\(\int f^{a+b x^3} x^4 \, dx\) [108]

Optimal result

Integrand size = 13, antiderivative size = 34 \[ \int f^{a+b x^3} x^4 \, dx=-\frac {f^a x^5 \Gamma \left (\frac {5}{3},-b x^3 \log (f)\right )}{3 \left (-b x^3 \log (f)\right )^{5/3}} \]

[Out]

-1/3*f^a*x^5*GAMMA(5/3,-b*x^3*ln(f))/(-b*x^3*ln(f))^(5/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, 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} \[ \int f^{a+b x^3} x^4 \, dx=-\frac {x^5 f^a \Gamma \left (\frac {5}{3},-b x^3 \log (f)\right )}{3 \left (-b x^3 \log (f)\right )^{5/3}} \]

[In]

Int[f^(a + b*x^3)*x^4,x]

[Out]

-1/3*(f^a*x^5*Gamma[5/3, -(b*x^3*Log[f])])/(-(b*x^3*Log[f]))^(5/3)

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 {f^a x^5 \Gamma \left (\frac {5}{3},-b x^3 \log (f)\right )}{3 \left (-b x^3 \log (f)\right )^{5/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^3} x^4 \, dx=-\frac {f^a x^5 \Gamma \left (\frac {5}{3},-b x^3 \log (f)\right )}{3 \left (-b x^3 \log (f)\right )^{5/3}} \]

[In]

Integrate[f^(a + b*x^3)*x^4,x]

[Out]

-1/3*(f^a*x^5*Gamma[5/3, -(b*x^3*Log[f])])/(-(b*x^3*Log[f]))^(5/3)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(105\) vs. \(2(28)=56\).

Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.12

method	result	size
meijerg	\(\frac {f^{a} \left (-\frac {2 x^{2} \left (-b \right )^{\frac {5}{3}} \ln \left (f \right )^{\frac {2}{3}} \Gamma \left (\frac {2}{3}\right )}{3 b \left (-b \,x^{3} \ln \left (f \right )\right )^{\frac {2}{3}}}+\frac {x^{2} \left (-b \right )^{\frac {5}{3}} \ln \left (f \right )^{\frac {2}{3}} {\mathrm e}^{b \,x^{3} \ln \left (f \right )}}{b}+\frac {2 x^{2} \left (-b \right )^{\frac {5}{3}} \ln \left (f \right )^{\frac {2}{3}} \Gamma \left (\frac {2}{3}, -b \,x^{3} \ln \left (f \right )\right )}{3 b \left (-b \,x^{3} \ln \left (f \right )\right )^{\frac {2}{3}}}\right )}{3 \left (-b \right )^{\frac {5}{3}} \ln \left (f \right )^{\frac {5}{3}}}\)	\(106\)

[In]

int(f^(b*x^3+a)*x^4,x,method=_RETURNVERBOSE)

[Out]

1/3*f^a/(-b)^(5/3)/ln(f)^(5/3)*(-2/3*x^2*(-b)^(5/3)*ln(f)^(2/3)/b*GAMMA(2/3)/(-b*x^3*ln(f))^(2/3)+x^2*(-b)^(5/
3)*ln(f)^(2/3)/b*exp(b*x^3*ln(f))+2/3*x^2*(-b)^(5/3)*ln(f)^(2/3)/b/(-b*x^3*ln(f))^(2/3)*GAMMA(2/3,-b*x^3*ln(f)
))

Fricas [A] (verification not implemented)

none

Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.44 \[ \int f^{a+b x^3} x^4 \, dx=\frac {3 \, b f^{b x^{3} + a} x^{2} \log \left (f\right ) - 2 \, \left (-b \log \left (f\right )\right )^{\frac {1}{3}} f^{a} \Gamma \left (\frac {2}{3}, -b x^{3} \log \left (f\right )\right )}{9 \, b^{2} \log \left (f\right )^{2}} \]

[In]

integrate(f^(b*x^3+a)*x^4,x, algorithm="fricas")

[Out]

1/9*(3*b*f^(b*x^3 + a)*x^2*log(f) - 2*(-b*log(f))^(1/3)*f^a*gamma(2/3, -b*x^3*log(f)))/(b^2*log(f)^2)

Sympy [F]

\[ \int f^{a+b x^3} x^4 \, dx=\int f^{a + b x^{3}} x^{4}\, dx \]

[In]

integrate(f**(b*x**3+a)*x**4,x)

[Out]

Integral(f**(a + b*x**3)*x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int f^{a+b x^3} x^4 \, dx=-\frac {f^{a} x^{5} \Gamma \left (\frac {5}{3}, -b x^{3} \log \left (f\right )\right )}{3 \, \left (-b x^{3} \log \left (f\right )\right )^{\frac {5}{3}}} \]

[In]

integrate(f^(b*x^3+a)*x^4,x, algorithm="maxima")

[Out]

-1/3*f^a*x^5*gamma(5/3, -b*x^3*log(f))/(-b*x^3*log(f))^(5/3)

Giac [F]

\[ \int f^{a+b x^3} x^4 \, dx=\int { f^{b x^{3} + a} x^{4} \,d x } \]

[In]

integrate(f^(b*x^3+a)*x^4,x, algorithm="giac")

[Out]

integrate(f^(b*x^3 + a)*x^4, x)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.09 \[ \int f^{a+b x^3} x^4 \, dx=\frac {2\,f^a\,x^5\,\Gamma \left (\frac {2}{3}\right )}{9\,{\left (-b\,x^3\,\ln \left (f\right )\right )}^{5/3}}-\frac {2\,f^a\,x^5\,\Gamma \left (\frac {2}{3},-b\,x^3\,\ln \left (f\right )\right )}{9\,{\left (-b\,x^3\,\ln \left (f\right )\right )}^{5/3}}+\frac {f^a\,f^{b\,x^3}\,x^2}{3\,b\,\ln \left (f\right )} \]

[In]

int(f^(a + b*x^3)*x^4,x)

[Out]

(2*f^a*x^5*gamma(2/3))/(9*(-b*x^3*log(f))^(5/3)) - (2*f^a*x^5*igamma(2/3, -b*x^3*log(f)))/(9*(-b*x^3*log(f))^(
5/3)) + (f^a*f^(b*x^3)*x^2)/(3*b*log(f))