\(\int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx\) [222]

Optimal result

Integrand size = 15, antiderivative size = 68 \[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=-\frac {b f^{\frac {c}{a+b x}}}{a}-\frac {f^{\frac {c}{a+b x}}}{x}-\frac {b c f^{\frac {c}{a}} \operatorname {ExpIntegralEi}\left (-\frac {b c x \log (f)}{a (a+b x)}\right ) \log (f)}{a^2} \]

[Out]

-b*f^(c/(b*x+a))/a-f^(c/(b*x+a))/x-b*c*f^(c/a)*Ei(-b*c*x*ln(f)/a/(b*x+a))*ln(f)/a^2

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {2255, 6874, 2254, 2241, 2260, 2209, 2240} \[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=-\frac {b c \log (f) f^{\frac {c}{a}} \operatorname {ExpIntegralEi}\left (-\frac {b c x \log (f)}{a (a+b x)}\right )}{a^2}-\frac {b f^{\frac {c}{a+b x}}}{a}-\frac {f^{\frac {c}{a+b x}}}{x} \]

[In]

Int[f^(c/(a + b*x))/x^2,x]

[Out]

-((b*f^(c/(a + b*x)))/a) - f^(c/(a + b*x))/x - (b*c*f^(c/a)*ExpIntegralEi[-((b*c*x*Log[f])/(a*(a + b*x)))]*Log
[f])/a^2

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

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 2254

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

Rule 2255

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(
F^(a + b/(c + d*x))/(f*(m + 1))), x] + Dist[b*d*(Log[F]/(f*(m + 1))), Int[(e + f*x)^(m + 1)*(F^(a + b/(c + d*x
))/(c + d*x)^2), x], x] /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && ILtQ[m, -1]

Rule 2260

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> Dist[-
d/(f*(d*g - c*h)), Subst[Int[F^(a - b*(h/(d*g - c*h)) + d*b*(x/(d*g - c*h)))/x, x], x, (g + h*x)/(c + d*x)], x
] /; FreeQ[{F, a, b, c, d, e, f}, x] && EqQ[d*e - c*f, 0]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = -\frac {f^{\frac {c}{a+b x}}}{x}-(b c \log (f)) \int \frac {f^{\frac {c}{a+b x}}}{x (a+b x)^2} \, dx \\ & = -\frac {f^{\frac {c}{a+b x}}}{x}-(b c \log (f)) \int \left (\frac {f^{\frac {c}{a+b x}}}{a^2 x}-\frac {b f^{\frac {c}{a+b x}}}{a (a+b x)^2}-\frac {b f^{\frac {c}{a+b x}}}{a^2 (a+b x)}\right ) \, dx \\ & = -\frac {f^{\frac {c}{a+b x}}}{x}-\frac {(b c \log (f)) \int \frac {f^{\frac {c}{a+b x}}}{x} \, dx}{a^2}+\frac {\left (b^2 c \log (f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{a+b x} \, dx}{a^2}+\frac {\left (b^2 c \log (f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{(a+b x)^2} \, dx}{a} \\ & = -\frac {b f^{\frac {c}{a+b x}}}{a}-\frac {f^{\frac {c}{a+b x}}}{x}-\frac {b c \text {Ei}\left (\frac {c \log (f)}{a+b x}\right ) \log (f)}{a^2}-\frac {(b c \log (f)) \int \frac {f^{\frac {c}{a+b x}}}{x (a+b x)} \, dx}{a}-\frac {\left (b^2 c \log (f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{a+b x} \, dx}{a^2} \\ & = -\frac {b f^{\frac {c}{a+b x}}}{a}-\frac {f^{\frac {c}{a+b x}}}{x}-\frac {(b c \log (f)) \text {Subst}\left (\int \frac {f^{\frac {c}{a}-\frac {b c x}{a}}}{x} \, dx,x,\frac {x}{a+b x}\right )}{a^2} \\ & = -\frac {b f^{\frac {c}{a+b x}}}{a}-\frac {f^{\frac {c}{a+b x}}}{x}-\frac {b c f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a (a+b x)}\right ) \log (f)}{a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=-\frac {b f^{\frac {c}{a+b x}}}{a}-\frac {f^{\frac {c}{a+b x}}}{x}-\frac {b c f^{\frac {c}{a}} \operatorname {ExpIntegralEi}\left (-\frac {b c x \log (f)}{a^2+a b x}\right ) \log (f)}{a^2} \]

[In]

Integrate[f^(c/(a + b*x))/x^2,x]

[Out]

-((b*f^(c/(a + b*x)))/a) - f^(c/(a + b*x))/x - (b*c*f^(c/a)*ExpIntegralEi[-((b*c*x*Log[f])/(a^2 + a*b*x))]*Log
[f])/a^2

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18

method	result	size
risch	\(\frac {\ln \left (f \right ) b c \,f^{\frac {c}{b x +a}}}{a^{2} \left (\frac {c \ln \left (f \right )}{b x +a}-\frac {c \ln \left (f \right )}{a}\right )}+\frac {\ln \left (f \right ) b c \,f^{\frac {c}{a}} \operatorname {Ei}_{1}\left (-\frac {c \ln \left (f \right )}{b x +a}+\frac {c \ln \left (f \right )}{a}\right )}{a^{2}}\)	\(80\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=-\frac {b c f^{\frac {c}{a}} x {\rm Ei}\left (-\frac {b c x \log \left (f\right )}{a b x + a^{2}}\right ) \log \left (f\right ) + {\left (a b x + a^{2}\right )} f^{\frac {c}{b x + a}}}{a^{2} x} \]

[In]

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

[Out]

-(b*c*f^(c/a)*x*Ei(-b*c*x*log(f)/(a*b*x + a^2))*log(f) + (a*b*x + a^2)*f^(c/(b*x + a)))/(a^2*x)

Sympy [F]

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

[In]

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

[Out]

Integral(f**(c/(a + b*x))/x**2, x)

Maxima [F]

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

[In]

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

[Out]

integrate(f^(c/(b*x + a))/x^2, x)

Giac [F]

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

[In]

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

[Out]

integrate(f^(c/(b*x + a))/x^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=\int \frac {f^{\frac {c}{a+b\,x}}}{x^2} \,d x \]

[In]

int(f^(c/(a + b*x))/x^2,x)

[Out]

int(f^(c/(a + b*x))/x^2, x)