\(\int f^{a+b x^n} x^{-1+n} \, dx\) [185]

Optimal result

Integrand size = 15, antiderivative size = 20 \[ \int f^{a+b x^n} x^{-1+n} \, dx=\frac {f^{a+b x^n}}{b n \log (f)} \]

[Out]

f^(a+b*x^n)/b/n/ln(f)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2240} \[ \int f^{a+b x^n} x^{-1+n} \, dx=\frac {f^{a+b x^n}}{b n \log (f)} \]

[In]

Int[f^(a + b*x^n)*x^(-1 + n),x]

[Out]

f^(a + b*x^n)/(b*n*Log[f])

Rule 2240

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

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x^n}}{b n \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^n} x^{-1+n} \, dx=\frac {f^{a+b x^n}}{b n \log (f)} \]

[In]

Integrate[f^(a + b*x^n)*x^(-1 + n),x]

[Out]

f^(a + b*x^n)/(b*n*Log[f])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
risch	\(\frac {f^{a +b \,x^{n}}}{b n \ln \left (f \right )}\)	\(21\)
norman	\(\frac {{\mathrm e}^{\left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right ) \ln \left (f \right )}}{\ln \left (f \right ) b n}\)	\(25\)
meijerg	\(-\frac {f^{a} \left (-\frac {\left (-1\right )^{-\frac {1}{n}-\frac {n -1}{n}}}{\Gamma \left (2-\frac {1}{n}-\frac {n -1}{n}\right )}+\frac {\left (-1\right )^{-\frac {1}{n}-\frac {n -1}{n}} {\mathrm e}^{b \,x^{n} \ln \left (f \right )}}{\Gamma \left (2-\frac {1}{n}-\frac {n -1}{n}\right )}\right )}{b \ln \left (f \right ) n}\)	\(96\)

[In]

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

[Out]

f^(a+b*x^n)/b/n/ln(f)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int f^{a+b x^n} x^{-1+n} \, dx=\frac {e^{\left (b x^{n} \log \left (f\right ) + a \log \left (f\right )\right )}}{b n \log \left (f\right )} \]

[In]

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

[Out]

e^(b*x^n*log(f) + a*log(f))/(b*n*log(f))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (14) = 28\).

Time = 0.95 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.20 \[ \int f^{a+b x^n} x^{-1+n} \, dx=\begin {cases} \log {\left (x \right )} & \text {for}\: b = 0 \wedge f = 1 \wedge n = 0 \\\frac {f^{a} x x^{n - 1}}{n} & \text {for}\: b = 0 \\f^{a + b} \log {\left (x \right )} & \text {for}\: n = 0 \\\frac {x x^{n - 1}}{n} & \text {for}\: f = 1 \\\frac {f^{a + b x^{n}}}{b n \log {\left (f \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((log(x), Eq(b, 0) & Eq(f, 1) & Eq(n, 0)), (f**a*x*x**(n - 1)/n, Eq(b, 0)), (f**(a + b)*log(x), Eq(n,
0)), (x*x**(n - 1)/n, Eq(f, 1)), (f**(a + b*x**n)/(b*n*log(f)), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^n} x^{-1+n} \, dx=\frac {f^{b x^{n} + a}}{b n \log \left (f\right )} \]

[In]

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

[Out]

f^(b*x^n + a)/(b*n*log(f))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^n} x^{-1+n} \, dx=\frac {f^{b x^{n} + a}}{b n \log \left (f\right )} \]

[In]

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

[Out]

f^(b*x^n + a)/(b*n*log(f))

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^n} x^{-1+n} \, dx=\frac {f^{a+b\,x^n}}{b\,n\,\ln \left (f\right )} \]

[In]

int(f^(a + b*x^n)*x^(n - 1),x)

[Out]

f^(a + b*x^n)/(b*n*log(f))