Integrand size = 21, antiderivative size = 131 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {1}{8} a \left (4 a^2+3 b^2\right ) x-\frac {b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}+\frac {a \left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d}-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d} \]
1/8*a*(4*a^2+3*b^2)*x-1/60*b*(27*a^2+8*b^2)*cos(d*x+c)^3/d+1/8*a*(4*a^2+3* b^2)*cos(d*x+c)*sin(d*x+c)/d-7/20*a*b*cos(d*x+c)^3*(a+b*sin(d*x+c))/d-1/5* b*cos(d*x+c)^3*(a+b*sin(d*x+c))^2/d
Time = 0.81 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-60 b \left (6 a^2+b^2\right ) \cos (c+d x)-10 \left (12 a^2 b+b^3\right ) \cos (3 (c+d x))+6 b^3 \cos (5 (c+d x))+15 a \left (4 \left (4 a^2+3 b^2\right ) (c+d x)+8 a^2 \sin (2 (c+d x))-3 b^2 \sin (4 (c+d x))\right )}{480 d} \]
(-60*b*(6*a^2 + b^2)*Cos[c + d*x] - 10*(12*a^2*b + b^3)*Cos[3*(c + d*x)] + 6*b^3*Cos[5*(c + d*x)] + 15*a*(4*(4*a^2 + 3*b^2)*(c + d*x) + 8*a^2*Sin[2* (c + d*x)] - 3*b^2*Sin[4*(c + d*x)]))/(480*d)
Time = 0.60 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3171, 3042, 3341, 3042, 3148, 3042, 3115, 24}
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.
| \(\displaystyle \int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^2 (a+b \sin (c+d x))^3dx\) |
| \(\Big \downarrow \) 3171 |
| \(\displaystyle \frac {1}{5} \int \cos ^2(c+d x) (a+b \sin (c+d x)) \left (5 a^2+7 b \sin (c+d x) a+2 b^2\right )dx-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \cos (c+d x)^2 (a+b \sin (c+d x)) \left (5 a^2+7 b \sin (c+d x) a+2 b^2\right )dx-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3341 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \cos ^2(c+d x) \left (5 a \left (4 a^2+3 b^2\right )+b \left (27 a^2+8 b^2\right ) \sin (c+d x)\right )dx-\frac {7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \cos (c+d x)^2 \left (5 a \left (4 a^2+3 b^2\right )+b \left (27 a^2+8 b^2\right ) \sin (c+d x)\right )dx-\frac {7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a \left (4 a^2+3 b^2\right ) \int \cos ^2(c+d x)dx-\frac {b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{3 d}\right )-\frac {7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a \left (4 a^2+3 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{3 d}\right )-\frac {7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a \left (4 a^2+3 b^2\right ) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{3 d}\right )-\frac {7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a \left (4 a^2+3 b^2\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{3 d}\right )-\frac {7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
-1/5*(b*Cos[c + d*x]^3*(a + b*Sin[c + d*x])^2)/d + ((-7*a*b*Cos[c + d*x]^3 *(a + b*Sin[c + d*x]))/(4*d) + (-1/3*(b*(27*a^2 + 8*b^2)*Cos[c + d*x]^3)/d + 5*a*(4*a^2 + 3*b^2)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4)/5
3.5.7.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[1/(m + p) Int[(g*Cos[e + f*x])^p* (a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1) *Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[1/(m + p + 1) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Sim p[a*c*(m + p + 1) + b*d*m + (a*d*m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && !LtQ[p, -1] && IntegerQ[2*m] && !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && S implerQ[c + d*x, a + b*x])
Time = 0.76 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )+3 a \,b^{2} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+b^{3} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )}{d}\) | \(123\) |
| default | \(\frac {a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )+3 a \,b^{2} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+b^{3} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )}{d}\) | \(123\) |
| parallelrisch | \(\frac {240 a^{3} d x +180 a \,b^{2} d x +6 \cos \left (5 d x +5 c \right ) b^{3}-45 \sin \left (4 d x +4 c \right ) a \,b^{2}-120 \cos \left (3 d x +3 c \right ) a^{2} b -10 \cos \left (3 d x +3 c \right ) b^{3}+120 \sin \left (2 d x +2 c \right ) a^{3}-360 \cos \left (d x +c \right ) a^{2} b -60 \cos \left (d x +c \right ) b^{3}-480 a^{2} b -64 b^{3}}{480 d}\) | \(128\) |
| risch | \(\frac {a^{3} x}{2}+\frac {3 a \,b^{2} x}{8}-\frac {3 b \cos \left (d x +c \right ) a^{2}}{4 d}-\frac {b^{3} \cos \left (d x +c \right )}{8 d}+\frac {b^{3} \cos \left (5 d x +5 c \right )}{80 d}-\frac {3 \sin \left (4 d x +4 c \right ) a \,b^{2}}{32 d}-\frac {b \cos \left (3 d x +3 c \right ) a^{2}}{4 d}-\frac {b^{3} \cos \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (2 d x +2 c \right ) a^{3}}{4 d}\) | \(131\) |
| norman | \(\frac {\left (\frac {1}{2} a^{3}+\frac {3}{8} a \,b^{2}\right ) x +\left (5 a^{3}+\frac {15}{4} a \,b^{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (5 a^{3}+\frac {15}{4} a \,b^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a^{3}+\frac {3}{8} a \,b^{2}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5}{2} a^{3}+\frac {15}{8} a \,b^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5}{2} a^{3}+\frac {15}{8} a \,b^{2}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {30 a^{2} b +4 b^{3}}{15 d}-\frac {6 a^{2} b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (6 a^{2} b +2 b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (12 a^{2} b -2 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (12 a^{2} b +4 b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a \left (4 a^{2}-3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {a \left (4 a^{2}-3 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (4 a^{2}+9 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a \left (4 a^{2}+9 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(388\) |
1/d*(a^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)-a^2*b*cos(d*x+c)^3+3*a* b^2*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+c)*sin(d*x+c)+1/8*d*x+1/8*c) +b^3*(-1/5*cos(d*x+c)^3*sin(d*x+c)^2-2/15*cos(d*x+c)^3))
Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {24 \, b^{3} \cos \left (d x + c\right )^{5} - 40 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (4 \, a^{3} + 3 \, a b^{2}\right )} d x - 15 \, {\left (6 \, a b^{2} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(24*b^3*cos(d*x + c)^5 - 40*(3*a^2*b + b^3)*cos(d*x + c)^3 + 15*(4*a ^3 + 3*a*b^2)*d*x - 15*(6*a*b^2*cos(d*x + c)^3 - (4*a^3 + 3*a*b^2)*cos(d*x + c))*sin(d*x + c))/d
Time = 0.26 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.80 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\begin {cases} \frac {a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {a^{2} b \cos ^{3}{\left (c + d x \right )}}{d} + \frac {3 a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {3 a b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 b^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{3} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((a**3*x*sin(c + d*x)**2/2 + a**3*x*cos(c + d*x)**2/2 + a**3*sin( c + d*x)*cos(c + d*x)/(2*d) - a**2*b*cos(c + d*x)**3/d + 3*a*b**2*x*sin(c + d*x)**4/8 + 3*a*b**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*a*b**2*x*co s(c + d*x)**4/8 + 3*a*b**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - 3*a*b**2*s in(c + d*x)*cos(c + d*x)**3/(8*d) - b**3*sin(c + d*x)**2*cos(c + d*x)**3/( 3*d) - 2*b**3*cos(c + d*x)**5/(15*d), Ne(d, 0)), (x*(a + b*sin(c))**3*cos( c)**2, True))
Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.71 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {480 \, a^{2} b \cos \left (d x + c\right )^{3} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 45 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2} - 32 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{480 \, d} \]
-1/480*(480*a^2*b*cos(d*x + c)^3 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^ 3 - 45*(4*d*x + 4*c - sin(4*d*x + 4*c))*a*b^2 - 32*(3*cos(d*x + c)^5 - 5*c os(d*x + c)^3)*b^3)/d
Time = 0.52 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {b^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {3 \, a b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{3} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {1}{8} \, {\left (4 \, a^{3} + 3 \, a b^{2}\right )} x - \frac {{\left (12 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {{\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{8 \, d} \]
1/80*b^3*cos(5*d*x + 5*c)/d - 3/32*a*b^2*sin(4*d*x + 4*c)/d + 1/4*a^3*sin( 2*d*x + 2*c)/d + 1/8*(4*a^3 + 3*a*b^2)*x - 1/48*(12*a^2*b + b^3)*cos(3*d*x + 3*c)/d - 1/8*(6*a^2*b + b^3)*cos(d*x + c)/d
Time = 6.09 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.72 \[ \int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^2+3\,b^2\right )}{4\,\left (a^3+\frac {3\,a\,b^2}{4}\right )}\right )\,\left (4\,a^2+3\,b^2\right )}{4\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a\,b^2}{4}-a^3\right )+2\,a^2\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^3+\frac {9\,a\,b^2}{2}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {3\,a\,b^2}{4}-a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (2\,a^3+\frac {9\,a\,b^2}{2}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,a^2\,b+\frac {4\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,a^2\,b-\frac {4\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (12\,a^2\,b+4\,b^3\right )+\frac {4\,b^3}{15}+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a\,\left (4\,a^2+3\,b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \]
(a*atan((a*tan(c/2 + (d*x)/2)*(4*a^2 + 3*b^2))/(4*((3*a*b^2)/4 + a^3)))*(4 *a^2 + 3*b^2))/(4*d) - (tan(c/2 + (d*x)/2)*((3*a*b^2)/4 - a^3) + 2*a^2*b - tan(c/2 + (d*x)/2)^3*((9*a*b^2)/2 + 2*a^3) - tan(c/2 + (d*x)/2)^9*((3*a*b ^2)/4 - a^3) + tan(c/2 + (d*x)/2)^7*((9*a*b^2)/2 + 2*a^3) + tan(c/2 + (d*x )/2)^2*(4*a^2*b + (4*b^3)/3) + tan(c/2 + (d*x)/2)^4*(8*a^2*b - (4*b^3)/3) + tan(c/2 + (d*x)/2)^6*(12*a^2*b + 4*b^3) + (4*b^3)/15 + 6*a^2*b*tan(c/2 + (d*x)/2)^8)/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan (c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (a*(4*a^2 + 3*b^2)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d)