\(\int \frac {f^{a+b x^n}}{x^2} \, dx\) [180]

Optimal result

Integrand size = 13, antiderivative size = 37 \[ \int \frac {f^{a+b x^n}}{x^2} \, dx=-\frac {f^a \Gamma \left (-\frac {1}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{\frac {1}{n}}}{n x} \]

[Out]

-f^a*GAMMA(-1/n,-b*x^n*ln(f))*(-b*x^n*ln(f))^(1/n)/n/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2250} \[ \int \frac {f^{a+b x^n}}{x^2} \, dx=-\frac {f^a \left (-b \log (f) x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b x^n \log (f)\right )}{n x} \]

[In]

Int[f^(a + b*x^n)/x^2,x]

[Out]

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

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {f^a \Gamma \left (-\frac {1}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{\frac {1}{n}}}{n x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^n}}{x^2} \, dx=-\frac {f^a \Gamma \left (-\frac {1}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{\frac {1}{n}}}{n x} \]

[In]

Integrate[f^(a + b*x^n)/x^2,x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.06 (sec) , antiderivative size = 195, normalized size of antiderivative = 5.27

method	result	size
meijerg	\(\frac {f^{a} \ln \left (f \right )^{\frac {1}{n}} \left (-b \right )^{\frac {1}{n}} \left (-\frac {n \left (-b \right )^{-\frac {1}{n}} \ln \left (f \right )^{-\frac {1}{n}} \left (x^{n} \ln \left (f \right ) b n +n -1\right ) \Gamma \left (1+\frac {1}{n}\right ) \Gamma \left (\frac {n -1}{n}+1\right ) L_{\frac {1}{n}}^{\left (\frac {n -1}{n}\right )}\left (b \,x^{n} \ln \left (f \right )\right )}{x \left (n -1\right ) \Gamma \left (\frac {1}{n}+\frac {n -1}{n}+1\right )}+\frac {n^{2} x^{n -1} \left (-b \right )^{-\frac {1}{n}} \ln \left (f \right )^{1-\frac {1}{n}} b L_{\frac {1}{n}}^{\left (\frac {n -1}{n}+1\right )}\left (b \,x^{n} \ln \left (f \right )\right ) \Gamma \left (1+\frac {1}{n}\right ) \Gamma \left (\frac {n -1}{n}+1\right )}{\left (n -1\right ) \Gamma \left (\frac {1}{n}+\frac {n -1}{n}+1\right )}\right )}{n}\)	\(195\)

[In]

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

[Out]

f^a*ln(f)^(1/n)*(-b)^(1/n)/n*(-n/x*(-b)^(-1/n)*ln(f)^(-1/n)*(x^n*ln(f)*b*n+n-1)/(n-1)/GAMMA(1/n+(n-1)/n+1)*GAM
MA(1+1/n)*GAMMA((n-1)/n+1)*LaguerreL(1/n,(n-1)/n,b*x^n*ln(f))+n^2*x^(n-1)*(-b)^(-1/n)*ln(f)^(1-1/n)*b/(n-1)*La
guerreL(1/n,(n-1)/n+1,b*x^n*ln(f))*GAMMA(1+1/n)*GAMMA((n-1)/n+1)/GAMMA(1/n+(n-1)/n+1))

Fricas [F]

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

[In]

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

[Out]

integral(f^(b*x^n + a)/x^2, x)

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b*x**n)/x**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^n}}{x^2} \, dx=-\frac {\left (-b x^{n} \log \left (f\right )\right )^{\left (\frac {1}{n}\right )} f^{a} \Gamma \left (-\frac {1}{n}, -b x^{n} \log \left (f\right )\right )}{n x} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(b*x^n + a)/x^2, x)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.41 \[ \int \frac {f^{a+b x^n}}{x^2} \, dx=-\frac {f^a\,{\mathrm {e}}^{\frac {b\,x^n\,\ln \left (f\right )}{2}}\,{\mathrm {M}}_{\frac {1}{2\,n}+\frac {1}{2},-\frac {1}{2\,n}}\left (b\,x^n\,\ln \left (f\right )\right )\,{\left (b\,x^n\,\ln \left (f\right )\right )}^{\frac {1}{2\,n}-\frac {1}{2}}}{x} \]

[In]

int(f^(a + b*x^n)/x^2,x)

[Out]

-(f^a*exp((b*x^n*log(f))/2)*whittakerM(1/(2*n) + 1/2, -1/(2*n), b*x^n*log(f))*(b*x^n*log(f))^(1/(2*n) - 1/2))/
x