Integrand size = 13, antiderivative size = 101 \[ \int \frac {\left (a+b x^2\right )^8}{x^{13}} \, dx=-\frac {a^8}{12 x^{12}}-\frac {4 a^7 b}{5 x^{10}}-\frac {7 a^6 b^2}{2 x^8}-\frac {28 a^5 b^3}{3 x^6}-\frac {35 a^4 b^4}{2 x^4}-\frac {28 a^3 b^5}{x^2}+4 a b^7 x^2+\frac {b^8 x^4}{4}+28 a^2 b^6 \log (x) \] Output:
-1/12*a^8/x^12-4/5*a^7*b/x^10-7/2*a^6*b^2/x^8-28/3*a^5*b^3/x^6-35/2*a^4*b^ 4/x^4-28*a^3*b^5/x^2+4*a*b^7*x^2+1/4*b^8*x^4+28*a^2*b^6*ln(x)
Time = 0.00 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^8}{x^{13}} \, dx=-\frac {a^8}{12 x^{12}}-\frac {4 a^7 b}{5 x^{10}}-\frac {7 a^6 b^2}{2 x^8}-\frac {28 a^5 b^3}{3 x^6}-\frac {35 a^4 b^4}{2 x^4}-\frac {28 a^3 b^5}{x^2}+4 a b^7 x^2+\frac {b^8 x^4}{4}+28 a^2 b^6 \log (x) \] Input:
Integrate[(a + b*x^2)^8/x^13,x]
Output:
-1/12*a^8/x^12 - (4*a^7*b)/(5*x^10) - (7*a^6*b^2)/(2*x^8) - (28*a^5*b^3)/( 3*x^6) - (35*a^4*b^4)/(2*x^4) - (28*a^3*b^5)/x^2 + 4*a*b^7*x^2 + (b^8*x^4) /4 + 28*a^2*b^6*Log[x]
Time = 0.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^8}{x^{13}} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^8}{x^{14}}dx^2\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \int \left (\frac {a^8}{x^{14}}+\frac {8 b a^7}{x^{12}}+\frac {28 b^2 a^6}{x^{10}}+\frac {56 b^3 a^5}{x^8}+\frac {70 b^4 a^4}{x^6}+\frac {56 b^5 a^3}{x^4}+\frac {28 b^6 a^2}{x^2}+8 b^7 a+b^8 x^2\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a^8}{6 x^{12}}-\frac {8 a^7 b}{5 x^{10}}-\frac {7 a^6 b^2}{x^8}-\frac {56 a^5 b^3}{3 x^6}-\frac {35 a^4 b^4}{x^4}-\frac {56 a^3 b^5}{x^2}+28 a^2 b^6 \log \left (x^2\right )+8 a b^7 x^2+\frac {b^8 x^4}{2}\right )\) |
Input:
Int[(a + b*x^2)^8/x^13,x]
Output:
(-1/6*a^8/x^12 - (8*a^7*b)/(5*x^10) - (7*a^6*b^2)/x^8 - (56*a^5*b^3)/(3*x^ 6) - (35*a^4*b^4)/x^4 - (56*a^3*b^5)/x^2 + 8*a*b^7*x^2 + (b^8*x^4)/2 + 28* a^2*b^6*Log[x^2])/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 = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {a^{8}}{12 x^{12}}-\frac {4 a^{7} b}{5 x^{10}}-\frac {7 a^{6} b^{2}}{2 x^{8}}-\frac {28 a^{5} b^{3}}{3 x^{6}}-\frac {35 a^{4} b^{4}}{2 x^{4}}-\frac {28 a^{3} b^{5}}{x^{2}}+4 a \,b^{7} x^{2}+\frac {b^{8} x^{4}}{4}+28 a^{2} b^{6} \ln \left (x \right )\) | \(90\) |
norman | \(\frac {-\frac {1}{12} a^{8}+\frac {1}{4} b^{8} x^{16}+4 a \,b^{7} x^{14}-28 a^{3} b^{5} x^{10}-\frac {35}{2} a^{4} b^{4} x^{8}-\frac {28}{3} a^{5} b^{3} x^{6}-\frac {7}{2} a^{6} b^{2} x^{4}-\frac {4}{5} a^{7} b \,x^{2}}{x^{12}}+28 a^{2} b^{6} \ln \left (x \right )\) | \(92\) |
parallelrisch | \(\frac {15 b^{8} x^{16}+240 a \,b^{7} x^{14}+1680 a^{2} b^{6} \ln \left (x \right ) x^{12}-1680 a^{3} b^{5} x^{10}-1050 a^{4} b^{4} x^{8}-560 a^{5} b^{3} x^{6}-210 a^{6} b^{2} x^{4}-48 a^{7} b \,x^{2}-5 a^{8}}{60 x^{12}}\) | \(95\) |
risch | \(\frac {b^{8} x^{4}}{4}+4 a \,b^{7} x^{2}+16 a^{2} b^{6}+\frac {-28 a^{3} b^{5} x^{10}-\frac {35}{2} a^{4} b^{4} x^{8}-\frac {28}{3} a^{5} b^{3} x^{6}-\frac {7}{2} a^{6} b^{2} x^{4}-\frac {4}{5} a^{7} b \,x^{2}-\frac {1}{12} a^{8}}{x^{12}}+28 a^{2} b^{6} \ln \left (x \right )\) | \(100\) |
Input:
int((b*x^2+a)^8/x^13,x,method=_RETURNVERBOSE)
Output:
-1/12*a^8/x^12-4/5*a^7*b/x^10-7/2*a^6*b^2/x^8-28/3*a^5*b^3/x^6-35/2*a^4/x^ 4*b^4-28*a^3*b^5/x^2+4*a*b^7*x^2+1/4*b^8*x^4+28*a^2*b^6*ln(x)
Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^8}{x^{13}} \, dx=\frac {15 \, b^{8} x^{16} + 240 \, a b^{7} x^{14} + 1680 \, a^{2} b^{6} x^{12} \log \left (x\right ) - 1680 \, a^{3} b^{5} x^{10} - 1050 \, a^{4} b^{4} x^{8} - 560 \, a^{5} b^{3} x^{6} - 210 \, a^{6} b^{2} x^{4} - 48 \, a^{7} b x^{2} - 5 \, a^{8}}{60 \, x^{12}} \] Input:
integrate((b*x^2+a)^8/x^13,x, algorithm="fricas")
Output:
1/60*(15*b^8*x^16 + 240*a*b^7*x^14 + 1680*a^2*b^6*x^12*log(x) - 1680*a^3*b ^5*x^10 - 1050*a^4*b^4*x^8 - 560*a^5*b^3*x^6 - 210*a^6*b^2*x^4 - 48*a^7*b* x^2 - 5*a^8)/x^12
Time = 0.33 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^8}{x^{13}} \, dx=28 a^{2} b^{6} \log {\left (x \right )} + 4 a b^{7} x^{2} + \frac {b^{8} x^{4}}{4} + \frac {- 5 a^{8} - 48 a^{7} b x^{2} - 210 a^{6} b^{2} x^{4} - 560 a^{5} b^{3} x^{6} - 1050 a^{4} b^{4} x^{8} - 1680 a^{3} b^{5} x^{10}}{60 x^{12}} \] Input:
integrate((b*x**2+a)**8/x**13,x)
Output:
28*a**2*b**6*log(x) + 4*a*b**7*x**2 + b**8*x**4/4 + (-5*a**8 - 48*a**7*b*x **2 - 210*a**6*b**2*x**4 - 560*a**5*b**3*x**6 - 1050*a**4*b**4*x**8 - 1680 *a**3*b**5*x**10)/(60*x**12)
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^8}{x^{13}} \, dx=\frac {1}{4} \, b^{8} x^{4} + 4 \, a b^{7} x^{2} + 14 \, a^{2} b^{6} \log \left (x^{2}\right ) - \frac {1680 \, a^{3} b^{5} x^{10} + 1050 \, a^{4} b^{4} x^{8} + 560 \, a^{5} b^{3} x^{6} + 210 \, a^{6} b^{2} x^{4} + 48 \, a^{7} b x^{2} + 5 \, a^{8}}{60 \, x^{12}} \] Input:
integrate((b*x^2+a)^8/x^13,x, algorithm="maxima")
Output:
1/4*b^8*x^4 + 4*a*b^7*x^2 + 14*a^2*b^6*log(x^2) - 1/60*(1680*a^3*b^5*x^10 + 1050*a^4*b^4*x^8 + 560*a^5*b^3*x^6 + 210*a^6*b^2*x^4 + 48*a^7*b*x^2 + 5* a^8)/x^12
Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^8}{x^{13}} \, dx=\frac {1}{4} \, b^{8} x^{4} + 4 \, a b^{7} x^{2} + 14 \, a^{2} b^{6} \log \left (x^{2}\right ) - \frac {2058 \, a^{2} b^{6} x^{12} + 1680 \, a^{3} b^{5} x^{10} + 1050 \, a^{4} b^{4} x^{8} + 560 \, a^{5} b^{3} x^{6} + 210 \, a^{6} b^{2} x^{4} + 48 \, a^{7} b x^{2} + 5 \, a^{8}}{60 \, x^{12}} \] Input:
integrate((b*x^2+a)^8/x^13,x, algorithm="giac")
Output:
1/4*b^8*x^4 + 4*a*b^7*x^2 + 14*a^2*b^6*log(x^2) - 1/60*(2058*a^2*b^6*x^12 + 1680*a^3*b^5*x^10 + 1050*a^4*b^4*x^8 + 560*a^5*b^3*x^6 + 210*a^6*b^2*x^4 + 48*a^7*b*x^2 + 5*a^8)/x^12
Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2\right )^8}{x^{13}} \, dx=\frac {b^8\,x^4}{4}-\frac {\frac {a^8}{12}+\frac {4\,a^7\,b\,x^2}{5}+\frac {7\,a^6\,b^2\,x^4}{2}+\frac {28\,a^5\,b^3\,x^6}{3}+\frac {35\,a^4\,b^4\,x^8}{2}+28\,a^3\,b^5\,x^{10}}{x^{12}}+4\,a\,b^7\,x^2+28\,a^2\,b^6\,\ln \left (x\right ) \] Input:
int((a + b*x^2)^8/x^13,x)
Output:
(b^8*x^4)/4 - (a^8/12 + (4*a^7*b*x^2)/5 + (7*a^6*b^2*x^4)/2 + (28*a^5*b^3* x^6)/3 + (35*a^4*b^4*x^8)/2 + 28*a^3*b^5*x^10)/x^12 + 4*a*b^7*x^2 + 28*a^2 *b^6*log(x)
Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^8}{x^{13}} \, dx=\frac {1680 \,\mathrm {log}\left (x \right ) a^{2} b^{6} x^{12}-5 a^{8}-48 a^{7} b \,x^{2}-210 a^{6} b^{2} x^{4}-560 a^{5} b^{3} x^{6}-1050 a^{4} b^{4} x^{8}-1680 a^{3} b^{5} x^{10}+240 a \,b^{7} x^{14}+15 b^{8} x^{16}}{60 x^{12}} \] Input:
int((b*x^2+a)^8/x^13,x)
Output:
(1680*log(x)*a**2*b**6*x**12 - 5*a**8 - 48*a**7*b*x**2 - 210*a**6*b**2*x** 4 - 560*a**5*b**3*x**6 - 1050*a**4*b**4*x**8 - 1680*a**3*b**5*x**10 + 240* a*b**7*x**14 + 15*b**8*x**16)/(60*x**12)