\(\int \frac {f^{a+b x^3}}{x^3} \, dx\) [113]

Optimal result

Integrand size = 13, antiderivative size = 34 \[ \int \frac {f^{a+b x^3}}{x^3} \, dx=-\frac {f^a \Gamma \left (-\frac {2}{3},-b x^3 \log (f)\right ) \left (-b x^3 \log (f)\right )^{2/3}}{3 x^2} \]

[Out]

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

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

[In]

Int[f^(a + b*x^3)/x^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^3}}{x^3} \, dx=-\frac {f^a \Gamma \left (-\frac {2}{3},-b x^3 \log (f)\right ) \left (-b x^3 \log (f)\right )^{2/3}}{3 x^2} \]

[In]

Integrate[f^(a + b*x^3)/x^3,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(101\) vs. \(2(28)=56\).

Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.00

method	result	size
meijerg	\(-\frac {f^{a} b \ln \left (f \right )^{\frac {2}{3}} \left (\frac {x \ln \left (f \right )^{\frac {1}{3}} b \pi \sqrt {3}}{\left (-b \right )^{\frac {2}{3}} \Gamma \left (\frac {2}{3}\right ) \left (-b \,x^{3} \ln \left (f \right )\right )^{\frac {1}{3}}}-\frac {3 \,{\mathrm e}^{b \,x^{3} \ln \left (f \right )}}{2 x^{2} \left (-b \right )^{\frac {2}{3}} \ln \left (f \right )^{\frac {2}{3}}}-\frac {3 x \ln \left (f \right )^{\frac {1}{3}} b \Gamma \left (\frac {1}{3}, -b \,x^{3} \ln \left (f \right )\right )}{2 \left (-b \right )^{\frac {2}{3}} \left (-b \,x^{3} \ln \left (f \right )\right )^{\frac {1}{3}}}\right )}{3 \left (-b \right )^{\frac {1}{3}}}\)	\(102\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {f^{a+b x^3}}{x^3} \, dx=\frac {\left (-b \log \left (f\right )\right )^{\frac {2}{3}} f^{a} x^{2} \Gamma \left (\frac {1}{3}, -b x^{3} \log \left (f\right )\right ) - f^{b x^{3} + a}}{2 \, x^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b*x**3)/x**3, x)

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(b*x^3 + a)/x^3, x)

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.06 \[ \int \frac {f^{a+b x^3}}{x^3} \, dx=\frac {f^a\,\Gamma \left (\frac {1}{3},-b\,x^3\,\ln \left (f\right )\right )\,{\left (-b\,x^3\,\ln \left (f\right )\right )}^{2/3}}{2\,x^2}-\frac {f^a\,f^{b\,x^3}}{2\,x^2}-\frac {\pi \,\sqrt {3}\,f^a\,{\left (-b\,x^3\,\ln \left (f\right )\right )}^{2/3}}{3\,x^2\,\Gamma \left (\frac {2}{3}\right )} \]

[In]

int(f^(a + b*x^3)/x^3,x)

[Out]

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