\(\int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx\) [8]

Optimal result

Integrand size = 35, antiderivative size = 251 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=-\frac {2 a^2 (2 B (3+n)+A (5+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {a^3 \left (2 B \left (9+13 n+4 n^2\right )+A \left (25+30 n+8 n^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+n,2+n,\sin (e+f x)\right ) \sqrt {1-\sin (e+f x)} (d \sin (e+f x))^{1+n}}{d f (5+2 n) \left (3+5 n+2 n^2\right ) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n)} \] Output:

-2*a^2*(2*B*(3+n)+A*(5+2*n))*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(3+2*n)/( 
5+2*n)/(a+a*sin(f*x+e))^(1/2)+a^3*(2*B*(4*n^2+13*n+9)+A*(8*n^2+30*n+25))*c 
os(f*x+e)*hypergeom([1/2, 1+n],[2+n],sin(f*x+e))*(1-sin(f*x+e))^(1/2)*(d*s 
in(f*x+e))^(1+n)/d/f/(5+2*n)/(2*n^2+5*n+3)/(a-a*sin(f*x+e))/(a+a*sin(f*x+e 
))^(1/2)-2*a*B*cos(f*x+e)*(d*sin(f*x+e))^(1+n)*(a+a*sin(f*x+e))^(1/2)/d/f/ 
(5+2*n)
 

Mathematica [A] (warning: unable to verify)

Time = 25.27 (sec) , antiderivative size = 478, normalized size of antiderivative = 1.90 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\frac {2^{1+n} \sec \left (\frac {1}{2} (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n (a (1+\sin (e+f x)))^{3/2} \tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}\right )^n \left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^n \left (\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {7}{2}+n,\frac {3+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{1+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(3 A+2 B) \operatorname {Hypergeometric2F1}\left (\frac {2+n}{2},\frac {7}{2}+n,\frac {4+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{2+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {2 (2 A+3 B) \operatorname {Hypergeometric2F1}\left (\frac {3+n}{2},\frac {7}{2}+n,\frac {5+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{3+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {2 (2 A+3 B) \operatorname {Hypergeometric2F1}\left (\frac {7}{2}+n,\frac {4+n}{2},\frac {6+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{4+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(3 A+2 B) \operatorname {Hypergeometric2F1}\left (\frac {7}{2}+n,\frac {5+n}{2},\frac {7+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{5+n}+\frac {A \operatorname {Hypergeometric2F1}\left (\frac {7}{2}+n,\frac {6+n}{2},\frac {8+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{6+n}\right )\right )\right )\right )\right )}{f \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \] Input:

Integrate[(d*Sin[e + f*x])^n*(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x 
]),x]
 

Output:

(2^(1 + n)*Sec[(e + f*x)/2]*(d*Sin[e + f*x])^n*(a*(1 + Sin[e + f*x]))^(3/2 
)*Tan[(e + f*x)/2]*(Tan[(e + f*x)/2]/(1 + Tan[(e + f*x)/2]^2))^n*(1 + Tan[ 
(e + f*x)/2]^2)^n*((A*Hypergeometric2F1[(1 + n)/2, 7/2 + n, (3 + n)/2, -Ta 
n[(e + f*x)/2]^2])/(1 + n) + Tan[(e + f*x)/2]*(((3*A + 2*B)*Hypergeometric 
2F1[(2 + n)/2, 7/2 + n, (4 + n)/2, -Tan[(e + f*x)/2]^2])/(2 + n) + Tan[(e 
+ f*x)/2]*((2*(2*A + 3*B)*Hypergeometric2F1[(3 + n)/2, 7/2 + n, (5 + n)/2, 
 -Tan[(e + f*x)/2]^2])/(3 + n) + Tan[(e + f*x)/2]*((2*(2*A + 3*B)*Hypergeo 
metric2F1[7/2 + n, (4 + n)/2, (6 + n)/2, -Tan[(e + f*x)/2]^2])/(4 + n) + T 
an[(e + f*x)/2]*(((3*A + 2*B)*Hypergeometric2F1[7/2 + n, (5 + n)/2, (7 + n 
)/2, -Tan[(e + f*x)/2]^2])/(5 + n) + (A*Hypergeometric2F1[7/2 + n, (6 + n) 
/2, (8 + n)/2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2])/(6 + n)))))))/(f*Sqr 
t[Sec[(e + f*x)/2]^2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sin[e + f*x] 
^n)
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.90, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3455, 27, 3042, 3460, 3042, 3255, 77, 75}

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.

\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x)) (d \sin (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x)) (d \sin (e+f x))^ndx\)

\(\Big \downarrow \) 3455

\(\displaystyle \frac {2 \int \frac {1}{2} (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a} \left (a d \left (2 B (n+1)+2 A \left (n+\frac {5}{2}\right )\right )+a d (2 B (n+3)+A (2 n+5)) \sin (e+f x)\right )dx}{d (2 n+5)}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a} (a d (2 B (n+1)+A (2 n+5))+a d (2 B (n+3)+A (2 n+5)) \sin (e+f x))dx}{d (2 n+5)}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a} (a d (2 B (n+1)+A (2 n+5))+a d (2 B (n+3)+A (2 n+5)) \sin (e+f x))dx}{d (2 n+5)}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {\frac {a d \left (A \left (8 n^2+30 n+25\right )+2 B \left (4 n^2+13 n+9\right )\right ) \int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a}dx}{2 n+3}-\frac {2 a^2 (A (2 n+5)+2 B (n+3)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a d \left (A \left (8 n^2+30 n+25\right )+2 B \left (4 n^2+13 n+9\right )\right ) \int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a}dx}{2 n+3}-\frac {2 a^2 (A (2 n+5)+2 B (n+3)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)}\)

\(\Big \downarrow \) 3255

\(\displaystyle \frac {\frac {a^3 d \left (A \left (8 n^2+30 n+25\right )+2 B \left (4 n^2+13 n+9\right )\right ) \cos (e+f x) \int \frac {(d \sin (e+f x))^n}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f (2 n+3) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 (A (2 n+5)+2 B (n+3)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)}\)

\(\Big \downarrow \) 77

\(\displaystyle \frac {\frac {a^3 d \left (A \left (8 n^2+30 n+25\right )+2 B \left (4 n^2+13 n+9\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f (2 n+3) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 (A (2 n+5)+2 B (n+3)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {-\frac {2 a^2 d \left (A \left (8 n^2+30 n+25\right )+2 B \left (4 n^2+13 n+9\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 (A (2 n+5)+2 B (n+3)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)}\)

Input:

Int[(d*Sin[e + f*x])^n*(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]),x]
 

Output:

(-2*a*B*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n)*Sqrt[a + a*Sin[e + f*x]])/(d 
*f*(5 + 2*n)) + ((-2*a^2*d*(2*B*(9 + 13*n + 4*n^2) + A*(25 + 30*n + 8*n^2) 
)*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, 1 - Sin[e + f*x]]*(d*Sin[e 
+ f*x])^n)/(f*(3 + 2*n)*Sin[e + f*x]^n*Sqrt[a + a*Sin[e + f*x]]) - (2*a^2* 
(2*B*(3 + n) + A*(5 + 2*n))*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(f*(3 + 
 2*n)*Sqrt[a + a*Sin[e + f*x]]))/(d*(5 + 2*n))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 77

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((-b)*(c/ 
d))^IntPart[m]*((b*x)^FracPart[m]/((-d)*(x/c))^FracPart[m])   Int[((-d)*(x/ 
c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && 
 !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0]
 

rule 3042

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

rule 3255

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + 
 f*x]]*Sqrt[a - b*Sin[e + f*x]]))   Subst[Int[(c + d*x)^n/Sqrt[a - b*x], x] 
, x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 
 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !IntegerQ[2*n]
 

rule 3455

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 
 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1 
] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3460

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] 
 /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
 b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]
 

Maple [F]

\[\int \left (d \sin \left (f x +e \right )\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (A +B \sin \left (f x +e \right )\right )d x\]

Input:

int((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)),x)
 

Output:

int((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)),x)
 

Fricas [F]

\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \] Input:

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)),x, algo 
rithm="fricas")
 

Output:

integral(-(B*a*cos(f*x + e)^2 - (A + B)*a*sin(f*x + e) - (A + B)*a)*sqrt(a 
*sin(f*x + e) + a)*(d*sin(f*x + e))^n, x)
 

Sympy [F]

\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (d \sin {\left (e + f x \right )}\right )^{n} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \] Input:

integrate((d*sin(f*x+e))**n*(a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e)),x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**(3/2)*(d*sin(e + f*x))**n*(A + B*sin(e + 
f*x)), x)
 

Maxima [F]

\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \] Input:

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)),x, algo 
rithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e)) 
^n, x)
 

Giac [F]

\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \] Input:

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)),x, algo 
rithm="giac")
 

Output:

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e)) 
^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \] Input:

int((d*sin(e + f*x))^n*(A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2),x)
 

Output:

int((d*sin(e + f*x))^n*(A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=d^{n} \sqrt {a}\, a \left (\left (\int \sin \left (f x +e \right )^{n} \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b +\left (\int \sin \left (f x +e \right )^{n} \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a +\left (\int \sin \left (f x +e \right )^{n} \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b +\left (\int \sin \left (f x +e \right )^{n} \sqrt {\sin \left (f x +e \right )+1}d x \right ) a \right ) \] Input:

int((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)),x)
 

Output:

d**n*sqrt(a)*a*(int(sin(e + f*x)**n*sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2 
,x)*b + int(sin(e + f*x)**n*sqrt(sin(e + f*x) + 1)*sin(e + f*x),x)*a + int 
(sin(e + f*x)**n*sqrt(sin(e + f*x) + 1)*sin(e + f*x),x)*b + int(sin(e + f* 
x)**n*sqrt(sin(e + f*x) + 1),x)*a)