\(\int \frac {f^{a+b x^2}}{x^2} \, dx\) [89]

Optimal result

Integrand size = 13, antiderivative size = 49 \[ \int \frac {f^{a+b x^2}}{x^2} \, dx=-\frac {f^{a+b x^2}}{x}+\sqrt {b} f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \sqrt {\log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2245, 2235} \[ \int \frac {f^{a+b x^2}}{x^2} \, dx=\sqrt {\pi } \sqrt {b} f^a \sqrt {\log (f)} \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )-\frac {f^{a+b x^2}}{x} \]

[In]

Int[f^(a + b*x^2)/x^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^2}}{x^2} \, dx=-\frac {f^{a+b x^2}}{x}+\sqrt {b} f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \sqrt {\log (f)} \]

[In]

Integrate[f^(a + b*x^2)/x^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90

method	result	size
risch	\(-\frac {f^{a} f^{b \,x^{2}}}{x}+\frac {f^{a} \ln \left (f \right ) b \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{\sqrt {-b \ln \left (f \right )}}\)	\(44\)
meijerg	\(-\frac {f^{a} b \sqrt {\ln \left (f \right )}\, \left (-\frac {2 \,{\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{x \sqrt {-b}\, \sqrt {\ln \left (f \right )}}+\frac {2 \sqrt {b}\, \sqrt {\pi }\, \operatorname {erfi}\left (x \sqrt {b}\, \sqrt {\ln \left (f \right )}\right )}{\sqrt {-b}}\right )}{2 \sqrt {-b}}\)	\(62\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \frac {f^{a+b x^2}}{x^2} \, dx=-\frac {\sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} x \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) + f^{b x^{2} + a}}{x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b*x**2)/x**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.57 \[ \int \frac {f^{a+b x^2}}{x^2} \, dx=-\frac {\sqrt {-b x^{2} \log \left (f\right )} f^{a} \Gamma \left (-\frac {1}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, x} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(b*x^2 + a)/x^2, x)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int \frac {f^{a+b x^2}}{x^2} \, dx=\frac {b\,f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (f\right )}{\sqrt {b\,\ln \left (f\right )}}\right )\,\ln \left (f\right )}{\sqrt {b\,\ln \left (f\right )}}-\frac {f^a\,f^{b\,x^2}}{x} \]

[In]

int(f^(a + b*x^2)/x^2,x)

[Out]

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