\(\int f^{a+b x^n} \, dx\) [178]

Optimal result

Integrand size = 9, antiderivative size = 35 \[ \int f^{a+b x^n} \, dx=-\frac {f^a x \Gamma \left (\frac {1}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{-1/n}}{n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 35, 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 = {2239} \[ \int f^{a+b x^n} \, dx=-\frac {x f^a \left (-b \log (f) x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-b x^n \log (f)\right )}{n} \]

[In]

Int[f^(a + b*x^n),x]

[Out]

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

Rule 2239

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(-F^a)*(c + d*x)*(Gamma[1/n, (-b)*(c + d
*x)^n*Log[F]]/(d*n*((-b)*(c + d*x)^n*Log[F])^(1/n))), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^n} \, dx=-\frac {f^a x \Gamma \left (\frac {1}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{-1/n}}{n} \]

[In]

Integrate[f^(a + b*x^n),x]

[Out]

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

Maple [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 201, normalized size of antiderivative = 5.74

method	result	size
meijerg	\(\frac {f^{a} \left (-b \right )^{-\frac {1}{n}} \ln \left (f \right )^{-\frac {1}{n}} \left (\frac {n x \left (-b \right )^{\frac {1}{n}} \ln \left (f \right )^{\frac {1}{n}} \left (x^{n} \ln \left (f \right ) b n +n +1\right ) \Gamma \left (1-\frac {1}{n}\right ) \Gamma \left (\frac {1+n}{n}+1\right ) L_{-\frac {1}{n}}^{\left (\frac {1+n}{n}\right )}\left (b \,x^{n} \ln \left (f \right )\right )}{\left (1+n \right ) \Gamma \left (-\frac {1}{n}+\frac {1+n}{n}+1\right )}-\frac {n^{2} x^{1+n} \left (-b \right )^{\frac {1}{n}} \ln \left (f \right )^{1+\frac {1}{n}} b L_{-\frac {1}{n}}^{\left (\frac {1+n}{n}+1\right )}\left (b \,x^{n} \ln \left (f \right )\right ) \Gamma \left (1-\frac {1}{n}\right ) \Gamma \left (\frac {1+n}{n}+1\right )}{\left (1+n \right ) \Gamma \left (-\frac {1}{n}+\frac {1+n}{n}+1\right )}\right )}{n}\)	\(201\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(f^(b*x^n + a), x)

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b*x**n), x)

Maxima [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^n} \, dx=-\frac {f^{a} x \Gamma \left (\frac {1}{n}, -b x^{n} \log \left (f\right )\right )}{\left (-b x^{n} \log \left (f\right )\right )^{\left (\frac {1}{n}\right )} n} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(b*x^n + a), x)

Mupad [F(-1)]

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

[In]

int(f^(a + b*x^n),x)

[Out]

int(f^(a + b*x^n), x)