3.2 \(\int F^{c (a+b x)} \sin ^3(d+e x) \, dx\)
Optimal. Leaf size=199 \[ \frac{b c \log (F) \sin ^3(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2}+\frac{6 b c e^2 \log (F) \sin (d+e x) F^{c (a+b x)}}{10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)+9 e^4}-\frac{6 e^3 \cos (d+e x) F^{c (a+b x)}}{10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)+9 e^4}-\frac{3 e \sin ^2(d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2} \]
[Out]
(-6*e^3*F^(c*(a + b*x))*Cos[d + e*x])/(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4) + (6*b*c*e^2*F^(c*(
a + b*x))*Log[F]*Sin[d + e*x])/(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4) - (3*e*F^(c*(a + b*x))*Cos
[d + e*x]*Sin[d + e*x]^2)/(9*e^2 + b^2*c^2*Log[F]^2) + (b*c*F^(c*(a + b*x))*Log[F]*Sin[d + e*x]^3)/(9*e^2 + b^
2*c^2*Log[F]^2)
________________________________________________________________________________________
Rubi [A] time = 0.0698005, antiderivative size = 199, normalized size of antiderivative = 1.,
number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used =
{4434, 4432} \[ \frac{b c \log (F) \sin ^3(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2}+\frac{6 b c e^2 \log (F) \sin (d+e x) F^{c (a+b x)}}{10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)+9 e^4}-\frac{6 e^3 \cos (d+e x) F^{c (a+b x)}}{10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)+9 e^4}-\frac{3 e \sin ^2(d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+9 e^2} \]
Antiderivative was successfully verified.
[In]
Int[F^(c*(a + b*x))*Sin[d + e*x]^3,x]
[Out]
(-6*e^3*F^(c*(a + b*x))*Cos[d + e*x])/(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4) + (6*b*c*e^2*F^(c*(
a + b*x))*Log[F]*Sin[d + e*x])/(9*e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4) - (3*e*F^(c*(a + b*x))*Cos
[d + e*x]*Sin[d + e*x]^2)/(9*e^2 + b^2*c^2*Log[F]^2) + (b*c*F^(c*(a + b*x))*Log[F]*Sin[d + e*x]^3)/(9*e^2 + b^
2*c^2*Log[F]^2)
Rule 4434
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*
x))*Sin[d + e*x]^n)/(e^2*n^2 + b^2*c^2*Log[F]^2), x] + (Dist[(n*(n - 1)*e^2)/(e^2*n^2 + b^2*c^2*Log[F]^2), Int
[F^(c*(a + b*x))*Sin[d + e*x]^(n - 2), x], x] - Simp[(e*n*F^(c*(a + b*x))*Cos[d + e*x]*Sin[d + e*x]^(n - 1))/(
e^2*n^2 + b^2*c^2*Log[F]^2), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 + b^2*c^2*Log[F]^2, 0] && GtQ[
n, 1]
Rule 4432
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*S
in[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] - Simp[(e*F^(c*(a + b*x))*Cos[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]
Rubi steps
\begin{align*} \int F^{c (a+b x)} \sin ^3(d+e x) \, dx &=-\frac{3 e F^{c (a+b x)} \cos (d+e x) \sin ^2(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac{b c F^{c (a+b x)} \log (F) \sin ^3(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac{\left (6 e^2\right ) \int F^{c (a+b x)} \sin (d+e x) \, dx}{9 e^2+b^2 c^2 \log ^2(F)}\\ &=-\frac{6 e^3 F^{c (a+b x)} \cos (d+e x)}{9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}+\frac{6 b c e^2 F^{c (a+b x)} \log (F) \sin (d+e x)}{9 e^4+10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}-\frac{3 e F^{c (a+b x)} \cos (d+e x) \sin ^2(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}+\frac{b c F^{c (a+b x)} \log (F) \sin ^3(d+e x)}{9 e^2+b^2 c^2 \log ^2(F)}\\ \end{align*}
Mathematica [A] time = 0.68481, size = 154, normalized size = 0.77 \[ \frac{F^{c (a+b x)} \left (-3 e \cos (d+e x) \left (b^2 c^2 \log ^2(F)+9 e^2\right )+3 \cos (3 (d+e x)) \left (b^2 c^2 e \log ^2(F)+e^3\right )-2 b c \log (F) \sin (d+e x) \left (\cos (2 (d+e x)) \left (b^2 c^2 \log ^2(F)+e^2\right )-b^2 c^2 \log ^2(F)-13 e^2\right )\right )}{4 \left (10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)+9 e^4\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[F^(c*(a + b*x))*Sin[d + e*x]^3,x]
[Out]
(F^(c*(a + b*x))*(-3*e*Cos[d + e*x]*(9*e^2 + b^2*c^2*Log[F]^2) + 3*Cos[3*(d + e*x)]*(e^3 + b^2*c^2*e*Log[F]^2)
- 2*b*c*Log[F]*(-13*e^2 - b^2*c^2*Log[F]^2 + Cos[2*(d + e*x)]*(e^2 + b^2*c^2*Log[F]^2))*Sin[d + e*x]))/(4*(9*
e^4 + 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4))
________________________________________________________________________________________
Maple [A] time = 0.138, size = 336, normalized size = 1.7 \begin{align*} -{\frac{3\,{F}^{ac}e{{\rm e}^{bcx\ln \left ( F \right ) }}}{4\,{e}^{2}+4\,{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}} \left ( 1+ \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2} \right ) ^{-1}}+{\frac{3\,{F}^{ac}e{{\rm e}^{bcx\ln \left ( F \right ) }}}{4\,{e}^{2}+4\,{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}} \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2} \left ( 1+ \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2} \right ) ^{-1}}+{\frac{3\,{F}^{ac}\ln \left ( F \right ) bc{{\rm e}^{bcx\ln \left ( F \right ) }}}{2\,{e}^{2}+2\,{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}}\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2} \right ) ^{-1}}+{\frac{3\,{F}^{ac}e{{\rm e}^{bcx\ln \left ( F \right ) }}}{36\,{e}^{2}+4\,{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}} \left ( 1+ \left ( \tan \left ({\frac{3\,ex}{2}}+{\frac{3\,d}{2}} \right ) \right ) ^{2} \right ) ^{-1}}-{\frac{3\,{F}^{ac}e{{\rm e}^{bcx\ln \left ( F \right ) }}}{36\,{e}^{2}+4\,{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}} \left ( \tan \left ({\frac{3\,ex}{2}}+{\frac{3\,d}{2}} \right ) \right ) ^{2} \left ( 1+ \left ( \tan \left ({\frac{3\,ex}{2}}+{\frac{3\,d}{2}} \right ) \right ) ^{2} \right ) ^{-1}}-{\frac{{F}^{ac}\ln \left ( F \right ) bc{{\rm e}^{bcx\ln \left ( F \right ) }}}{2\,{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}+18\,{e}^{2}}\tan \left ({\frac{3\,ex}{2}}+{\frac{3\,d}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{3\,ex}{2}}+{\frac{3\,d}{2}} \right ) \right ) ^{2} \right ) ^{-1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(F^(c*(b*x+a))*sin(e*x+d)^3,x)
[Out]
-3/4*F^(a*c)/(1+tan(1/2*d+1/2*e*x)^2)/(e^2+b^2*c^2*ln(F)^2)*e*exp(b*c*x*ln(F))+3/4*F^(a*c)/(1+tan(1/2*d+1/2*e*
x)^2)/(e^2+b^2*c^2*ln(F)^2)*e*exp(b*c*x*ln(F))*tan(1/2*d+1/2*e*x)^2+3/2*F^(a*c)/(1+tan(1/2*d+1/2*e*x)^2)*ln(F)
*b*c/(e^2+b^2*c^2*ln(F)^2)*exp(b*c*x*ln(F))*tan(1/2*d+1/2*e*x)+3/4*F^(a*c)/(1+tan(3/2*e*x+3/2*d)^2)/(9*e^2+b^2
*c^2*ln(F)^2)*e*exp(b*c*x*ln(F))-3/4*F^(a*c)/(1+tan(3/2*e*x+3/2*d)^2)/(9*e^2+b^2*c^2*ln(F)^2)*e*exp(b*c*x*ln(F
))*tan(3/2*e*x+3/2*d)^2-1/2*F^(a*c)/(1+tan(3/2*e*x+3/2*d)^2)*ln(F)*b*c/(9*e^2+b^2*c^2*ln(F)^2)*exp(b*c*x*ln(F)
)*tan(3/2*e*x+3/2*d)
________________________________________________________________________________________
Maxima [B] time = 1.33787, size = 1098, normalized size = 5.52 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*sin(e*x+d)^3,x, algorithm="maxima")
[Out]
-1/8*((F^(a*c)*b^3*c^3*log(F)^3*sin(3*d) - 3*F^(a*c)*b^2*c^2*e*cos(3*d)*log(F)^2 + F^(a*c)*b*c*e^2*log(F)*sin(
3*d) - 3*F^(a*c)*e^3*cos(3*d))*F^(b*c*x)*cos(3*e*x) - (F^(a*c)*b^3*c^3*log(F)^3*sin(3*d) + 3*F^(a*c)*b^2*c^2*e
*cos(3*d)*log(F)^2 + F^(a*c)*b*c*e^2*log(F)*sin(3*d) + 3*F^(a*c)*e^3*cos(3*d))*F^(b*c*x)*cos(3*e*x + 6*d) + 3*
(F^(a*c)*b^3*c^3*log(F)^3*sin(3*d) + F^(a*c)*b^2*c^2*e*cos(3*d)*log(F)^2 + 9*F^(a*c)*b*c*e^2*log(F)*sin(3*d) +
9*F^(a*c)*e^3*cos(3*d))*F^(b*c*x)*cos(e*x + 4*d) - 3*(F^(a*c)*b^3*c^3*log(F)^3*sin(3*d) - F^(a*c)*b^2*c^2*e*c
os(3*d)*log(F)^2 + 9*F^(a*c)*b*c*e^2*log(F)*sin(3*d) - 9*F^(a*c)*e^3*cos(3*d))*F^(b*c*x)*cos(e*x - 2*d) + (F^(
a*c)*b^3*c^3*cos(3*d)*log(F)^3 + 3*F^(a*c)*b^2*c^2*e*log(F)^2*sin(3*d) + F^(a*c)*b*c*e^2*cos(3*d)*log(F) + 3*F
^(a*c)*e^3*sin(3*d))*F^(b*c*x)*sin(3*e*x) + (F^(a*c)*b^3*c^3*cos(3*d)*log(F)^3 - 3*F^(a*c)*b^2*c^2*e*log(F)^2*
sin(3*d) + F^(a*c)*b*c*e^2*cos(3*d)*log(F) - 3*F^(a*c)*e^3*sin(3*d))*F^(b*c*x)*sin(3*e*x + 6*d) - 3*(F^(a*c)*b
^3*c^3*cos(3*d)*log(F)^3 - F^(a*c)*b^2*c^2*e*log(F)^2*sin(3*d) + 9*F^(a*c)*b*c*e^2*cos(3*d)*log(F) - 9*F^(a*c)
*e^3*sin(3*d))*F^(b*c*x)*sin(e*x + 4*d) - 3*(F^(a*c)*b^3*c^3*cos(3*d)*log(F)^3 + F^(a*c)*b^2*c^2*e*log(F)^2*si
n(3*d) + 9*F^(a*c)*b*c*e^2*cos(3*d)*log(F) + 9*F^(a*c)*e^3*sin(3*d))*F^(b*c*x)*sin(e*x - 2*d))/(b^4*c^4*cos(3*
d)^2*log(F)^4 + b^4*c^4*log(F)^4*sin(3*d)^2 + 9*(cos(3*d)^2 + sin(3*d)^2)*e^4 + 10*(b^2*c^2*cos(3*d)^2*log(F)^
2 + b^2*c^2*log(F)^2*sin(3*d)^2)*e^2)
________________________________________________________________________________________
Fricas [A] time = 0.507948, size = 386, normalized size = 1.94 \begin{align*} \frac{{\left (3 \, e^{3} \cos \left (e x + d\right )^{3} - 9 \, e^{3} \cos \left (e x + d\right ) + 3 \,{\left (b^{2} c^{2} e \cos \left (e x + d\right )^{3} - b^{2} c^{2} e \cos \left (e x + d\right )\right )} \log \left (F\right )^{2} -{\left ({\left (b^{3} c^{3} \cos \left (e x + d\right )^{2} - b^{3} c^{3}\right )} \log \left (F\right )^{3} +{\left (b c e^{2} \cos \left (e x + d\right )^{2} - 7 \, b c e^{2}\right )} \log \left (F\right )\right )} \sin \left (e x + d\right )\right )} F^{b c x + a c}}{b^{4} c^{4} \log \left (F\right )^{4} + 10 \, b^{2} c^{2} e^{2} \log \left (F\right )^{2} + 9 \, e^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*sin(e*x+d)^3,x, algorithm="fricas")
[Out]
(3*e^3*cos(e*x + d)^3 - 9*e^3*cos(e*x + d) + 3*(b^2*c^2*e*cos(e*x + d)^3 - b^2*c^2*e*cos(e*x + d))*log(F)^2 -
((b^3*c^3*cos(e*x + d)^2 - b^3*c^3)*log(F)^3 + (b*c*e^2*cos(e*x + d)^2 - 7*b*c*e^2)*log(F))*sin(e*x + d))*F^(b
*c*x + a*c)/(b^4*c^4*log(F)^4 + 10*b^2*c^2*e^2*log(F)^2 + 9*e^4)
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F**(c*(b*x+a))*sin(e*x+d)**3,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [C] time = 1.29847, size = 1770, normalized size = 8.89 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*sin(e*x+d)^3,x, algorithm="giac")
[Out]
-1/4*(2*b*c*log(abs(F))*sin(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c + 3*x*e + 3*d)
/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 6*e)^2) - (pi*b*c*sgn(F) - pi*b*c + 6*e)*cos(1/2*pi*b*c*
x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c + 3*x*e + 3*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn
(F) - pi*b*c + 6*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 3/4*(2*b*c*log(abs(F))*sin(1/2*pi*b*c*x*sgn(
F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c + x*e + d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b
*c + 2*e)^2) - (pi*b*c*sgn(F) - pi*b*c + 2*e)*cos(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2
*pi*a*c + x*e + d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 2*e)^2))*e^(b*c*x*log(abs(F)) + a*c*lo
g(abs(F))) - 3/4*(2*b*c*log(abs(F))*sin(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c -
x*e - d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 2*e)^2) - (pi*b*c*sgn(F) - pi*b*c - 2*e)*cos(1/2
*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c - x*e - d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c
*sgn(F) - pi*b*c - 2*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 1/4*(2*b*c*log(abs(F))*sin(1/2*pi*b*c*x*
sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c - 3*x*e - 3*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F
) - pi*b*c - 6*e)^2) - (pi*b*c*sgn(F) - pi*b*c - 6*e)*cos(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(
F) - 1/2*pi*a*c - 3*x*e - 3*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 6*e)^2))*e^(b*c*x*log(abs(
F)) + a*c*log(abs(F))) + 1/2*(-2*I*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*
a*c + 3*I*x*e + 3*I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(abs(F)) + 48*I*e) - 2*I*e^(-1/2*I*pi*b*c*x
*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c - 3*I*x*e - 3*I*d)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*
b*c + 16*b*c*log(abs(F)) - 48*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 1/2*(6*I*e^(1/2*I*pi*b*c*x*sgn(F
) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c + I*x*e + I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*
c*log(abs(F)) + 16*I*e) + 6*I*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c
- I*x*e - I*d)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*b*c*log(abs(F)) - 16*I*e))*e^(b*c*x*log(abs(F)) + a*c*log
(abs(F))) + 1/2*(-6*I*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c - I*x*e -
I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(abs(F)) - 16*I*e) - 6*I*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*p
i*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c + I*x*e + I*d)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*b*c*log(abs(
F)) + 16*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 1/2*(2*I*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x +
1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c - 3*I*x*e - 3*I*d)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(abs(F)) - 4
8*I*e) + 2*I*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c + 3*I*x*e + 3*I*d
)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*b*c*log(abs(F)) + 48*I*e))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F)))