\(\int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx\) [398]

Optimal result

Integrand size = 21, antiderivative size = 116 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx=\frac {d F^{a+\frac {b}{c+d x}}}{f (d e-c f)}-\frac {F^{a+\frac {b}{c+d x}}}{f (e+f x)}-\frac {b d F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log (F)}{(d e-c f)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2255, 6874, 2240, 2241, 2254, 2260, 2209} \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx=-\frac {b d \log (F) F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{(d e-c f)^2}+\frac {d F^{a+\frac {b}{c+d x}}}{f (d e-c f)}-\frac {F^{a+\frac {b}{c+d x}}}{f (e+f x)} \]

[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 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^{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)) \text {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] (verified)

Time = 0.22 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx=\frac {d F^{a+\frac {b}{c+d x}}}{f (d e-c f)}-\frac {F^{a+\frac {b}{c+d x}}}{f (e+f x)}-\frac {b d F^{a+\frac {b f}{-d e+c f}} \operatorname {ExpIntegralEi}\left (-\frac {b f \log (F)}{-d e+c f}+\frac {b \log (F)}{c+d x}\right ) \log (F)}{(d e-c f)^2} \]

[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

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.65

method	result	size
risch	\(\frac {d \ln \left (F \right ) b \,F^{a} F^{\frac {b}{d x +c}}}{\left (c f -d e \right )^{2} \left (\frac {b \ln \left (F \right )}{d x +c}+a \ln \left (F \right )-\frac {\ln \left (F \right ) a c f}{c f -d e}+\frac {\ln \left (F \right ) a d e}{c f -d e}-\frac {\ln \left (F \right ) b f}{c f -d e}\right )}+\frac {d \ln \left (F \right ) b \,F^{\frac {a c f -a d e +b f}{c f -d e}} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (F \right )}{d x +c}-a \ln \left (F \right )-\frac {-\ln \left (F \right ) a c f +a e d \ln \left (F \right )-\ln \left (F \right ) b f}{c f -d e}\right )}{\left (c f -d e \right )^{2}}\)	\(191\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.54 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx=-\frac {{\left (b d f x + b d e\right )} F^{\frac {a d e - {\left (a c + b\right )} f}{d e - c f}} {\rm Ei}\left (\frac {{\left (b d f x + b d e\right )} \log \left (F\right )}{c d e - c^{2} f + {\left (d^{2} e - c d f\right )} x}\right ) \log \left (F\right ) - {\left (c d e - c^{2} f + {\left (d^{2} e - c d f\right )} x\right )} F^{\frac {a d x + a c + b}{d x + c}}}{d^{2} e^{3} - 2 \, c d e^{2} f + c^{2} e f^{2} + {\left (d^{2} e^{2} f - 2 \, c d e f^{2} + c^{2} f^{3}\right )} x} \]

[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)

Sympy [F]

\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx=\int \frac {F^{a + \frac {b}{c + d x}}}{\left (e + f x\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{{\left (f x + e\right )}^{2}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{{\left (f x + e\right )}^{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx=\int \frac {F^{a+\frac {b}{c+d\,x}}}{{\left (e+f\,x\right )}^2} \,d x \]

[In]

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

[Out]

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