Integrand size = 13, antiderivative size = 132 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=-\frac {105 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{32 b^{9/2} \log ^{\frac {9}{2}}(f)}+\frac {105 f^{a+\frac {b}{x^2}}}{16 b^4 x \log ^4(f)}-\frac {35 f^{a+\frac {b}{x^2}}}{8 b^3 x^3 \log ^3(f)}+\frac {7 f^{a+\frac {b}{x^2}}}{4 b^2 x^5 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)} \]
105/16*f^(a+b/x^2)/b^4/x/ln(f)^4-35/8*f^(a+b/x^2)/b^3/x^3/ln(f)^3+7/4*f^(a +b/x^2)/b^2/x^5/ln(f)^2-1/2*f^(a+b/x^2)/b/x^7/ln(f)-105/32*f^a*erfi(b^(1/2 )*ln(f)^(1/2)/x)*Pi^(1/2)/b^(9/2)/ln(f)^(9/2)
Time = 0.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\frac {f^a \left (-105 \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )+\frac {2 \sqrt {b} f^{\frac {b}{x^2}} \sqrt {\log (f)} \left (105 x^6-70 b x^4 \log (f)+28 b^2 x^2 \log ^2(f)-8 b^3 \log ^3(f)\right )}{x^7}\right )}{32 b^{9/2} \log ^{\frac {9}{2}}(f)} \]
(f^a*(-105*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x] + (2*Sqrt[b]*f^(b/x^2)* Sqrt[Log[f]]*(105*x^6 - 70*b*x^4*Log[f] + 28*b^2*x^2*Log[f]^2 - 8*b^3*Log[ f]^3))/x^7))/(32*b^(9/2)*Log[f]^(9/2))
Time = 0.46 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.27, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2641, 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.
\(\displaystyle \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {7 \int \frac {f^{a+\frac {b}{x^2}}}{x^8}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {7 \left (-\frac {5 \int \frac {f^{a+\frac {b}{x^2}}}{x^6}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \int \frac {f^{a+\frac {b}{x^2}}}{x^4}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \left (-\frac {\int \frac {f^{a+\frac {b}{x^2}}}{x^2}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\) |
\(\Big \downarrow \) 2640 |
\(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \left (\frac {\int f^{a+\frac {b}{x^2}}d\frac {1}{x}}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \left (\frac {\sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\) |
-1/2*f^(a + b/x^2)/(b*x^7*Log[f]) - (7*(-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])))/(2*b*Log[f])
3.2.51.3.1 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 = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.78
method | result | size |
meijerg | \(\frac {f^{a} \sqrt {-b}\, \left (-\frac {\left (-b \right )^{\frac {9}{2}} \sqrt {\ln \left (f \right )}\, \left (-\frac {72 b^{3} \ln \left (f \right )^{3}}{x^{6}}+\frac {252 b^{2} \ln \left (f \right )^{2}}{x^{4}}-\frac {630 b \ln \left (f \right )}{x^{2}}+945\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{72 x \,b^{4}}+\frac {105 \left (-b \right )^{\frac {9}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{16 b^{\frac {9}{2}}}\right )}{2 b^{5} \ln \left (f \right )^{\frac {9}{2}}}\) | \(103\) |
risch | \(-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 x^{7} b \ln \left (f \right )}+\frac {7 f^{a} f^{\frac {b}{x^{2}}}}{4 \ln \left (f \right )^{2} b^{2} x^{5}}-\frac {35 f^{a} f^{\frac {b}{x^{2}}}}{8 \ln \left (f \right )^{3} b^{3} x^{3}}+\frac {105 f^{a} f^{\frac {b}{x^{2}}}}{16 \ln \left (f \right )^{4} b^{4} x}-\frac {105 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{32 \ln \left (f \right )^{4} b^{4} \sqrt {-b \ln \left (f \right )}}\) | \(124\) |
1/2*f^a/b^5/ln(f)^(9/2)*(-b)^(1/2)*(-1/72/x*(-b)^(9/2)*ln(f)^(1/2)*(-72*b^ 3*ln(f)^3/x^6+252*b^2*ln(f)^2/x^4-630*b*ln(f)/x^2+945)/b^4*exp(b*ln(f)/x^2 )+105/16*(-b)^(9/2)/b^(9/2)*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x))
Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\frac {105 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} x^{7} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) + 2 \, {\left (105 \, b x^{6} \log \left (f\right ) - 70 \, b^{2} x^{4} \log \left (f\right )^{2} + 28 \, b^{3} x^{2} \log \left (f\right )^{3} - 8 \, b^{4} \log \left (f\right )^{4}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{32 \, b^{5} x^{7} \log \left (f\right )^{5}} \]
1/32*(105*sqrt(pi)*sqrt(-b*log(f))*f^a*x^7*erf(sqrt(-b*log(f))/x) + 2*(105 *b*x^6*log(f) - 70*b^2*x^4*log(f)^2 + 28*b^3*x^2*log(f)^3 - 8*b^4*log(f)^4 )*f^((a*x^2 + b)/x^2))/(b^5*x^7*log(f)^5)
Timed out. \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.21 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\frac {f^{a} \Gamma \left (\frac {9}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, x^{9} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {9}{2}}} \]
\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{10}} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.92 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=-\frac {\frac {f^a\,\left (105\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (f\right )}{x\,\sqrt {b\,\ln \left (f\right )}}\right )-\frac {210\,f^{\frac {b}{x^2}}\,\sqrt {b\,\ln \left (f\right )}}{x}\right )}{32\,\sqrt {b\,\ln \left (f\right )}}-\frac {7\,b^2\,f^a\,f^{\frac {b}{x^2}}\,{\ln \left (f\right )}^2}{4\,x^5}+\frac {b^3\,f^a\,f^{\frac {b}{x^2}}\,{\ln \left (f\right )}^3}{2\,x^7}+\frac {35\,b\,f^a\,f^{\frac {b}{x^2}}\,\ln \left (f\right )}{8\,x^3}}{b^4\,{\ln \left (f\right )}^4} \]