\(\int f^{a+b x^2} \, dx\) [88]

Optimal result

Integrand size = 9, antiderivative size = 37 \[ \int f^{a+b x^2} \, dx=\frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{2 \sqrt {b} \sqrt {\log (f)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2235} \[ \int f^{a+b x^2} \, dx=\frac {\sqrt {\pi } f^a \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{2 \sqrt {b} \sqrt {\log (f)}} \]

[In]

Int[f^(a + b*x^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{2 \sqrt {b} \sqrt {\log (f)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^2} \, dx=\frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{2 \sqrt {b} \sqrt {\log (f)}} \]

[In]

Integrate[f^(a + b*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70

method	result	size
meijerg	\(\frac {f^{a} \operatorname {erfi}\left (x \sqrt {b}\, \sqrt {\ln \left (f \right )}\right ) \sqrt {\pi }}{2 \sqrt {b}\, \sqrt {\ln \left (f \right )}}\)	\(26\)
risch	\(\frac {f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{2 \sqrt {-b \ln \left (f \right )}}\)	\(26\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int f^{a+b x^2} \, dx=-\frac {\sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right )}{2 \, b \log \left (f\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b*x**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int f^{a+b x^2} \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right )}{2 \, \sqrt {-b \log \left (f\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int f^{a+b x^2} \, dx=-\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (f\right )} x\right )}{2 \, \sqrt {-b \log \left (f\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int f^{a+b x^2} \, dx=\frac {f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (f\right )}{\sqrt {b\,\ln \left (f\right )}}\right )}{2\,\sqrt {b\,\ln \left (f\right )}} \]

[In]

int(f^(a + b*x^2),x)

[Out]

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