3.145 \(\int f^{a+\frac {b}{x^2}} x^2 \, dx\)

Optimal. Leaf size=73 \[ -\frac {2}{3} \sqrt {\pi } b^{3/2} f^a \log ^{\frac {3}{2}}(f) \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )+\frac {2}{3} b x \log (f) f^{a+\frac {b}{x^2}}+\frac {1}{3} x^3 f^{a+\frac {b}{x^2}} \]

[Out]

1/3*f^(a+b/x^2)*x^3+2/3*b*f^(a+b/x^2)*x*ln(f)-2/3*b^(3/2)*f^a*erfi(b^(1/2)*ln(f)^(1/2)/x)*ln(f)^(3/2)*Pi^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2214, 2206, 2211, 2204} \[ -\frac {2}{3} \sqrt {\pi } b^{3/2} f^a \log ^{\frac {3}{2}}(f) \text {Erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )+\frac {1}{3} x^3 f^{a+\frac {b}{x^2}}+\frac {2}{3} b x \log (f) f^{a+\frac {b}{x^2}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)*x^2,x]

[Out]

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

Rule 2204

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 2206

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 2211

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 2214

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*} \int f^{a+\frac {b}{x^2}} x^2 \, dx &=\frac {1}{3} f^{a+\frac {b}{x^2}} x^3+\frac {1}{3} (2 b \log (f)) \int f^{a+\frac {b}{x^2}} \, dx\\ &=\frac {1}{3} f^{a+\frac {b}{x^2}} x^3+\frac {2}{3} b f^{a+\frac {b}{x^2}} x \log (f)+\frac {1}{3} \left (4 b^2 \log ^2(f)\right ) \int \frac {f^{a+\frac {b}{x^2}}}{x^2} \, dx\\ &=\frac {1}{3} f^{a+\frac {b}{x^2}} x^3+\frac {2}{3} b f^{a+\frac {b}{x^2}} x \log (f)-\frac {1}{3} \left (4 b^2 \log ^2(f)\right ) \operatorname {Subst}\left (\int f^{a+b x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{3} f^{a+\frac {b}{x^2}} x^3+\frac {2}{3} b f^{a+\frac {b}{x^2}} x \log (f)-\frac {2}{3} b^{3/2} f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \log ^{\frac {3}{2}}(f)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 60, normalized size = 0.82 \[ \frac {1}{3} f^a \left (x f^{\frac {b}{x^2}} \left (2 b \log (f)+x^2\right )-2 \sqrt {\pi } b^{3/2} \log ^{\frac {3}{2}}(f) \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)*x^2,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 56, normalized size = 0.77 \[ \frac {2}{3} \, \sqrt {\pi } \sqrt {-b \log \relax (f)} b f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \relax (f)}}{x}\right ) \log \relax (f) + \frac {1}{3} \, {\left (x^{3} + 2 \, b x \log \relax (f)\right )} f^{\frac {a x^{2} + b}{x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(pi)*sqrt(-b*log(f))*b*f^a*erf(sqrt(-b*log(f))/x)*log(f) + 1/3*(x^3 + 2*b*x*log(f))*f^((a*x^2 + b)/x^2
)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(a + b/x^2)*x^2, x)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 67, normalized size = 0.92 \[ -\frac {2 \sqrt {\pi }\, b^{2} f^{a} \erf \left (\frac {\sqrt {-b \ln \relax (f )}}{x}\right ) \ln \relax (f )^{2}}{3 \sqrt {-b \ln \relax (f )}}+\frac {2 b x \,f^{a} f^{\frac {b}{x^{2}}} \ln \relax (f )}{3}+\frac {x^{3} f^{a} f^{\frac {b}{x^{2}}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^2)*x^2,x)

[Out]

1/3*f^a*x^3*f^(b/x^2)+2/3*f^a*ln(f)*b*x*f^(b/x^2)-2/3*f^a*ln(f)^2*b^2*Pi^(1/2)/(-b*ln(f))^(1/2)*erf((-b*ln(f))
^(1/2)/x)

________________________________________________________________________________________

maxima [A]  time = 1.27, size = 28, normalized size = 0.38 \[ \frac {1}{2} \, f^{a} x^{3} \left (-\frac {b \log \relax (f)}{x^{2}}\right )^{\frac {3}{2}} \Gamma \left (-\frac {3}{2}, -\frac {b \log \relax (f)}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f^a*x^3*(-b*log(f)/x^2)^(3/2)*gamma(-3/2, -b*log(f)/x^2)

________________________________________________________________________________________

mupad [B]  time = 3.61, size = 71, normalized size = 0.97 \[ x^3\,\left (\frac {f^a\,f^{\frac {b}{x^2}}}{3}+\frac {2\,b\,f^a\,f^{\frac {b}{x^2}}\,\ln \relax (f)}{3\,x^2}\right )-\frac {2\,b^2\,f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \relax (f)}{x\,\sqrt {b\,\ln \relax (f)}}\right )\,{\ln \relax (f)}^2}{3\,\sqrt {b\,\ln \relax (f)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^2)*x^2,x)

[Out]

x^3*((f^a*f^(b/x^2))/3 + (2*b*f^a*f^(b/x^2)*log(f))/(3*x^2)) - (2*b^2*f^a*pi^(1/2)*erfi((b*log(f))/(x*(b*log(f
))^(1/2)))*log(f)^2)/(3*(b*log(f))^(1/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b/x**2)*x**2, x)

________________________________________________________________________________________