\(\int f^{a+b x^n} x^m \, dx\) [174]

Optimal result

Integrand size = 13, antiderivative size = 46 \[ \int f^{a+b x^n} x^m \, dx=-\frac {f^a x^{1+m} \Gamma \left (\frac {1+m}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{-\frac {1+m}{n}}}{n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2250} \[ \int f^{a+b x^n} x^m \, dx=-\frac {f^a x^{m+1} \left (-b \log (f) x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-b x^n \log (f)\right )}{n} \]

[In]

Int[f^(a + b*x^n)*x^m,x]

[Out]

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

Rule 2250

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^n} x^m \, dx=-\frac {f^a x^{1+m} \Gamma \left (\frac {1+m}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{-\frac {1+m}{n}}}{n} \]

[In]

Integrate[f^(a + b*x^n)*x^m,x]

[Out]

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

Maple [C] (verified)

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

Time = 0.10 (sec) , antiderivative size = 280, normalized size of antiderivative = 6.09

method	result	size
meijerg	\(\frac {f^{a} \left (-b \right )^{-\frac {m}{n}-\frac {1}{n}} \ln \left (f \right )^{-\frac {m}{n}-\frac {1}{n}} \left (\frac {n \,x^{1+m} \left (-b \right )^{\frac {m}{n}+\frac {1}{n}} \ln \left (f \right )^{\frac {m}{n}+\frac {1}{n}} \left (x^{n} \ln \left (f \right ) b n +m +n +1\right ) L_{-\frac {1+m}{n}}^{\left (\frac {1+m +n}{n}\right )}\left (b \,x^{n} \ln \left (f \right )\right ) \Gamma \left (-\frac {1+m}{n}+1\right ) \Gamma \left (\frac {1+m +n}{n}+1\right )}{\left (1+m \right ) \left (1+m +n \right ) \Gamma \left (-\frac {1+m}{n}+\frac {1+m +n}{n}+1\right )}-\frac {n^{2} x^{1+m +n} \left (-b \right )^{\frac {m}{n}+\frac {1}{n}} \ln \left (f \right )^{1+\frac {m}{n}+\frac {1}{n}} b L_{-\frac {1+m}{n}}^{\left (\frac {1+m +n}{n}+1\right )}\left (b \,x^{n} \ln \left (f \right )\right ) \Gamma \left (-\frac {1+m}{n}+1\right ) \Gamma \left (\frac {1+m +n}{n}+1\right )}{\left (1+m \right ) \left (1+m +n \right ) \Gamma \left (-\frac {1+m}{n}+\frac {1+m +n}{n}+1\right )}\right )}{n}\)	\(280\)

[In]

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

[Out]

f^a*(-b)^(-m/n-1/n)*ln(f)^(-m/n-1/n)/n*(n/(1+m)*x^(1+m)*(-b)^(m/n+1/n)*ln(f)^(m/n+1/n)*(x^n*ln(f)*b*n+m+n+1)/(
1+m+n)*LaguerreL(-(1+m)/n,(1+m+n)/n,b*x^n*ln(f))*GAMMA(-(1+m)/n+1)*GAMMA((1+m+n)/n+1)/GAMMA(-(1+m)/n+(1+m+n)/n
+1)-n^2/(1+m)*x^(1+m+n)*(-b)^(m/n+1/n)*ln(f)^(1+m/n+1/n)*b/(1+m+n)*LaguerreL(-(1+m)/n,(1+m+n)/n+1,b*x^n*ln(f))
*GAMMA(-(1+m)/n+1)*GAMMA((1+m+n)/n+1)/GAMMA(-(1+m)/n+(1+m+n)/n+1))

Fricas [F]

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

[In]

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

[Out]

integral(f^(b*x^n + a)*x^m, x)

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b*x**n)*x**m, x)

Maxima [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int f^{a+b x^n} x^m \, dx=-\frac {f^{a} x^{m + 1} \Gamma \left (\frac {m + 1}{n}, -b x^{n} \log \left (f\right )\right )}{\left (-b x^{n} \log \left (f\right )\right )^{\frac {m + 1}{n}} n} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(b*x^n + a)*x^m, x)

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.72 \[ \int f^{a+b x^n} x^m \, dx=\frac {f^a\,f^{b\,x^n}\,x^{m+1}\,{\mathrm {e}}^{-\frac {b\,x^n\,\ln \left (f\right )}{2}}\,{\mathrm {M}}_{1-\frac {m+n+1}{2\,n},\frac {m+n+1}{2\,n}-\frac {1}{2}}\left (b\,x^n\,\ln \left (f\right )\right )}{\left (m+1\right )\,{\left (b\,x^n\,\ln \left (f\right )\right )}^{\frac {m+n+1}{2\,n}}} \]

[In]

int(f^(a + b*x^n)*x^m,x)

[Out]

(f^a*f^(b*x^n)*x^(m + 1)*exp(-(b*x^n*log(f))/2)*whittakerM(1 - (m + n + 1)/(2*n), (m + n + 1)/(2*n) - 1/2, b*x
^n*log(f)))/((m + 1)*(b*x^n*log(f))^((m + n + 1)/(2*n)))