Integrand size = 21, antiderivative size = 71 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx=-\frac {F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{c+d x}\right )}{f}+\frac {F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{f} \] Output:
-F^a*Ei(b*ln(F)/(d*x+c))/f+F^(a-b*f/(-c*f+d*e))*Ei(b*d*(f*x+e)*ln(F)/(-c*f +d*e)/(d*x+c))/f
Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx=\frac {F^a \left (-\operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{c+d x}\right )+F^{\frac {b f}{-d e+c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )\right )}{f} \] Input:
Integrate[F^(a + b/(c + d*x))/(e + f*x),x]
Output:
(F^a*(-ExpIntegralEi[(b*Log[F])/(c + d*x)] + F^((b*f)/(-(d*e) + c*f))*ExpI ntegralEi[(b*d*(e + f*x)*Log[F])/((d*e - c*f)*(c + d*x))]))/f
Time = 1.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2652, 2639, 2658, 2609}
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.
\(\displaystyle \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx\) |
\(\Big \downarrow \) 2652 |
\(\displaystyle \frac {d \int \frac {F^{a+\frac {b}{c+d x}}}{c+d x}dx}{f}-\frac {(d e-c f) \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x) (e+f x)}dx}{f}\) |
\(\Big \downarrow \) 2639 |
\(\displaystyle -\frac {(d e-c f) \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x) (e+f x)}dx}{f}-\frac {F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{c+d x}\right )}{f}\) |
\(\Big \downarrow \) 2658 |
\(\displaystyle \frac {\int \frac {F^{a+\frac {b d (e+f x)}{(d e-c f) (c+d x)}-\frac {b f}{d e-c f}} (c+d x)}{e+f x}d\frac {e+f x}{c+d x}}{f}-\frac {F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{c+d x}\right )}{f}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{f}-\frac {F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{c+d x}\right )}{f}\) |
Input:
Int[F^(a + b/(c + d*x))/(e + f*x),x]
Output:
-((F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)])/f) + (F^(a - (b*f)/(d*e - c*f) )*ExpIntegralEi[(b*d*(e + f*x)*Log[F])/((d*e - c*f)*(c + d*x))])/f
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F reeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
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] /; Free Q[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]
Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/((e_.) + (f_.)*(x_)), x_Symbo l] :> Simp[d/f Int[F^(a + b/(c + d*x))/(c + d*x), x], x] - Simp[(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]
Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> Simp[-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]
Time = 0.24 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.49
method | result | size |
risch | \(-\frac {F^{\frac {a c f -a d e +b f}{c f -d e}} \operatorname {expIntegral}_{1}\left (-\frac {b \ln \left (F \right )}{d x +c}-\ln \left (F \right ) a -\frac {-\ln \left (F \right ) a c f +\ln \left (F \right ) a d e -\ln \left (F \right ) b f}{c f -d e}\right )}{f}+\frac {F^{a} \operatorname {expIntegral}_{1}\left (-\frac {b \ln \left (F \right )}{d x +c}\right )}{f}\) | \(106\) |
Input:
int(F^(a+b/(d*x+c))/(f*x+e),x,method=_RETURNVERBOSE)
Output:
-1/f*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))+1/f*F^a*Ei(1,-b*ln(F)/(d*x+c))
Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.25 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx=\frac {F^{\frac {a d e - {\left (a c + b\right )} f}{d e - c f}} {\rm Ei}\left (\frac {{\left (b d f x + b d e\right )} \log \left (F\right )}{c d e - c^{2} f + {\left (d^{2} e - c d f\right )} x}\right ) - F^{a} {\rm Ei}\left (\frac {b \log \left (F\right )}{d x + c}\right )}{f} \] Input:
integrate(F^(a+b/(d*x+c))/(f*x+e),x, algorithm="fricas")
Output:
(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)) - F^a*Ei(b*log(F)/(d*x + c)))/f
\[ \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx=\int \frac {F^{a + \frac {b}{c + d x}}}{e + f x}\, dx \] Input:
integrate(F**(a+b/(d*x+c))/(f*x+e),x)
Output:
Integral(F**(a + b/(c + d*x))/(e + f*x), x)
\[ \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{f x + e} \,d x } \] Input:
integrate(F^(a+b/(d*x+c))/(f*x+e),x, algorithm="maxima")
Output:
integrate(F^(a + b/(d*x + c))/(f*x + e), x)
\[ \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{f x + e} \,d x } \] Input:
integrate(F^(a+b/(d*x+c))/(f*x+e),x, algorithm="giac")
Output:
integrate(F^(a + b/(d*x + c))/(f*x + e), x)
Timed out. \[ \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx=\int \frac {F^{a+\frac {b}{c+d\,x}}}{e+f\,x} \,d x \] Input:
int(F^(a + b/(c + d*x))/(e + f*x),x)
Output:
int(F^(a + b/(c + d*x))/(e + f*x), x)
\[ \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx=\int \frac {f^{\frac {a d x +a c +b}{d x +c}}}{f x +e}d x \] Input:
int(F^(a+b/(d*x+c))/(f*x+e),x)
Output:
int(f**((a*c + a*d*x + b)/(c + d*x))/(e + f*x),x)