Integrand size = 26, antiderivative size = 159 \[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^2} \, dx=\frac {d F^{e+\frac {b f}{d}-\frac {(b c-a d) f}{d (c+d x)}}}{h (d g-c h)}-\frac {F^{e+\frac {f (a+b x)}{c+d x}}}{h (g+h x)}+\frac {(b c-a d) f F^{e+\frac {f (b g-a h)}{d g-c h}} \operatorname {ExpIntegralEi}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right ) \log (F)}{(d g-c h)^2} \]
d*F^(e+b*f/d-(-a*d+b*c)*f/d/(d*x+c))/h/(-c*h+d*g)-F^(e+f*(b*x+a)/(d*x+c))/ h/(h*x+g)+(-a*d+b*c)*f*F^(e+f*(-a*h+b*g)/(-c*h+d*g))*Ei(-(-a*d+b*c)*f*(h*x +g)*ln(F)/(-c*h+d*g)/(d*x+c))*ln(F)/(-c*h+d*g)^2
\[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^2} \, dx=\int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^2} \, dx \]
Time = 2.24 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.13, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2662, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {F^{\frac {f (a+b x)}{c+d x}+e}}{(g+h x)^2} \, dx\) |
| \(\Big \downarrow \) 2662 |
| \(\displaystyle \frac {f \log (F) (b c-a d) \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(c+d x)^2 (g+h x)}dx}{h}-\frac {F^{\frac {f (a+b x)}{c+d x}+e}}{h (g+h x)}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {f \log (F) (b c-a d) \int \left (-\frac {d h F^{e+\frac {f (a+b x)}{c+d x}}}{(d g-c h)^2 (c+d x)}+\frac {h^2 F^{e+\frac {f (a+b x)}{c+d x}}}{(d g-c h)^2 (g+h x)}+\frac {d F^{e+\frac {f (a+b x)}{c+d x}}}{(d g-c h) (c+d x)^2}\right )dx}{h}-\frac {F^{\frac {f (a+b x)}{c+d x}+e}}{h (g+h x)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f \log (F) (b c-a d) \left (\frac {h F^{\frac {f (b g-a h)}{d g-c h}+e} \operatorname {ExpIntegralEi}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right )}{(d g-c h)^2}+\frac {d F^{-\frac {f (b c-a d)}{d (c+d x)}+\frac {b f}{d}+e}}{f \log (F) (b c-a d) (d g-c h)}\right )}{h}-\frac {F^{\frac {f (a+b x)}{c+d x}+e}}{h (g+h x)}\) |
-(F^(e + (f*(a + b*x))/(c + d*x))/(h*(g + h*x))) + ((b*c - a*d)*f*((F^(e + (f*(b*g - a*h))/(d*g - c*h))*h*ExpIntegralEi[-(((b*c - a*d)*f*(g + h*x)*L og[F])/((d*g - c*h)*(c + d*x)))])/(d*g - c*h)^2 + (d*F^(e + (b*f)/d - ((b* c - a*d)*f)/(d*(c + d*x))))/((b*c - a*d)*f*(d*g - c*h)*Log[F]))*Log[F])/h
3.5.23.3.1 Defintions of rubi rules used
Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_), x_Symbol] :> Simp[(g + h*x)^(m + 1)*(F^(e + f*((a + b* x)/(c + d*x)))/(h*(m + 1))), x] - Simp[f*(b*c - a*d)*(Log[F]/(h*(m + 1))) Int[(g + h*x)^(m + 1)*(F^(e + f*((a + b*x)/(c + d*x)))/(c + d*x)^2), x], x ] /; FreeQ[{F, a, b, c, d, e, f, g, h}, x] && NeQ[b*c - a*d, 0] && NeQ[d*g - c*h, 0] && ILtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(581\) vs. \(2(159)=318\).
Time = 0.53 (sec) , antiderivative size = 582, normalized size of antiderivative = 3.66
| method | result | size |
| risch | \(\frac {\ln \left (F \right ) F^{\frac {f \left (a d -c b \right )}{d \left (d x +c \right )}} F^{\frac {b f +d e}{d}} a d f}{\left (c h -d g \right )^{2} \left (\frac {f \ln \left (F \right ) a}{d x +c}-\frac {f \ln \left (F \right ) c b}{d \left (d x +c \right )}+\frac {\ln \left (F \right ) b f}{d}+\ln \left (F \right ) e -\frac {\ln \left (F \right ) a f h}{c h -d g}+\frac {\ln \left (F \right ) b f g}{c h -d g}-\frac {\ln \left (F \right ) c e h}{c h -d g}+\frac {\ln \left (F \right ) d e g}{c h -d g}\right )}-\frac {\ln \left (F \right ) F^{\frac {f \left (a d -c b \right )}{d \left (d x +c \right )}} F^{\frac {b f +d e}{d}} b c f}{\left (c h -d g \right )^{2} \left (\frac {f \ln \left (F \right ) a}{d x +c}-\frac {f \ln \left (F \right ) c b}{d \left (d x +c \right )}+\frac {\ln \left (F \right ) b f}{d}+\ln \left (F \right ) e -\frac {\ln \left (F \right ) a f h}{c h -d g}+\frac {\ln \left (F \right ) b f g}{c h -d g}-\frac {\ln \left (F \right ) c e h}{c h -d g}+\frac {\ln \left (F \right ) d e g}{c h -d g}\right )}+\frac {\ln \left (F \right ) F^{\frac {a f h -b f g +c e h -d e g}{c h -d g}} \operatorname {Ei}_{1}\left (-\frac {\left (a d f -b c f \right ) \ln \left (F \right )}{d \left (d x +c \right )}-\frac {\left (b f +d e \right ) \ln \left (F \right )}{d}-\frac {-\ln \left (F \right ) a f h +\ln \left (F \right ) b f g -\ln \left (F \right ) c e h +\ln \left (F \right ) d e g}{c h -d g}\right ) a d f}{\left (c h -d g \right )^{2}}-\frac {\ln \left (F \right ) F^{\frac {a f h -b f g +c e h -d e g}{c h -d g}} \operatorname {Ei}_{1}\left (-\frac {\left (a d f -b c f \right ) \ln \left (F \right )}{d \left (d x +c \right )}-\frac {\left (b f +d e \right ) \ln \left (F \right )}{d}-\frac {-\ln \left (F \right ) a f h +\ln \left (F \right ) b f g -\ln \left (F \right ) c e h +\ln \left (F \right ) d e g}{c h -d g}\right ) b c f}{\left (c h -d g \right )^{2}}\) | \(582\) |
ln(F)/(c*h-d*g)^2*F^(f*(a*d-b*c)/d/(d*x+c))*F^((b*f+d*e)/d)/(f*ln(F)/(d*x+ c)*a-f*ln(F)/d/(d*x+c)*c*b+ln(F)/d*b*f+ln(F)*e-1/(c*h-d*g)*ln(F)*a*f*h+1/( c*h-d*g)*ln(F)*b*f*g-1/(c*h-d*g)*ln(F)*c*e*h+1/(c*h-d*g)*ln(F)*d*e*g)*a*d* f-ln(F)/(c*h-d*g)^2*F^(f*(a*d-b*c)/d/(d*x+c))*F^((b*f+d*e)/d)/(f*ln(F)/(d* x+c)*a-f*ln(F)/d/(d*x+c)*c*b+ln(F)/d*b*f+ln(F)*e-1/(c*h-d*g)*ln(F)*a*f*h+1 /(c*h-d*g)*ln(F)*b*f*g-1/(c*h-d*g)*ln(F)*c*e*h+1/(c*h-d*g)*ln(F)*d*e*g)*b* c*f+ln(F)/(c*h-d*g)^2*F^((a*f*h-b*f*g+c*e*h-d*e*g)/(c*h-d*g))*Ei(1,-(a*d*f -b*c*f)*ln(F)/d/(d*x+c)-(b*f+d*e)*ln(F)/d-(-ln(F)*a*f*h+ln(F)*b*f*g-ln(F)* c*e*h+ln(F)*d*e*g)/(c*h-d*g))*a*d*f-ln(F)/(c*h-d*g)^2*F^((a*f*h-b*f*g+c*e* h-d*e*g)/(c*h-d*g))*Ei(1,-(a*d*f-b*c*f)*ln(F)/d/(d*x+c)-(b*f+d*e)*ln(F)/d- (-ln(F)*a*f*h+ln(F)*b*f*g-ln(F)*c*e*h+ln(F)*d*e*g)/(c*h-d*g))*b*c*f
Time = 0.33 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.38 \[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^2} \, dx=\frac {{\left ({\left (b c - a d\right )} f h x + {\left (b c - a d\right )} f g\right )} F^{\frac {{\left (d e + b f\right )} g - {\left (c e + a f\right )} h}{d g - c h}} {\rm Ei}\left (-\frac {{\left ({\left (b c - a d\right )} f h x + {\left (b c - a d\right )} f g\right )} \log \left (F\right )}{c d g - c^{2} h + {\left (d^{2} g - c d h\right )} x}\right ) \log \left (F\right ) + {\left (c d g - c^{2} h + {\left (d^{2} g - c d h\right )} x\right )} F^{\frac {c e + a f + {\left (d e + b f\right )} x}{d x + c}}}{d^{2} g^{3} - 2 \, c d g^{2} h + c^{2} g h^{2} + {\left (d^{2} g^{2} h - 2 \, c d g h^{2} + c^{2} h^{3}\right )} x} \]
(((b*c - a*d)*f*h*x + (b*c - a*d)*f*g)*F^(((d*e + b*f)*g - (c*e + a*f)*h)/ (d*g - c*h))*Ei(-((b*c - a*d)*f*h*x + (b*c - a*d)*f*g)*log(F)/(c*d*g - c^2 *h + (d^2*g - c*d*h)*x))*log(F) + (c*d*g - c^2*h + (d^2*g - c*d*h)*x)*F^(( c*e + a*f + (d*e + b*f)*x)/(d*x + c)))/(d^2*g^3 - 2*c*d*g^2*h + c^2*g*h^2 + (d^2*g^2*h - 2*c*d*g*h^2 + c^2*h^3)*x)
\[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^2} \, dx=\int \frac {F^{e + \frac {f \left (a + b x\right )}{c + d x}}}{\left (g + h x\right )^{2}}\, dx \]
\[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^2} \, dx=\int { \frac {F^{e + \frac {{\left (b x + a\right )} f}{d x + c}}}{{\left (h x + g\right )}^{2}} \,d x } \]
\[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^2} \, dx=\int { \frac {F^{e + \frac {{\left (b x + a\right )} f}{d x + c}}}{{\left (h x + g\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^2} \, dx=\int \frac {F^{e+\frac {f\,\left (a+b\,x\right )}{c+d\,x}}}{{\left (g+h\,x\right )}^2} \,d x \]