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

Optimal. Leaf size=45 \[ \frac {x^n f^{a+b x^n}}{b n \log (f)}-\frac {f^{a+b x^n}}{b^2 n \log ^2(f)} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2213, 2209} \[ \frac {x^n f^{a+b x^n}}{b n \log (f)}-\frac {f^{a+b x^n}}{b^2 n \log ^2(f)} \]

Antiderivative was successfully verified.

[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 2209

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 2213

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*} \int f^{a+b x^n} x^{-1+2 n} \, dx &=\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*}

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Mathematica [C]  time = 0.01, size = 25, normalized size = 0.56 \[ -\frac {f^a \Gamma \left (2,-b x^n \log (f)\right )}{b^2 n \log ^2(f)} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.42, size = 33, normalized size = 0.73 \[ \frac {{\left (b x^{n} \log \relax (f) - 1\right )} e^{\left (b x^{n} \log \relax (f) + a \log \relax (f)\right )}}{b^{2} n \log \relax (f)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{b x^{n} + a} x^{2 \, n - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 56, normalized size = 1.24 \[ \frac {{\mathrm e}^{n \ln \relax (x )} {\mathrm e}^{\left (b \,{\mathrm e}^{n \ln \relax (x )}+a \right ) \ln \relax (f )}}{b n \ln \relax (f )}-\frac {{\mathrm e}^{\left (b \,{\mathrm e}^{n \ln \relax (x )}+a \right ) \ln \relax (f )}}{b^{2} n \ln \relax (f )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.78, size = 34, normalized size = 0.76 \[ \frac {{\left (b f^{a} x^{n} \log \relax (f) - f^{a}\right )} f^{b x^{n}}}{b^{2} n \log \relax (f)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int f^{a+b\,x^n}\,x^{2\,n-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^n)*x^(2*n - 1),x)

[Out]

int(f^(a + b*x^n)*x^(2*n - 1), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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