3.6.29 \(\int \cos ^5(c+d x) (a+b \tan (c+d x))^2 \, dx\) [529]
Optimal. Leaf size=114 \[ -\frac {3 a b \cos ^5(c+d x)}{20 d}+\frac {\left (4 a^2+b^2\right ) \sin (c+d x)}{4 d}-\frac {\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{6 d}+\frac {\left (4 a^2+b^2\right ) \sin ^5(c+d x)}{20 d}-\frac {b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d} \]
[Out]
-3/20*a*b*cos(d*x+c)^5/d+1/4*(4*a^2+b^2)*sin(d*x+c)/d-1/6*(4*a^2+b^2)*sin(d*x+c)^3/d+1/20*(4*a^2+b^2)*sin(d*x+
c)^5/d-1/4*b*cos(d*x+c)^5*(a+b*tan(d*x+c))/d
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Rubi [A]
time = 0.09, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3589, 3567,
2713} \begin {gather*} \frac {\left (4 a^2+b^2\right ) \sin ^5(c+d x)}{20 d}-\frac {\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{6 d}+\frac {\left (4 a^2+b^2\right ) \sin (c+d x)}{4 d}-\frac {3 a b \cos ^5(c+d x)}{20 d}-\frac {b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]
[Out]
(-3*a*b*Cos[c + d*x]^5)/(20*d) + ((4*a^2 + b^2)*Sin[c + d*x])/(4*d) - ((4*a^2 + b^2)*Sin[c + d*x]^3)/(6*d) + (
(4*a^2 + b^2)*Sin[c + d*x]^5)/(20*d) - (b*Cos[c + d*x]^5*(a + b*Tan[c + d*x]))/(4*d)
Rule 2713
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Rule 3567
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] + Dist[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 3589
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[b*(d*Sec
[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]
Rubi steps
\begin {align*} \int \cos ^5(c+d x) (a+b \tan (c+d x))^2 \, dx &=-\frac {b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d}-\frac {1}{4} \int \cos ^5(c+d x) \left (-4 a^2-b^2-3 a b \tan (c+d x)\right ) \, dx\\ &=-\frac {3 a b \cos ^5(c+d x)}{20 d}-\frac {b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d}-\frac {1}{4} \left (-4 a^2-b^2\right ) \int \cos ^5(c+d x) \, dx\\ &=-\frac {3 a b \cos ^5(c+d x)}{20 d}-\frac {b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d}-\frac {\left (4 a^2+b^2\right ) \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=-\frac {3 a b \cos ^5(c+d x)}{20 d}+\frac {\left (4 a^2+b^2\right ) \sin (c+d x)}{4 d}-\frac {\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{6 d}+\frac {\left (4 a^2+b^2\right ) \sin ^5(c+d x)}{20 d}-\frac {b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d}\\ \end {align*}
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Mathematica [A]
time = 0.24, size = 116, normalized size = 1.02 \begin {gather*} \frac {-60 a b \cos (c+d x)-30 a b \cos (3 (c+d x))-6 a b \cos (5 (c+d x))+150 a^2 \sin (c+d x)+30 b^2 \sin (c+d x)+25 a^2 \sin (3 (c+d x))-5 b^2 \sin (3 (c+d x))+3 a^2 \sin (5 (c+d x))-3 b^2 \sin (5 (c+d x))}{240 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]
[Out]
(-60*a*b*Cos[c + d*x] - 30*a*b*Cos[3*(c + d*x)] - 6*a*b*Cos[5*(c + d*x)] + 150*a^2*Sin[c + d*x] + 30*b^2*Sin[c
+ d*x] + 25*a^2*Sin[3*(c + d*x)] - 5*b^2*Sin[3*(c + d*x)] + 3*a^2*Sin[5*(c + d*x)] - 3*b^2*Sin[5*(c + d*x)])/
(240*d)
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Maple [A]
time = 0.20, size = 88, normalized size = 0.77
| | |
| method |
result |
size |
| | |
| derivativedivides |
\(\frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{15}\right )-\frac {2 a b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) |
\(88\) |
| default |
\(\frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{15}\right )-\frac {2 a b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) |
\(88\) |
| risch |
\(-\frac {a b \cos \left (d x +c \right )}{4 d}+\frac {5 a^{2} \sin \left (d x +c \right )}{8 d}+\frac {b^{2} \sin \left (d x +c \right )}{8 d}-\frac {a b \cos \left (5 d x +5 c \right )}{40 d}+\frac {\sin \left (5 d x +5 c \right ) a^{2}}{80 d}-\frac {\sin \left (5 d x +5 c \right ) b^{2}}{80 d}-\frac {a b \cos \left (3 d x +3 c \right )}{8 d}+\frac {5 \sin \left (3 d x +3 c \right ) a^{2}}{48 d}-\frac {\sin \left (3 d x +3 c \right ) b^{2}}{48 d}\) |
\(143\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^5*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
[Out]
1/d*(b^2*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(cos(d*x+c)^2+2)*sin(d*x+c))-2/5*a*b*cos(d*x+c)^5+1/5*a^2*(8/3+cos
(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))
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Maxima [A]
time = 0.28, size = 77, normalized size = 0.68 \begin {gather*} -\frac {6 \, a b \cos \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2} + {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} b^{2}}{15 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
[Out]
-1/15*(6*a*b*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^2 + (3*sin(d*x + c)^5
- 5*sin(d*x + c)^3)*b^2)/d
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Fricas [A]
time = 0.37, size = 74, normalized size = 0.65 \begin {gather*} -\frac {6 \, a b \cos \left (d x + c\right )^{5} - {\left (3 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
[Out]
-1/15*(6*a*b*cos(d*x + c)^5 - (3*(a^2 - b^2)*cos(d*x + c)^4 + (4*a^2 + b^2)*cos(d*x + c)^2 + 8*a^2 + 2*b^2)*si
n(d*x + c))/d
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \cos ^{5}{\left (c + d x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**5*(a+b*tan(d*x+c))**2,x)
[Out]
Integral((a + b*tan(c + d*x))**2*cos(c + d*x)**5, x)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 28204 vs.
\(2 (104) = 208\).
time = 50.38, size = 28204, normalized size = 247.40 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="giac")
[Out]
-1/960*(45*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^
2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2
*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^10*tan(1/2*c)^10 + 45*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(
1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*
tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^10*tan(1/2*c)^10
+ 45*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 + 2*
tan(1/2*c) - 1)*tan(1/2*d*x)^10*tan(1/2*c)^10 + 45*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*t
an(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*tan(1/2*d*x)^10*tan(1/2*c)^10 + 60*pi*a*b*sgn(ta
n(1/2*d*x)^2*tan(1/2*c)^2 - tan(1/2*d*x)^2 - 4*tan(1/2*d*x)*tan(1/2*c) - tan(1/2*c)^2 + 1)*tan(1/2*d*x)^10*tan
(1/2*c)^10 + 225*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1
/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + t
an(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^10*tan(1/2*c)^8 + 225*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 -
2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^
2 - 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^10*tan(1/2*
c)^8 + 225*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^
2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2
*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^8*tan(1/2*c)^10 + 225*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(
1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*
tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^8*tan(1/2*c)^10 -
90*a*b*arctan((tan(1/2*d*x)*tan(1/2*c) + tan(1/2*d*x) - tan(1/2*c) + 1)/(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*
x) + tan(1/2*c) + 1))*tan(1/2*d*x)^10*tan(1/2*c)^10 - 90*a*b*arctan((tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) +
tan(1/2*c) + 1)/(tan(1/2*d*x)*tan(1/2*c) + tan(1/2*d*x) - tan(1/2*c) + 1))*tan(1/2*d*x)^10*tan(1/2*c)^10 + 90*
a*b*arctan((tan(1/2*d*x)*tan(1/2*c) + tan(1/2*d*x) + tan(1/2*c) - 1)/(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) -
tan(1/2*c) - 1))*tan(1/2*d*x)^10*tan(1/2*c)^10 + 90*a*b*arctan((tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(
1/2*c) - 1)/(tan(1/2*d*x)*tan(1/2*c) + tan(1/2*d*x) + tan(1/2*c) - 1))*tan(1/2*d*x)^10*tan(1/2*c)^10 + 225*pi*
a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 + 2*tan(1/2*
c) - 1)*tan(1/2*d*x)^10*tan(1/2*c)^8 + 225*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c
) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*tan(1/2*d*x)^10*tan(1/2*c)^8 + 300*pi*a*b*sgn(tan(1/2*d*
x)^2*tan(1/2*c)^2 - tan(1/2*d*x)^2 - 4*tan(1/2*d*x)*tan(1/2*c) - tan(1/2*c)^2 + 1)*tan(1/2*d*x)^10*tan(1/2*c)^
8 + 225*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 +
2*tan(1/2*c) - 1)*tan(1/2*d*x)^8*tan(1/2*c)^10 + 225*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^
2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*tan(1/2*d*x)^8*tan(1/2*c)^10 + 300*pi*a*b*sgn
(tan(1/2*d*x)^2*tan(1/2*c)^2 - tan(1/2*d*x)^2 - 4*tan(1/2*d*x)*tan(1/2*c) - tan(1/2*c)^2 + 1)*tan(1/2*d*x)^8*t
an(1/2*c)^10 + 384*a*b*tan(1/2*d*x)^10*tan(1/2*c)^10 + 450*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*
d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(
1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^10*tan(1/2*c)^6 + 450
*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(
1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*
tan(1/2*d*x) - 1)*tan(1/2*d*x)^10*tan(1/2*c)^6 + 1125*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^
2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d
*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^8*tan(1/2*c)^8 + 1125*pi*a
*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c
) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1
/2*d*x) - 1)*tan(1/2*d*x)^8*tan(1/2*c)^8 - 450*a*b*arctan((tan(1/2*d*x)*tan(1/2*c) + tan(1/2*d*x) - tan(1/2*c)
+ 1)/(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) + tan(1/2*c) + 1))*tan(1/2*d*x)^10*tan(1/2*c)^8 - 450*a*b*arctan
((tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) + tan(...
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Mupad [B]
time = 3.77, size = 115, normalized size = 1.01 \begin {gather*} \frac {2\,\left (\frac {3\,\sin \left (c+d\,x\right )\,a^2\,{\cos \left (c+d\,x\right )}^4}{2}+2\,\sin \left (c+d\,x\right )\,a^2\,{\cos \left (c+d\,x\right )}^2+4\,\sin \left (c+d\,x\right )\,a^2-3\,a\,b\,{\cos \left (c+d\,x\right )}^5-\frac {3\,\sin \left (c+d\,x\right )\,b^2\,{\cos \left (c+d\,x\right )}^4}{2}+\frac {\sin \left (c+d\,x\right )\,b^2\,{\cos \left (c+d\,x\right )}^2}{2}+\sin \left (c+d\,x\right )\,b^2\right )}{15\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^5*(a + b*tan(c + d*x))^2,x)
[Out]
(2*(4*a^2*sin(c + d*x) + b^2*sin(c + d*x) + 2*a^2*cos(c + d*x)^2*sin(c + d*x) + (3*a^2*cos(c + d*x)^4*sin(c +
d*x))/2 + (b^2*cos(c + d*x)^2*sin(c + d*x))/2 - (3*b^2*cos(c + d*x)^4*sin(c + d*x))/2 - 3*a*b*cos(c + d*x)^5))
/(15*d)
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