3.95 \(\int f^{a+b x^3} x^m \, dx\)

Optimal. Leaf size=46 \[ -\frac {1}{3} f^a x^{m+1} \left (-b x^3 \log (f)\right )^{\frac {1}{3} (-m-1)} \Gamma \left (\frac {m+1}{3},-b x^3 \log (f)\right ) \]

[Out]

-1/3*f^a*x^(1+m)*GAMMA(1/3+1/3*m,-b*x^3*ln(f))*(-b*x^3*ln(f))^(-1/3-1/3*m)

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Rubi [A]  time = 0.02, 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} \[ -\frac {1}{3} f^a x^{m+1} \left (-b x^3 \log (f)\right )^{\frac {1}{3} (-m-1)} \text {Gamma}\left (\frac {m+1}{3},-b x^3 \log (f)\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^3)*x^m,x]

[Out]

-(f^a*x^(1 + m)*Gamma[(1 + m)/3, -(b*x^3*Log[f])]*(-(b*x^3*Log[f]))^((-1 - m)/3))/3

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^3} x^m \, dx &=-\frac {1}{3} f^a x^{1+m} \Gamma \left (\frac {1+m}{3},-b x^3 \log (f)\right ) \left (-b x^3 \log (f)\right )^{\frac {1}{3} (-1-m)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 46, normalized size = 1.00 \[ -\frac {1}{3} f^a x^{m+1} \left (-b x^3 \log (f)\right )^{\frac {1}{3} (-m-1)} \Gamma \left (\frac {m+1}{3},-b x^3 \log (f)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^3)*x^m,x]

[Out]

-1/3*(f^a*x^(1 + m)*Gamma[(1 + m)/3, -(b*x^3*Log[f])]*(-(b*x^3*Log[f]))^((-1 - m)/3))

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fricas [A]  time = 0.45, size = 40, normalized size = 0.87 \[ \frac {e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-b \log \relax (f)\right ) + a \log \relax (f)\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -b x^{3} \log \relax (f)\right )}{3 \, b \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*e^(-1/3*(m - 2)*log(-b*log(f)) + a*log(f))*gamma(1/3*m + 1/3, -b*x^3*log(f))/(b*log(f))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{b x^{3} + a} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(b*x^3 + a)*x^m, x)

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maple [B]  time = 0.06, size = 140, normalized size = 3.04 \[ \frac {\left (\frac {3 \left (\frac {m}{3}+\frac {1}{3}\right ) x^{m +1} \left (-b \right )^{\frac {m}{3}+\frac {1}{3}} \left (-b \,x^{3} \ln \relax (f )\right )^{-\frac {m}{3}-\frac {1}{3}} \ln \relax (f )^{\frac {m}{3}+\frac {1}{3}} \Gamma \left (\frac {m}{3}+\frac {1}{3}\right )}{m +1}+\frac {3 \left (-\frac {m}{3}-\frac {1}{3}\right ) x^{m +1} \left (-b \right )^{\frac {m}{3}+\frac {1}{3}} \left (-b \,x^{3} \ln \relax (f )\right )^{-\frac {m}{3}-\frac {1}{3}} \ln \relax (f )^{\frac {m}{3}+\frac {1}{3}} \Gamma \left (\frac {m}{3}+\frac {1}{3}, -b \,x^{3} \ln \relax (f )\right )}{m +1}\right ) f^{a} \left (-b \right )^{-\frac {m}{3}-\frac {1}{3}} \ln \relax (f )^{-\frac {m}{3}-\frac {1}{3}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^3+a)*x^m,x)

[Out]

1/3*f^a*(-b)^(-1/3*m-1/3)*ln(f)^(-1/3*m-1/3)*(3/(m+1)*x^(m+1)*(-b)^(1/3*m+1/3)*ln(f)^(1/3*m+1/3)*(1/3*m+1/3)*(
-b*x^3*ln(f))^(-1/3*m-1/3)*GAMMA(1/3*m+1/3)+3/(m+1)*x^(m+1)*(-b)^(1/3*m+1/3)*ln(f)^(1/3*m+1/3)*(-1/3*m-1/3)*(-
b*x^3*ln(f))^(-1/3*m-1/3)*GAMMA(1/3*m+1/3,-b*x^3*ln(f)))

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maxima [A]  time = 1.32, size = 38, normalized size = 0.83 \[ -\frac {1}{3} \, \left (-b x^{3} \log \relax (f)\right )^{-\frac {1}{3} \, m - \frac {1}{3}} f^{a} x^{m + 1} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -b x^{3} \log \relax (f)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-b*x^3*log(f))^(-1/3*m - 1/3)*f^a*x^(m + 1)*gamma(1/3*m + 1/3, -b*x^3*log(f))

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mupad [B]  time = 3.39, size = 56, normalized size = 1.22 \[ \frac {f^a\,x^{m+1}\,{\mathrm {e}}^{\frac {b\,x^3\,\ln \relax (f)}{2}}\,{\mathrm {M}}_{\frac {1}{3}-\frac {m}{6},\frac {m}{6}+\frac {1}{6}}\left (b\,x^3\,\ln \relax (f)\right )}{\left (m+1\right )\,{\left (b\,x^3\,\ln \relax (f)\right )}^{\frac {m}{6}+\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^3)*x^m,x)

[Out]

(f^a*x^(m + 1)*exp((b*x^3*log(f))/2)*whittakerM(1/3 - m/6, m/6 + 1/6, b*x^3*log(f)))/((m + 1)*(b*x^3*log(f))^(
m/6 + 2/3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + b x^{3}} x^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**3+a)*x**m,x)

[Out]

Integral(f**(a + b*x**3)*x**m, x)

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