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

Optimal. Leaf size=68 \[ -\frac{b c \log (f) f^{\frac{c}{a}} \text{Ei}\left (-\frac{b c x \log (f)}{a (a+b x)}\right )}{a^2}-\frac{b f^{\frac{c}{a+b x}}}{a}-\frac{f^{\frac{c}{a+b x}}}{x} \]

[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

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Rubi [A]  time = 0.39663, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {2223, 6742, 2222, 2210, 2228, 2178, 2209} \[ -\frac{b c \log (f) f^{\frac{c}{a}} \text{Ei}\left (-\frac{b c x \log (f)}{a (a+b x)}\right )}{a^2}-\frac{b f^{\frac{c}{a+b x}}}{a}-\frac{f^{\frac{c}{a+b x}}}{x} \]

Antiderivative was successfully verified.

[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 2223

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 6742

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

Rule 2222

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 2210

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 2228

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 2178

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

Rule 2209

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]

Rubi steps

\begin{align*} \int \frac{f^{\frac{c}{a+b x}}}{x^2} \, dx &=-\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)) \operatorname{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]  time = 0.102447, size = 68, normalized size = 1. \[ -\frac{b c \log (f) f^{\frac{c}{a}} \text{Ei}\left (-\frac{b c x \log (f)}{a^2+b x a}\right )}{a^2}-\frac{b f^{\frac{c}{a+b x}}}{a}-\frac{f^{\frac{c}{a+b x}}}{x} \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.093, size = 80, normalized size = 1.2 \begin{align*}{\frac{cb\ln \left ( f \right ) }{{a}^{2}}{f}^{{\frac{c}{bx+a}}} \left ({\frac{c\ln \left ( f \right ) }{bx+a}}-{\frac{c\ln \left ( f \right ) }{a}} \right ) ^{-1}}+{\frac{cb\ln \left ( f \right ) }{{a}^{2}}{f}^{{\frac{c}{a}}}{\it Ei} \left ( 1,-{\frac{c\ln \left ( f \right ) }{bx+a}}+{\frac{c\ln \left ( f \right ) }{a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{\frac{c}{b x + a}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(c/(b*x + a))/x^2, x)

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Fricas [A]  time = 1.54048, size = 131, normalized size = 1.93 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{\frac{c}{a + b x}}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(c/(a + b*x))/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{\frac{c}{b x + a}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(c/(b*x + a))/x^2, x)