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

Optimal. Leaf size=159 \[ \frac {f \log (F) (b c-a d) F^{\frac {f (b g-a h)}{d g-c h}+e} \text {Ei}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right )}{(d g-c h)^2}+\frac {d F^{-\frac {f (b c-a d)}{d (c+d x)}+\frac {b f}{d}+e}}{h (d g-c h)}-\frac {F^{\frac {f (a+b x)}{c+d x}+e}}{h (g+h x)} \]

[Out]

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

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Rubi [A]  time = 2.56, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2232, 6742, 2230, 2209, 2210, 2231, 2233, 2178} \[ \frac {f \log (F) (b c-a d) F^{\frac {f (b g-a h)}{d g-c h}+e} \text {Ei}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right )}{(d g-c h)^2}+\frac {d F^{-\frac {f (b c-a d)}{d (c+d x)}+\frac {b f}{d}+e}}{h (d g-c h)}-\frac {F^{\frac {f (a+b x)}{c+d x}+e}}{h (g+h x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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 2230

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :>
Int[(g + h*x)^m*F^((d*e + b*f)/d - (f*(b*c - a*d))/(d*(c + d*x))), x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m},
x] && NeQ[b*c - a*d, 0] && EqQ[d*g - c*h, 0]

Rule 2231

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))/((g_.) + (h_.)*(x_)), x_Symbol] :> Dist[d
/h, Int[F^(e + (f*(a + b*x))/(c + d*x))/(c + d*x), x], x] - Dist[(d*g - c*h)/h, Int[F^(e + (f*(a + b*x))/(c +
d*x))/((c + d*x)*(g + h*x)), 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]

Rule 2232

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_), x_Symbol] :> S
imp[((g + h*x)^(m + 1)*F^(e + (f*(a + b*x))/(c + d*x)))/(h*(m + 1)), x] - Dist[(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]

Rule 2233

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))/(((g_.) + (h_.)*(x_))*((i_.) + (j_.)*(x_)
)), x_Symbol] :> -Dist[d/(h*(d*i - c*j)), Subst[Int[F^(e + (f*(b*i - a*j))/(d*i - c*j) - ((b*c - a*d)*f*x)/(d*
i - c*j))/x, x], x, (i + j*x)/(c + d*x)], x] /; FreeQ[{F, a, b, c, d, e, f, g, h}, x] && EqQ[d*g - c*h, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [F]  time = 1.19, size = 0, normalized size = 0.00 \[ \int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{(g+h x)^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 220, normalized size = 1.38 \[ \frac {{\left ({\left (b c - a d\right )} f h x + {\left (b c - a d\right )} f g\right )} F^{\frac {{\left (d e + b f\right )} g - {\left (c e + a f\right )} h}{d g - c h}} {\rm Ei}\left (-\frac {{\left ({\left (b c - a d\right )} f h x + {\left (b c - a d\right )} f g\right )} \log \relax (F)}{c d g - c^{2} h + {\left (d^{2} g - c d h\right )} x}\right ) \log \relax (F) + {\left (c d g - c^{2} h + {\left (d^{2} g - c d h\right )} x\right )} F^{\frac {c e + a f + {\left (d e + b f\right )} x}{d x + c}}}{d^{2} g^{3} - 2 \, c d g^{2} h + c^{2} g h^{2} + {\left (d^{2} g^{2} h - 2 \, c d g h^{2} + c^{2} h^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{e + \frac {{\left (b x + a\right )} f}{d x + c}}}{{\left (h x + g\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.21, size = 580, normalized size = 3.65 \[ \frac {a d f \,F^{\frac {b f +d e}{d}} F^{\frac {\left (a d -b c \right ) f}{\left (d x +c \right ) d}} \ln \relax (F )}{\left (c h -d g \right )^{2} \left (-\frac {a f h \ln \relax (F )}{c h -d g}+\frac {b f g \ln \relax (F )}{c h -d g}-\frac {c e h \ln \relax (F )}{c h -d g}+\frac {d e g \ln \relax (F )}{c h -d g}+\frac {a f \ln \relax (F )}{d x +c}-\frac {b c f \ln \relax (F )}{\left (d x +c \right ) d}+\frac {b f \ln \relax (F )}{d}+e \ln \relax (F )\right )}+\frac {a d f \,F^{\frac {a f h -b f g +c e h -d e g}{c h -d g}} \Ei \left (1, -\frac {\left (a d -b c \right ) f \ln \relax (F )}{\left (d x +c \right ) d}-\frac {\left (b f +d e \right ) \ln \relax (F )}{d}-\frac {-a f h \ln \relax (F )+b f g \ln \relax (F )-c e h \ln \relax (F )+d e g \ln \relax (F )}{c h -d g}\right ) \ln \relax (F )}{\left (c h -d g \right )^{2}}-\frac {b c f \,F^{\frac {b f +d e}{d}} F^{\frac {\left (a d -b c \right ) f}{\left (d x +c \right ) d}} \ln \relax (F )}{\left (c h -d g \right )^{2} \left (-\frac {a f h \ln \relax (F )}{c h -d g}+\frac {b f g \ln \relax (F )}{c h -d g}-\frac {c e h \ln \relax (F )}{c h -d g}+\frac {d e g \ln \relax (F )}{c h -d g}+\frac {a f \ln \relax (F )}{d x +c}-\frac {b c f \ln \relax (F )}{\left (d x +c \right ) d}+\frac {b f \ln \relax (F )}{d}+e \ln \relax (F )\right )}-\frac {b c f \,F^{\frac {a f h -b f g +c e h -d e g}{c h -d g}} \Ei \left (1, -\frac {\left (a d -b c \right ) f \ln \relax (F )}{\left (d x +c \right ) d}-\frac {\left (b f +d e \right ) \ln \relax (F )}{d}-\frac {-a f h \ln \relax (F )+b f g \ln \relax (F )-c e h \ln \relax (F )+d e g \ln \relax (F )}{c h -d g}\right ) \ln \relax (F )}{\left (c h -d g \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{e + \frac {{\left (b x + a\right )} f}{d x + c}}}{{\left (h x + g\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {F^{e+\frac {f\,\left (a+b\,x\right )}{c+d\,x}}}{{\left (g+h\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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