Integrand size = 13, antiderivative size = 109 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\frac {15 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{16 b^{7/2} \log ^{\frac {7}{2}}(f)}-\frac {15 f^{a+\frac {b}{x^2}}}{8 b^3 x \log ^3(f)}+\frac {5 f^{a+\frac {b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)} \]
-15/8*f^(a+b/x^2)/b^3/x/ln(f)^3+5/4*f^(a+b/x^2)/b^2/x^3/ln(f)^2-1/2*f^(a+b /x^2)/b/x^5/ln(f)+15/16*f^a*erfi(b^(1/2)*ln(f)^(1/2)/x)*Pi^(1/2)/b^(7/2)/l n(f)^(7/2)
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.79 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\frac {15 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{16 b^{7/2} \log ^{\frac {7}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}} \left (15 x^4-10 b x^2 \log (f)+4 b^2 \log ^2(f)\right )}{8 b^3 x^5 \log ^3(f)} \]
(15*f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x])/(16*b^(7/2)*Log[f]^(7/2)) - (f^(a + b/x^2)*(15*x^4 - 10*b*x^2*Log[f] + 4*b^2*Log[f]^2))/(8*b^3*x^5* Log[f]^3)
Time = 0.39 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.22, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2641, 2641, 2641, 2640, 2633}
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.
\(\displaystyle \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {5 \int \frac {f^{a+\frac {b}{x^2}}}{x^6}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {5 \left (-\frac {3 \int \frac {f^{a+\frac {b}{x^2}}}{x^4}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {5 \left (-\frac {3 \left (-\frac {\int \frac {f^{a+\frac {b}{x^2}}}{x^2}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\) |
\(\Big \downarrow \) 2640 |
\(\displaystyle -\frac {5 \left (-\frac {3 \left (\frac {\int f^{a+\frac {b}{x^2}}d\frac {1}{x}}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle -\frac {5 \left (-\frac {3 \left (\frac {\sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\) |
-1/2*f^(a + b/x^2)/(b*x^5*Log[f]) - (5*(-1/2*f^(a + b/x^2)/(b*x^3*Log[f]) - (3*((f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x])/(4*b^(3/2)*Log[f]^(3/2 )) - f^(a + b/x^2)/(2*b*x*Log[f])))/(2*b*Log[f])))/(2*b*Log[f])
3.2.50.3.1 Defintions of rubi rules used
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ F, a, b, c, d}, x] && PosQ[b]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[1/(d*(m + 1)) Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1)]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[(c + d*x)^(m - n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*L og[F])), x] - Simp[(m - n + 1)/(b*n*Log[F]) Int[(c + d*x)^(m - n)*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/ n)] && LtQ[0, (m + 1)/n, 5] && IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n , 0])
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83
method | result | size |
meijerg | \(-\frac {f^{a} \sqrt {-b}\, \left (\frac {\left (-b \right )^{\frac {7}{2}} \sqrt {\ln \left (f \right )}\, \left (\frac {28 b^{2} \ln \left (f \right )^{2}}{x^{4}}-\frac {70 b \ln \left (f \right )}{x^{2}}+105\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{28 x \,b^{3}}-\frac {15 \left (-b \right )^{\frac {7}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{8 b^{\frac {7}{2}}}\right )}{2 \ln \left (f \right )^{\frac {7}{2}} b^{4}}\) | \(91\) |
risch | \(-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 x^{5} b \ln \left (f \right )}+\frac {5 f^{a} f^{\frac {b}{x^{2}}}}{4 \ln \left (f \right )^{2} b^{2} x^{3}}-\frac {15 f^{a} f^{\frac {b}{x^{2}}}}{8 \ln \left (f \right )^{3} b^{3} x}+\frac {15 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{16 \ln \left (f \right )^{3} b^{3} \sqrt {-b \ln \left (f \right )}}\) | \(102\) |
-1/2*f^a/ln(f)^(7/2)/b^4*(-b)^(1/2)*(1/28/x*(-b)^(7/2)*ln(f)^(1/2)*(28*b^2 *ln(f)^2/x^4-70*b*ln(f)/x^2+105)/b^3*exp(b*ln(f)/x^2)-15/8*(-b)^(7/2)/b^(7 /2)*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x))
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.81 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=-\frac {15 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} x^{5} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) + 2 \, {\left (15 \, b x^{4} \log \left (f\right ) - 10 \, b^{2} x^{2} \log \left (f\right )^{2} + 4 \, b^{3} \log \left (f\right )^{3}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{16 \, b^{4} x^{5} \log \left (f\right )^{4}} \]
-1/16*(15*sqrt(pi)*sqrt(-b*log(f))*f^a*x^5*erf(sqrt(-b*log(f))/x) + 2*(15* b*x^4*log(f) - 10*b^2*x^2*log(f)^2 + 4*b^3*log(f)^3)*f^((a*x^2 + b)/x^2))/ (b^4*x^5*log(f)^4)
Timed out. \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.26 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\frac {f^{a} \Gamma \left (\frac {7}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, x^{7} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {7}{2}}} \]
\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{8}} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\frac {5\,f^a\,f^{\frac {b}{x^2}}}{4\,b^2\,x^3\,{\ln \left (f\right )}^2}-\frac {f^a\,f^{\frac {b}{x^2}}}{2\,b\,x^5\,\ln \left (f\right )}-\frac {15\,f^a\,f^{\frac {b}{x^2}}}{8\,b^3\,x\,{\ln \left (f\right )}^3}+\frac {15\,f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (f\right )}{x\,\sqrt {b\,\ln \left (f\right )}}\right )}{16\,b^3\,{\ln \left (f\right )}^3\,\sqrt {b\,\ln \left (f\right )}} \]