3.223 \(\int \frac {f^{\frac {c}{a+b x}}}{x^3} \, dx\)

Optimal. Leaf size=166 \[ \frac {b^2 c^2 \log ^2(f) f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a (a+b x)}\right )}{2 a^4}+\frac {b^2 c \log (f) f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a (a+b x)}\right )}{a^3}+\frac {b^2 c \log (f) f^{\frac {c}{a+b x}}}{2 a^3}+\frac {b^2 f^{\frac {c}{a+b x}}}{2 a^2}+\frac {b c \log (f) f^{\frac {c}{a+b x}}}{2 a^2 x}-\frac {f^{\frac {c}{a+b x}}}{2 x^2} \]

[Out]

1/2*b^2*f^(c/(b*x+a))/a^2-1/2*f^(c/(b*x+a))/x^2+1/2*b^2*c*f^(c/(b*x+a))*ln(f)/a^3+1/2*b*c*f^(c/(b*x+a))*ln(f)/
a^2/x+b^2*c*f^(c/a)*Ei(-b*c*x*ln(f)/a/(b*x+a))*ln(f)/a^3+1/2*b^2*c^2*f^(c/a)*Ei(-b*c*x*ln(f)/a/(b*x+a))*ln(f)^
2/a^4

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Rubi [A]  time = 0.72, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 18, 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} \[ \frac {b^2 c^2 \log ^2(f) f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a (a+b x)}\right )}{2 a^4}+\frac {b^2 c \log (f) f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a (a+b x)}\right )}{a^3}+\frac {b^2 f^{\frac {c}{a+b x}}}{2 a^2}+\frac {b^2 c \log (f) f^{\frac {c}{a+b x}}}{2 a^3}+\frac {b c \log (f) f^{\frac {c}{a+b x}}}{2 a^2 x}-\frac {f^{\frac {c}{a+b x}}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[f^(c/(a + b*x))/x^3,x]

[Out]

(b^2*f^(c/(a + b*x)))/(2*a^2) - f^(c/(a + b*x))/(2*x^2) + (b^2*c*f^(c/(a + b*x))*Log[f])/(2*a^3) + (b*c*f^(c/(
a + b*x))*Log[f])/(2*a^2*x) + (b^2*c*f^(c/a)*ExpIntegralEi[-((b*c*x*Log[f])/(a*(a + b*x)))]*Log[f])/a^3 + (b^2
*c^2*f^(c/a)*ExpIntegralEi[-((b*c*x*Log[f])/(a*(a + b*x)))]*Log[f]^2)/(2*a^4)

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^3} \, dx &=-\frac {f^{\frac {c}{a+b x}}}{2 x^2}-\frac {1}{2} (b c \log (f)) \int \frac {f^{\frac {c}{a+b x}}}{x^2 (a+b x)^2} \, dx\\ &=-\frac {f^{\frac {c}{a+b x}}}{2 x^2}-\frac {1}{2} (b c \log (f)) \int \left (\frac {f^{\frac {c}{a+b x}}}{a^2 x^2}-\frac {2 b f^{\frac {c}{a+b x}}}{a^3 x}+\frac {b^2 f^{\frac {c}{a+b x}}}{a^2 (a+b x)^2}+\frac {2 b^2 f^{\frac {c}{a+b x}}}{a^3 (a+b x)}\right ) \, dx\\ &=-\frac {f^{\frac {c}{a+b x}}}{2 x^2}-\frac {(b c \log (f)) \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx}{2 a^2}+\frac {\left (b^2 c \log (f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{x} \, dx}{a^3}-\frac {\left (b^3 c \log (f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{a+b x} \, dx}{a^3}-\frac {\left (b^3 c \log (f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{(a+b x)^2} \, dx}{2 a^2}\\ &=\frac {b^2 f^{\frac {c}{a+b x}}}{2 a^2}-\frac {f^{\frac {c}{a+b x}}}{2 x^2}+\frac {b c f^{\frac {c}{a+b x}} \log (f)}{2 a^2 x}+\frac {b^2 c \text {Ei}\left (\frac {c \log (f)}{a+b x}\right ) \log (f)}{a^3}+\frac {\left (b^2 c \log (f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{x (a+b x)} \, dx}{a^2}+\frac {\left (b^3 c \log (f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{a+b x} \, dx}{a^3}+\frac {\left (b^2 c^2 \log ^2(f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{x (a+b x)^2} \, dx}{2 a^2}\\ &=\frac {b^2 f^{\frac {c}{a+b x}}}{2 a^2}-\frac {f^{\frac {c}{a+b x}}}{2 x^2}+\frac {b c f^{\frac {c}{a+b x}} \log (f)}{2 a^2 x}+\frac {\left (b^2 c \log (f)\right ) \operatorname {Subst}\left (\int \frac {f^{\frac {c}{a}-\frac {b c x}{a}}}{x} \, dx,x,\frac {x}{a+b x}\right )}{a^3}+\frac {\left (b^2 c^2 \log ^2(f)\right ) \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}{2 a^2}\\ &=\frac {b^2 f^{\frac {c}{a+b x}}}{2 a^2}-\frac {f^{\frac {c}{a+b x}}}{2 x^2}+\frac {b c f^{\frac {c}{a+b x}} \log (f)}{2 a^2 x}+\frac {b^2 c f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a (a+b x)}\right ) \log (f)}{a^3}+\frac {\left (b^2 c^2 \log ^2(f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{x} \, dx}{2 a^4}-\frac {\left (b^3 c^2 \log ^2(f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{a+b x} \, dx}{2 a^4}-\frac {\left (b^3 c^2 \log ^2(f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{(a+b x)^2} \, dx}{2 a^3}\\ &=\frac {b^2 f^{\frac {c}{a+b x}}}{2 a^2}-\frac {f^{\frac {c}{a+b x}}}{2 x^2}+\frac {b^2 c f^{\frac {c}{a+b x}} \log (f)}{2 a^3}+\frac {b c f^{\frac {c}{a+b x}} \log (f)}{2 a^2 x}+\frac {b^2 c f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a (a+b x)}\right ) \log (f)}{a^3}+\frac {b^2 c^2 \text {Ei}\left (\frac {c \log (f)}{a+b x}\right ) \log ^2(f)}{2 a^4}+\frac {\left (b^2 c^2 \log ^2(f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{x (a+b x)} \, dx}{2 a^3}+\frac {\left (b^3 c^2 \log ^2(f)\right ) \int \frac {f^{\frac {c}{a+b x}}}{a+b x} \, dx}{2 a^4}\\ &=\frac {b^2 f^{\frac {c}{a+b x}}}{2 a^2}-\frac {f^{\frac {c}{a+b x}}}{2 x^2}+\frac {b^2 c f^{\frac {c}{a+b x}} \log (f)}{2 a^3}+\frac {b c f^{\frac {c}{a+b x}} \log (f)}{2 a^2 x}+\frac {b^2 c f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a (a+b x)}\right ) \log (f)}{a^3}+\frac {\left (b^2 c^2 \log ^2(f)\right ) \operatorname {Subst}\left (\int \frac {f^{\frac {c}{a}-\frac {b c x}{a}}}{x} \, dx,x,\frac {x}{a+b x}\right )}{2 a^4}\\ &=\frac {b^2 f^{\frac {c}{a+b x}}}{2 a^2}-\frac {f^{\frac {c}{a+b x}}}{2 x^2}+\frac {b^2 c f^{\frac {c}{a+b x}} \log (f)}{2 a^3}+\frac {b c f^{\frac {c}{a+b x}} \log (f)}{2 a^2 x}+\frac {b^2 c f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a (a+b x)}\right ) \log (f)}{a^3}+\frac {b^2 c^2 f^{\frac {c}{a}} \text {Ei}\left (-\frac {b c x \log (f)}{a (a+b x)}\right ) \log ^2(f)}{2 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 115, normalized size = 0.69 \[ \frac {b^2 (2 a+c \log (f)) f^{\frac {c}{a+b x}}}{2 a^3}+\frac {b^2 c \log (f) f^{\frac {c}{a}} (2 a+c \log (f)) \text {Ei}\left (-\frac {b c x \log (f)}{a^2+b x a}\right )-\frac {a^2 f^{\frac {c}{a+b x}} \left (a^2+b^2 x^2-b c x \log (f)\right )}{x^2}}{2 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(c/(a + b*x))/x^3,x]

[Out]

(b^2*f^(c/(a + b*x))*(2*a + c*Log[f]))/(2*a^3) + (b^2*c*f^(c/a)*ExpIntegralEi[-((b*c*x*Log[f])/(a^2 + a*b*x))]
*Log[f]*(2*a + c*Log[f]) - (a^2*f^(c/(a + b*x))*(a^2 + b^2*x^2 - b*c*x*Log[f]))/x^2)/(2*a^4)

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fricas [A]  time = 0.44, size = 110, normalized size = 0.66 \[ \frac {{\left (b^{2} c^{2} x^{2} \log \relax (f)^{2} + 2 \, a b^{2} c x^{2} \log \relax (f)\right )} f^{\frac {c}{a}} {\rm Ei}\left (-\frac {b c x \log \relax (f)}{a b x + a^{2}}\right ) + {\left (a^{2} b^{2} x^{2} - a^{4} + {\left (a b^{2} c x^{2} + a^{2} b c x\right )} \log \relax (f)\right )} f^{\frac {c}{b x + a}}}{2 \, a^{4} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b^2*c^2*x^2*log(f)^2 + 2*a*b^2*c*x^2*log(f))*f^(c/a)*Ei(-b*c*x*log(f)/(a*b*x + a^2)) + (a^2*b^2*x^2 - a^
4 + (a*b^2*c*x^2 + a^2*b*c*x)*log(f))*f^(c/(b*x + a)))/(a^4*x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{\frac {c}{b x + a}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(c/(b*x + a))/x^3, x)

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maple [A]  time = 0.13, size = 226, normalized size = 1.36 \[ -\frac {b^{2} c^{2} f^{\frac {c}{a}} \Ei \left (1, -\frac {c \ln \relax (f )}{b x +a}+\frac {c \ln \relax (f )}{a}\right ) \ln \relax (f )^{2}}{2 a^{4}}-\frac {b^{2} c \,f^{\frac {c}{a}} \Ei \left (1, -\frac {c \ln \relax (f )}{b x +a}+\frac {c \ln \relax (f )}{a}\right ) \ln \relax (f )}{a^{3}}-\frac {b^{2} c^{2} f^{\frac {c}{b x +a}} \ln \relax (f )^{2}}{2 \left (\frac {c \ln \relax (f )}{b x +a}-\frac {c \ln \relax (f )}{a}\right )^{2} a^{4}}-\frac {b^{2} c^{2} f^{\frac {c}{b x +a}} \ln \relax (f )^{2}}{2 \left (\frac {c \ln \relax (f )}{b x +a}-\frac {c \ln \relax (f )}{a}\right ) a^{4}}-\frac {b^{2} c \,f^{\frac {c}{b x +a}} \ln \relax (f )}{\left (\frac {c \ln \relax (f )}{b x +a}-\frac {c \ln \relax (f )}{a}\right ) a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(1/(b*x+a)*c)/x^3,x)

[Out]

-ln(f)*b^2*c/a^3*f^(1/(b*x+a)*c)/(1/(b*x+a)*c*ln(f)-1/a*c*ln(f))-ln(f)*b^2*c/a^3*f^(1/a*c)*Ei(1,-1/(b*x+a)*c*l
n(f)+1/a*c*ln(f))-1/2*ln(f)^2*b^2*c^2/a^4*f^(1/(b*x+a)*c)/(1/(b*x+a)*c*ln(f)-1/a*c*ln(f))^2-1/2*ln(f)^2*b^2*c^
2/a^4*f^(1/(b*x+a)*c)/(1/(b*x+a)*c*ln(f)-1/a*c*ln(f))-1/2*ln(f)^2*b^2*c^2/a^4*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 \[ \int \frac {f^{\frac {c}{b x + a}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(c/(b*x + a))/x^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {f^{\frac {c}{a+b\,x}}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c/(a + b*x))/x^3,x)

[Out]

int(f^(c/(a + b*x))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{\frac {c}{a + b x}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(c/(a + b*x))/x**3, x)

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