3.2.45 \(\int f^{a+\frac {b}{x^2}} x^2 \, dx\) [145]

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

[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)

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Rubi [A]
time = 0.05, 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 = {2245, 2237, 2242, 2235} \begin {gather*} -\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}} \end {gather*}

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 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*} \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 ) \text {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*}

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Mathematica [A]
time = 0.02, size = 60, normalized size = 0.82 \begin {gather*} \frac {1}{3} f^a \left (-2 b^{3/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \log ^{\frac {3}{2}}(f)+f^{\frac {b}{x^2}} x \left (x^2+2 b \log (f)\right )\right ) \end {gather*}

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

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Maple [A]
time = 0.03, size = 67, normalized size = 0.92

method	result	size
risch	\(\frac {f^{a} x^{3} f^{\frac {b}{x^{2}}}}{3}+\frac {2 f^{a} \ln \left (f \right ) b x \,f^{\frac {b}{x^{2}}}}{3}-\frac {2 f^{a} \ln \left (f \right )^{2} b^{2} \sqrt {\pi }\, \erf \left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{3 \sqrt {-b \ln \left (f \right )}}\)	\(67\)
meijerg	\(\frac {f^{a} b \ln \left (f \right )^{\frac {3}{2}} \sqrt {-b}\, \left (-\frac {2 x^{3} \left (\frac {2 b \ln \left (f \right )}{x^{2}}+1\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{3 \left (-b \right )^{\frac {3}{2}} \ln \left (f \right )^{\frac {3}{2}}}+\frac {4 b^{\frac {3}{2}} \sqrt {\pi }\, \erfi \left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{3 \left (-b \right )^{\frac {3}{2}}}\right )}{2}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [A]
time = 0.32, size = 28, normalized size = 0.38 \begin {gather*} \frac {1}{2} \, f^{a} x^{3} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {3}{2}} \Gamma \left (-\frac {3}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right ) \end {gather*}

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)

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Fricas [A]
time = 0.38, size = 56, normalized size = 0.77 \begin {gather*} \frac {2}{3} \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) \log \left (f\right ) + \frac {1}{3} \, {\left (x^{3} + 2 \, b x \log \left (f\right )\right )} f^{\frac {a x^{2} + b}{x^{2}}} \end {gather*}

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
)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int f^{a + \frac {b}{x^{2}}} x^{2}\, dx \end {gather*}

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

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))

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