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

Optimal. Leaf size=104 \[ \frac {F^{\frac {f (b g-a h)}{d g-c h}+e} \text {Ei}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right )}{h}-\frac {F^{\frac {b f}{d}+e} \text {Ei}\left (-\frac {(b c-a d) f \log (F)}{d (c+d x)}\right )}{h} \]

[Out]

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

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Rubi [A]  time = 1.05, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2231, 2230, 2210, 2233, 2178} \[ \frac {F^{\frac {f (b g-a h)}{d g-c h}+e} \text {Ei}\left (-\frac {(b c-a d) f (g+h x) \log (F)}{(d g-c h) (c+d x)}\right )}{h}-\frac {F^{\frac {b f}{d}+e} \text {Ei}\left (-\frac {(b c-a d) f \log (F)}{d (c+d x)}\right )}{h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.32, size = 103, normalized size = 0.99 \[ \frac {F^{\frac {b f}{d}+e} \left (F^{\frac {f h (b c-a d)}{d (d g-c h)}} \text {Ei}\left (\frac {(b c-a d) f (g+h x) \log (F)}{(c h-d g) (c+d x)}\right )-\text {Ei}\left (\frac {(a d f-b c f) \log (F)}{d (c+d x)}\right )\right )}{h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 135, normalized size = 1.30 \[ -\frac {F^{\frac {d e + b f}{d}} {\rm Ei}\left (-\frac {{\left (b c - a d\right )} f \log \relax (F)}{d^{2} x + c d}\right ) - F^{\frac {{\left (d e + b f\right )} g - {\left (c e + a f\right )} h}{d g - c h}} {\rm Ei}\left (-\frac {{\left ({\left (b c - a d\right )} f h x + {\left (b c - a d\right )} f g\right )} \log \relax (F)}{c d g - c^{2} h + {\left (d^{2} g - c d h\right )} x}\right )}{h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.26, size = 432, normalized size = 4.15 \[ \frac {a d \,F^{\frac {b f +d e}{d}} \Ei \left (1, -\frac {\left (a d -b c \right ) f \ln \relax (F )}{\left (d x +c \right ) d}-\frac {\left (b f +d e \right ) \ln \relax (F )}{d}-\frac {-b f \ln \relax (F )-d e \ln \relax (F )}{d}\right )}{\left (a d -b c \right ) h}-\frac {a d \,F^{\frac {a f h -b f g +c e h -d e g}{c h -d g}} \Ei \left (1, -\frac {\left (a d -b c \right ) f \ln \relax (F )}{\left (d x +c \right ) d}-\frac {\left (b f +d e \right ) \ln \relax (F )}{d}-\frac {-a f h \ln \relax (F )+b f g \ln \relax (F )-c e h \ln \relax (F )+d e g \ln \relax (F )}{c h -d g}\right )}{\left (a d -b c \right ) h}-\frac {b c \,F^{\frac {b f +d e}{d}} \Ei \left (1, -\frac {\left (a d -b c \right ) f \ln \relax (F )}{\left (d x +c \right ) d}-\frac {\left (b f +d e \right ) \ln \relax (F )}{d}-\frac {-b f \ln \relax (F )-d e \ln \relax (F )}{d}\right )}{\left (a d -b c \right ) h}+\frac {b c \,F^{\frac {a f h -b f g +c e h -d e g}{c h -d g}} \Ei \left (1, -\frac {\left (a d -b c \right ) f \ln \relax (F )}{\left (d x +c \right ) d}-\frac {\left (b f +d e \right ) \ln \relax (F )}{d}-\frac {-a f h \ln \relax (F )+b f g \ln \relax (F )-c e h \ln \relax (F )+d e g \ln \relax (F )}{c h -d g}\right )}{\left (a d -b c \right ) h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d/h/(a*d-b*c)*F^((b*f+d*e)/d)*Ei(1,-f*(a*d-b*c)*ln(F)/d/(d*x+c)-(b*f+d*e)*ln(F)/d-(-b*f*ln(F)-d*e*ln(F))/d)*a-
1/h/(a*d-b*c)*F^((b*f+d*e)/d)*Ei(1,-f*(a*d-b*c)*ln(F)/d/(d*x+c)-(b*f+d*e)*ln(F)/d-(-b*f*ln(F)-d*e*ln(F))/d)*b*
c-d/h/(a*d-b*c)*F^((a*f*h-b*f*g+c*e*h-d*e*g)/(c*h-d*g))*Ei(1,-f*(a*d-b*c)*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+1/h/(a*d-b*c)*F^((a*f*h-b*f*g+c*e*h-d*e*g)/(c*h-
d*g))*Ei(1,-f*(a*d-b*c)*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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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