {-40 a^3 b x^2+12 a^2 b^2 x^4-4 a b^3 x^6+b^4 x^8-\frac {2 a^6}{\left (a+b x^2\right )^2}+\frac {24 a^5}{a+b x^2}+60 a^4 \log \left (a+b x^2\right )}{8 b^7} \] Input:
Integrate[x^13/(a + b*x^2)^3,x]
Output:
(-40*a^3*b*x^2 + 12*a^2*b^2*x^4 - 4*a*b^3*x^6 + b^4*x^8 - (2*a^6)/(a + b*x ^2)^2 + (24*a^5)/(a + b*x^2) + 60*a^4*Log[a + b*x^2])/(8*b^7)
Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {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.
\(\displaystyle \int \frac {x^{13}}{\left (a+b x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {x^{12}}{\left (b x^2+a\right )^3}dx^2\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \int \left (\frac {a^6}{b^6 \left (b x^2+a\right )^3}-\frac {6 a^5}{b^6 \left (b x^2+a\right )^2}+\frac {15 a^4}{b^6 \left (b x^2+a\right )}-\frac {10 a^3}{b^6}+\frac {6 x^2 a^2}{b^5}-\frac {3 x^4 a}{b^4}+\frac {x^6}{b^3}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a^6}{2 b^7 \left (a+b x^2\right )^2}+\frac {6 a^5}{b^7 \left (a+b x^2\right )}+\frac {15 a^4 \log \left (a+b x^2\right )}{b^7}-\frac {10 a^3 x^2}{b^6}+\frac {3 a^2 x^4}{b^5}-\frac {a x^6}{b^4}+\frac {x^8}{4 b^3}\right )\) |
Input:
Int[x^13/(a + b*x^2)^3,x]
Output:
((-10*a^3*x^2)/b^6 + (3*a^2*x^4)/b^5 - (a*x^6)/b^4 + x^8/(4*b^3) - a^6/(2* b^7*(a + b*x^2)^2) + (6*a^5)/(b^7*(a + b*x^2)) + (15*a^4*Log[a + b*x^2])/b ^7)/2
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]
Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87
method | result | size |
norman | \(\frac {\frac {15 a^{5} x^{2}}{b^{6}}+\frac {x^{12}}{8 b}-\frac {a \,x^{10}}{4 b^{2}}+\frac {5 a^{2} x^{8}}{8 b^{3}}+\frac {45 a^{6}}{4 b^{7}}-\frac {5 a^{3} x^{6}}{2 b^{4}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {15 a^{4} \ln \left (b \,x^{2}+a \right )}{2 b^{7}}\) | \(87\) |
risch | \(\frac {x^{8}}{8 b^{3}}-\frac {a \,x^{6}}{2 b^{4}}+\frac {3 a^{2} x^{4}}{2 b^{5}}-\frac {5 a^{3} x^{2}}{b^{6}}+\frac {3 a^{5} x^{2}+\frac {11 a^{6}}{4 b}}{b^{6} \left (b \,x^{2}+a \right )^{2}}+\frac {15 a^{4} \ln \left (b \,x^{2}+a \right )}{2 b^{7}}\) | \(87\) |
default | \(-\frac {-\frac {1}{8} b^{3} x^{8}+\frac {1}{2} a \,b^{2} x^{6}-\frac {3}{2} a^{2} b \,x^{4}+5 a^{3} x^{2}}{b^{6}}+\frac {a^{4} \left (-\frac {a^{2}}{2 b \left (b \,x^{2}+a \right )^{2}}+\frac {6 a}{b \left (b \,x^{2}+a \right )}+\frac {15 \ln \left (b \,x^{2}+a \right )}{b}\right )}{2 b^{6}}\) | \(96\) |
parallelrisch | \(\frac {b^{6} x^{12}-2 a \,b^{5} x^{10}+5 a^{2} b^{4} x^{8}-20 a^{3} x^{6} b^{3}+60 \ln \left (b \,x^{2}+a \right ) x^{4} a^{4} b^{2}+120 \ln \left (b \,x^{2}+a \right ) x^{2} a^{5} b +120 a^{5} b \,x^{2}+60 \ln \left (b \,x^{2}+a \right ) a^{6}+90 a^{6}}{8 b^{7} \left (b \,x^{2}+a \right )^{2}}\) | \(117\) |
Input:
int(x^13/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
Output:
(15*a^5/b^6*x^2+1/8/b*x^12-1/4*a/b^2*x^10+5/8*a^2/b^3*x^8+45/4*a^6/b^7-5/2 *a^3/b^4*x^6)/(b*x^2+a)^2+15/2*a^4*ln(b*x^2+a)/b^7
Time = 0.06 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.25 \[ \int \frac {x^{13}}{\left (a+b x^2\right )^3} \, dx=\frac {b^{6} x^{12} - 2 \, a b^{5} x^{10} + 5 \, a^{2} b^{4} x^{8} - 20 \, a^{3} b^{3} x^{6} - 68 \, a^{4} b^{2} x^{4} - 16 \, a^{5} b x^{2} + 22 \, a^{6} + 60 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{2} + a^{6}\right )} \log \left (b x^{2} + a\right )}{8 \, {\left (b^{9} x^{4} + 2 \, a b^{8} x^{2} + a^{2} b^{7}\right )}} \] Input:
integrate(x^13/(b*x^2+a)^3,x, algorithm="fricas")
Output:
1/8*(b^6*x^12 - 2*a*b^5*x^10 + 5*a^2*b^4*x^8 - 20*a^3*b^3*x^6 - 68*a^4*b^2 *x^4 - 16*a^5*b*x^2 + 22*a^6 + 60*(a^4*b^2*x^4 + 2*a^5*b*x^2 + a^6)*log(b* x^2 + a))/(b^9*x^4 + 2*a*b^8*x^2 + a^2*b^7)
Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04 \[ \int \frac {x^{13}}{\left (a+b x^2\right )^3} \, dx=\frac {15 a^{4} \log {\left (a + b x^{2} \right )}}{2 b^{7}} - \frac {5 a^{3} x^{2}}{b^{6}} + \frac {3 a^{2} x^{4}}{2 b^{5}} - \frac {a x^{6}}{2 b^{4}} + \frac {11 a^{6} + 12 a^{5} b x^{2}}{4 a^{2} b^{7} + 8 a b^{8} x^{2} + 4 b^{9} x^{4}} + \frac {x^{8}}{8 b^{3}} \] Input:
integrate(x**13/(b*x**2+a)**3,x)
Output:
15*a**4*log(a + b*x**2)/(2*b**7) - 5*a**3*x**2/b**6 + 3*a**2*x**4/(2*b**5) - a*x**6/(2*b**4) + (11*a**6 + 12*a**5*b*x**2)/(4*a**2*b**7 + 8*a*b**8*x* *2 + 4*b**9*x**4) + x**8/(8*b**3)
Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99 \[ \int \frac {x^{13}}{\left (a+b x^2\right )^3} \, dx=\frac {12 \, a^{5} b x^{2} + 11 \, a^{6}}{4 \, {\left (b^{9} x^{4} + 2 \, a b^{8} x^{2} + a^{2} b^{7}\right )}} + \frac {15 \, a^{4} \log \left (b x^{2} + a\right )}{2 \, b^{7}} + \frac {b^{3} x^{8} - 4 \, a b^{2} x^{6} + 12 \, a^{2} b x^{4} - 40 \, a^{3} x^{2}}{8 \, b^{6}} \] Input:
integrate(x^13/(b*x^2+a)^3,x, algorithm="maxima")
Output:
1/4*(12*a^5*b*x^2 + 11*a^6)/(b^9*x^4 + 2*a*b^8*x^2 + a^2*b^7) + 15/2*a^4*l og(b*x^2 + a)/b^7 + 1/8*(b^3*x^8 - 4*a*b^2*x^6 + 12*a^2*b*x^4 - 40*a^3*x^2 )/b^6
Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02 \[ \int \frac {x^{13}}{\left (a+b x^2\right )^3} \, dx=\frac {15 \, a^{4} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{7}} - \frac {45 \, a^{4} b^{2} x^{4} + 78 \, a^{5} b x^{2} + 34 \, a^{6}}{4 \, {\left (b x^{2} + a\right )}^{2} b^{7}} + \frac {b^{9} x^{8} - 4 \, a b^{8} x^{6} + 12 \, a^{2} b^{7} x^{4} - 40 \, a^{3} b^{6} x^{2}}{8 \, b^{12}} \] Input:
integrate(x^13/(b*x^2+a)^3,x, algorithm="giac")
Output:
15/2*a^4*log(abs(b*x^2 + a))/b^7 - 1/4*(45*a^4*b^2*x^4 + 78*a^5*b*x^2 + 34 *a^6)/((b*x^2 + a)^2*b^7) + 1/8*(b^9*x^8 - 4*a*b^8*x^6 + 12*a^2*b^7*x^4 - 40*a^3*b^6*x^2)/b^12
Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {x^{13}}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {11\,a^6}{4\,b}+3\,a^5\,x^2}{a^2\,b^6+2\,a\,b^7\,x^2+b^8\,x^4}+\frac {x^8}{8\,b^3}-\frac {a\,x^6}{2\,b^4}+\frac {15\,a^4\,\ln \left (b\,x^2+a\right )}{2\,b^7}+\frac {3\,a^2\,x^4}{2\,b^5}-\frac {5\,a^3\,x^2}{b^6} \] Input:
int(x^13/(a + b*x^2)^3,x)
Output:
((11*a^6)/(4*b) + 3*a^5*x^2)/(a^2*b^6 + b^8*x^4 + 2*a*b^7*x^2) + x^8/(8*b^ 3) - (a*x^6)/(2*b^4) + (15*a^4*log(a + b*x^2))/(2*b^7) + (3*a^2*x^4)/(2*b^ 5) - (5*a^3*x^2)/b^6
Time = 0.19 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.29 \[ \int \frac {x^{13}}{\left (a+b x^2\right )^3} \, dx=\frac {60 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{6}+120 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{5} b \,x^{2}+60 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{4} b^{2} x^{4}+30 a^{6}-60 a^{4} b^{2} x^{4}-20 a^{3} b^{3} x^{6}+5 a^{2} b^{4} x^{8}-2 a \,b^{5} x^{10}+b^{6} x^{12}}{8 b^{7} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^13/(b*x^2+a)^3,x)
Output:
(60*log(a + b*x**2)*a**6 + 120*log(a + b*x**2)*a**5*b*x**2 + 60*log(a + b* x**2)*a**4*b**2*x**4 + 30*a**6 - 60*a**4*b**2*x**4 - 20*a**3*b**3*x**6 + 5 *a**2*b**4*x**8 - 2*a*b**5*x**10 + b**6*x**12)/(8*b**7*(a**2 + 2*a*b*x**2 + b**2*x**4))