\(\int \cos ^7(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [50]

Optimal result

Integrand size = 24, antiderivative size = 106 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {3 i a^3 \cos ^5(c+d x)}{35 d}+\frac {3 a^3 \sin (c+d x)}{7 d}-\frac {2 a^3 \sin ^3(c+d x)}{7 d}+\frac {3 a^3 \sin ^5(c+d x)}{35 d}-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^2}{7 d} \] Output:

-3/35*I*a^3*cos(d*x+c)^5/d+3/7*a^3*sin(d*x+c)/d-2/7*a^3*sin(d*x+c)^3/d+3/3 
5*a^3*sin(d*x+c)^5/d-2/7*I*a*cos(d*x+c)^7*(a+I*a*tan(d*x+c))^2/d
 

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.70 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 (-i \cos (3 (c+d x))+\sin (3 (c+d x))) \left (35 \sqrt {\cos ^2(c+d x)}+\left (8+84 \sqrt {\cos ^2(c+d x)}\right ) \cos (2 (c+d x))+\left (8-15 \sqrt {\cos ^2(c+d x)}\right ) \cos (4 (c+d x))-8 i \sin (2 (c+d x))-56 i \sqrt {\cos ^2(c+d x)} \sin (2 (c+d x))-8 i \sin (4 (c+d x))+20 i \sqrt {\cos ^2(c+d x)} \sin (4 (c+d x))\right )}{280 d \sqrt {\cos ^2(c+d x)}} \] Input:

Integrate[Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^3,x]
 

Output:

(a^3*((-I)*Cos[3*(c + d*x)] + Sin[3*(c + d*x)])*(35*Sqrt[Cos[c + d*x]^2] + 
 (8 + 84*Sqrt[Cos[c + d*x]^2])*Cos[2*(c + d*x)] + (8 - 15*Sqrt[Cos[c + d*x 
]^2])*Cos[4*(c + d*x)] - (8*I)*Sin[2*(c + d*x)] - (56*I)*Sqrt[Cos[c + d*x] 
^2]*Sin[2*(c + d*x)] - (8*I)*Sin[4*(c + d*x)] + (20*I)*Sqrt[Cos[c + d*x]^2 
]*Sin[4*(c + d*x)]))/(280*d*Sqrt[Cos[c + d*x]^2])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3977, 3042, 3967, 3042, 3113, 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.

\(\displaystyle \int \cos ^7(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{\sec (c+d x)^7}dx\)

\(\Big \downarrow \) 3977

\(\displaystyle \frac {3}{7} a^2 \int \cos ^5(c+d x) (i \tan (c+d x) a+a)dx-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{7} a^2 \int \frac {i \tan (c+d x) a+a}{\sec (c+d x)^5}dx-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3967

\(\displaystyle \frac {3}{7} a^2 \left (a \int \cos ^5(c+d x)dx-\frac {i a \cos ^5(c+d x)}{5 d}\right )-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{7} a^2 \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {i a \cos ^5(c+d x)}{5 d}\right )-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3113

\(\displaystyle \frac {3}{7} a^2 \left (-\frac {a \int \left (\sin ^4(c+d x)-2 \sin ^2(c+d x)+1\right )d(-\sin (c+d x))}{d}-\frac {i a \cos ^5(c+d x)}{5 d}\right )-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3}{7} a^2 \left (-\frac {a \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}-\frac {i a \cos ^5(c+d x)}{5 d}\right )-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

Input:

Int[Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^3,x]
 

Output:

(3*a^2*(((-1/5*I)*a*Cos[c + d*x]^5)/d - (a*(-Sin[c + d*x] + (2*Sin[c + d*x 
]^3)/3 - Sin[c + d*x]^5/5))/d))/7 - (((2*I)/7)*a*Cos[c + d*x]^7*(a + I*a*T 
an[c + d*x])^2)/d
 

Defintions of rubi rules used

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 3113

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] 
 && IGtQ[(n - 1)/2, 0]
 

rule 3967

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a   Int[(d 
*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] 
|| NeQ[a^2 + b^2, 0])
 

rule 3977

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( 
n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m))   Int[(d*Sec[e + f*x]) 
^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] 
&& EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || 
 (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & 
& LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) 
&& IntegerQ[2*m]
 

Maple [A] (verified)

Time = 77.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.80

method	result	size
risch	\(-\frac {i a^{3} {\mathrm e}^{7 i \left (d x +c \right )}}{112 d}-\frac {i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}}{20 d}-\frac {i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{8 d}-\frac {3 i a^{3} \cos \left (d x +c \right )}{16 d}+\frac {5 a^{3} \sin \left (d x +c \right )}{16 d}\)	\(85\)
derivativedivides	\(\frac {-i a^{3} \left (-\frac {\cos \left (d x +c \right )^{5} \sin \left (d x +c \right )^{2}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )-3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{6}}{7}+\frac {\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {3 i a^{3} \cos \left (d x +c \right )^{7}}{7}+\frac {a^{3} \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\)	\(146\)
default	\(\frac {-i a^{3} \left (-\frac {\cos \left (d x +c \right )^{5} \sin \left (d x +c \right )^{2}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )-3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{6}}{7}+\frac {\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {3 i a^{3} \cos \left (d x +c \right )^{7}}{7}+\frac {a^{3} \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\)	\(146\)

Input:

int(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

-1/112*I/d*a^3*exp(7*I*(d*x+c))-1/20*I/d*a^3*exp(5*I*(d*x+c))-1/8*I/d*a^3* 
exp(3*I*(d*x+c))-3/16*I/d*a^3*cos(d*x+c)+5/16*a^3*sin(d*x+c)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.72 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {{\left (-5 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 28 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 70 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 140 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 35 i \, a^{3}\right )} e^{\left (-i \, d x - i \, c\right )}}{560 \, d} \] Input:

integrate(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

1/560*(-5*I*a^3*e^(8*I*d*x + 8*I*c) - 28*I*a^3*e^(6*I*d*x + 6*I*c) - 70*I* 
a^3*e^(4*I*d*x + 4*I*c) - 140*I*a^3*e^(2*I*d*x + 2*I*c) + 35*I*a^3)*e^(-I* 
d*x - I*c)/d
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.79 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^3 \, dx=\begin {cases} \frac {\left (- 10240 i a^{3} d^{4} e^{8 i c} e^{7 i d x} - 57344 i a^{3} d^{4} e^{6 i c} e^{5 i d x} - 143360 i a^{3} d^{4} e^{4 i c} e^{3 i d x} - 286720 i a^{3} d^{4} e^{2 i c} e^{i d x} + 71680 i a^{3} d^{4} e^{- i d x}\right ) e^{- i c}}{1146880 d^{5}} & \text {for}\: d^{5} e^{i c} \neq 0 \\\frac {x \left (a^{3} e^{8 i c} + 4 a^{3} e^{6 i c} + 6 a^{3} e^{4 i c} + 4 a^{3} e^{2 i c} + a^{3}\right ) e^{- i c}}{16} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**7*(a+I*a*tan(d*x+c))**3,x)
 

Output:

Piecewise(((-10240*I*a**3*d**4*exp(8*I*c)*exp(7*I*d*x) - 57344*I*a**3*d**4 
*exp(6*I*c)*exp(5*I*d*x) - 143360*I*a**3*d**4*exp(4*I*c)*exp(3*I*d*x) - 28 
6720*I*a**3*d**4*exp(2*I*c)*exp(I*d*x) + 71680*I*a**3*d**4*exp(-I*d*x))*ex 
p(-I*c)/(1146880*d**5), Ne(d**5*exp(I*c), 0)), (x*(a**3*exp(8*I*c) + 4*a** 
3*exp(6*I*c) + 6*a**3*exp(4*I*c) + 4*a**3*exp(2*I*c) + a**3)*exp(-I*c)/16, 
 True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.16 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {15 i \, a^{3} \cos \left (d x + c\right )^{7} + i \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{3} + {\left (15 \, \sin \left (d x + c\right )^{7} - 42 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3}\right )} a^{3} + {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{3}}{35 \, d} \] Input:

integrate(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

-1/35*(15*I*a^3*cos(d*x + c)^7 + I*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a 
^3 + (15*sin(d*x + c)^7 - 42*sin(d*x + c)^5 + 35*sin(d*x + c)^3)*a^3 + (5* 
sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))* 
a^3)/d
 

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 465 vs. \(2 (90) = 180\).

Time = 0.53 (sec) , antiderivative size = 465, normalized size of antiderivative = 4.39 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {19635 \, a^{3} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 39270 \, a^{3} e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 19635 \, a^{3} e^{\left (i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 19635 \, a^{3} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 39270 \, a^{3} e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 19635 \, a^{3} e^{\left (i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 19635 \, a^{3} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 39270 \, a^{3} e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 19635 \, a^{3} e^{\left (i \, d x - i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 19635 \, a^{3} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 39270 \, a^{3} e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 19635 \, a^{3} e^{\left (i \, d x - i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 640 i \, a^{3} e^{\left (12 i \, d x + 10 i \, c\right )} - 4864 i \, a^{3} e^{\left (10 i \, d x + 8 i \, c\right )} - 16768 i \, a^{3} e^{\left (8 i \, d x + 6 i \, c\right )} - 39424 i \, a^{3} e^{\left (6 i \, d x + 4 i \, c\right )} - 40320 i \, a^{3} e^{\left (4 i \, d x + 2 i \, c\right )} - 8960 i \, a^{3} e^{\left (2 i \, d x\right )} + 4480 i \, a^{3} e^{\left (-2 i \, c\right )}}{71680 \, {\left (d e^{\left (5 i \, d x + 3 i \, c\right )} + 2 \, d e^{\left (3 i \, d x + i \, c\right )} + d e^{\left (i \, d x - i \, c\right )}\right )}} \] Input:

integrate(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

1/71680*(19635*a^3*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 39270* 
a^3*e^(3*I*d*x + I*c)*log(I*e^(I*d*x + I*c) + 1) + 19635*a^3*e^(I*d*x - I* 
c)*log(I*e^(I*d*x + I*c) + 1) + 19635*a^3*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d 
*x + I*c) - 1) + 39270*a^3*e^(3*I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1) + 
19635*a^3*e^(I*d*x - I*c)*log(I*e^(I*d*x + I*c) - 1) - 19635*a^3*e^(5*I*d* 
x + 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 39270*a^3*e^(3*I*d*x + I*c)*log(- 
I*e^(I*d*x + I*c) + 1) - 19635*a^3*e^(I*d*x - I*c)*log(-I*e^(I*d*x + I*c) 
+ 1) - 19635*a^3*e^(5*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 39270*a 
^3*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) - 19635*a^3*e^(I*d*x - I* 
c)*log(-I*e^(I*d*x + I*c) - 1) - 640*I*a^3*e^(12*I*d*x + 10*I*c) - 4864*I* 
a^3*e^(10*I*d*x + 8*I*c) - 16768*I*a^3*e^(8*I*d*x + 6*I*c) - 39424*I*a^3*e 
^(6*I*d*x + 4*I*c) - 40320*I*a^3*e^(4*I*d*x + 2*I*c) - 8960*I*a^3*e^(2*I*d 
*x) + 4480*I*a^3*e^(-2*I*c))/(d*e^(5*I*d*x + 3*I*c) + 2*d*e^(3*I*d*x + I*c 
) + d*e^(I*d*x - I*c))
 

Mupad [B] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.26 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {2\,a^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {17\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {17\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{2}+\frac {31\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{2}-\frac {5\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{2}+\frac {\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,35{}\mathrm {i}}{8}-\frac {\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,35{}\mathrm {i}}{8}+\frac {\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,119{}\mathrm {i}}{8}-\frac {\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,15{}\mathrm {i}}{8}\right )}{35\,d\,\left (\cos \left (3\,c+3\,d\,x\right )-\sin \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}\right )} \] Input:

int(cos(c + d*x)^7*(a + a*tan(c + d*x)*1i)^3,x)
 

Output:

-(2*a^3*cos(c/2 + (d*x)/2)*((cos(c/2 + (d*x)/2)*35i)/8 - (cos((3*c)/2 + (3 
*d*x)/2)*35i)/8 + (cos((5*c)/2 + (5*d*x)/2)*119i)/8 - (cos((7*c)/2 + (7*d* 
x)/2)*15i)/8 + (17*sin(c/2 + (d*x)/2))/2 - (17*sin((3*c)/2 + (3*d*x)/2))/2 
 + (31*sin((5*c)/2 + (5*d*x)/2))/2 - (5*sin((7*c)/2 + (7*d*x)/2))/2))/(35* 
d*(cos(3*c + 3*d*x) - sin(3*c + 3*d*x)*1i))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^{3} \left (20 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} i -53 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} i +46 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i -13 \cos \left (d x +c \right ) i -20 \sin \left (d x +c \right )^{7}+63 \sin \left (d x +c \right )^{5}-70 \sin \left (d x +c \right )^{3}+35 \sin \left (d x +c \right )+13 i \right )}{35 d} \] Input:

int(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^3,x)
 

Output:

(a**3*(20*cos(c + d*x)*sin(c + d*x)**6*i - 53*cos(c + d*x)*sin(c + d*x)**4 
*i + 46*cos(c + d*x)*sin(c + d*x)**2*i - 13*cos(c + d*x)*i - 20*sin(c + d* 
x)**7 + 63*sin(c + d*x)**5 - 70*sin(c + d*x)**3 + 35*sin(c + d*x) + 13*i)) 
/(35*d)