3.398 \(\int \frac{F^{a+\frac{b}{c+d x}}}{(e+f x)^2} \, dx\)

Optimal. Leaf size=116 \[ -\frac{b d \log (F) F^{a-\frac{b f}{d e-c f}} \text{Ei}\left (\frac{b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{(d e-c f)^2}+\frac{d F^{a+\frac{b}{c+d x}}}{f (d e-c f)}-\frac{F^{a+\frac{b}{c+d x}}}{f (e+f x)} \]

[Out]

(d*F^(a + b/(c + d*x)))/(f*(d*e - c*f)) - F^(a + b/(c + d*x))/(f*(e + f*x)) - (b*d*F^(a - (b*f)/(d*e - c*f))*E
xpIntegralEi[(b*d*(e + f*x)*Log[F])/((d*e - c*f)*(c + d*x))]*Log[F])/(d*e - c*f)^2

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Rubi [A]  time = 1.01075, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2223, 6742, 2209, 2210, 2222, 2228, 2178} \[ -\frac{b d \log (F) F^{a-\frac{b f}{d e-c f}} \text{Ei}\left (\frac{b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{(d e-c f)^2}+\frac{d F^{a+\frac{b}{c+d x}}}{f (d e-c f)}-\frac{F^{a+\frac{b}{c+d x}}}{f (e+f x)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x))/(e + f*x)^2,x]

[Out]

(d*F^(a + b/(c + d*x)))/(f*(d*e - c*f)) - F^(a + b/(c + d*x))/(f*(e + f*x)) - (b*d*F^(a - (b*f)/(d*e - c*f))*E
xpIntegralEi[(b*d*(e + f*x)*Log[F])/((d*e - c*f)*(c + d*x))]*Log[F])/(d*e - c*f)^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 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 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

Rubi steps

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

Mathematica [A]  time = 0.339725, size = 116, normalized size = 1. \[ -\frac{b d \log (F) F^{a+\frac{b f}{c f-d e}} \text{Ei}\left (\frac{b \log (F)}{c+d x}-\frac{b f \log (F)}{c f-d e}\right )}{(d e-c f)^2}+\frac{d F^{a+\frac{b}{c+d x}}}{f (d e-c f)}-\frac{F^{a+\frac{b}{c+d x}}}{f (e+f x)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x))/(e + f*x)^2,x]

[Out]

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

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Maple [A]  time = 0.128, size = 191, normalized size = 1.7 \begin{align*}{\frac{\ln \left ( F \right ) bd{F}^{a}}{ \left ( cf-de \right ) ^{2}}{F}^{{\frac{b}{dx+c}}} \left ({\frac{b\ln \left ( F \right ) }{dx+c}}+\ln \left ( F \right ) a-{\frac{\ln \left ( F \right ) acf}{cf-de}}+{\frac{\ln \left ( F \right ) ade}{cf-de}}-{\frac{\ln \left ( F \right ) bf}{cf-de}} \right ) ^{-1}}+{\frac{\ln \left ( F \right ) bd}{ \left ( cf-de \right ) ^{2}}{F}^{{\frac{acf-ade+bf}{cf-de}}}{\it Ei} \left ( 1,-{\frac{b\ln \left ( F \right ) }{dx+c}}-\ln \left ( F \right ) a-{\frac{-\ln \left ( F \right ) acf+\ln \left ( F \right ) ade-\ln \left ( F \right ) bf}{cf-de}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(F^(a + b/(d*x + c))/(f*x + e)^2, x)

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Fricas [A]  time = 1.6438, size = 375, normalized size = 3.23 \begin{align*} -\frac{{\left (b d f x + b d e\right )} F^{\frac{a d e -{\left (a c + b\right )} f}{d e - c f}}{\rm Ei}\left (\frac{{\left (b d f x + b d e\right )} \log \left (F\right )}{c d e - c^{2} f +{\left (d^{2} e - c d f\right )} x}\right ) \log \left (F\right ) -{\left (c d e - c^{2} f +{\left (d^{2} e - c d f\right )} x\right )} F^{\frac{a d x + a c + b}{d x + c}}}{d^{2} e^{3} - 2 \, c d e^{2} f + c^{2} e f^{2} +{\left (d^{2} e^{2} f - 2 \, c d e f^{2} + c^{2} f^{3}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b*d*f*x + b*d*e)*F^((a*d*e - (a*c + b)*f)/(d*e - c*f))*Ei((b*d*f*x + b*d*e)*log(F)/(c*d*e - c^2*f + (d^2*e
- c*d*f)*x))*log(F) - (c*d*e - c^2*f + (d^2*e - c*d*f)*x)*F^((a*d*x + a*c + b)/(d*x + c)))/(d^2*e^3 - 2*c*d*e^
2*f + c^2*e*f^2 + (d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(F^(a + b/(d*x + c))/(f*x + e)^2, x)