\(\int f^{a+b x^n} x^{-1-\frac {n}{2}} \, dx\) [192]

Optimal result

Integrand size = 19, antiderivative size = 66 \[ \int f^{a+b x^n} x^{-1-\frac {n}{2}} \, dx=-\frac {2 f^{a+b x^n} x^{-n/2}}{n}+\frac {2 \sqrt {b} f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x^{n/2} \sqrt {\log (f)}\right ) \sqrt {\log (f)}}{n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2246, 2242, 2235} \[ \int f^{a+b x^n} x^{-1-\frac {n}{2}} \, dx=\frac {2 \sqrt {\pi } \sqrt {b} f^a \sqrt {\log (f)} \text {erfi}\left (\sqrt {b} \sqrt {\log (f)} x^{n/2}\right )}{n}-\frac {2 x^{-n/2} f^{a+b x^n}}{n} \]

[In]

Int[f^(a + b*x^n)*x^(-1 - n/2),x]

[Out]

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

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 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 2246

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)^Simplify[m + n]*F^(a +
b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && IntegerQ[2*Simplify[(m + 1)/n]] && LtQ[-4, Simpl
ify[(m + 1)/n], 5] &&  !RationalQ[m] && SumSimplerQ[m, n]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.59 \[ \int f^{a+b x^n} x^{-1-\frac {n}{2}} \, dx=-\frac {f^a x^{-n/2} \Gamma \left (-\frac {1}{2},-b x^n \log (f)\right ) \sqrt {-b x^n \log (f)}}{n} \]

[In]

Integrate[f^(a + b*x^n)*x^(-1 - n/2),x]

[Out]

-((f^a*Gamma[-1/2, -(b*x^n*Log[f])]*Sqrt[-(b*x^n*Log[f])])/(n*x^(n/2)))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {2 f^{a} x^{-\frac {n}{2}} f^{b \,x^{n}}}{n}+\frac {2 f^{a} \ln \left (f \right ) b \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x^{\frac {n}{2}}\right )}{n \sqrt {-b \ln \left (f \right )}}\)	\(59\)
meijerg	\(\frac {f^{a} \sqrt {-b}\, \sqrt {\ln \left (f \right )}\, \left (-\frac {2 x^{-\frac {n}{2}} {\mathrm e}^{b \,x^{n} \ln \left (f \right )}}{\sqrt {-b}\, \sqrt {\ln \left (f \right )}}+\frac {2 \sqrt {b}\, \sqrt {\pi }\, \operatorname {erfi}\left (x^{\frac {n}{2}} \sqrt {b}\, \sqrt {\ln \left (f \right )}\right )}{\sqrt {-b}}\right )}{n}\)	\(69\)

[In]

int(f^(a+b*x^n)*x^(-1-1/2*n),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.26 \[ \int f^{a+b x^n} x^{-1-\frac {n}{2}} \, dx=-\frac {2 \, {\left (\sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x x^{-\frac {1}{2} \, n - 1}}\right ) + x x^{-\frac {1}{2} \, n - 1} e^{\left (\frac {a x^{2} x^{-n - 2} \log \left (f\right ) + b \log \left (f\right )}{x^{2} x^{-n - 2}}\right )}\right )}}{n} \]

[In]

integrate(f^(a+b*x^n)*x^(-1-1/2*n),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int f^{a+b x^n} x^{-1-\frac {n}{2}} \, dx=\int f^{a + b x^{n}} x^{- \frac {n}{2} - 1}\, dx \]

[In]

integrate(f**(a+b*x**n)*x**(-1-1/2*n),x)

[Out]

Integral(f**(a + b*x**n)*x**(-n/2 - 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.53 \[ \int f^{a+b x^n} x^{-1-\frac {n}{2}} \, dx=-\frac {\sqrt {-b x^{n} \log \left (f\right )} f^{a} \Gamma \left (-\frac {1}{2}, -b x^{n} \log \left (f\right )\right )}{n x^{\frac {1}{2} \, n}} \]

[In]

integrate(f^(a+b*x^n)*x^(-1-1/2*n),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int f^{a+b x^n} x^{-1-\frac {n}{2}} \, dx=\int { f^{b x^{n} + a} x^{-\frac {1}{2} \, n - 1} \,d x } \]

[In]

integrate(f^(a+b*x^n)*x^(-1-1/2*n),x, algorithm="giac")

[Out]

integrate(f^(b*x^n + a)*x^(-1/2*n - 1), x)

Mupad [F(-1)]

Timed out. \[ \int f^{a+b x^n} x^{-1-\frac {n}{2}} \, dx=\int \frac {f^{a+b\,x^n}}{x^{\frac {n}{2}+1}} \,d x \]

[In]

int(f^(a + b*x^n)/x^(n/2 + 1),x)

[Out]

int(f^(a + b*x^n)/x^(n/2 + 1), x)