\(\int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx\) [300]

Optimal result

Integrand size = 27, antiderivative size = 169 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {2 \sqrt {e} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {2 \sqrt {e} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d (a+a \cos (c+d x)+a \sin (c+d x))} \] Output:

2*e^(1/2)*arcsinh((e*cos(d*x+c))^(1/2)/e^(1/2))*(1+cos(d*x+c))^(1/2)*(a+a* 
sin(d*x+c))^(1/2)/d/(a+a*cos(d*x+c)+a*sin(d*x+c))+2*e^(1/2)*arctan(e^(1/2) 
*sin(d*x+c)/(e*cos(d*x+c))^(1/2)/(1+cos(d*x+c))^(1/2))*(1+cos(d*x+c))^(1/2 
)*(a+a*sin(d*x+c))^(1/2)/d/(a+a*cos(d*x+c)+a*sin(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \sqrt [4]{2} (e \cos (c+d x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{4},\frac {7}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{3 d e \sqrt [4]{1+\sin (c+d x)} \sqrt {a (1+\sin (c+d x))}} \] Input:

Integrate[Sqrt[e*Cos[c + d*x]]/Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(-2*2^(1/4)*(e*Cos[c + d*x])^(3/2)*Hypergeometric2F1[3/4, 3/4, 7/4, (1 - S 
in[c + d*x])/2])/(3*d*e*(1 + Sin[c + d*x])^(1/4)*Sqrt[a*(1 + Sin[c + d*x]) 
])
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3163, 3042, 25, 3254, 216, 3312, 63, 222}

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.

\(\displaystyle \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a \sin (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a \sin (c+d x)+a}}dx\)

\(\Big \downarrow \) 3163

\(\displaystyle \frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\cos (c+d x)+1}}{\sqrt {e \cos (c+d x)}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}+\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\)

\(\Big \downarrow \) 3254

\(\displaystyle \frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {2 e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x)}{\cos (c+d x)+1}+1}d\left (-\frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3312

\(\displaystyle \frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 63

\(\displaystyle \frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\sqrt {\cos (c+d x)+1}}d\sqrt {e \cos (c+d x)}}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\)

Input:

Int[Sqrt[e*Cos[c + d*x]]/Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(2*Sqrt[e]*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sq 
rt[a + a*Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*Sin[c + d*x])) + (2*Sqr 
t[e]*ArcTan[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + 
d*x]])]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a + a*Cos[c + 
 d*x] + a*Sin[c + d*x]))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 63

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b   S 
ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x 
] && GtQ[c, 0]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042

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

rule 3163

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[g*Sqrt[1 + Cos[e + f*x]]*(Sqrt[a + b*Sin[e + f*x 
]]/(a + a*Cos[e + f*x] + b*Sin[e + f*x]))   Int[Sqrt[1 + Cos[e + f*x]]/Sqrt 
[g*Cos[e + f*x]], x], x] - Simp[g*Sqrt[1 + Cos[e + f*x]]*(Sqrt[a + b*Sin[e 
+ f*x]]/(b + b*Cos[e + f*x] + a*Sin[e + f*x]))   Int[Sin[e + f*x]/(Sqrt[g*C 
os[e + f*x]]*Sqrt[1 + Cos[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, g}, x] & 
& EqQ[a^2 - b^2, 0]
 

rule 3254

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(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 3312

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( 
c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f)   Su 
bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, 
b, c, d, e, f, m, n}, x]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 4.01 (sec) , antiderivative size = 827, normalized size of antiderivative = 4.89

method	result	size
default	\(\text {Expression too large to display}\)	\(827\)

Input:

int((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/d/(2*2^(1/2)+3)^(1/2)/(1+2^(1/2))*(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2) 
/e/(2*cos(1/2*d*x+1/2*c)^2-1)*(cos(1/2*d*x+1/2*c)+1)^2*(-2*e^(1/2)*(e*(2*c 
os(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(1+2^(1/2))*arctanh 
(2^(1/2)*e^(1/2)*cos(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)+1)/(e*(2*cos(1/2*d 
*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))-2^(1/2)*e^(1/2)*arctanh(2^ 
(1/2)*e^(1/2)*cos(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)+1)/(e*(2*cos(1/2*d*x+ 
1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))*(e*(2*cos(1/2*d*x+1/2*c)^2-1) 
/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(1+2^(1/2))+2*e*2^(1/2)*((2^(1/2)*cos(1/2 
*d*x+1/2*c)-2^(1/2)+2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1))^(1/2)* 
(-2*(2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)-2*cos(1/2*d*x+1/2*c)+1)/(cos(1/2*d 
*x+1/2*c)+1))^(1/2)*EllipticF((1+2^(1/2))*(csc(1/2*d*x+1/2*c)-cot(1/2*d*x+ 
1/2*c)),-2*2^(1/2)+3)*(1+2^(1/2))-8*((2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)+2 
*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1))^(1/2)*(-2*(2^(1/2)*cos(1/2* 
d*x+1/2*c)-2^(1/2)-2*cos(1/2*d*x+1/2*c)+1)/(cos(1/2*d*x+1/2*c)+1))^(1/2)*E 
llipticPi((2*2^(1/2)+3)^(1/2)*(csc(1/2*d*x+1/2*c)-cot(1/2*d*x+1/2*c)),-1/( 
2*2^(1/2)+3),(-2*2^(1/2)+3)^(1/2)/(2*2^(1/2)+3)^(1/2))*e-4*e*2^(1/2)*((2^( 
1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)+2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c 
)+1))^(1/2)*(-2*(2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)-2*cos(1/2*d*x+1/2*c)+1 
)/(cos(1/2*d*x+1/2*c)+1))^(1/2)*EllipticPi((2*2^(1/2)+3)^(1/2)*(csc(1/2*d* 
x+1/2*c)-cot(1/2*d*x+1/2*c)),-1/(2*2^(1/2)+3),(-2*2^(1/2)+3)^(1/2)/(2*2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (147) = 294\).

Time = 0.15 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.89 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {4 \, \sqrt {2} \sqrt {\frac {e}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {\frac {e}{a}} \sin \left (d x + c\right )}{e \cos \left (d x + c\right )^{2} + e \cos \left (d x + c\right ) \sin \left (d x + c\right ) + e \cos \left (d x + c\right )}\right ) + \sqrt {2} \sqrt {\frac {e}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {\frac {e}{a}} {\left (\cos \left (d x + c\right ) + 1\right )} + 2 \, e \cos \left (d x + c\right )^{2} + 3 \, e \cos \left (d x + c\right ) + {\left (2 \, e \cos \left (d x + c\right ) + e\right )} \sin \left (d x + c\right ) + e}{\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1}\right ) - \sqrt {2} \sqrt {\frac {e}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {\frac {e}{a}} {\left (\cos \left (d x + c\right ) + 1\right )} - 2 \, e \cos \left (d x + c\right )^{2} - 3 \, e \cos \left (d x + c\right ) - {\left (2 \, e \cos \left (d x + c\right ) + e\right )} \sin \left (d x + c\right ) - e}{\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1}\right )}{4 \, d} \] Input:

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas 
")
 

Output:

1/4*(4*sqrt(2)*sqrt(e/a)*arctan(sqrt(2)*sqrt(e*cos(d*x + c))*sqrt(a*sin(d* 
x + c) + a)*sqrt(e/a)*sin(d*x + c)/(e*cos(d*x + c)^2 + e*cos(d*x + c)*sin( 
d*x + c) + e*cos(d*x + c))) + sqrt(2)*sqrt(e/a)*log((2*sqrt(2)*sqrt(e*cos( 
d*x + c))*sqrt(a*sin(d*x + c) + a)*sqrt(e/a)*(cos(d*x + c) + 1) + 2*e*cos( 
d*x + c)^2 + 3*e*cos(d*x + c) + (2*e*cos(d*x + c) + e)*sin(d*x + c) + e)/( 
cos(d*x + c) + sin(d*x + c) + 1)) - sqrt(2)*sqrt(e/a)*log(-(2*sqrt(2)*sqrt 
(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)*sqrt(e/a)*(cos(d*x + c) + 1) - 2 
*e*cos(d*x + c)^2 - 3*e*cos(d*x + c) - (2*e*cos(d*x + c) + e)*sin(d*x + c) 
 - e)/(cos(d*x + c) + sin(d*x + c) + 1)))/d
 

Sympy [F]

\[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sqrt {e \cos {\left (c + d x \right )}}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \] Input:

integrate((e*cos(d*x+c))**(1/2)/(a+a*sin(d*x+c))**(1/2),x)
 

Output:

Integral(sqrt(e*cos(c + d*x))/sqrt(a*(sin(c + d*x) + 1)), x)
 

Maxima [F]

\[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\sqrt {e \cos \left (d x + c\right )}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \] Input:

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima 
")
 

Output:

integrate(sqrt(e*cos(d*x + c))/sqrt(a*sin(d*x + c) + a), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sqrt {e\,\cos \left (c+d\,x\right )}}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \] Input:

int((e*cos(c + d*x))^(1/2)/(a + a*sin(c + d*x))^(1/2),x)
 

Output:

int((e*cos(c + d*x))^(1/2)/(a + a*sin(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {e}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\sin \left (d x +c \right )+1}d x \right )}{a} \] Input:

int((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(1/2),x)
 

Output:

(sqrt(e)*sqrt(a)*int((sqrt(sin(c + d*x) + 1)*sqrt(cos(c + d*x)))/(sin(c + 
d*x) + 1),x))/a