\(\int \frac {f^{a+b x^n}}{x^4} \, dx\) [182]

Optimal result

Integrand size = 13, antiderivative size = 39 \[ \int \frac {f^{a+b x^n}}{x^4} \, dx=-\frac {f^a \Gamma \left (-\frac {3}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{3/n}}{n x^3} \]

[Out]

-f^a*GAMMA(-3/n,-b*x^n*ln(f))*(-b*x^n*ln(f))^(3/n)/n/x^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, 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} \[ \int \frac {f^{a+b x^n}}{x^4} \, dx=-\frac {f^a \left (-b \log (f) x^n\right )^{3/n} \Gamma \left (-\frac {3}{n},-b x^n \log (f)\right )}{n x^3} \]

[In]

Int[f^(a + b*x^n)/x^4,x]

[Out]

-((f^a*Gamma[-3/n, -(b*x^n*Log[f])]*(-(b*x^n*Log[f]))^(3/n))/(n*x^3))

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 {3}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{3/n}}{n x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^n}}{x^4} \, dx=-\frac {f^a \Gamma \left (-\frac {3}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{3/n}}{n x^3} \]

[In]

Integrate[f^(a + b*x^n)/x^4,x]

[Out]

-((f^a*Gamma[-3/n, -(b*x^n*Log[f])]*(-(b*x^n*Log[f]))^(3/n))/(n*x^3))

Maple [C] (verified)

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

Time = 0.16 (sec) , antiderivative size = 212, normalized size of antiderivative = 5.44

method	result	size
meijerg	\(\frac {f^{a} \ln \left (f \right )^{\frac {3}{n}} \left (-b \right )^{\frac {3}{n}} \left (-\frac {n \left (-b \right )^{-\frac {3}{n}} \ln \left (f \right )^{-\frac {3}{n}} \left (x^{n} \ln \left (f \right ) b n +n -3\right ) \Gamma \left (1+\frac {3}{n}\right ) \Gamma \left (\frac {n -3}{n}+1\right ) L_{\frac {3}{n}}^{\left (\frac {n -3}{n}\right )}\left (b \,x^{n} \ln \left (f \right )\right )}{3 x^{3} \left (n -3\right ) \Gamma \left (\frac {3}{n}+\frac {n -3}{n}+1\right )}+\frac {n^{2} x^{n -3} \left (-b \right )^{-\frac {3}{n}} \ln \left (f \right )^{1-\frac {3}{n}} b L_{\frac {3}{n}}^{\left (\frac {n -3}{n}+1\right )}\left (b \,x^{n} \ln \left (f \right )\right ) \Gamma \left (1+\frac {3}{n}\right ) \Gamma \left (\frac {n -3}{n}+1\right )}{3 \left (n -3\right ) \Gamma \left (\frac {3}{n}+\frac {n -3}{n}+1\right )}\right )}{n}\)	\(212\)

[In]

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

[Out]

f^a*ln(f)^(3/n)*(-b)^(3/n)/n*(-1/3*n/x^3*(-b)^(-3/n)*ln(f)^(-3/n)*(x^n*ln(f)*b*n+n-3)/(n-3)/GAMMA(3/n+(n-3)/n+
1)*GAMMA(1+3/n)*GAMMA((n-3)/n+1)*LaguerreL(3/n,(n-3)/n,b*x^n*ln(f))+1/3*n^2*x^(n-3)*(-b)^(-3/n)*ln(f)^(1-3/n)*
b/(n-3)*LaguerreL(3/n,(n-3)/n+1,b*x^n*ln(f))*GAMMA(1+3/n)*GAMMA((n-3)/n+1)/GAMMA(3/n+(n-3)/n+1))

Fricas [F]

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

[In]

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

[Out]

integral(f^(b*x^n + a)/x^4, x)

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b*x**n)/x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^n}}{x^4} \, dx=-\frac {\left (-b x^{n} \log \left (f\right )\right )^{\frac {3}{n}} f^{a} \Gamma \left (-\frac {3}{n}, -b x^{n} \log \left (f\right )\right )}{n x^{3}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(b*x^n + a)/x^4, x)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.33 \[ \int \frac {f^{a+b x^n}}{x^4} \, dx=-\frac {f^a\,{\mathrm {e}}^{\frac {b\,x^n\,\ln \left (f\right )}{2}}\,{\mathrm {M}}_{\frac {3}{2\,n}+\frac {1}{2},-\frac {3}{2\,n}}\left (b\,x^n\,\ln \left (f\right )\right )\,{\left (b\,x^n\,\ln \left (f\right )\right )}^{\frac {3}{2\,n}-\frac {1}{2}}}{3\,x^3} \]

[In]

int(f^(a + b*x^n)/x^4,x)

[Out]

-(f^a*exp((b*x^n*log(f))/2)*whittakerM(3/(2*n) + 1/2, -3/(2*n), b*x^n*log(f))*(b*x^n*log(f))^(3/(2*n) - 1/2))/
(3*x^3)