Integrand size = 13, antiderivative size = 86 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=-\frac {3 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}+\frac {3 f^{a+\frac {b}{x^2}}}{4 b^2 x \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)} \] Output:
-3/8*f^a*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x)/b^(5/2)/ln(f)^(5/2)+3/4*f^(a +b/x^2)/b^2/x/ln(f)^2-1/2*f^(a+b/x^2)/b/x^3/ln(f)
Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=-\frac {3 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}+\frac {f^{a+\frac {b}{x^2}} \left (3 x^2-2 b \log (f)\right )}{4 b^2 x^3 \log ^2(f)} \] Input:
Integrate[f^(a + b/x^2)/x^6,x]
Output:
(-3*f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x])/(8*b^(5/2)*Log[f]^(5/2)) + (f^(a + b/x^2)*(3*x^2 - 2*b*Log[f]))/(4*b^2*x^3*Log[f]^2)
Time = 0.52 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {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.
\(\displaystyle \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {3 \int \frac {f^{a+\frac {b}{x^2}}}{x^4}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle -\frac {3 \left (-\frac {\int \frac {f^{a+\frac {b}{x^2}}}{x^2}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\) |
\(\Big \downarrow \) 2640 |
\(\displaystyle -\frac {3 \left (\frac {\int f^{a+\frac {b}{x^2}}d\frac {1}{x}}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle -\frac {3 \left (\frac {\sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\) |
Input:
Int[f^(a + b/x^2)/x^6,x]
Output:
-1/2*f^(a + b/x^2)/(b*x^3*Log[f]) - (3*((f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[L og[f]])/x])/(4*b^(3/2)*Log[f]^(3/2)) - f^(a + b/x^2)/(2*b*x*Log[f])))/(2*b *Log[f])
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 = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92
method | result | size |
meijerg | \(\frac {f^{a} \sqrt {-b}\, \left (-\frac {\left (-b \right )^{\frac {5}{2}} \sqrt {\ln \left (f \right )}\, \left (-\frac {10 b \ln \left (f \right )}{x^{2}}+15\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{10 x \,b^{2}}+\frac {3 \left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{4 b^{\frac {5}{2}}}\right )}{2 \ln \left (f \right )^{\frac {5}{2}} b^{3}}\) | \(79\) |
risch | \(-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 x^{3} b \ln \left (f \right )}+\frac {3 f^{a} f^{\frac {b}{x^{2}}}}{4 \ln \left (f \right )^{2} b^{2} x}-\frac {3 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{8 \ln \left (f \right )^{2} b^{2} \sqrt {-b \ln \left (f \right )}}\) | \(80\) |
Input:
int(f^(a+b/x^2)/x^6,x,method=_RETURNVERBOSE)
Output:
1/2*f^a/ln(f)^(5/2)/b^3*(-b)^(1/2)*(-1/10/x*(-b)^(5/2)*ln(f)^(1/2)*(-10*b* ln(f)/x^2+15)/b^2*exp(b*ln(f)/x^2)+3/4*(-b)^(5/2)/b^(5/2)*Pi^(1/2)*erfi(b^ (1/2)*ln(f)^(1/2)/x))
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=\frac {3 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} x^{3} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) + 2 \, {\left (3 \, b x^{2} \log \left (f\right ) - 2 \, b^{2} \log \left (f\right )^{2}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{8 \, b^{3} x^{3} \log \left (f\right )^{3}} \] Input:
integrate(f^(a+b/x^2)/x^6,x, algorithm="fricas")
Output:
1/8*(3*sqrt(pi)*sqrt(-b*log(f))*f^a*x^3*erf(sqrt(-b*log(f))/x) + 2*(3*b*x^ 2*log(f) - 2*b^2*log(f)^2)*f^((a*x^2 + b)/x^2))/(b^3*x^3*log(f)^3)
Timed out. \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=\text {Timed out} \] Input:
integrate(f**(a+b/x**2)/x**6,x)
Output:
Timed out
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.33 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=\frac {f^{a} \Gamma \left (\frac {5}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, x^{5} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {5}{2}}} \] Input:
integrate(f^(a+b/x^2)/x^6,x, algorithm="maxima")
Output:
1/2*f^a*gamma(5/2, -b*log(f)/x^2)/(x^5*(-b*log(f)/x^2)^(5/2))
\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{6}} \,d x } \] Input:
integrate(f^(a+b/x^2)/x^6,x, algorithm="giac")
Output:
integrate(f^(a + b/x^2)/x^6, x)
Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=-\frac {f^a\,\left (3\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (f\right )}{x\,\sqrt {b\,\ln \left (f\right )}}\right )-\frac {6\,f^{\frac {b}{x^2}}\,\sqrt {b\,\ln \left (f\right )}}{x}\right )}{8\,b^2\,{\ln \left (f\right )}^2\,\sqrt {b\,\ln \left (f\right )}}-\frac {f^a\,f^{\frac {b}{x^2}}}{2\,b\,x^3\,\ln \left (f\right )} \] Input:
int(f^(a + b/x^2)/x^6,x)
Output:
- (f^a*(3*pi^(1/2)*erfi((b*log(f))/(x*(b*log(f))^(1/2))) - (6*f^(b/x^2)*(b *log(f))^(1/2))/x))/(8*b^2*log(f)^2*(b*log(f))^(1/2)) - (f^a*f^(b/x^2))/(2 *b*x^3*log(f))
\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=\int \frac {f^{\frac {a \,x^{2}+b}{x^{2}}}}{x^{6}}d x \] Input:
int(f^(a+b/x^2)/x^6,x)
Output:
int(f**((a*x**2 + b)/x**2)/x**6,x)