\(\int f^{a+\frac {b}{x^2}} x^4 \, dx\) [144]

Optimal result

Integrand size = 13, antiderivative size = 96 \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {1}{5} f^{a+\frac {b}{x^2}} x^5+\frac {2}{15} b f^{a+\frac {b}{x^2}} x^3 \log (f)+\frac {4}{15} b^2 f^{a+\frac {b}{x^2}} x \log ^2(f)-\frac {4}{15} b^{5/2} f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \log ^{\frac {5}{2}}(f) \]

[Out]

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*erfi(b^(1/2)*
ln(f)^(1/2)/x)*ln(f)^(5/2)*Pi^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2245, 2237, 2242, 2235} \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=-\frac {4}{15} \sqrt {\pi } b^{5/2} f^a \log ^{\frac {5}{2}}(f) \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )+\frac {4}{15} b^2 x \log ^2(f) f^{a+\frac {b}{x^2}}+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}+\frac {2}{15} b x^3 \log (f) f^{a+\frac {b}{x^2}} \]

[In]

Int[f^(a + b/x^2)*x^4,x]

[Out]

(f^(a + b/x^2)*x^5)/5 + (2*b*f^(a + b/x^2)*x^3*Log[f])/15 + (4*b^2*f^(a + b/x^2)*x*Log[f]^2)/15 - (4*b^(5/2)*f
^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x]*Log[f]^(5/2))/15

Rule 2235

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]

Rule 2237

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(c + d*x)*(F^(a + b*(c + d*x)^n)/d), x]
- Dist[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]

Rule 2242

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[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
)]

Rule 2245

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] - Dist[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]))

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} f^{a+\frac {b}{x^2}} x^5+\frac {1}{5} (2 b \log (f)) \int f^{a+\frac {b}{x^2}} x^2 \, 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 {1}{15} \left (4 b^2 \log ^2(f)\right ) \int f^{a+\frac {b}{x^2}} \, 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 {1}{15} \left (8 b^3 \log ^3(f)\right ) \int \frac {f^{a+\frac {b}{x^2}}}{x^2} \, 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 {1}{15} \left (8 b^3 \log ^3(f)\right ) \text {Subst}\left (\int f^{a+b x^2} \, dx,x,\frac {1}{x}\right ) \\ & = \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) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77 \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {1}{15} f^a \left (-4 b^{5/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \log ^{\frac {5}{2}}(f)+f^{\frac {b}{x^2}} x \left (3 x^4+2 b x^2 \log (f)+4 b^2 \log ^2(f)\right )\right ) \]

[In]

Integrate[f^(a + b/x^2)*x^4,x]

[Out]

(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

Maple [A] (verified)

Time = 0.09 (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	\(\frac {f^{a} x^{5} f^{\frac {b}{x^{2}}}}{5}+\frac {2 f^{a} \ln \left (f \right ) b \,x^{3} f^{\frac {b}{x^{2}}}}{15}+\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} x \,f^{\frac {b}{x^{2}}}}{15}-\frac {4 f^{a} \ln \left (f \right )^{3} b^{3} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{15 \sqrt {-b \ln \left (f \right )}}\)	\(89\)

[In]

int(f^(a+b/x^2)*x^4,x,method=_RETURNVERBOSE)

[Out]

-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)*e
xp(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))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77 \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {4}{15} \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b^{2} f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) \log \left (f\right )^{2} + \frac {1}{15} \, {\left (3 \, x^{5} + 2 \, b x^{3} \log \left (f\right ) + 4 \, b^{2} x \log \left (f\right )^{2}\right )} f^{\frac {a x^{2} + b}{x^{2}}} \]

[In]

integrate(f^(a+b/x^2)*x^4,x, algorithm="fricas")

[Out]

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)

Sympy [F]

\[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\int f^{a + \frac {b}{x^{2}}} x^{4}\, dx \]

[In]

integrate(f**(a+b/x**2)*x**4,x)

[Out]

Integral(f**(a + b/x**2)*x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.29 \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {1}{2} \, f^{a} x^{5} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {5}{2}} \Gamma \left (-\frac {5}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right ) \]

[In]

integrate(f^(a+b/x^2)*x^4,x, algorithm="maxima")

[Out]

1/2*f^a*x^5*(-b*log(f)/x^2)^(5/2)*gamma(-5/2, -b*log(f)/x^2)

Giac [F]

\[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\int { f^{a + \frac {b}{x^{2}}} x^{4} \,d x } \]

[In]

integrate(f^(a+b/x^2)*x^4,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)*x^4, x)

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.11 \[ \int f^{a+\frac {b}{x^2}} x^4 \, dx=\frac {f^a\,f^{\frac {b}{x^2}}\,x^5}{5}+\frac {4\,f^a\,x^5\,\sqrt {\pi }\,{\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}^{5/2}}{15}-\frac {4\,f^a\,x^5\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \left (f\right )}{x^2}}\right )\,{\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}^{5/2}}{15}+\frac {4\,b^2\,f^a\,f^{\frac {b}{x^2}}\,x\,{\ln \left (f\right )}^2}{15}+\frac {2\,b\,f^a\,f^{\frac {b}{x^2}}\,x^3\,\ln \left (f\right )}{15} \]

[In]

int(f^(a + b/x^2)*x^4,x)

[Out]

(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