3.1.11 \(\int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx\) [11]

3.1.11.1 Optimal result

Integrand size = 19, antiderivative size = 127 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {5 a x}{16}+\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d}-\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d} \]

output

5/16*a*x+a*arctanh(sin(d*x+c))/d-a*sin(d*x+c)/d-5/16*a*cos(d*x+c)*sin(d*x+ 
c)/d-1/3*a*sin(d*x+c)^3/d-5/24*a*cos(d*x+c)*sin(d*x+c)^3/d-1/5*a*sin(d*x+c 
)^5/d-1/6*a*cos(d*x+c)*sin(d*x+c)^5/d
 

3.1.11.2 Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.68 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {a \left (960 \text {arctanh}(\sin (c+d x))-960 \sin (c+d x)-320 \sin ^3(c+d x)-192 \sin ^5(c+d x)+5 (60 c+60 d x-45 \sin (2 (c+d x))+9 \sin (4 (c+d x))-\sin (6 (c+d x)))\right )}{960 d} \]

input

Integrate[(a + a*Sec[c + d*x])*Sin[c + d*x]^6,x]
 

output

(a*(960*ArcTanh[Sin[c + d*x]] - 960*Sin[c + d*x] - 320*Sin[c + d*x]^3 - 19 
2*Sin[c + d*x]^5 + 5*(60*c + 60*d*x - 45*Sin[2*(c + d*x)] + 9*Sin[4*(c + d 
*x)] - Sin[6*(c + d*x)])))/(960*d)
 

3.1.11.3 Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.842, Rules used = {3042, 4360, 25, 25, 3042, 3317, 3042, 3072, 254, 2009, 3115, 3042, 3115, 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.

\(\displaystyle \int \sin ^6(c+d x) (a \sec (c+d x)+a) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^6 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 4360

\(\displaystyle \int -\left (\sin ^5(c+d x) \tan (c+d x) (a (-\cos (c+d x))-a)\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -\left ((\cos (c+d x) a+a) \sin ^5(c+d x) \tan (c+d x)\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \sin ^5(c+d x) \tan (c+d x) (a \cos (c+d x)+a)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^6 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3317

\(\displaystyle a \int \sin ^6(c+d x)dx+a \int \sin ^5(c+d x) \tan (c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \sin (c+d x)^6dx+a \int \sin (c+d x)^5 \tan (c+d x)dx\)

\(\Big \downarrow \) 3072

\(\displaystyle \frac {a \int \frac {\sin ^6(c+d x)}{1-\sin ^2(c+d x)}d\sin (c+d x)}{d}+a \int \sin (c+d x)^6dx\)

\(\Big \downarrow \) 254

\(\displaystyle \frac {a \int \left (-\sin ^4(c+d x)-\sin ^2(c+d x)+\frac {1}{1-\sin ^2(c+d x)}-1\right )d\sin (c+d x)}{d}+a \int \sin (c+d x)^6dx\)

\(\Big \downarrow \) 2009

\(\displaystyle a \int \sin (c+d x)^6dx+\frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3115

\(\displaystyle a \left (\frac {5}{6} \int \sin ^4(c+d x)dx-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {5}{6} \int \sin (c+d x)^4dx-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3115

\(\displaystyle a \left (\frac {5}{6} \left (\frac {3}{4} \int \sin ^2(c+d x)dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {5}{6} \left (\frac {3}{4} \int \sin (c+d x)^2dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3115

\(\displaystyle a \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {a \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+a \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )\)

input

Int[(a + a*Sec[c + d*x])*Sin[c + d*x]^6,x]
 

output

(a*(ArcTanh[Sin[c + d*x]] - Sin[c + d*x] - Sin[c + d*x]^3/3 - Sin[c + d*x] 
^5/5))/d + a*(-1/6*(Cos[c + d*x]*Sin[c + d*x]^5)/d + (5*(-1/4*(Cos[c + d*x 
]*Sin[c + d*x]^3)/d + (3*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/6)
 

3.1.11.3.1 Defintions of rubi rules used

rule 24

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

rule 25

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

rule 254

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, 
 a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
 

rule 2009

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

rule 3042

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

rule 3072

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f   Subst[Int[ 
(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x 
]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
 

rule 3115

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]
 

rule 3317

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[(g*Co 
s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d   Int[(g*Cos[e + f*x])^ 
p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
 

rule 4360

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 

3.1.11.4 Maple [A] (verified)

Time = 1.55 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(96\)
default	\(\frac {a \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(96\)
parts	\(\frac {a \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}+\frac {a \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\)	\(98\)
parallelrisch	\(-\frac {a \left (-60 d x +\sin \left (6 d x +6 c \right )+192 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-192 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+264 \sin \left (d x +c \right )+45 \sin \left (2 d x +2 c \right )-28 \sin \left (3 d x +3 c \right )-9 \sin \left (4 d x +4 c \right )+\frac {12 \sin \left (5 d x +5 c \right )}{5}\right )}{192 d}\)	\(101\)
risch	\(\frac {5 a x}{16}+\frac {11 i a \,{\mathrm e}^{i \left (d x +c \right )}}{16 d}-\frac {11 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \sin \left (6 d x +6 c \right )}{192 d}-\frac {a \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 a \sin \left (4 d x +4 c \right )}{64 d}+\frac {7 a \sin \left (3 d x +3 c \right )}{48 d}-\frac {15 a \sin \left (2 d x +2 c \right )}{64 d}\)	\(150\)
norman	\(\frac {\frac {5 a x}{16}-\frac {21 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {389 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}-\frac {853 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}-\frac {523 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 d}-\frac {73 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{8 d}-\frac {11 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {15 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {75 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16}+\frac {25 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}+\frac {75 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16}+\frac {15 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {5 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(248\)

input

int((a+a*sec(d*x+c))*sin(d*x+c)^6,x,method=_RETURNVERBOSE)
 

output

1/d*(a*(-1/5*sin(d*x+c)^5-1/3*sin(d*x+c)^3-sin(d*x+c)+ln(sec(d*x+c)+tan(d* 
x+c)))+a*(-1/6*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+ 
5/16*d*x+5/16*c))
 

3.1.11.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {75 \, a d x + 120 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 120 \, a \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (40 \, a \cos \left (d x + c\right )^{5} + 48 \, a \cos \left (d x + c\right )^{4} - 130 \, a \cos \left (d x + c\right )^{3} - 176 \, a \cos \left (d x + c\right )^{2} + 165 \, a \cos \left (d x + c\right ) + 368 \, a\right )} \sin \left (d x + c\right )}{240 \, d} \]

input

integrate((a+a*sec(d*x+c))*sin(d*x+c)^6,x, algorithm="fricas")
 

output

1/240*(75*a*d*x + 120*a*log(sin(d*x + c) + 1) - 120*a*log(-sin(d*x + c) + 
1) - (40*a*cos(d*x + c)^5 + 48*a*cos(d*x + c)^4 - 130*a*cos(d*x + c)^3 - 1 
76*a*cos(d*x + c)^2 + 165*a*cos(d*x + c) + 368*a)*sin(d*x + c))/d
 

3.1.11.6 Sympy [F]

\[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=a \left (\int \sin ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin ^{6}{\left (c + d x \right )}\, dx\right ) \]

input

integrate((a+a*sec(d*x+c))*sin(d*x+c)**6,x)
 

output

a*(Integral(sin(c + d*x)**6*sec(c + d*x), x) + Integral(sin(c + d*x)**6, x 
))
 

3.1.11.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=-\frac {32 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{960 \, d} \]

input

integrate((a+a*sec(d*x+c))*sin(d*x+c)^6,x, algorithm="maxima")
 

output

-1/960*(32*(6*sin(d*x + c)^5 + 10*sin(d*x + c)^3 - 15*log(sin(d*x + c) + 1 
) + 15*log(sin(d*x + c) - 1) + 30*sin(d*x + c))*a - 5*(4*sin(2*d*x + 2*c)^ 
3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a)/d
 

3.1.11.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.15 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {75 \, {\left (d x + c\right )} a + 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (165 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1095 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 3138 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5118 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1945 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 315 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

input

integrate((a+a*sec(d*x+c))*sin(d*x+c)^6,x, algorithm="giac")
 

output

1/240*(75*(d*x + c)*a + 240*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 240*a*l 
og(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(165*a*tan(1/2*d*x + 1/2*c)^11 + 109 
5*a*tan(1/2*d*x + 1/2*c)^9 + 3138*a*tan(1/2*d*x + 1/2*c)^7 + 5118*a*tan(1/ 
2*d*x + 1/2*c)^5 + 1945*a*tan(1/2*d*x + 1/2*c)^3 + 315*a*tan(1/2*d*x + 1/2 
*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d
 

3.1.11.9 Mupad [B] (verification not implemented)

Time = 13.66 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {5\,a\,x}{16}+\frac {2\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {15\,a\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {7\,a\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {3\,a\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}-\frac {a\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}-\frac {a\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}-\frac {11\,a\,\sin \left (c+d\,x\right )}{8\,d} \]

input

int(sin(c + d*x)^6*(a + a/cos(c + d*x)),x)
 

output

(5*a*x)/16 + (2*a*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (15*a* 
sin(2*c + 2*d*x))/(64*d) + (7*a*sin(3*c + 3*d*x))/(48*d) + (3*a*sin(4*c + 
4*d*x))/(64*d) - (a*sin(5*c + 5*d*x))/(80*d) - (a*sin(6*c + 6*d*x))/(192*d 
) - (11*a*sin(c + d*x))/(8*d)