\(\int \frac {f^{a+b x^3}}{x^4} \, dx\) [103]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int \frac {f^{a+b x^3}}{x^4} \, dx=-\frac {f^{a+b x^3}}{3 x^3}+\frac {1}{3} b f^a \operatorname {ExpIntegralEi}\left (b x^3 \log (f)\right ) \log (f) \]

[Out]

-1/3*f^(b*x^3+a)/x^3+1/3*b*f^a*Ei(b*x^3*ln(f))*ln(f)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2245, 2241} \[ \int \frac {f^{a+b x^3}}{x^4} \, dx=\frac {1}{3} b f^a \log (f) \operatorname {ExpIntegralEi}\left (b x^3 \log (f)\right )-\frac {f^{a+b x^3}}{3 x^3} \]

[In]

Int[f^(a + b*x^3)/x^4,x]

[Out]

-1/3*f^(a + b*x^3)/x^3 + (b*f^a*ExpIntegralEi[b*x^3*Log[f]]*Log[f])/3

Rule 2241

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[F^a*(ExpIntegralEi[
b*(c + d*x)^n*Log[F]]/(f*n)), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2245

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))), x] - Dist[b*n*(Log[F]/(m + 1)), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rubi steps \begin{align*} \text {integral}& = -\frac {f^{a+b x^3}}{3 x^3}+(b \log (f)) \int \frac {f^{a+b x^3}}{x} \, dx \\ & = -\frac {f^{a+b x^3}}{3 x^3}+\frac {1}{3} b f^a \text {Ei}\left (b x^3 \log (f)\right ) \log (f) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int \frac {f^{a+b x^3}}{x^4} \, dx=\frac {1}{3} f^a \left (-\frac {f^{b x^3}}{x^3}+b \operatorname {ExpIntegralEi}\left (b x^3 \log (f)\right ) \log (f)\right ) \]

[In]

Integrate[f^(a + b*x^3)/x^4,x]

[Out]

(f^a*(-(f^(b*x^3)/x^3) + b*ExpIntegralEi[b*x^3*Log[f]]*Log[f]))/3

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(96\) vs. \(2(31)=62\).

Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.77

method	result	size
meijerg	\(-\frac {f^{a} b \ln \left (f \right ) \left (\frac {1}{b \,x^{3} \ln \left (f \right )}+1-3 \ln \left (x \right )-\ln \left (-b \right )-\ln \left (\ln \left (f \right )\right )-\frac {2+2 b \,x^{3} \ln \left (f \right )}{2 b \,x^{3} \ln \left (f \right )}+\frac {{\mathrm e}^{b \,x^{3} \ln \left (f \right )}}{b \,x^{3} \ln \left (f \right )}+\ln \left (-b \,x^{3} \ln \left (f \right )\right )+\operatorname {Ei}_{1}\left (-b \,x^{3} \ln \left (f \right )\right )\right )}{3}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^3}}{x^4} \, dx=\frac {b f^{a} x^{3} {\rm Ei}\left (b x^{3} \log \left (f\right )\right ) \log \left (f\right ) - f^{b x^{3} + a}}{3 \, x^{3}} \]

[In]

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

[Out]

1/3*(b*f^a*x^3*Ei(b*x^3*log(f))*log(f) - f^(b*x^3 + a))/x^3

Sympy [F]

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

[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 = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.51 \[ \int \frac {f^{a+b x^3}}{x^4} \, dx=\frac {1}{3} \, b f^{a} \Gamma \left (-1, -b x^{3} \log \left (f\right )\right ) \log \left (f\right ) \]

[In]

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

[Out]

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

Giac [F]

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

[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 = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int \frac {f^{a+b x^3}}{x^4} \, dx=-\frac {f^a\,\left (f^{b\,x^3}+b\,x^3\,\ln \left (f\right )\,\mathrm {expint}\left (-b\,x^3\,\ln \left (f\right )\right )\right )}{3\,x^3} \]

[In]

int(f^(a + b*x^3)/x^4,x)

[Out]

-(f^a*(f^(b*x^3) + b*x^3*log(f)*expint(-b*x^3*log(f))))/(3*x^3)