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

Optimal. Leaf size=46 \[ -\frac {f^a x^{m+1} \left (-b \log (f) x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-b x^n \log (f)\right )}{n} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2218} \[ -\frac {f^a x^{m+1} \left (-b \log (f) x^n\right )^{-\frac {m+1}{n}} \text {Gamma}\left (\frac {m+1}{n},-b \log (f) x^n\right )}{n} \]

Antiderivative was successfully verified.

[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 2218

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

Rubi steps

\begin {align*} \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}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 46, normalized size = 1.00 \[ -\frac {f^a x^{m+1} \left (-b \log (f) x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-b x^n \log (f)\right )}{n} \]

Antiderivative was successfully verified.

[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)))

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (f^{b x^{n} + a} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(f^(b*x^n + a)*x^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(b*x^n + a)*x^m, x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int f^{b \,x^{n}+a} x^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.17, size = 47, normalized size = 1.02 \[ -\frac {f^{a} x^{m + 1} \Gamma \left (\frac {m + 1}{n}, -b x^{n} \log \relax (f)\right )}{\left (-b x^{n} \log \relax (f)\right )^{\frac {m + 1}{n}} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.76, size = 79, normalized size = 1.72 \[ \frac {f^a\,f^{b\,x^n}\,x^{m+1}\,{\mathrm {e}}^{-\frac {b\,x^n\,\ln \relax (f)}{2}}\,{\mathrm {M}}_{1-\frac {m+n+1}{2\,n},\frac {m+n+1}{2\,n}-\frac {1}{2}}\left (b\,x^n\,\ln \relax (f)\right )}{\left (m+1\right )\,{\left (b\,x^n\,\ln \relax (f)\right )}^{\frac {m+n+1}{2\,n}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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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**m,x)

[Out]

Timed out

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