\(\int f^{a+b x^3} x^m \, dx\) [95]

Optimal result

Integrand size = 13, antiderivative size = 46 \[ \int f^{a+b x^3} x^m \, dx=-\frac {1}{3} f^a x^{1+m} \Gamma \left (\frac {1+m}{3},-b x^3 \log (f)\right ) \left (-b x^3 \log (f)\right )^{\frac {1}{3} (-1-m)} \]

[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)

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} \[ \int f^{a+b x^3} x^m \, dx=-\frac {1}{3} f^a x^{m+1} \left (-b x^3 \log (f)\right )^{\frac {1}{3} (-m-1)} \Gamma \left (\frac {m+1}{3},-b x^3 \log (f)\right ) \]

[In]

Int[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))

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 {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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^3} x^m \, dx=-\frac {1}{3} f^a x^{1+m} \Gamma \left (\frac {1+m}{3},-b x^3 \log (f)\right ) \left (-b x^3 \log (f)\right )^{\frac {1}{3} (-1-m)} \]

[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))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(139\) vs. \(2(38)=76\).

Time = 0.05 (sec) , antiderivative size = 140, normalized size of antiderivative = 3.04

method	result	size
meijerg	\(\frac {f^{a} \left (-b \right )^{-\frac {1}{3}-\frac {m}{3}} \ln \left (f \right )^{-\frac {1}{3}-\frac {m}{3}} \left (\frac {3 x^{1+m} \left (-b \right )^{\frac {1}{3}+\frac {m}{3}} \ln \left (f \right )^{\frac {1}{3}+\frac {m}{3}} \left (\frac {1}{3}+\frac {m}{3}\right ) \left (-b \,x^{3} \ln \left (f \right )\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1}{3}+\frac {m}{3}\right )}{1+m}+\frac {3 x^{1+m} \left (-b \right )^{\frac {1}{3}+\frac {m}{3}} \ln \left (f \right )^{\frac {1}{3}+\frac {m}{3}} \left (-\frac {1}{3}-\frac {m}{3}\right ) \left (-b \,x^{3} \ln \left (f \right )\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1}{3}+\frac {m}{3}, -b \,x^{3} \ln \left (f \right )\right )}{1+m}\right )}{3}\)	\(140\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int f^{a+b x^3} x^m \, dx=\frac {e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-b \log \left (f\right )\right ) + a \log \left (f\right )\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -b x^{3} \log \left (f\right )\right )}{3 \, b \log \left (f\right )} \]

[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))

Sympy [F]

\[ \int f^{a+b x^3} x^m \, dx=\int f^{a + b x^{3}} x^{m}\, dx \]

[In]

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

[Out]

Integral(f**(a + b*x**3)*x**m, x)

Maxima [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int f^{a+b x^3} x^m \, dx=-\frac {1}{3} \, \left (-b x^{3} \log \left (f\right )\right )^{-\frac {1}{3} \, m - \frac {1}{3}} f^{a} x^{m + 1} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -b x^{3} \log \left (f\right )\right ) \]

[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))

Giac [F]

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

[In]

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

[Out]

integrate(f^(b*x^3 + a)*x^m, x)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.22 \[ \int f^{a+b x^3} x^m \, dx=\frac {f^a\,x^{m+1}\,{\mathrm {e}}^{\frac {b\,x^3\,\ln \left (f\right )}{2}}\,{\mathrm {M}}_{\frac {1}{3}-\frac {m}{6},\frac {m}{6}+\frac {1}{6}}\left (b\,x^3\,\ln \left (f\right )\right )}{\left (m+1\right )\,{\left (b\,x^3\,\ln \left (f\right )\right )}^{\frac {m}{6}+\frac {2}{3}}} \]

[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))