\(\int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx\) [48]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 165 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 a b}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 \left (3 a^2-2 b^2\right ) \cos (c+d x)}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 \left (3 a^2-2 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d e^4 \sqrt {\sin (c+d x)}} \] Output:

-2/5*(b+a*cos(d*x+c))*(a+b*cos(d*x+c))/d/e/(e*sin(d*x+c))^(5/2)-2/5*a*b/d/ 
e^3/(e*sin(d*x+c))^(1/2)-2/5*(3*a^2-2*b^2)*cos(d*x+c)/d/e^3/(e*sin(d*x+c)) 
^(1/2)+2/5*(3*a^2-2*b^2)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*s 
in(d*x+c))^(1/2)/d/e^4/sin(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.66 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=-\frac {8 a b+\left (7 a^2+2 b^2\right ) \cos (c+d x)-3 a^2 \cos (3 (c+d x))+2 b^2 \cos (3 (c+d x))-4 \left (3 a^2-2 b^2\right ) E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sin ^{\frac {5}{2}}(c+d x)}{10 d e (e \sin (c+d x))^{5/2}} \] Input:

Integrate[(a + b*Cos[c + d*x])^2/(e*Sin[c + d*x])^(7/2),x]
 

Output:

-1/10*(8*a*b + (7*a^2 + 2*b^2)*Cos[c + d*x] - 3*a^2*Cos[3*(c + d*x)] + 2*b 
^2*Cos[3*(c + d*x)] - 4*(3*a^2 - 2*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, 2 
]*Sin[c + d*x]^(5/2))/(d*e*(e*Sin[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3170, 27, 3042, 3148, 3042, 3116, 3042, 3121, 3042, 3119}

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.

\(\displaystyle \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{7/2}}dx\)

\(\Big \downarrow \) 3170

\(\displaystyle -\frac {2 \int -\frac {3 a^2+b \cos (c+d x) a-2 b^2}{2 (e \sin (c+d x))^{3/2}}dx}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 a^2+b \cos (c+d x) a-2 b^2}{(e \sin (c+d x))^{3/2}}dx}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {3 a^2-b \sin \left (c+d x-\frac {\pi }{2}\right ) a-2 b^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3148

\(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{(e \sin (c+d x))^{3/2}}dx-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{(e \sin (c+d x))^{3/2}}dx-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \left (-\frac {\int \sqrt {e \sin (c+d x)}dx}{e^2}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \left (-\frac {\int \sqrt {e \sin (c+d x)}dx}{e^2}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \left (-\frac {\sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \left (-\frac {\sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \left (-\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\)

Input:

Int[(a + b*Cos[c + d*x])^2/(e*Sin[c + d*x])^(7/2),x]
 

Output:

(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x]))/(5*d*e*(e*Sin[c + d*x])^(5/ 
2)) + ((-2*a*b)/(d*e*Sqrt[e*Sin[c + d*x]]) + (3*a^2 - 2*b^2)*((-2*Cos[c + 
d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqr 
t[e*Sin[c + d*x]])/(d*e^2*Sqrt[Sin[c + d*x]])))/(5*e^2)
 

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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3116
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3148
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + 
 Simp[a   Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && 
 (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
 

rule 3170
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x 
])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Simp[1/(g^2*(p + 1)) 
  Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + 
a^2*(p + 2) + a*b*(m + p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g 
}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[2*m, 2* 
p] || IntegerQ[m])
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(350\) vs. \(2(147)=294\).

Time = 3.34 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.13

method result size
default \(\frac {-\frac {4 a b}{5 e \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticE}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}-4 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticE}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}-3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}+2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}+6 a^{2} \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )-4 b^{2} \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )-8 a^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+2 b^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) \(351\)
parts \(\frac {a^{2} \left (6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticE}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+6 \sin \left (d x +c \right )^{5}-4 \sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )\right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {2 b^{2} \left (2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticE}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+2 \sin \left (d x +c \right )^{5}-3 \sin \left (d x +c \right )^{3}+\sin \left (d x +c \right )\right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {4 a b}{5 \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}} e d}\) \(360\)

Input:

int((a+cos(d*x+c)*b)^2/(e*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

(-4/5*a*b/e/(e*sin(d*x+c))^(5/2)+1/5/e^3*(6*(1-sin(d*x+c))^(1/2)*(2+2*sin( 
d*x+c))^(1/2)*sin(d*x+c)^(7/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2)) 
*a^2-4*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(7/2)*Ellipt 
icE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-3*(1-sin(d*x+c))^(1/2)*(2+2*sin( 
d*x+c))^(1/2)*sin(d*x+c)^(7/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2)) 
*a^2+2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(7/2)*Ellipt 
icF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2+6*a^2*cos(d*x+c)^4*sin(d*x+c)-4* 
b^2*cos(d*x+c)^4*sin(d*x+c)-8*a^2*cos(d*x+c)^2*sin(d*x+c)+2*b^2*cos(d*x+c) 
^2*sin(d*x+c))/sin(d*x+c)^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.31 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=-\frac {2 \, {\left ({\left ({\left (3 i \, a^{2} - 2 i \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {-\frac {1}{2} i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left ({\left (-3 i \, a^{2} + 2 i \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {\frac {1}{2} i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left ({\left (3 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - 2 \, a b - {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{5 \, {\left (d e^{4} \cos \left (d x + c\right )^{2} - d e^{4}\right )} \sin \left (d x + c\right )} \] Input:

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

-2/5*(((3*I*a^2 - 2*I*b^2)*cos(d*x + c)^2 - 3*I*a^2 + 2*I*b^2)*sqrt(-1/2*I 
*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + 
 c) + I*sin(d*x + c))) + ((-3*I*a^2 + 2*I*b^2)*cos(d*x + c)^2 + 3*I*a^2 - 
2*I*b^2)*sqrt(1/2*I*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInve 
rse(4, 0, cos(d*x + c) - I*sin(d*x + c))) + ((3*a^2 - 2*b^2)*cos(d*x + c)^ 
3 - 2*a*b - (4*a^2 - b^2)*cos(d*x + c))*sqrt(e*sin(d*x + c)))/((d*e^4*cos( 
d*x + c)^2 - d*e^4)*sin(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**2/(e*sin(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

integrate((b*cos(d*x + c) + a)^2/(e*sin(d*x + c))^(7/2), x)
 

Giac [F]

\[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

integrate((b*cos(d*x + c) + a)^2/(e*sin(d*x + c))^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}} \,d x \] Input:

int((a + b*cos(c + d*x))^2/(e*sin(c + d*x))^(7/2),x)
                                                                                    
                                                                                    
 

Output:

int((a + b*cos(c + d*x))^2/(e*sin(c + d*x))^(7/2), x)
 

Reduce [F]

\[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=\frac {\sqrt {e}\, \left (-4 \sqrt {\sin \left (d x +c \right )}\, a b +5 \left (\int \frac {\sqrt {\sin \left (d x +c \right )}}{\sin \left (d x +c \right )^{4}}d x \right ) \sin \left (d x +c \right )^{3} a^{2} d +5 \left (\int \frac {\sqrt {\sin \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\sin \left (d x +c \right )^{4}}d x \right ) \sin \left (d x +c \right )^{3} b^{2} d \right )}{5 \sin \left (d x +c \right )^{3} d \,e^{4}} \] Input:

int((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),x)
 

Output:

(sqrt(e)*( - 4*sqrt(sin(c + d*x))*a*b + 5*int(sqrt(sin(c + d*x))/sin(c + d 
*x)**4,x)*sin(c + d*x)**3*a**2*d + 5*int((sqrt(sin(c + d*x))*cos(c + d*x)* 
*2)/sin(c + d*x)**4,x)*sin(c + d*x)**3*b**2*d))/(5*sin(c + d*x)**3*d*e**4)