\(\int \frac {f^{a+b x^3}}{x^2} \, dx\) [112]

Optimal result

Integrand size = 13, antiderivative size = 34 \[ \int \frac {f^{a+b x^3}}{x^2} \, dx=-\frac {f^a \Gamma \left (-\frac {1}{3},-b x^3 \log (f)\right ) \sqrt [3]{-b x^3 \log (f)}}{3 x} \]

[Out]

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

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} \[ \int \frac {f^{a+b x^3}}{x^2} \, dx=-\frac {f^a \sqrt [3]{-b x^3 \log (f)} \Gamma \left (-\frac {1}{3},-b x^3 \log (f)\right )}{3 x} \]

[In]

Int[f^(a + b*x^3)/x^2,x]

[Out]

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

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 {1}{3},-b x^3 \log (f)\right ) \sqrt [3]{-b x^3 \log (f)}}{3 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^3}}{x^2} \, dx=-\frac {f^a \Gamma \left (-\frac {1}{3},-b x^3 \log (f)\right ) \sqrt [3]{-b x^3 \log (f)}}{3 x} \]

[In]

Integrate[f^(a + b*x^3)/x^2,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(99\) vs. \(2(28)=56\).

Time = 0.02 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.94

method	result	size
meijerg	\(\frac {f^{a} \left (-b \right )^{\frac {1}{3}} \ln \left (f \right )^{\frac {1}{3}} \left (\frac {3 x^{2} \ln \left (f \right )^{\frac {2}{3}} b \Gamma \left (\frac {2}{3}\right )}{\left (-b \right )^{\frac {1}{3}} \left (-b \,x^{3} \ln \left (f \right )\right )^{\frac {2}{3}}}-\frac {3 \,{\mathrm e}^{b \,x^{3} \ln \left (f \right )}}{x \left (-b \right )^{\frac {1}{3}} \ln \left (f \right )^{\frac {1}{3}}}-\frac {3 x^{2} \ln \left (f \right )^{\frac {2}{3}} b \Gamma \left (\frac {2}{3}, -b \,x^{3} \ln \left (f \right )\right )}{\left (-b \right )^{\frac {1}{3}} \left (-b \,x^{3} \ln \left (f \right )\right )^{\frac {2}{3}}}\right )}{3}\)	\(100\)

[In]

int(f^(b*x^3+a)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/3*f^a*(-b)^(1/3)*ln(f)^(1/3)*(3*x^2/(-b)^(1/3)*ln(f)^(2/3)*b*GAMMA(2/3)/(-b*x^3*ln(f))^(2/3)-3/x/(-b)^(1/3)/
ln(f)^(1/3)*exp(b*x^3*ln(f))-3*x^2/(-b)^(1/3)*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 = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {f^{a+b x^3}}{x^2} \, dx=\frac {\left (-b \log \left (f\right )\right )^{\frac {1}{3}} f^{a} x \Gamma \left (\frac {2}{3}, -b x^{3} \log \left (f\right )\right ) - f^{b x^{3} + a}}{x} \]

[In]

integrate(f^(b*x^3+a)/x^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {f^{a+b x^3}}{x^2} \, dx=\int \frac {f^{a + b x^{3}}}{x^{2}}\, dx \]

[In]

integrate(f**(b*x**3+a)/x**2,x)

[Out]

Integral(f**(a + b*x**3)/x**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {f^{a+b x^3}}{x^2} \, dx=-\frac {\left (-b x^{3} \log \left (f\right )\right )^{\frac {1}{3}} f^{a} \Gamma \left (-\frac {1}{3}, -b x^{3} \log \left (f\right )\right )}{3 \, x} \]

[In]

integrate(f^(b*x^3+a)/x^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {f^{a+b x^3}}{x^2} \, dx=\int { \frac {f^{b x^{3} + a}}{x^{2}} \,d x } \]

[In]

integrate(f^(b*x^3+a)/x^2,x, algorithm="giac")

[Out]

integrate(f^(b*x^3 + a)/x^2, x)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.85 \[ \int \frac {f^{a+b x^3}}{x^2} \, dx=\frac {f^a\,\Gamma \left (\frac {2}{3},-b\,x^3\,\ln \left (f\right )\right )\,{\left (-b\,x^3\,\ln \left (f\right )\right )}^{1/3}}{x}-\frac {f^a\,\Gamma \left (\frac {2}{3}\right )\,{\left (-b\,x^3\,\ln \left (f\right )\right )}^{1/3}}{x}-\frac {f^a\,f^{b\,x^3}}{x} \]

[In]

int(f^(a + b*x^3)/x^2,x)

[Out]

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