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/(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

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Rubi [A]  time = 0.40, antiderivative size = 68, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 6742

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

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*}

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Mathematica [A]  time = 0.11, size = 68, normalized size = 1.00 \[ -\frac {b c \log (f) f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a^2+b x a}\right )}{a^2}-\frac {b f^{\frac {c}{a+b x}}}{a}-\frac {f^{\frac {c}{a+b x}}}{x} \]

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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fricas [A]  time = 0.45, size = 60, normalized size = 0.88 \[ -\frac {b c f^{\frac {c}{a}} x {\rm Ei}\left (-\frac {b c x \log \relax (f)}{a b x + a^{2}}\right ) \log \relax (f) + {\left (a b x + a^{2}\right )} f^{\frac {c}{b x + a}}}{a^{2} x} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{\frac {c}{b x + a}}}{x^{2}}\,{d x} \]

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)

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maple [A]  time = 0.12, size = 80, normalized size = 1.18 \[ \frac {b c \,f^{\frac {c}{a}} \Ei \left (1, -\frac {c \ln \relax (f )}{b x +a}+\frac {c \ln \relax (f )}{a}\right ) \ln \relax (f )}{a^{2}}+\frac {b c \,f^{\frac {c}{b x +a}} \ln \relax (f )}{\left (\frac {c \ln \relax (f )}{b x +a}-\frac {c \ln \relax (f )}{a}\right ) a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(1/(b*x+a)*c)/x^2,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{\frac {c}{b x + a}}}{x^{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {f^{\frac {c}{a+b\,x}}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c/(a + b*x))/x^2,x)

[Out]

int(f^(c/(a + b*x))/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{\frac {c}{a + b x}}}{x^{2}}\, dx \]

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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