\(\int f^{a+b x^n} x^{-1+3 n} \, dx\) [183]

Optimal result

Integrand size = 17, antiderivative size = 71 \[ \int f^{a+b x^n} x^{-1+3 n} \, dx=\frac {2 f^{a+b x^n}}{b^3 n \log ^3(f)}-\frac {2 f^{a+b x^n} x^n}{b^2 n \log ^2(f)}+\frac {f^{a+b x^n} x^{2 n}}{b n \log (f)} \]

[Out]

2*f^(a+b*x^n)/b^3/n/ln(f)^3-2*f^(a+b*x^n)*x^n/b^2/n/ln(f)^2+f^(a+b*x^n)*x^(2*n)/b/n/ln(f)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2244, 2240} \[ \int f^{a+b x^n} x^{-1+3 n} \, dx=\frac {2 f^{a+b x^n}}{b^3 n \log ^3(f)}-\frac {2 x^n f^{a+b x^n}}{b^2 n \log ^2(f)}+\frac {x^{2 n} f^{a+b x^n}}{b n \log (f)} \]

[In]

Int[f^(a + b*x^n)*x^(-1 + 3*n),x]

[Out]

(2*f^(a + b*x^n))/(b^3*n*Log[f]^3) - (2*f^(a + b*x^n)*x^n)/(b^2*n*Log[f]^2) + (f^(a + b*x^n)*x^(2*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^{2 n}}{b n \log (f)}-\frac {2 \int f^{a+b x^n} x^{-1+2 n} \, dx}{b \log (f)} \\ & = -\frac {2 f^{a+b x^n} x^n}{b^2 n \log ^2(f)}+\frac {f^{a+b x^n} x^{2 n}}{b n \log (f)}+\frac {2 \int f^{a+b x^n} x^{-1+n} \, dx}{b^2 \log ^2(f)} \\ & = \frac {2 f^{a+b x^n}}{b^3 n \log ^3(f)}-\frac {2 f^{a+b x^n} x^n}{b^2 n \log ^2(f)}+\frac {f^{a+b x^n} x^{2 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 = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.34 \[ \int f^{a+b x^n} x^{-1+3 n} \, dx=\frac {f^a \Gamma \left (3,-b x^n \log (f)\right )}{b^3 n \log ^3(f)} \]

[In]

Integrate[f^(a + b*x^n)*x^(-1 + 3*n),x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {\left (b^{2} x^{2 n} \ln \left (f \right )^{2}-2 b \,x^{n} \ln \left (f \right )+2\right ) f^{a +b \,x^{n}}}{b^{3} \ln \left (f \right )^{3} n}\)	\(44\)
meijerg	\(-\frac {f^{a} \left (2-\frac {\left (3 b^{2} x^{2 n} \ln \left (f \right )^{2}-6 b \,x^{n} \ln \left (f \right )+6\right ) {\mathrm e}^{b \,x^{n} \ln \left (f \right )}}{3}\right )}{\ln \left (f \right )^{3} b^{3} n}\)	\(52\)

[In]

int(f^(a+b*x^n)*x^(-1+3*n),x,method=_RETURNVERBOSE)

[Out]

(b^2*(x^n)^2*ln(f)^2-2*b*x^n*ln(f)+2)/b^3/ln(f)^3/n*f^(a+b*x^n)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.66 \[ \int f^{a+b x^n} x^{-1+3 n} \, dx=\frac {{\left (b^{2} x^{2 \, n} \log \left (f\right )^{2} - 2 \, b x^{n} \log \left (f\right ) + 2\right )} e^{\left (b x^{n} \log \left (f\right ) + a \log \left (f\right )\right )}}{b^{3} n \log \left (f\right )^{3}} \]

[In]

integrate(f^(a+b*x^n)*x^(-1+3*n),x, algorithm="fricas")

[Out]

(b^2*x^(2*n)*log(f)^2 - 2*b*x^n*log(f) + 2)*e^(b*x^n*log(f) + a*log(f))/(b^3*n*log(f)^3)

Sympy [A] (verification not implemented)

Time = 2.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int f^{a+b x^n} x^{-1+3 n} \, dx=\begin {cases} \log {\left (x \right )} & \text {for}\: b = 0 \wedge f = 1 \wedge n = 0 \\\frac {f^{a} x x^{3 n - 1}}{3 n} & \text {for}\: b = 0 \\f^{a + b} \log {\left (x \right )} & \text {for}\: n = 0 \\\frac {x x^{3 n - 1}}{3 n} & \text {for}\: f = 1 \\\frac {f^{a + b x^{n}} x^{2 n}}{b n \log {\left (f \right )}} - \frac {2 f^{a + b x^{n}} x^{n}}{b^{2} n \log {\left (f \right )}^{2}} + \frac {2 f^{a + b x^{n}}}{b^{3} n \log {\left (f \right )}^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(f**(a+b*x**n)*x**(-1+3*n),x)

[Out]

Piecewise((log(x), Eq(b, 0) & Eq(f, 1) & Eq(n, 0)), (f**a*x*x**(3*n - 1)/(3*n), Eq(b, 0)), (f**(a + b)*log(x),
Eq(n, 0)), (x*x**(3*n - 1)/(3*n), Eq(f, 1)), (f**(a + b*x**n)*x**(2*n)/(b*n*log(f)) - 2*f**(a + b*x**n)*x**n/
(b**2*n*log(f)**2) + 2*f**(a + b*x**n)/(b**3*n*log(f)**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.72 \[ \int f^{a+b x^n} x^{-1+3 n} \, dx=\frac {{\left (b^{2} f^{a} x^{2 \, n} \log \left (f\right )^{2} - 2 \, b f^{a} x^{n} \log \left (f\right ) + 2 \, f^{a}\right )} f^{b x^{n}}}{b^{3} n \log \left (f\right )^{3}} \]

[In]

integrate(f^(a+b*x^n)*x^(-1+3*n),x, algorithm="maxima")

[Out]

(b^2*f^a*x^(2*n)*log(f)^2 - 2*b*f^a*x^n*log(f) + 2*f^a)*f^(b*x^n)/(b^3*n*log(f)^3)

Giac [F]

\[ \int f^{a+b x^n} x^{-1+3 n} \, dx=\int { f^{b x^{n} + a} x^{3 \, n - 1} \,d x } \]

[In]

integrate(f^(a+b*x^n)*x^(-1+3*n),x, algorithm="giac")

[Out]

integrate(f^(b*x^n + a)*x^(3*n - 1), x)

Mupad [F(-1)]

Timed out. \[ \int f^{a+b x^n} x^{-1+3 n} \, dx=\int f^{a+b\,x^n}\,x^{3\,n-1} \,d x \]

[In]

int(f^(a + b*x^n)*x^(3*n - 1),x)

[Out]

int(f^(a + b*x^n)*x^(3*n - 1), x)