\(\int f^{a+b x^3} x \, dx\) [110]

Optimal result

Integrand size = 11, antiderivative size = 34 \[ \int f^{a+b x^3} x \, dx=-\frac {f^a x^2 \Gamma \left (\frac {2}{3},-b x^3 \log (f)\right )}{3 \left (-b x^3 \log (f)\right )^{2/3}} \]

[Out]

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

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2250} \[ \int f^{a+b x^3} x \, dx=-\frac {x^2 f^a \Gamma \left (\frac {2}{3},-b x^3 \log (f)\right )}{3 \left (-b x^3 \log (f)\right )^{2/3}} \]

[In]

Int[f^(a + b*x^3)*x,x]

[Out]

-1/3*(f^a*x^2*Gamma[2/3, -(b*x^3*Log[f])])/(-(b*x^3*Log[f]))^(2/3)

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^2 \Gamma \left (\frac {2}{3},-b x^3 \log (f)\right )}{3 \left (-b x^3 \log (f)\right )^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^3} x \, dx=-\frac {f^a x^2 \Gamma \left (\frac {2}{3},-b x^3 \log (f)\right )}{3 \left (-b x^3 \log (f)\right )^{2/3}} \]

[In]

Integrate[f^(a + b*x^3)*x,x]

[Out]

-1/3*(f^a*x^2*Gamma[2/3, -(b*x^3*Log[f])])/(-(b*x^3*Log[f]))^(2/3)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(74\) vs. \(2(28)=56\).

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.21

method	result	size
meijerg	\(\frac {f^{a} \left (\frac {x^{2} \left (-b \right )^{\frac {2}{3}} \ln \left (f \right )^{\frac {2}{3}} \Gamma \left (\frac {2}{3}\right )}{\left (-b \,x^{3} \ln \left (f \right )\right )^{\frac {2}{3}}}-\frac {x^{2} \left (-b \right )^{\frac {2}{3}} \ln \left (f \right )^{\frac {2}{3}} \Gamma \left (\frac {2}{3}, -b \,x^{3} \ln \left (f \right )\right )}{\left (-b \,x^{3} \ln \left (f \right )\right )^{\frac {2}{3}}}\right )}{3 \left (-b \right )^{\frac {2}{3}} \ln \left (f \right )^{\frac {2}{3}}}\)	\(75\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int f^{a+b x^3} x \, dx=\frac {\left (-b \log \left (f\right )\right )^{\frac {1}{3}} f^{a} \Gamma \left (\frac {2}{3}, -b x^{3} \log \left (f\right )\right )}{3 \, b \log \left (f\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b*x**3)*x, x)

Maxima [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int f^{a+b x^3} x \, dx=-\frac {f^{a} x^{2} \Gamma \left (\frac {2}{3}, -b x^{3} \log \left (f\right )\right )}{3 \, \left (-b x^{3} \log \left (f\right )\right )^{\frac {2}{3}}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(b*x^3 + a)*x, x)

Mupad [F(-1)]

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

[In]

int(f^(a + b*x^3)*x,x)

[Out]

int(f^(a + b*x^3)*x, x)