3.37 \(\int \frac{\csc (e+f x) \sqrt{c+d \sin (e+f x)}}{\sqrt{a+a \sin (e+f x)}} \, dx\)
Optimal. Leaf size=140 \[ \frac{\sqrt{2} \sqrt{c-d} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{a} f}-\frac{2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{a} f} \]
[Out]
(-2*Sqrt[c]*ArcTanh[(Sqrt[a]*Sqrt[c]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(Sqrt
[a]*f) + (Sqrt[2]*Sqrt[c - d]*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqr
t[c + d*Sin[e + f*x]])])/(Sqrt[a]*f)
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Rubi [A] time = 0.50238, antiderivative size = 140, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{2944, 2782, 208, 2943, 206} \[ \frac{\sqrt{2} \sqrt{c-d} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{a} f}-\frac{2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{a} f} \]
Antiderivative was successfully verified.
[In]
Int[(Csc[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/Sqrt[a + a*Sin[e + f*x]],x]
[Out]
(-2*Sqrt[c]*ArcTanh[(Sqrt[a]*Sqrt[c]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(Sqrt
[a]*f) + (Sqrt[2]*Sqrt[c - d]*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqr
t[c + d*Sin[e + f*x]])])/(Sqrt[a]*f)
Rule 2944
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/(sin[(e_.) + (f_.)*(x_)]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x
_)]]), x_Symbol] :> Dist[(b*c - a*d)/c, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] + Di
st[a/c, Int[Sqrt[c + d*Sin[e + f*x]]/(Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && EqQ[c^2 - d^2, 0]
Rule 2782
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rule 2943
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/(sin[(e_.) + (f_.)*(x_)]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x
_)]]), x_Symbol] :> Dist[(-2*a)/f, Subst[Int[1/(1 - a*c*x^2), x], x, Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sq
rt[c + d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[
b*c + a*d, 0]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{\csc (e+f x) \sqrt{c+d \sin (e+f x)}}{\sqrt{a+a \sin (e+f x)}} \, dx &=\frac{c \int \frac{\csc (e+f x) \sqrt{a+a \sin (e+f x)}}{\sqrt{c+d \sin (e+f x)}} \, dx}{a}+(-c+d) \int \frac{1}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx\\ &=-\frac{(2 c) \operatorname{Subst}\left (\int \frac{1}{1-a c x^2} \, dx,x,\frac{\cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{f}+\frac{(2 a (c-d)) \operatorname{Subst}\left (\int \frac{1}{2 a^2-(a c-a d) x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{f}\\ &=-\frac{2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{a} f}+\frac{\sqrt{2} \sqrt{c-d} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{a} f}\\ \end{align*}
Mathematica [C] time = 35.0772, size = 472502, normalized size = 3375.01
\[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Csc[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/Sqrt[a + a*Sin[e + f*x]],x]
[Out]
Result too large to show
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Maple [B] time = 0.315, size = 347, normalized size = 2.5 \begin{align*} -{\frac{ \left ( -1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) \right ) \sqrt{2}}{2\,f\sin \left ( fx+e \right ) }\sqrt{c+d\sin \left ( fx+e \right ) } \left ( \sqrt{2\,c-2\,d}\ln \left ( -2\,{\frac{1}{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) } \left ( \sqrt{2\,c-2\,d}\sqrt{2}\sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sin \left ( fx+e \right ) +c\cos \left ( fx+e \right ) -d\cos \left ( fx+e \right ) +c\sin \left ( fx+e \right ) -d\sin \left ( fx+e \right ) -c+d \right ) } \right ) \sqrt{c}+\ln \left ( -{\frac{1}{\sin \left ( fx+e \right ) } \left ( -\sqrt{c}\sqrt{2}\sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sin \left ( fx+e \right ) +c\cos \left ( fx+e \right ) -d\sin \left ( fx+e \right ) -c \right ){\frac{1}{\sqrt{c}}}} \right ) c-c\ln \left ( 2\,{\frac{1}{-1+\cos \left ( fx+e \right ) } \left ( -\sqrt{c}\sqrt{2}\sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sin \left ( fx+e \right ) +d\cos \left ( fx+e \right ) -c\sin \left ( fx+e \right ) -d \right ) } \right ) \right ){\frac{1}{\sqrt{c}}}{\frac{1}{\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }}}{\frac{1}{\sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*sin(f*x+e))^(1/2)/sin(f*x+e)/(a+a*sin(f*x+e))^(1/2),x)
[Out]
-1/2/f/c^(1/2)*(-1+cos(f*x+e)-sin(f*x+e))*(c+d*sin(f*x+e))^(1/2)*((2*c-2*d)^(1/2)*ln(-2*((2*c-2*d)^(1/2)*2^(1/
2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)
/(-1+cos(f*x+e)-sin(f*x+e)))*c^(1/2)+ln(-1/c^(1/2)*(-c^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*s
in(f*x+e)+c*cos(f*x+e)-d*sin(f*x+e)-c)/sin(f*x+e))*c-c*ln(2*(-c^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1)
)^(1/2)*sin(f*x+e)+d*cos(f*x+e)-c*sin(f*x+e)-d)/(-1+cos(f*x+e))))/(a*(1+sin(f*x+e)))^(1/2)/sin(f*x+e)*2^(1/2)/
((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \sin \left (f x + e\right ) + c}}{\sqrt{a \sin \left (f x + e\right ) + a} \sin \left (f x + e\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(1/2)/sin(f*x+e)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate(sqrt(d*sin(f*x + e) + c)/(sqrt(a*sin(f*x + e) + a)*sin(f*x + e)), x)
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Fricas [B] time = 5.66698, size = 6734, normalized size = 48.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(1/2)/sin(f*x+e)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
[1/4*(sqrt(2)*sqrt((c - d)/a)*log(((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^3 + 4*sqrt(2)*((c - 3*d)*cos(f*x + e)^
2 - (3*c - d)*cos(f*x + e) + ((c - 3*d)*cos(f*x + e) + 4*c - 4*d)*sin(f*x + e) - 4*c + 4*d)*sqrt(a*sin(f*x + e
) + a)*sqrt(d*sin(f*x + e) + c)*sqrt((c - d)/a) - (13*c^2 - 22*c*d - 3*d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4
*d^2 - 2*(9*c^2 - 14*c*d + 9*d^2)*cos(f*x + e) + ((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d
^2 + 2*(7*c^2 - 18*c*d + 7*d^2)*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + (cos(f*x + e)
^2 - 2*cos(f*x + e) - 4)*sin(f*x + e) - 2*cos(f*x + e) - 4)) + sqrt(c/a)*log(((c^4 - 28*c^3*d + 70*c^2*d^2 - 2
8*c*d^3 + d^4)*cos(f*x + e)^5 - (31*c^4 - 196*c^3*d + 154*c^2*d^2 - 4*c*d^3 - d^4)*cos(f*x + e)^4 + c^4 + 4*c^
3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 - 2*(81*c^4 - 252*c^3*d + 150*c^2*d^2 - 28*c*d^3 + d^4)*cos(f*x + e)^3 + 2*(79
*c^4 - 100*c^3*d + 74*c^2*d^2 - 4*c*d^3 - d^4)*cos(f*x + e)^2 - 8*((c^3 - 7*c^2*d + 7*c*d^2 - d^3)*cos(f*x + e
)^4 - 2*(5*c^3 - 14*c^2*d + 5*c*d^2)*cos(f*x + e)^3 + 51*c^3 - 59*c^2*d + 17*c*d^2 - d^3 - 2*(18*c^3 - 33*c^2*
d + 12*c*d^2 - d^3)*cos(f*x + e)^2 + 2*(13*c^3 - 14*c^2*d + 5*c*d^2)*cos(f*x + e) + ((c^3 - 7*c^2*d + 7*c*d^2
- d^3)*cos(f*x + e)^3 - 51*c^3 + 59*c^2*d - 17*c*d^2 + d^3 + (11*c^3 - 35*c^2*d + 17*c*d^2 - d^3)*cos(f*x + e)
^2 - (25*c^3 - 31*c^2*d + 7*c*d^2 - d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x +
e) + c)*sqrt(c/a) + (289*c^4 - 476*c^3*d + 230*c^2*d^2 - 28*c*d^3 + d^4)*cos(f*x + e) + ((c^4 - 28*c^3*d + 70
*c^2*d^2 - 28*c*d^3 + d^4)*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 + 32*(c^4 - 7*c^3*d + 7*
c^2*d^2 - c*d^3)*cos(f*x + e)^3 - 2*(65*c^4 - 140*c^3*d + 38*c^2*d^2 - 12*c*d^3 + d^4)*cos(f*x + e)^2 - 32*(9*
c^4 - 15*c^3*d + 7*c^2*d^2 - c*d^3)*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e)^5 + cos(f*x + e)^4 - 2*cos(f*x +
e)^3 - 2*cos(f*x + e)^2 + (cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sin(f*x + e) + cos(f*x + e) + 1)))/f, 1/4*(
sqrt(2)*sqrt((c - d)/a)*log(((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^3 + 4*sqrt(2)*((c - 3*d)*cos(f*x + e)^2 - (3
*c - d)*cos(f*x + e) + ((c - 3*d)*cos(f*x + e) + 4*c - 4*d)*sin(f*x + e) - 4*c + 4*d)*sqrt(a*sin(f*x + e) + a)
*sqrt(d*sin(f*x + e) + c)*sqrt((c - d)/a) - (13*c^2 - 22*c*d - 3*d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^2 -
2*(9*c^2 - 14*c*d + 9*d^2)*cos(f*x + e) + ((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^2 + 2
*(7*c^2 - 18*c*d + 7*d^2)*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + (cos(f*x + e)^2 - 2
*cos(f*x + e) - 4)*sin(f*x + e) - 2*cos(f*x + e) - 4)) + 2*sqrt(-c/a)*arctan(-1/4*((c^2 - 6*c*d + d^2)*cos(f*x
+ e)^2 - 9*c^2 + 6*c*d - d^2 + 8*(c^2 - c*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*
sqrt(-c/a)/((c^2*d - c*d^2)*cos(f*x + e)^3 - (c^3 - 3*c^2*d)*cos(f*x + e)*sin(f*x + e) + (2*c^3 - c^2*d + c*d^
2)*cos(f*x + e))))/f, 1/4*(2*sqrt(2)*sqrt(-(c - d)/a)*arctan(1/4*sqrt(2)*sqrt(a*sin(f*x + e) + a)*((c - 3*d)*s
in(f*x + e) - 3*c + d)*sqrt(d*sin(f*x + e) + c)*sqrt(-(c - d)/a)/((c*d - d^2)*cos(f*x + e)*sin(f*x + e) + (c^2
- c*d)*cos(f*x + e))) + sqrt(c/a)*log(((c^4 - 28*c^3*d + 70*c^2*d^2 - 28*c*d^3 + d^4)*cos(f*x + e)^5 - (31*c^
4 - 196*c^3*d + 154*c^2*d^2 - 4*c*d^3 - d^4)*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 - 2*(8
1*c^4 - 252*c^3*d + 150*c^2*d^2 - 28*c*d^3 + d^4)*cos(f*x + e)^3 + 2*(79*c^4 - 100*c^3*d + 74*c^2*d^2 - 4*c*d^
3 - d^4)*cos(f*x + e)^2 - 8*((c^3 - 7*c^2*d + 7*c*d^2 - d^3)*cos(f*x + e)^4 - 2*(5*c^3 - 14*c^2*d + 5*c*d^2)*c
os(f*x + e)^3 + 51*c^3 - 59*c^2*d + 17*c*d^2 - d^3 - 2*(18*c^3 - 33*c^2*d + 12*c*d^2 - d^3)*cos(f*x + e)^2 + 2
*(13*c^3 - 14*c^2*d + 5*c*d^2)*cos(f*x + e) + ((c^3 - 7*c^2*d + 7*c*d^2 - d^3)*cos(f*x + e)^3 - 51*c^3 + 59*c^
2*d - 17*c*d^2 + d^3 + (11*c^3 - 35*c^2*d + 17*c*d^2 - d^3)*cos(f*x + e)^2 - (25*c^3 - 31*c^2*d + 7*c*d^2 - d^
3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(c/a) + (289*c^4 - 476*c^
3*d + 230*c^2*d^2 - 28*c*d^3 + d^4)*cos(f*x + e) + ((c^4 - 28*c^3*d + 70*c^2*d^2 - 28*c*d^3 + d^4)*cos(f*x + e
)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 + 32*(c^4 - 7*c^3*d + 7*c^2*d^2 - c*d^3)*cos(f*x + e)^3 - 2*(6
5*c^4 - 140*c^3*d + 38*c^2*d^2 - 12*c*d^3 + d^4)*cos(f*x + e)^2 - 32*(9*c^4 - 15*c^3*d + 7*c^2*d^2 - c*d^3)*co
s(f*x + e))*sin(f*x + e))/(cos(f*x + e)^5 + cos(f*x + e)^4 - 2*cos(f*x + e)^3 - 2*cos(f*x + e)^2 + (cos(f*x +
e)^4 - 2*cos(f*x + e)^2 + 1)*sin(f*x + e) + cos(f*x + e) + 1)))/f, 1/2*(sqrt(2)*sqrt(-(c - d)/a)*arctan(1/4*sq
rt(2)*sqrt(a*sin(f*x + e) + a)*((c - 3*d)*sin(f*x + e) - 3*c + d)*sqrt(d*sin(f*x + e) + c)*sqrt(-(c - d)/a)/((
c*d - d^2)*cos(f*x + e)*sin(f*x + e) + (c^2 - c*d)*cos(f*x + e))) + sqrt(-c/a)*arctan(-1/4*((c^2 - 6*c*d + d^2
)*cos(f*x + e)^2 - 9*c^2 + 6*c*d - d^2 + 8*(c^2 - c*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x +
e) + c)*sqrt(-c/a)/((c^2*d - c*d^2)*cos(f*x + e)^3 - (c^3 - 3*c^2*d)*cos(f*x + e)*sin(f*x + e) + (2*c^3 - c^2
*d + c*d^2)*cos(f*x + e))))/f]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c + d \sin{\left (e + f x \right )}}}{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )} \sin{\left (e + f x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))**(1/2)/sin(f*x+e)/(a+a*sin(f*x+e))**(1/2),x)
[Out]
Integral(sqrt(c + d*sin(e + f*x))/(sqrt(a*(sin(e + f*x) + 1))*sin(e + f*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \sin \left (f x + e\right ) + c}}{\sqrt{a \sin \left (f x + e\right ) + a} \sin \left (f x + e\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(1/2)/sin(f*x+e)/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(d*sin(f*x + e) + c)/(sqrt(a*sin(f*x + e) + a)*sin(f*x + e)), x)