3.5.22 \(\int \frac {F^{e+\frac {f (a+b x)}{c+d x}}}{g+h x} \, dx\) [422]

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

[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

________________________________________________________________________________________

Rubi [A]
time = 0.70, 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 = {2263, 2262, 2241, 2265, 2209} \begin {gather*} \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} \end {gather*}

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 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2241

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 2262

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 2263

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 2265

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 {\text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.22, size = 103, normalized size = 0.99 \begin {gather*} \frac {F^{e+\frac {b f}{d}} \left (-\text {Ei}\left (\frac {(-b c f+a d f) \log (F)}{d (c+d x)}\right )+F^{\frac {(b c-a d) f h}{d (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 )\right )}{h} \end {gather*}

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(431\) vs. \(2(104)=208\).
time = 0.17, size = 432, normalized size = 4.15

method	result	size
risch	\(\frac {d \,F^{\frac {b f +e d}{d}} \expIntegral \left (1, -\frac {f \left (a d -c b \right ) \ln \left (F \right )}{d \left (d x +c \right )}-\frac {\left (b f +e d \right ) \ln \left (F \right )}{d}-\frac {-\ln \left (F \right ) b f -d e \ln \left (F \right )}{d}\right ) a}{h \left (a d -c b \right )}-\frac {F^{\frac {b f +e d}{d}} \expIntegral \left (1, -\frac {f \left (a d -c b \right ) \ln \left (F \right )}{d \left (d x +c \right )}-\frac {\left (b f +e d \right ) \ln \left (F \right )}{d}-\frac {-\ln \left (F \right ) b f -d e \ln \left (F \right )}{d}\right ) c b}{h \left (a d -c b \right )}-\frac {d \,F^{\frac {a f h -b f g +c e h -d e g}{c h -d g}} \expIntegral \left (1, -\frac {f \left (a d -c b \right ) \ln \left (F \right )}{d \left (d x +c \right )}-\frac {\left (b f +e d \right ) \ln \left (F \right )}{d}-\frac {-\ln \left (F \right ) a f h +\ln \left (F \right ) b f g -\ln \left (F \right ) c e h +\ln \left (F \right ) d e g}{c h -d g}\right ) a}{h \left (a d -c b \right )}+\frac {F^{\frac {a f h -b f g +c e h -d e g}{c h -d g}} \expIntegral \left (1, -\frac {f \left (a d -c b \right ) \ln \left (F \right )}{d \left (d x +c \right )}-\frac {\left (b f +e d \right ) \ln \left (F \right )}{d}-\frac {-\ln \left (F \right ) a f h +\ln \left (F \right ) b f g -\ln \left (F \right ) c e h +\ln \left (F \right ) d e g}{c h -d g}\right ) c b}{h \left (a d -c b \right )}\)	\(432\)

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,method=_RETURNVERBOSE)

[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-(-ln(F)*b*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-(-ln(F)*b*f-d*e*ln(F))/d)*c*
b-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))*c*b

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 137, normalized size = 1.32 \begin {gather*} -\frac {F^{\frac {b f + d e}{d}} {\rm Ei}\left (-\frac {{\left (b c - a d\right )} f \log \left (F\right )}{d^{2} x + c d}\right ) - F^{\frac {b f g - a f h + {\left (d g - c h\right )} e}{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 )}{h} \end {gather*}

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

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________