\(\int f^{a+b x^n} x^{-1+2 n} \, dx\) [184]

Optimal result

Integrand size = 17, antiderivative size = 45 \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=-\frac {f^{a+b x^n}}{b^2 n \log ^2(f)}+\frac {f^{a+b x^n} x^n}{b n \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2244, 2240} \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=\frac {x^n f^{a+b x^n}}{b n \log (f)}-\frac {f^{a+b x^n}}{b^2 n \log ^2(f)} \]

[In]

Int[f^(a + b*x^n)*x^(-1 + 2*n),x]

[Out]

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

Rule 2244

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), 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]] && Lt
Q[0, Simplify[(m + 1)/n], 5] &&  !RationalQ[m] && SumSimplerQ[m, -n]

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x^n} x^n}{b n \log (f)}-\frac {\int f^{a+b x^n} x^{-1+n} \, dx}{b \log (f)} \\ & = -\frac {f^{a+b x^n}}{b^2 n \log ^2(f)}+\frac {f^{a+b x^n} x^n}{b n \log (f)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.56 \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=-\frac {f^a \Gamma \left (2,-b x^n \log (f)\right )}{b^2 n \log ^2(f)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {\left (b \,x^{n} \ln \left (f \right )-1\right ) f^{a +b \,x^{n}}}{\ln \left (f \right )^{2} b^{2} n}\)	\(30\)
meijerg	\(\frac {f^{a} \left (1-\frac {\left (2-2 b \,x^{n} \ln \left (f \right )\right ) {\mathrm e}^{b \,x^{n} \ln \left (f \right )}}{2}\right )}{\ln \left (f \right )^{2} b^{2} n}\)	\(37\)
norman	\(\frac {{\mathrm e}^{n \ln \left (x \right )} {\mathrm e}^{\left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right ) \ln \left (f \right )}}{\ln \left (f \right ) b n}-\frac {{\mathrm e}^{\left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right ) \ln \left (f \right )}}{\ln \left (f \right )^{2} b^{2} n}\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.73 \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=\frac {{\left (b x^{n} \log \left (f\right ) - 1\right )} e^{\left (b x^{n} \log \left (f\right ) + a \log \left (f\right )\right )}}{b^{2} n \log \left (f\right )^{2}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (36) = 72\).

Time = 1.53 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.62 \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=\begin {cases} \log {\left (x \right )} & \text {for}\: b = 0 \wedge f = 1 \wedge n = 0 \\\frac {f^{a} x x^{2 n - 1}}{2 n} & \text {for}\: b = 0 \\f^{a + b} \log {\left (x \right )} & \text {for}\: n = 0 \\\frac {x x^{2 n - 1}}{2 n} & \text {for}\: f = 1 \\\frac {f^{a + b x^{n}} x^{n}}{b n \log {\left (f \right )}} - \frac {f^{a + b x^{n}}}{b^{2} n \log {\left (f \right )}^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((log(x), Eq(b, 0) & Eq(f, 1) & Eq(n, 0)), (f**a*x*x**(2*n - 1)/(2*n), Eq(b, 0)), (f**(a + b)*log(x),
Eq(n, 0)), (x*x**(2*n - 1)/(2*n), Eq(f, 1)), (f**(a + b*x**n)*x**n/(b*n*log(f)) - f**(a + b*x**n)/(b**2*n*log
(f)**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.76 \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=\frac {{\left (b f^{a} x^{n} \log \left (f\right ) - f^{a}\right )} f^{b x^{n}}}{b^{2} n \log \left (f\right )^{2}} \]

[In]

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

[Out]

(b*f^a*x^n*log(f) - f^a)*f^(b*x^n)/(b^2*n*log(f)^2)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(f^(a + b*x^n)*x^(2*n - 1),x)

[Out]

int(f^(a + b*x^n)*x^(2*n - 1), x)