\(\int \frac {f^{a+b x^2}}{x^4} \, dx\) [90]

Optimal result

Integrand size = 13, antiderivative size = 73 \[ \int \frac {f^{a+b x^2}}{x^4} \, dx=-\frac {f^{a+b x^2}}{3 x^3}-\frac {2 b f^{a+b x^2} \log (f)}{3 x}+\frac {2}{3} b^{3/2} f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \log ^{\frac {3}{2}}(f) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2245, 2235} \[ \int \frac {f^{a+b x^2}}{x^4} \, dx=\frac {2}{3} \sqrt {\pi } b^{3/2} f^a \log ^{\frac {3}{2}}(f) \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )-\frac {2 b \log (f) f^{a+b x^2}}{3 x}-\frac {f^{a+b x^2}}{3 x^3} \]

[In]

Int[f^(a + b*x^2)/x^4,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.85 \[ \int \frac {f^{a+b x^2}}{x^4} \, dx=\frac {1}{3} f^a \left (2 b^{3/2} \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \log ^{\frac {3}{2}}(f)-\frac {f^{b x^2} \left (1+2 b x^2 \log (f)\right )}{x^3}\right ) \]

[In]

Integrate[f^(a + b*x^2)/x^4,x]

[Out]

(f^a*(2*b^(3/2)*Sqrt[Pi]*Erfi[Sqrt[b]*x*Sqrt[Log[f]]]*Log[f]^(3/2) - (f^(b*x^2)*(1 + 2*b*x^2*Log[f]))/x^3))/3

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.92

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78 \[ \int \frac {f^{a+b x^2}}{x^4} \, dx=-\frac {2 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b f^{a} x^{3} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) \log \left (f\right ) + {\left (2 \, b x^{2} \log \left (f\right ) + 1\right )} f^{b x^{2} + a}}{3 \, x^{3}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b*x**2)/x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.38 \[ \int \frac {f^{a+b x^2}}{x^4} \, dx=-\frac {\left (-b x^{2} \log \left (f\right )\right )^{\frac {3}{2}} f^{a} \Gamma \left (-\frac {3}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, x^{3}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(b*x^2 + a)/x^4, x)

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int \frac {f^{a+b x^2}}{x^4} \, dx=\frac {2\,b^2\,f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (f\right )}{\sqrt {b\,\ln \left (f\right )}}\right )\,{\ln \left (f\right )}^2}{3\,\sqrt {b\,\ln \left (f\right )}}-\frac {\frac {f^a\,f^{b\,x^2}}{3}+\frac {2\,b\,f^a\,f^{b\,x^2}\,x^2\,\ln \left (f\right )}{3}}{x^3} \]

[In]

int(f^(a + b*x^2)/x^4,x)

[Out]

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