\(\int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [248]

Optimal result

Integrand size = 24, antiderivative size = 751 \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}-\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}+\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}+\frac {3 a^2 f \log (a+b \sin (c+d x))}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}+\frac {3 a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}-\frac {3 a f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}+\frac {3 a f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a f}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \] Output:

3/2*I*a^3*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2) 
^(5/2)/d-3/2*I*a*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a 
^2-b^2)^(3/2)/d-3/2*I*a^3*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/ 
2)))/b/(a^2-b^2)^(5/2)/d+3/2*I*a*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b 
^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d+3/2*a^2*f*ln(a+b*sin(d*x+c))/b/(a^2-b^2)^2 
/d^2-f*ln(a+b*sin(d*x+c))/b/(a^2-b^2)/d^2+3/2*a^3*f*polylog(2,I*b*exp(I*(d 
*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^2-3/2*a*f*polylog(2,I*b*ex 
p(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2-3/2*a^3*f*polylog( 
2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^2+3/2*a*f*po 
lylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2-1/2* 
a*(f*x+e)*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^2-1/2*a*f/b/(a^2-b^2)/d^ 
2/(a+b*sin(d*x+c))-3/2*a^2*(f*x+e)*cos(d*x+c)/(a^2-b^2)^2/d/(a+b*sin(d*x+c 
))+(f*x+e)*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2666\) vs. \(2(751)=1502\).

Time = 19.80 (sec) , antiderivative size = 2666, normalized size of antiderivative = 3.55 \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Result too large to show} \] Input:

Integrate[((e + f*x)*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

(-(a*d*e*Cos[c + d*x]) + a*c*f*Cos[c + d*x] - a*f*(c + d*x)*Cos[c + d*x])/ 
(2*(a - b)*(a + b)*d^2*(a + b*Sin[c + d*x])^2) + (-(a^3*f) + a*b^2*f - a^2 
*b*d*e*Cos[c + d*x] - 2*b^3*d*e*Cos[c + d*x] + a^2*b*c*f*Cos[c + d*x] + 2* 
b^3*c*f*Cos[c + d*x] - a^2*b*f*(c + d*x)*Cos[c + d*x] - 2*b^3*f*(c + d*x)* 
Cos[c + d*x])/(2*(a - b)^2*b*(a + b)^2*d^2*(a + b*Sin[c + d*x])) + (((-3*a 
*b*d*e*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 
 (3*a*b*c*f*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b 
^2] - ((a^2 + 2*b^2)*f*Log[Sec[(c + d*x)/2]^2])/(2*b) + (a^2*f*Log[Sec[(c 
+ d*x)/2]^2*(a + b*Sin[c + d*x])])/(2*b) + b*f*Log[Sec[(c + d*x)/2]^2*(a + 
 b*Sin[c + d*x])] + (((3*I)/2)*a*b*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - 
Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])])/Sqrt 
[-a^2 + b^2] - (((3*I)/2)*a*b*f*Log[1 - I*Tan[(c + d*x)/2]]*Log[-((b - Sqr 
t[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a - b + Sqrt[-a^2 + b^2]))])/Sqrt[- 
a^2 + b^2] + (((3*I)/2)*a*b*f*Log[1 - I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a 
^2 + b^2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^2 + b^2])])/Sqrt[-a^ 
2 + b^2] - (((3*I)/2)*a*b*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 
 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])])/Sqrt[-a^2 + b 
^2] + (((3*I)/2)*a*b*f*PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + 
 Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] - (((3*I)/2)*a*b*f*PolyLog[2, (a*(1 
 + I*Tan[(c + d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^...
 

Rubi [A] (verified)

Time = 3.25 (sec) , antiderivative size = 751, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {7293, 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.

\(\displaystyle \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 a f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {3 a f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{3/2}}-\frac {a f}{2 b d^2 \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {3 a^2 f \log (a+b \sin (c+d x))}{2 b d^2 \left (a^2-b^2\right )^2}-\frac {f \log (a+b \sin (c+d x))}{b d^2 \left (a^2-b^2\right )}-\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d \left (a^2-b^2\right )^{3/2}}+\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{2 b d \left (a^2-b^2\right )^{3/2}}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {a (e+f x) \cos (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}+\frac {(e+f x) \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {3 a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{5/2}}-\frac {3 a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{5/2}}+\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d \left (a^2-b^2\right )^{5/2}}-\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{2 b d \left (a^2-b^2\right )^{5/2}}\)

Input:

Int[((e + f*x)*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

(((3*I)/2)*a^3*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2 
])])/(b*(a^2 - b^2)^(5/2)*d) - (((3*I)/2)*a*(e + f*x)*Log[1 - (I*b*E^(I*(c 
 + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d) - (((3*I)/2)*a^3 
*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - 
 b^2)^(5/2)*d) + (((3*I)/2)*a*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + 
 Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d) + (3*a^2*f*Log[a + b*Sin[c + d 
*x]])/(2*b*(a^2 - b^2)^2*d^2) - (f*Log[a + b*Sin[c + d*x]])/(b*(a^2 - b^2) 
*d^2) + (3*a^3*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/ 
(2*b*(a^2 - b^2)^(5/2)*d^2) - (3*a*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - 
 Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(3/2)*d^2) - (3*a^3*f*PolyLog[2, (I*b 
*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(5/2)*d^2) + (3 
*a*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(2*b*(a^2 - 
b^2)^(3/2)*d^2) - (a*(e + f*x)*Cos[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sin[c 
 + d*x])^2) - (a*f)/(2*b*(a^2 - b^2)*d^2*(a + b*Sin[c + d*x])) - (3*a^2*(e 
 + f*x)*Cos[c + d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])) + ((e + f*x 
)*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x]))
 

Defintions of rubi rules used

rule 2009

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

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 6.60 (sec) , antiderivative size = 1084, normalized size of antiderivative = 1.44

method	result	size
risch	\(\text {Expression too large to display}\)	\(1084\)

Input:

int((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

I*(5*I*b^3*a*d*e*exp(I*(d*x+c))-2*I*b^2*f*exp(2*I*(d*x+c))*a^2-3*I*b^3*a*d 
*f*x*exp(3*I*(d*x+c))+4*I*b*a^3*d*f*x*exp(I*(d*x+c))+2*a^4*d*f*x*exp(2*I*( 
d*x+c))+5*exp(2*I*(d*x+c))*a^2*b^2*d*f*x+2*b^4*d*f*x*exp(2*I*(d*x+c))-3*I* 
b^3*a*d*e*exp(3*I*(d*x+c))+2*I*a^4*f*exp(2*I*(d*x+c))+5*I*b^3*a*d*f*x*exp( 
I*(d*x+c))+4*I*b*a^3*d*e*exp(I*(d*x+c))+2*a^4*d*e*exp(2*I*(d*x+c))+exp(3*I 
*(d*x+c))*a^3*b*f+5*exp(2*I*(d*x+c))*a^2*b^2*d*e-b^3*a*f*exp(3*I*(d*x+c))+ 
2*b^4*d*e*exp(2*I*(d*x+c))-a^2*b^2*d*f*x-2*b^4*d*f*x-b*a^3*f*exp(I*(d*x+c) 
)-a^2*b^2*d*e+b^3*a*f*exp(I*(d*x+c))-2*b^4*d*e)/(-I*b*exp(2*I*(d*x+c))+2*a 
*exp(I*(d*x+c))+I*b)^2/(a^2-b^2)^2/d^2/b-1/b/d^2/(-a^2+b^2)^2*a^2*f*ln(exp 
(I*(d*x+c)))+1/2/b/d^2/(-a^2+b^2)^2*a^2*f*ln(I*b*exp(2*I*(d*x+c))-2*a*exp( 
I*(d*x+c))-I*b)-2*b/d^2/(-a^2+b^2)^2*f*ln(exp(I*(d*x+c)))+b/d^2/(-a^2+b^2) 
^2*f*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-3*I*b/d/(-a^2+b^2)^(5 
/2)*a*e*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))+3*I*b/d^2/ 
(-a^2+b^2)^(5/2)*a*f*c*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1 
/2))-3/2*b/d/(-a^2+b^2)^(5/2)*a*f*ln((I*a+exp(I*(d*x+c))*b-(-a^2+b^2)^(1/2 
))/(I*a-(-a^2+b^2)^(1/2)))*x+3/2*b/d/(-a^2+b^2)^(5/2)*a*f*ln((I*a+exp(I*(d 
*x+c))*b+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*x-3/2*b/d^2/(-a^2+b^2)^ 
(5/2)*a*f*ln((I*a+exp(I*(d*x+c))*b-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2) 
))*c+3/2*b/d^2/(-a^2+b^2)^(5/2)*a*f*ln((I*a+exp(I*(d*x+c))*b+(-a^2+b^2)^(1 
/2))/(I*a+(-a^2+b^2)^(1/2)))*c+3/2*I*b/d^2/(-a^2+b^2)^(5/2)*a*f*dilog((...
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2429 vs. \(2 (651) = 1302\).

Time = 0.35 (sec) , antiderivative size = 2429, normalized size of antiderivative = 3.23 \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

-1/4*(3*(I*a*b^5*f*cos(d*x + c)^2 - 2*I*a^2*b^4*f*sin(d*x + c) - I*(a^3*b^ 
3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + 
 c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 
1) + 3*(-I*a*b^5*f*cos(d*x + c)^2 + 2*I*a^2*b^4*f*sin(d*x + c) + I*(a^3*b^ 
3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + 
 c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 
1) + 3*(-I*a*b^5*f*cos(d*x + c)^2 + 2*I*a^2*b^4*f*sin(d*x + c) + I*(a^3*b^ 
3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x 
+ c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 
 1) + 3*(I*a*b^5*f*cos(d*x + c)^2 - 2*I*a^2*b^4*f*sin(d*x + c) - I*(a^3*b^ 
3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x 
+ c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 
 1) + 3*((a^3*b^3 + a*b^5)*d*f*x + (a^3*b^3 + a*b^5)*c*f - (a*b^5*d*f*x + 
a*b^5*c*f)*cos(d*x + c)^2 + 2*(a^2*b^4*d*f*x + a^2*b^4*c*f)*sin(d*x + c))* 
sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d* 
x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - 3*((a^3*b^3 + 
a*b^5)*d*f*x + (a^3*b^3 + a*b^5)*c*f - (a*b^5*d*f*x + a*b^5*c*f)*cos(d*x + 
 c)^2 + 2*(a^2*b^4*d*f*x + a^2*b^4*c*f)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^ 
2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x 
 + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + 3*((a^3*b^3 + a*b^5)*d*f*x + (a...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))**3,x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [F]

\[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {{\left (f x + e\right )} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((f*x + e)*sin(d*x + c)/(b*sin(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Hanged} \] Input:

int((sin(c + d*x)*(e + f*x))/(a + b*sin(c + d*x))^3,x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int((f*x+e)*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)
 

Output:

( - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))* 
sin(c + d*x)**2*a**2*b**3*e - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)* 
a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**3*b**2*e - 12*sqrt(a**2 - b**2)* 
atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**4*b*e - 2*cos(c + d*x) 
*sin(c + d*x)*a**5*b*e - 2*cos(c + d*x)*sin(c + d*x)*a**3*b**3*e + 4*cos(c 
 + d*x)*sin(c + d*x)*a*b**5*e - 4*cos(c + d*x)*a**6*e + 2*cos(c + d*x)*a** 
4*b**2*e + 2*cos(c + d*x)*a**2*b**4*e + 4*int((sin(c + d*x)*x)/(sin(c + d* 
x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2*b + a**3),x)*s 
in(c + d*x)**2*a**7*b**2*d*f - 12*int((sin(c + d*x)*x)/(sin(c + d*x)**3*b* 
*3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2*b + a**3),x)*sin(c + d 
*x)**2*a**5*b**4*d*f + 12*int((sin(c + d*x)*x)/(sin(c + d*x)**3*b**3 + 3*s 
in(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2*b + a**3),x)*sin(c + d*x)**2*a 
**3*b**6*d*f - 4*int((sin(c + d*x)*x)/(sin(c + d*x)**3*b**3 + 3*sin(c + d* 
x)**2*a*b**2 + 3*sin(c + d*x)*a**2*b + a**3),x)*sin(c + d*x)**2*a*b**8*d*f 
 + 8*int((sin(c + d*x)*x)/(sin(c + d*x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 
 + 3*sin(c + d*x)*a**2*b + a**3),x)*sin(c + d*x)*a**8*b*d*f - 24*int((sin( 
c + d*x)*x)/(sin(c + d*x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d 
*x)*a**2*b + a**3),x)*sin(c + d*x)*a**6*b**3*d*f + 24*int((sin(c + d*x)*x) 
/(sin(c + d*x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2*b 
+ a**3),x)*sin(c + d*x)*a**4*b**5*d*f - 8*int((sin(c + d*x)*x)/(sin(c +...