Integrand size = 13, antiderivative size = 96 \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {1}{5} f^{a+\frac {b}{x^2}} x^5+\frac {2}{15} b f^{a+\frac {b}{x^2}} x^3 \log (f)+\frac {4}{15} b^2 f^{a+\frac {b}{x^2}} x \log ^2(f)-\frac {4}{15} b^{5/2} f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \log ^{\frac {5}{2}}(f) \] Output:
1/5*f^(a+b/x^2)*x^5+2/15*b*f^(a+b/x^2)*x^3*ln(f)+4/15*b^2*f^(a+b/x^2)*x*ln (f)^2-4/15*b^(5/2)*f^a*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x)*ln(f)^(5/2)
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77 \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {1}{15} f^a \left (-4 b^{5/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \log ^{\frac {5}{2}}(f)+f^{\frac {b}{x^2}} x \left (3 x^4+2 b x^2 \log (f)+4 b^2 \log ^2(f)\right )\right ) \] Input:
Integrate[f^(a + b/x^2)*x^4,x]
Output:
(f^a*(-4*b^(5/2)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x]*Log[f]^(5/2) + f^ (b/x^2)*x*(3*x^4 + 2*b*x^2*Log[f] + 4*b^2*Log[f]^2)))/15
Time = 0.55 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2643, 2643, 2635, 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.
\(\displaystyle \int x^4 f^{a+\frac {b}{x^2}} \, dx\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {2}{5} b \log (f) \int f^{a+\frac {b}{x^2}} x^2dx+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \int f^{a+\frac {b}{x^2}}dx+\frac {1}{3} x^3 f^{a+\frac {b}{x^2}}\right )+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}\) |
\(\Big \downarrow \) 2635 |
\(\displaystyle \frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \left (2 b \log (f) \int \frac {f^{a+\frac {b}{x^2}}}{x^2}dx+x f^{a+\frac {b}{x^2}}\right )+\frac {1}{3} x^3 f^{a+\frac {b}{x^2}}\right )+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}\) |
\(\Big \downarrow \) 2640 |
\(\displaystyle \frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \left (x f^{a+\frac {b}{x^2}}-2 b \log (f) \int f^{a+\frac {b}{x^2}}d\frac {1}{x}\right )+\frac {1}{3} x^3 f^{a+\frac {b}{x^2}}\right )+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \left (x f^{a+\frac {b}{x^2}}-\sqrt {\pi } \sqrt {b} f^a \sqrt {\log (f)} \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )\right )+\frac {1}{3} x^3 f^{a+\frac {b}{x^2}}\right )+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}\) |
Input:
Int[f^(a + b/x^2)*x^4,x]
Output:
(f^(a + b/x^2)*x^5)/5 + (2*b*Log[f]*((f^(a + b/x^2)*x^3)/3 + (2*b*(f^(a + b/x^2)*x - Sqrt[b]*f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x]*Sqrt[Log[f] ])*Log[f])/3))/5
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_)), x_Symbol] :> Simp[(c + d*x)*(F^(a + b*(c + d*x)^n)/d), x] - Simp[b*n*Log[F] Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n] && ILtQ[n, 0]
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 + 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))) , x] - Simp[b*n*(Log[F]/(m + 1)) 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[ -4, (m + 1)/n, 5] && IntegerQ[n] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))
Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.92
method | result | size |
meijerg | \(-\frac {f^{a} \ln \left (f \right )^{\frac {5}{2}} b^{2} \sqrt {-b}\, \left (-\frac {2 x^{5} \left (\frac {4 b^{2} \ln \left (f \right )^{2}}{3 x^{4}}+\frac {2 b \ln \left (f \right )}{3 x^{2}}+1\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{5 \left (-b \right )^{\frac {5}{2}} \ln \left (f \right )^{\frac {5}{2}}}+\frac {8 b^{\frac {5}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{15 \left (-b \right )^{\frac {5}{2}}}\right )}{2}\) | \(88\) |
risch | \(\frac {f^{a} f^{\frac {b}{x^{2}}} x^{5}}{5}+\frac {2 f^{a} \ln \left (f \right ) b \,x^{3} f^{\frac {b}{x^{2}}}}{15}+\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} x \,f^{\frac {b}{x^{2}}}}{15}-\frac {4 f^{a} \ln \left (f \right )^{3} b^{3} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{15 \sqrt {-b \ln \left (f \right )}}\) | \(89\) |
Input:
int(f^(a+b/x^2)*x^4,x,method=_RETURNVERBOSE)
Output:
-1/2*f^a*ln(f)^(5/2)*b^2*(-b)^(1/2)*(-2/5*x^5/(-b)^(5/2)/ln(f)^(5/2)*(4/3* b^2*ln(f)^2/x^4+2/3*b*ln(f)/x^2+1)*exp(b*ln(f)/x^2)+8/15/(-b)^(5/2)*b^(5/2 )*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x))
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77 \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {4}{15} \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b^{2} f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) \log \left (f\right )^{2} + \frac {1}{15} \, {\left (3 \, x^{5} + 2 \, b x^{3} \log \left (f\right ) + 4 \, b^{2} x \log \left (f\right )^{2}\right )} f^{\frac {a x^{2} + b}{x^{2}}} \] Input:
integrate(f^(a+b/x^2)*x^4,x, algorithm="fricas")
Output:
4/15*sqrt(pi)*sqrt(-b*log(f))*b^2*f^a*erf(sqrt(-b*log(f))/x)*log(f)^2 + 1/ 15*(3*x^5 + 2*b*x^3*log(f) + 4*b^2*x*log(f)^2)*f^((a*x^2 + b)/x^2)
\[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\int f^{a + \frac {b}{x^{2}}} x^{4}\, dx \] Input:
integrate(f**(a+b/x**2)*x**4,x)
Output:
Integral(f**(a + b/x**2)*x**4, x)
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.29 \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {1}{2} \, f^{a} x^{5} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {5}{2}} \Gamma \left (-\frac {5}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right ) \] Input:
integrate(f^(a+b/x^2)*x^4,x, algorithm="maxima")
Output:
1/2*f^a*x^5*(-b*log(f)/x^2)^(5/2)*gamma(-5/2, -b*log(f)/x^2)
\[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\int { f^{a + \frac {b}{x^{2}}} x^{4} \,d x } \] Input:
integrate(f^(a+b/x^2)*x^4,x, algorithm="giac")
Output:
integrate(f^(a + b/x^2)*x^4, x)
Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.11 \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {f^a\,f^{\frac {b}{x^2}}\,x^5}{5}+\frac {4\,f^a\,x^5\,\sqrt {\pi }\,{\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}^{5/2}}{15}-\frac {4\,f^a\,x^5\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \left (f\right )}{x^2}}\right )\,{\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}^{5/2}}{15}+\frac {4\,b^2\,f^a\,f^{\frac {b}{x^2}}\,x\,{\ln \left (f\right )}^2}{15}+\frac {2\,b\,f^a\,f^{\frac {b}{x^2}}\,x^3\,\ln \left (f\right )}{15} \] Input:
int(f^(a + b/x^2)*x^4,x)
Output:
(f^a*f^(b/x^2)*x^5)/5 + (4*f^a*x^5*pi^(1/2)*(-(b*log(f))/x^2)^(5/2))/15 - (4*f^a*x^5*pi^(1/2)*erfc((-(b*log(f))/x^2)^(1/2))*(-(b*log(f))/x^2)^(5/2)) /15 + (4*b^2*f^a*f^(b/x^2)*x*log(f)^2)/15 + (2*b*f^a*f^(b/x^2)*x^3*log(f)) /15
\[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {-8 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right )^{3} b^{3}+4 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right )^{2} b^{2} x^{2}+2 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right ) b \,x^{4}+3 f^{\frac {a \,x^{2}+b}{x^{2}}} x^{6}-16 \left (\int \frac {f^{\frac {a \,x^{2}+b}{x^{2}}}}{x^{4}}d x \right ) \mathrm {log}\left (f \right )^{4} b^{4} x}{15 x} \] Input:
int(f^(a+b/x^2)*x^4,x)
Output:
( - 8*f**((a*x**2 + b)/x**2)*log(f)**3*b**3 + 4*f**((a*x**2 + b)/x**2)*log (f)**2*b**2*x**2 + 2*f**((a*x**2 + b)/x**2)*log(f)*b*x**4 + 3*f**((a*x**2 + b)/x**2)*x**6 - 16*int(f**((a*x**2 + b)/x**2)/x**4,x)*log(f)**4*b**4*x)/ (15*x)