\(\int \frac {(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx\) [1380]

Optimal result

Integrand size = 33, antiderivative size = 501 \[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {a^2 \left (-a^2+b^2\right )^{3/4} g^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} f}-\frac {a^2 \left (-a^2+b^2\right )^{3/4} g^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} f}+\frac {2 a^2 g (g \cos (e+f x))^{3/2}}{3 b^3 f}-\frac {2 (g \cos (e+f x))^{7/2}}{7 b f g}+\frac {2 a^3 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \sqrt {\cos (e+f x)}}-\frac {6 a g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt {\cos (e+f x)}}-\frac {a^3 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^3 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 a g (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f} \] Output:

a^2*(-a^2+b^2)^(3/4)*g^(5/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2 
)^(1/4)/g^(1/2))/b^(9/2)/f-a^2*(-a^2+b^2)^(3/4)*g^(5/2)*arctanh(b^(1/2)*(g 
*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(9/2)/f+2/3*a^2*g*(g*cos(f* 
x+e))^(3/2)/b^3/f-2/7*(g*cos(f*x+e))^(7/2)/b/f/g+2*a^3*g^2*(g*cos(f*x+e))^ 
(1/2)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/b^4/f/cos(f*x+e)^(1/2)-6/5*a*g 
^2*(g*cos(f*x+e))^(1/2)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/b^2/f/cos(f* 
x+e)^(1/2)-a^3*(a^2-b^2)*g^3*cos(f*x+e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e 
),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))/b^5/(b-(-a^2+b^2)^(1/2))/f/(g*cos(f*x+ 
e))^(1/2)-a^3*(a^2-b^2)*g^3*cos(f*x+e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e) 
,2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))/b^5/(b+(-a^2+b^2)^(1/2))/f/(g*cos(f*x+e 
))^(1/2)-2/5*a*g*(g*cos(f*x+e))^(3/2)*sin(f*x+e)/b^2/f
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 26.51 (sec) , antiderivative size = 824, normalized size of antiderivative = 1.64 \[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {a (g \cos (e+f x))^{5/2} \left (-\frac {4 a b \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (e+f x)}{\sqrt {1-\cos ^2(e+f x)} (a+b \sin (e+f x))}-\frac {\left (5 a^2-3 b^2\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \sin ^2(e+f x)}{12 b^{3/2} \left (-a^2+b^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}\right )}{5 b^3 f \cos ^{\frac {5}{2}}(e+f x)}+\frac {(g \cos (e+f x))^{5/2} \sec ^2(e+f x) \left (-\frac {\left (-28 a^2+9 b^2\right ) \cos (e+f x)}{42 b^3}-\frac {\cos (3 (e+f x))}{14 b}-\frac {a \sin (2 (e+f x))}{5 b^2}\right )}{f} \] Input:

Integrate[((g*Cos[e + f*x])^(5/2)*Sin[e + f*x]^2)/(a + b*Sin[e + f*x]),x]
 

Output:

(a*(g*Cos[e + f*x])^(5/2)*((-4*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((a*Ap 
pellF1[3/4, 1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2) 
]*Cos[e + f*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I 
)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)* 
Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - ( 
1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]] + 
 Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f* 
x]] + I*b*Cos[e + f*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)))*Sin[e + f*x])/(Sqr 
t[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) - ((5*a^2 - 3*b^2)*(a + b*Sqrt 
[1 - Cos[e + f*x]^2])*(8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Cos[e + f*x]^ 
2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(3/2) + 3*Sqrt[2]*a*(a^ 
2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b 
^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2) 
^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos 
[e + f*x]] + b*Cos[e + f*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 
- b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]]))*Sin[e + f*x]^2)/(12*b^ 
(3/2)*(-a^2 + b^2)*(1 - Cos[e + f*x]^2)*(a + b*Sin[e + f*x]))))/(5*b^3*f*C 
os[e + f*x]^(5/2)) + ((g*Cos[e + f*x])^(5/2)*Sec[e + f*x]^2*(-1/42*((-28*a 
^2 + 9*b^2)*Cos[e + f*x])/b^3 - Cos[3*(e + f*x)]/(14*b) - (a*Sin[2*(e + f* 
x)])/(5*b^2)))/f
 

Rubi [A] (verified)

Time = 1.43 (sec) , antiderivative size = 501, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3042, 3377, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sin ^2(e+f x) (g \cos (e+f x))^{5/2}}{a+b \sin (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^2 (g \cos (e+f x))^{5/2}}{a+b \sin (e+f x)}dx\)

\(\Big \downarrow \) 3377

\(\displaystyle \int \left (\frac {a^2 (g \cos (e+f x))^{5/2}}{b^2 (a+b \sin (e+f x))}-\frac {a (g \cos (e+f x))^{5/2}}{b^2}+\frac {\sin (e+f x) (g \cos (e+f x))^{5/2}}{b}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 a^3 g^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b^4 f \sqrt {\cos (e+f x)}}+\frac {a^2 g^{5/2} \left (b^2-a^2\right )^{3/4} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f}-\frac {a^2 g^{5/2} \left (b^2-a^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f}+\frac {2 a^2 g (g \cos (e+f x))^{3/2}}{3 b^3 f}-\frac {a^3 g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^5 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^5 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}-\frac {6 a g^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{5 b^2 f \sqrt {\cos (e+f x)}}-\frac {2 a g \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b^2 f}-\frac {2 (g \cos (e+f x))^{7/2}}{7 b f g}\)

Input:

Int[((g*Cos[e + f*x])^(5/2)*Sin[e + f*x]^2)/(a + b*Sin[e + f*x]),x]
 

Output:

(a^2*(-a^2 + b^2)^(3/4)*g^(5/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a 
^2 + b^2)^(1/4)*Sqrt[g])])/(b^(9/2)*f) - (a^2*(-a^2 + b^2)^(3/4)*g^(5/2)*A 
rcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(9 
/2)*f) + (2*a^2*g*(g*Cos[e + f*x])^(3/2))/(3*b^3*f) - (2*(g*Cos[e + f*x])^ 
(7/2))/(7*b*f*g) + (2*a^3*g^2*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 
2])/(b^4*f*Sqrt[Cos[e + f*x]]) - (6*a*g^2*Sqrt[g*Cos[e + f*x]]*EllipticE[( 
e + f*x)/2, 2])/(5*b^2*f*Sqrt[Cos[e + f*x]]) - (a^3*(a^2 - b^2)*g^3*Sqrt[C 
os[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^ 
5*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - (a^3*(a^2 - b^2)*g^3*Sq 
rt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2]) 
/(b^5*(b + Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - (2*a*g*(g*Cos[e + f 
*x])^(3/2)*Sin[e + f*x])/(5*b^2*f)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3377

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a 
_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + 
 f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, 
 g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 
2, 0])
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 11.74 (sec) , antiderivative size = 1144, normalized size of antiderivative = 2.28

method	result	size
default	\(\text {Expression too large to display}\)	\(1144\)

Input:

int((g*cos(f*x+e))^(5/2)*sin(f*x+e)^2/(a+b*sin(f*x+e)),x,method=_RETURNVER 
BOSE)
 

Output:

(16*g^3*b*(1/168/b^4/g*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(24*sin(1/2*f*x 
+1/2*e)^6*b^2-36*sin(1/2*f*x+1/2*e)^4*b^2-14*sin(1/2*f*x+1/2*e)^2*a^2+18*s 
in(1/2*f*x+1/2*e)^2*b^2+7*a^2-3*b^2)-1/64*a^2/b^6*(a^2-b^2)/(g^2*(a^2-b^2) 
/b^2)^(1/4)*2^(1/2)*(ln((2*g*cos(1/2*f*x+1/2*e)^2-g-(g^2*(a^2-b^2)/b^2)^(1 
/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2))/ 
(2*g*cos(1/2*f*x+1/2*e)^2-g+(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2 
*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan(2^(1/2)/(g^2*( 
a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+1)-2*arctan(-2^(1/2 
)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+1)))+1/40*( 
g*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g^3*a*(128*cos(1 
/2*f*x+1/2*e)^7*b^4-256*cos(1/2*f*x+1/2*e)^5*b^4+160*cos(1/2*f*x+1/2*e)^3* 
b^4+80*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*Ellip 
ticE(cos(1/2*f*x+1/2*e),2^(1/2))*a^2*b^2-48*(sin(1/2*f*x+1/2*e)^2)^(1/2)*( 
1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*b^4- 
32*cos(1/2*f*x+1/2*e)*b^4+5*sum((a^2-b^2)/_alpha*(8*(g*(2*_alpha^2*b^2+a^2 
-2*b^2)/b^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2) 
^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*_alp 
ha^3*b^2-8*b^2*_alpha*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e) 
^2)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*( 
g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)+a^2*2^(1/2)*arctanh(1/2*g*(4*_a...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))^(5/2)*sin(f*x+e)^2/(a+b*sin(f*x+e)),x, algorithm= 
"fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))**(5/2)*sin(f*x+e)**2/(a+b*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{b \sin \left (f x + e\right ) + a} \,d x } \] Input:

integrate((g*cos(f*x+e))^(5/2)*sin(f*x+e)^2/(a+b*sin(f*x+e)),x, algorithm= 
"maxima")
 

Output:

integrate((g*cos(f*x + e))^(5/2)*sin(f*x + e)^2/(b*sin(f*x + e) + a), x)
 

Giac [F]

\[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{b \sin \left (f x + e\right ) + a} \,d x } \] Input:

integrate((g*cos(f*x+e))^(5/2)*sin(f*x+e)^2/(a+b*sin(f*x+e)),x, algorithm= 
"giac")
 

Output:

integrate((g*cos(f*x + e))^(5/2)*sin(f*x + e)^2/(b*sin(f*x + e) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^2\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}}{a+b\,\sin \left (e+f\,x\right )} \,d x \] Input:

int((sin(e + f*x)^2*(g*cos(e + f*x))^(5/2))/(a + b*sin(e + f*x)),x)
 

Output:

int((sin(e + f*x)^2*(g*cos(e + f*x))^(5/2))/(a + b*sin(e + f*x)), x)
 

Reduce [F]

\[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\sqrt {g}\, \left (\int \frac {\sqrt {\cos \left (f x +e \right )}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right ) b +a}d x \right ) g^{2} \] Input:

int((g*cos(f*x+e))^(5/2)*sin(f*x+e)^2/(a+b*sin(f*x+e)),x)
 

Output:

sqrt(g)*int((sqrt(cos(e + f*x))*cos(e + f*x)**2*sin(e + f*x)**2)/(sin(e + 
f*x)*b + a),x)*g**2