\(\int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx\) [1107]

Optimal result

Integrand size = 27, antiderivative size = 129 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a b x}{8}-\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a b \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d} \]

[Out]

1/8*a*b*x-1/105*(a^2+6*b^2)*cos(d*x+c)^5/d+1/8*a*b*cos(d*x+c)*sin(d*x+c)/d+1/12*a*b*cos(d*x+c)^3*sin(d*x+c)/d-
1/21*a*cos(d*x+c)^5*(a+b*sin(d*x+c))/d-1/7*cos(d*x+c)^5*(a+b*sin(d*x+c))^2/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2941, 2748, 2715, 8} \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}+\frac {a b \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac {a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a b x}{8} \]

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(a*b*x)/8 - ((a^2 + 6*b^2)*Cos[c + d*x]^5)/(105*d) + (a*b*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*b*Cos[c + d*x]
^3*Sin[c + d*x])/(12*d) - (a*Cos[c + d*x]^5*(a + b*Sin[c + d*x]))/(21*d) - (Cos[c + d*x]^5*(a + b*Sin[c + d*x]
)^2)/(7*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

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] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[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 2941

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
] + Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[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] && SimplerQ[c + d*x, a + b*x])

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{7} \int \cos ^4(c+d x) (2 b+2 a \sin (c+d x)) (a+b \sin (c+d x)) \, dx \\ & = -\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{42} \int \cos ^4(c+d x) \left (14 a b+2 \left (a^2+6 b^2\right ) \sin (c+d x)\right ) \, dx \\ & = -\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{3} (a b) \int \cos ^4(c+d x) \, dx \\ & = -\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}+\frac {a b \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{4} (a b) \int \cos ^2(c+d x) \, dx \\ & = -\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a b \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{8} (a b) \int 1 \, dx \\ & = \frac {a b x}{8}-\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a b \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.02 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {840 a b c+840 a b d x-105 \left (8 a^2+3 b^2\right ) \cos (c+d x)-105 \left (4 a^2+b^2\right ) \cos (3 (c+d x))-84 a^2 \cos (5 (c+d x))+21 b^2 \cos (5 (c+d x))+15 b^2 \cos (7 (c+d x))+210 a b \sin (2 (c+d x))-210 a b \sin (4 (c+d x))-70 a b \sin (6 (c+d x))}{6720 d} \]

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(840*a*b*c + 840*a*b*d*x - 105*(8*a^2 + 3*b^2)*Cos[c + d*x] - 105*(4*a^2 + b^2)*Cos[3*(c + d*x)] - 84*a^2*Cos[
5*(c + d*x)] + 21*b^2*Cos[5*(c + d*x)] + 15*b^2*Cos[7*(c + d*x)] + 210*a*b*Sin[2*(c + d*x)] - 210*a*b*Sin[4*(c
+ d*x)] - 70*a*b*Sin[6*(c + d*x)])/(6720*d)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+2 a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )}{d}\)	\(105\)
default	\(\frac {-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+2 a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )}{d}\)	\(105\)
parallelrisch	\(\frac {\left (-420 a^{2}-105 b^{2}\right ) \cos \left (3 d x +3 c \right )+\left (-84 a^{2}+21 b^{2}\right ) \cos \left (5 d x +5 c \right )+15 b^{2} \cos \left (7 d x +7 c \right )+210 a b \sin \left (2 d x +2 c \right )-210 a b \sin \left (4 d x +4 c \right )-70 a b \sin \left (6 d x +6 c \right )+\left (-840 a^{2}-315 b^{2}\right ) \cos \left (d x +c \right )+840 a b x d -1344 a^{2}-384 b^{2}}{6720 d}\)	\(136\)
risch	\(\frac {a b x}{8}-\frac {a^{2} \cos \left (d x +c \right )}{8 d}-\frac {3 b^{2} \cos \left (d x +c \right )}{64 d}+\frac {\cos \left (7 d x +7 c \right ) b^{2}}{448 d}-\frac {a b \sin \left (6 d x +6 c \right )}{96 d}-\frac {\cos \left (5 d x +5 c \right ) a^{2}}{80 d}+\frac {\cos \left (5 d x +5 c \right ) b^{2}}{320 d}-\frac {a b \sin \left (4 d x +4 c \right )}{32 d}-\frac {\cos \left (3 d x +3 c \right ) a^{2}}{16 d}-\frac {\cos \left (3 d x +3 c \right ) b^{2}}{64 d}+\frac {a b \sin \left (2 d x +2 c \right )}{32 d}\)	\(168\)
norman	\(\frac {-\frac {14 a^{2}+4 b^{2}}{35 d}+\frac {a b x}{8}-\frac {2 a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {\left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (6 a^{2}-4 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (8 a^{2}+8 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (22 a^{2}-8 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {11 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {31 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {31 a b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {11 a b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a b \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {7 a b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {21 a b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {35 a b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {35 a b x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {21 a b x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {7 a b x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {a b x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}\)	\(411\)

[In]

int(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/5*a^2*cos(d*x+c)^5+2*a*b*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1
/16*d*x+1/16*c)+b^2*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.66 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {120 \, b^{2} \cos \left (d x + c\right )^{7} - 168 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{5} + 105 \, a b d x - 35 \, {\left (8 \, a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/840*(120*b^2*cos(d*x + c)^7 - 168*(a^2 + b^2)*cos(d*x + c)^5 + 105*a*b*d*x - 35*(8*a*b*cos(d*x + c)^5 - 2*a*
b*cos(d*x + c)^3 - 3*a*b*cos(d*x + c))*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.73 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} - \frac {a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {a b x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {3 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a b x \cos ^{6}{\left (c + d x \right )}}{8} + \frac {a b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {a b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac {b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {2 b^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \sin {\left (c \right )} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)*(a+b*sin(d*x+c))**2,x)

[Out]

Piecewise((-a**2*cos(c + d*x)**5/(5*d) + a*b*x*sin(c + d*x)**6/8 + 3*a*b*x*sin(c + d*x)**4*cos(c + d*x)**2/8 +
3*a*b*x*sin(c + d*x)**2*cos(c + d*x)**4/8 + a*b*x*cos(c + d*x)**6/8 + a*b*sin(c + d*x)**5*cos(c + d*x)/(8*d)
+ a*b*sin(c + d*x)**3*cos(c + d*x)**3/(3*d) - a*b*sin(c + d*x)*cos(c + d*x)**5/(8*d) - b**2*sin(c + d*x)**2*co
s(c + d*x)**5/(5*d) - 2*b**2*cos(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)*cos(c)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.63 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {672 \, a^{2} \cos \left (d x + c\right )^{5} - 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 96 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} b^{2}}{3360 \, d} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3360*(672*a^2*cos(d*x + c)^5 - 35*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a*b - 96*(5*c
os(d*x + c)^7 - 7*cos(d*x + c)^5)*b^2)/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.09 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {1}{8} \, a b x + \frac {b^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {a b \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a b \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} - \frac {{\left (4 \, a^{2} - b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (4 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {{\left (8 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )}{64 \, d} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*a*b*x + 1/448*b^2*cos(7*d*x + 7*c)/d - 1/96*a*b*sin(6*d*x + 6*c)/d - 1/32*a*b*sin(4*d*x + 4*c)/d + 1/32*a*
b*sin(2*d*x + 2*c)/d - 1/320*(4*a^2 - b^2)*cos(5*d*x + 5*c)/d - 1/64*(4*a^2 + b^2)*cos(3*d*x + 3*c)/d - 1/64*(
8*a^2 + 3*b^2)*cos(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 14.49 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.98 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a\,b\,x}{8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (6\,a^2-4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (4\,a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {4\,a^2}{5}+\frac {4\,b^2}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (8\,a^2+8\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {22\,a^2}{5}-\frac {8\,b^2}{5}\right )+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\frac {2\,a^2}{5}+\frac {4\,b^2}{35}-\frac {11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {31\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}-\frac {31\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}+\frac {11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

[In]

int(cos(c + d*x)^4*sin(c + d*x)*(a + b*sin(c + d*x))^2,x)

[Out]

(a*b*x)/8 - (tan(c/2 + (d*x)/2)^8*(6*a^2 - 4*b^2) + tan(c/2 + (d*x)/2)^10*(4*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)
^2*((4*a^2)/5 + (4*b^2)/5) + tan(c/2 + (d*x)/2)^6*(8*a^2 + 8*b^2) + tan(c/2 + (d*x)/2)^4*((22*a^2)/5 - (8*b^2)
/5) + 2*a^2*tan(c/2 + (d*x)/2)^12 + (2*a^2)/5 + (4*b^2)/35 - (11*a*b*tan(c/2 + (d*x)/2)^3)/3 + (31*a*b*tan(c/2
+ (d*x)/2)^5)/12 - (31*a*b*tan(c/2 + (d*x)/2)^9)/12 + (11*a*b*tan(c/2 + (d*x)/2)^11)/3 - (a*b*tan(c/2 + (d*x)
/2)^13)/4 + (a*b*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)