Integrand size = 26, antiderivative size = 634 \[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^4} \, dx=\frac {d^3 F^{e+\frac {b f}{d}-\frac {(b c-a d) f}{d (c+d x)}}}{3 h (d g-c h)^3}-\frac {F^{e+\frac {f (a+b x)}{c+d x}}}{3 h (g+h x)^3}+\frac {5 d^2 (b c-a d) f F^{e+\frac {b f}{d}-\frac {(b c-a d) f}{d (c+d x)}} \log (F)}{6 (d g-c h)^4}-\frac {(b c-a d) f F^{e+\frac {f (a+b x)}{c+d x}} \log (F)}{6 (d g-c h)^2 (g+h x)^2}-\frac {2 d (b c-a d) f F^{e+\frac {f (a+b x)}{c+d x}} \log (F)}{3 (d g-c h)^3 (g+h x)}+\frac {d^2 (b c-a d) f F^{e+\frac {f (b g-a h)}{d g-c h}} \operatorname {ExpIntegralEi}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right ) \log (F)}{(d g-c h)^4}+\frac {d (b c-a d)^2 f^2 F^{e+\frac {b f}{d}-\frac {(b c-a d) f}{d (c+d x)}} h \log ^2(F)}{6 (d g-c h)^5}-\frac {(b c-a d)^2 f^2 F^{e+\frac {f (a+b x)}{c+d x}} h \log ^2(F)}{6 (d g-c h)^4 (g+h x)}+\frac {d (b c-a d)^2 f^2 F^{e+\frac {f (b g-a h)}{d g-c h}} h \operatorname {ExpIntegralEi}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right ) \log ^2(F)}{(d g-c h)^5}+\frac {(b c-a d)^3 f^3 F^{e+\frac {f (b g-a h)}{d g-c h}} h^2 \operatorname {ExpIntegralEi}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right ) \log ^3(F)}{6 (d g-c h)^6} \]
1/3*d^3*F^(e+b*f/d-(-a*d+b*c)*f/d/(d*x+c))/h/(-c*h+d*g)^3-1/3*F^(e+f*(b*x+ a)/(d*x+c))/h/(h*x+g)^3+5/6*d^2*(-a*d+b*c)*f*F^(e+b*f/d-(-a*d+b*c)*f/d/(d* x+c))*ln(F)/(-c*h+d*g)^4-1/6*(-a*d+b*c)*f*F^(e+f*(b*x+a)/(d*x+c))*ln(F)/(- c*h+d*g)^2/(h*x+g)^2-2/3*d*(-a*d+b*c)*f*F^(e+f*(b*x+a)/(d*x+c))*ln(F)/(-c* h+d*g)^3/(h*x+g)+d^2*(-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)^4+1/6*d*(-a*d+b* c)^2*f^2*F^(e+b*f/d-(-a*d+b*c)*f/d/(d*x+c))*h*ln(F)^2/(-c*h+d*g)^5-1/6*(-a *d+b*c)^2*f^2*F^(e+f*(b*x+a)/(d*x+c))*h*ln(F)^2/(-c*h+d*g)^4/(h*x+g)+d*(-a *d+b*c)^2*f^2*F^(e+f*(-a*h+b*g)/(-c*h+d*g))*h*Ei(-(-a*d+b*c)*f*(h*x+g)*ln( F)/(-c*h+d*g)/(d*x+c))*ln(F)^2/(-c*h+d*g)^5+1/6*(-a*d+b*c)^3*f^3*F^(e+f*(- a*h+b*g)/(-c*h+d*g))*h^2*Ei(-(-a*d+b*c)*f*(h*x+g)*ln(F)/(-c*h+d*g)/(d*x+c) )*ln(F)^3/(-c*h+d*g)^6
\[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^4} \, dx=\int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^4} \, dx \]
Time = 7.90 (sec) , antiderivative size = 612, normalized size of antiderivative = 0.97, 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.
| \(\displaystyle \int \frac {F^{\frac {f (a+b x)}{c+d x}+e}}{(g+h x)^4} \, dx\) |
| \(\Big \downarrow \) 2662 |
| \(\displaystyle \frac {f \log (F) (b c-a d) \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(c+d x)^2 (g+h x)^3}dx}{3 h}-\frac {F^{\frac {f (a+b x)}{c+d x}+e}}{3 h (g+h x)^3}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {f \log (F) (b c-a d) \int \left (-\frac {3 d^3 h F^{e+\frac {f (a+b x)}{c+d x}}}{(d g-c h)^4 (c+d x)}+\frac {3 d^2 h^2 F^{e+\frac {f (a+b x)}{c+d x}}}{(d g-c h)^4 (g+h x)}+\frac {d^3 F^{e+\frac {f (a+b x)}{c+d x}}}{(d g-c h)^3 (c+d x)^2}+\frac {2 d h^2 F^{e+\frac {f (a+b x)}{c+d x}}}{(d g-c h)^3 (g+h x)^2}+\frac {h^2 F^{e+\frac {f (a+b x)}{c+d x}}}{(d g-c h)^2 (g+h x)^3}\right )dx}{3 h}-\frac {F^{\frac {f (a+b x)}{c+d x}+e}}{3 h (g+h x)^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f \log (F) (b c-a d) \left (\frac {d^3 F^{-\frac {f (b c-a d)}{d (c+d x)}+\frac {b f}{d}+e}}{f \log (F) (b c-a d) (d g-c h)^3}+\frac {3 d^2 h F^{\frac {f (b g-a h)}{d g-c h}+e} \operatorname {ExpIntegralEi}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right )}{(d g-c h)^4}+\frac {5 d^2 h F^{-\frac {f (b c-a d)}{d (c+d x)}+\frac {b f}{d}+e}}{2 (d g-c h)^4}+\frac {f^2 h^3 \log ^2(F) (b c-a d)^2 F^{\frac {f (b g-a h)}{d g-c h}+e} \operatorname {ExpIntegralEi}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right )}{2 (d g-c h)^6}+\frac {3 d f h^2 \log (F) (b c-a d) F^{\frac {f (b g-a h)}{d g-c h}+e} \operatorname {ExpIntegralEi}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right )}{(d g-c h)^5}+\frac {d f h^2 \log (F) (b c-a d) F^{-\frac {f (b c-a d)}{d (c+d x)}+\frac {b f}{d}+e}}{2 (d g-c h)^5}-\frac {f h^2 \log (F) (b c-a d) F^{\frac {f (a+b x)}{c+d x}+e}}{2 (g+h x) (d g-c h)^4}-\frac {2 d h F^{\frac {f (a+b x)}{c+d x}+e}}{(g+h x) (d g-c h)^3}-\frac {h F^{\frac {f (a+b x)}{c+d x}+e}}{2 (g+h x)^2 (d g-c h)^2}\right )}{3 h}-\frac {F^{\frac {f (a+b x)}{c+d x}+e}}{3 h (g+h x)^3}\) |
-1/3*F^(e + (f*(a + b*x))/(c + d*x))/(h*(g + h*x)^3) + ((b*c - a*d)*f*Log[ F]*((5*d^2*F^(e + (b*f)/d - ((b*c - a*d)*f)/(d*(c + d*x)))*h)/(2*(d*g - c* h)^4) - (F^(e + (f*(a + b*x))/(c + d*x))*h)/(2*(d*g - c*h)^2*(g + h*x)^2) - (2*d*F^(e + (f*(a + b*x))/(c + d*x))*h)/((d*g - c*h)^3*(g + h*x)) + (3*d ^2*F^(e + (f*(b*g - a*h))/(d*g - c*h))*h*ExpIntegralEi[-(((b*c - a*d)*f*(g + h*x)*Log[F])/((d*g - c*h)*(c + d*x)))])/(d*g - c*h)^4 + (d^3*F^(e + (b* f)/d - ((b*c - a*d)*f)/(d*(c + d*x))))/((b*c - a*d)*f*(d*g - c*h)^3*Log[F] ) + (d*(b*c - a*d)*f*F^(e + (b*f)/d - ((b*c - a*d)*f)/(d*(c + d*x)))*h^2*L og[F])/(2*(d*g - c*h)^5) - ((b*c - a*d)*f*F^(e + (f*(a + b*x))/(c + d*x))* h^2*Log[F])/(2*(d*g - c*h)^4*(g + h*x)) + (3*d*(b*c - a*d)*f*F^(e + (f*(b* g - a*h))/(d*g - c*h))*h^2*ExpIntegralEi[-(((b*c - a*d)*f*(g + h*x)*Log[F] )/((d*g - c*h)*(c + d*x)))]*Log[F])/(d*g - c*h)^5 + ((b*c - a*d)^2*f^2*F^( e + (f*(b*g - a*h))/(d*g - c*h))*h^3*ExpIntegralEi[-(((b*c - a*d)*f*(g + h *x)*Log[F])/((d*g - c*h)*(c + d*x)))]*Log[F]^2)/(2*(d*g - c*h)^6)))/(3*h)
3.5.25.3.1 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. \(4679\) vs. \(2(618)=1236\).
Time = 1.23 (sec) , antiderivative size = 4680, normalized size of antiderivative = 7.38
ln(F)^2*d^3*f^2*h/(c*h-d*g)^5*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^2+1/6*ln(F)^3*d^3*f^3*h^2/(c*h-d*g )^6*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^3+ln(F)*d^3/(c*h-d*g)^4*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*f-ln(F)*d^2/(c*h-d*g)^4*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-1/6*ln(F)^3*f^3*h^ 2/(c*h-d*g)^6*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^3*c^3-1/3*ln(F)^3*f^3*h^2/(c*h-d*g)^6*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)^3*b^3*c^3-1/6*ln(F)^3*f^3*h^2/(c*h -d*g)^6*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)*l n(F)*b*f*g-1/(c*h-d*g)*ln(F)*c*e*h+1/(c*h-d*g)*ln(F)*d*e*g)^2*b^3*c^3-1...
Leaf count of result is larger than twice the leaf count of optimal. 2250 vs. \(2 (618) = 1236\).
Time = 0.43 (sec) , antiderivative size = 2250, normalized size of antiderivative = 3.55 \[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^4} \, dx=\text {Too large to display} \]
1/6*((((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^3*h^5*x^3 + 3 *(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^3*g*h^4*x^2 + 3*(b^ 3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^3*g^2*h^3*x + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^3*g^3*h^2)*log(F)^3 + 6*((b^2 *c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*f^2*g^4*h - (b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*f^2*g^3*h^2 + ((b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*f^2*g*h^4 - (b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*f^2*h^5)*x^3 + 3*((b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*f^2*g^2*h^3 - (b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c* d^3)*f^2*g*h^4)*x^2 + 3*((b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*f^2*g^3*h^2 - (b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*f^2*g^2*h^3)*x)*log(F)^2 + 6*(( b*c*d^4 - a*d^5)*f*g^5 - 2*(b*c^2*d^3 - a*c*d^4)*f*g^4*h + (b*c^3*d^2 - a* c^2*d^3)*f*g^3*h^2 + ((b*c*d^4 - a*d^5)*f*g^2*h^3 - 2*(b*c^2*d^3 - a*c*d^4 )*f*g*h^4 + (b*c^3*d^2 - a*c^2*d^3)*f*h^5)*x^3 + 3*((b*c*d^4 - a*d^5)*f*g^ 3*h^2 - 2*(b*c^2*d^3 - a*c*d^4)*f*g^2*h^3 + (b*c^3*d^2 - a*c^2*d^3)*f*g*h^ 4)*x^2 + 3*((b*c*d^4 - a*d^5)*f*g^4*h - 2*(b*c^2*d^3 - a*c*d^4)*f*g^3*h^2 + (b*c^3*d^2 - a*c^2*d^3)*f*g^2*h^3)*x)*log(F))*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)) + (6*c*d^5*g^5 - 24*c^2*d^4*g^4*h + 38*c ^3*d^3*g^3*h^2 - 30*c^4*d^2*g^2*h^3 + 12*c^5*d*g*h^4 - 2*c^6*h^5 + 2*(d^6* g^3*h^2 - 3*c*d^5*g^2*h^3 + 3*c^2*d^4*g*h^4 - c^3*d^3*h^5)*x^3 + 6*(d^6...
Timed out. \[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^4} \, dx=\text {Timed out} \]
\[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^4} \, dx=\int { \frac {F^{e + \frac {{\left (b x + a\right )} f}{d x + c}}}{{\left (h x + g\right )}^{4}} \,d x } \]
\[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^4} \, dx=\int { \frac {F^{e + \frac {{\left (b x + a\right )} f}{d x + c}}}{{\left (h x + g\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^4} \, dx=\int \frac {F^{e+\frac {f\,\left (a+b\,x\right )}{c+d\,x}}}{{\left (g+h\,x\right )}^4} \,d x \]