3.7.31 \(\int \cos ^5(c+d x) (a+b \sin (c+d x))^m \, dx\) [631]

Optimal. Leaf size=167 \[ \frac {\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{1+m}}{b^5 d (1+m)}-\frac {4 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{2+m}}{b^5 d (2+m)}+\frac {2 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{3+m}}{b^5 d (3+m)}-\frac {4 a (a+b \sin (c+d x))^{4+m}}{b^5 d (4+m)}+\frac {(a+b \sin (c+d x))^{5+m}}{b^5 d (5+m)} \]

[Out]

(a^2-b^2)^2*(a+b*sin(d*x+c))^(1+m)/b^5/d/(1+m)-4*a*(a^2-b^2)*(a+b*sin(d*x+c))^(2+m)/b^5/d/(2+m)+2*(3*a^2-b^2)*
(a+b*sin(d*x+c))^(3+m)/b^5/d/(3+m)-4*a*(a+b*sin(d*x+c))^(4+m)/b^5/d/(4+m)+(a+b*sin(d*x+c))^(5+m)/b^5/d/(5+m)

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Rubi [A]
time = 0.08, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2747, 711} \begin {gather*} \frac {\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{m+1}}{b^5 d (m+1)}-\frac {4 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{m+2}}{b^5 d (m+2)}+\frac {2 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{m+3}}{b^5 d (m+3)}-\frac {4 a (a+b \sin (c+d x))^{m+4}}{b^5 d (m+4)}+\frac {(a+b \sin (c+d x))^{m+5}}{b^5 d (m+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^m,x]

[Out]

((a^2 - b^2)^2*(a + b*Sin[c + d*x])^(1 + m))/(b^5*d*(1 + m)) - (4*a*(a^2 - b^2)*(a + b*Sin[c + d*x])^(2 + m))/
(b^5*d*(2 + m)) + (2*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(3 + m))/(b^5*d*(3 + m)) - (4*a*(a + b*Sin[c + d*x])^(
4 + m))/(b^5*d*(4 + m)) + (a + b*Sin[c + d*x])^(5 + m)/(b^5*d*(5 + m))

Rule 711

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+b \sin (c+d x))^m \, dx &=\frac {\text {Subst}\left (\int (a+x)^m \left (b^2-x^2\right )^2 \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\text {Subst}\left (\int \left (\left (a^2-b^2\right )^2 (a+x)^m-4 \left (a^3-a b^2\right ) (a+x)^{1+m}+2 \left (3 a^2-b^2\right ) (a+x)^{2+m}-4 a (a+x)^{3+m}+(a+x)^{4+m}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{1+m}}{b^5 d (1+m)}-\frac {4 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{2+m}}{b^5 d (2+m)}+\frac {2 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{3+m}}{b^5 d (3+m)}-\frac {4 a (a+b \sin (c+d x))^{4+m}}{b^5 d (4+m)}+\frac {(a+b \sin (c+d x))^{5+m}}{b^5 d (5+m)}\\ \end {align*}

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Mathematica [A]
time = 0.95, size = 330, normalized size = 1.98 \begin {gather*} \frac {(a+b \sin (c+d x))^{1+m} \left (192 a^4-544 a^2 b^2+712 b^4-144 a^2 b^2 m+758 b^4 m+16 a^2 b^2 m^2+281 b^4 m^2+46 b^4 m^3+3 b^4 m^4+4 b^2 \left (2+3 m+m^2\right ) \left (-12 a^2+b^2 \left (28+11 m+m^2\right )\right ) \cos (2 (c+d x))+b^4 \left (24+50 m+35 m^2+10 m^3+m^4\right ) \cos (4 (c+d x))-192 a^3 b \sin (c+d x)+496 a b^3 \sin (c+d x)-192 a^3 b m \sin (c+d x)+664 a b^3 m \sin (c+d x)+176 a b^3 m^2 \sin (c+d x)+8 a b^3 m^3 \sin (c+d x)+48 a b^3 \sin (3 (c+d x))+88 a b^3 m \sin (3 (c+d x))+48 a b^3 m^2 \sin (3 (c+d x))+8 a b^3 m^3 \sin (3 (c+d x))\right )}{8 b^5 d (1+m) (2+m) (3+m) (4+m) (5+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^m,x]

[Out]

((a + b*Sin[c + d*x])^(1 + m)*(192*a^4 - 544*a^2*b^2 + 712*b^4 - 144*a^2*b^2*m + 758*b^4*m + 16*a^2*b^2*m^2 +
281*b^4*m^2 + 46*b^4*m^3 + 3*b^4*m^4 + 4*b^2*(2 + 3*m + m^2)*(-12*a^2 + b^2*(28 + 11*m + m^2))*Cos[2*(c + d*x)
] + b^4*(24 + 50*m + 35*m^2 + 10*m^3 + m^4)*Cos[4*(c + d*x)] - 192*a^3*b*Sin[c + d*x] + 496*a*b^3*Sin[c + d*x]
- 192*a^3*b*m*Sin[c + d*x] + 664*a*b^3*m*Sin[c + d*x] + 176*a*b^3*m^2*Sin[c + d*x] + 8*a*b^3*m^3*Sin[c + d*x]
+ 48*a*b^3*Sin[3*(c + d*x)] + 88*a*b^3*m*Sin[3*(c + d*x)] + 48*a*b^3*m^2*Sin[3*(c + d*x)] + 8*a*b^3*m^3*Sin[3
*(c + d*x)]))/(8*b^5*d*(1 + m)*(2 + m)*(3 + m)*(4 + m)*(5 + m))

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Maple [F]
time = 0.15, size = 0, normalized size = 0.00 \[\int \left (\cos ^{5}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*sin(d*x+c))^m,x)

[Out]

int(cos(d*x+c)^5*(a+b*sin(d*x+c))^m,x)

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Maxima [A]
time = 0.30, size = 286, normalized size = 1.71 \begin {gather*} \frac {\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{m + 1}}{b {\left (m + 1\right )}} - \frac {2 \, {\left ({\left (m^{2} + 3 \, m + 2\right )} b^{3} \sin \left (d x + c\right )^{3} + {\left (m^{2} + m\right )} a b^{2} \sin \left (d x + c\right )^{2} - 2 \, a^{2} b m \sin \left (d x + c\right ) + 2 \, a^{3}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} b^{3}} + \frac {{\left ({\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} b^{5} \sin \left (d x + c\right )^{5} + {\left (m^{4} + 6 \, m^{3} + 11 \, m^{2} + 6 \, m\right )} a b^{4} \sin \left (d x + c\right )^{4} - 4 \, {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} a^{2} b^{3} \sin \left (d x + c\right )^{3} + 12 \, {\left (m^{2} + m\right )} a^{3} b^{2} \sin \left (d x + c\right )^{2} - 24 \, a^{4} b m \sin \left (d x + c\right ) + 24 \, a^{5}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{{\left (m^{5} + 15 \, m^{4} + 85 \, m^{3} + 225 \, m^{2} + 274 \, m + 120\right )} b^{5}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

((b*sin(d*x + c) + a)^(m + 1)/(b*(m + 1)) - 2*((m^2 + 3*m + 2)*b^3*sin(d*x + c)^3 + (m^2 + m)*a*b^2*sin(d*x +
c)^2 - 2*a^2*b*m*sin(d*x + c) + 2*a^3)*(b*sin(d*x + c) + a)^m/((m^3 + 6*m^2 + 11*m + 6)*b^3) + ((m^4 + 10*m^3
+ 35*m^2 + 50*m + 24)*b^5*sin(d*x + c)^5 + (m^4 + 6*m^3 + 11*m^2 + 6*m)*a*b^4*sin(d*x + c)^4 - 4*(m^3 + 3*m^2
+ 2*m)*a^2*b^3*sin(d*x + c)^3 + 12*(m^2 + m)*a^3*b^2*sin(d*x + c)^2 - 24*a^4*b*m*sin(d*x + c) + 24*a^5)*(b*sin
(d*x + c) + a)^m/((m^5 + 15*m^4 + 85*m^3 + 225*m^2 + 274*m + 120)*b^5))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (167) = 334\).
time = 0.41, size = 381, normalized size = 2.28 \begin {gather*} \frac {{\left (24 \, a^{5} - 80 \, a^{3} b^{2} + 120 \, a b^{4} + {\left (a b^{4} m^{4} + 6 \, a b^{4} m^{3} + 11 \, a b^{4} m^{2} + 6 \, a b^{4} m\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (a^{3} b^{2} + 3 \, a b^{4}\right )} m^{2} + 4 \, {\left (2 \, a b^{4} m^{3} - 3 \, {\left (a^{3} b^{2} - 3 \, a b^{4}\right )} m^{2} - {\left (3 \, a^{3} b^{2} - 7 \, a b^{4}\right )} m\right )} \cos \left (d x + c\right )^{2} - 24 \, {\left (a^{3} b^{2} - 5 \, a b^{4}\right )} m + {\left (64 \, b^{5} + {\left (b^{5} m^{4} + 10 \, b^{5} m^{3} + 35 \, b^{5} m^{2} + 50 \, b^{5} m + 24 \, b^{5}\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (3 \, a^{2} b^{3} + b^{5}\right )} m^{2} + 4 \, {\left (8 \, b^{5} + {\left (a^{2} b^{3} + b^{5}\right )} m^{3} + {\left (3 \, a^{2} b^{3} + 7 \, b^{5}\right )} m^{2} + 2 \, {\left (a^{2} b^{3} + 7 \, b^{5}\right )} m\right )} \cos \left (d x + c\right )^{2} - 24 \, {\left (a^{4} b - 3 \, a^{2} b^{3} - 2 \, b^{5}\right )} m\right )} \sin \left (d x + c\right )\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{b^{5} d m^{5} + 15 \, b^{5} d m^{4} + 85 \, b^{5} d m^{3} + 225 \, b^{5} d m^{2} + 274 \, b^{5} d m + 120 \, b^{5} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

(24*a^5 - 80*a^3*b^2 + 120*a*b^4 + (a*b^4*m^4 + 6*a*b^4*m^3 + 11*a*b^4*m^2 + 6*a*b^4*m)*cos(d*x + c)^4 + 8*(a^
3*b^2 + 3*a*b^4)*m^2 + 4*(2*a*b^4*m^3 - 3*(a^3*b^2 - 3*a*b^4)*m^2 - (3*a^3*b^2 - 7*a*b^4)*m)*cos(d*x + c)^2 -
24*(a^3*b^2 - 5*a*b^4)*m + (64*b^5 + (b^5*m^4 + 10*b^5*m^3 + 35*b^5*m^2 + 50*b^5*m + 24*b^5)*cos(d*x + c)^4 +
8*(3*a^2*b^3 + b^5)*m^2 + 4*(8*b^5 + (a^2*b^3 + b^5)*m^3 + (3*a^2*b^3 + 7*b^5)*m^2 + 2*(a^2*b^3 + 7*b^5)*m)*co
s(d*x + c)^2 - 24*(a^4*b - 3*a^2*b^3 - 2*b^5)*m)*sin(d*x + c))*(b*sin(d*x + c) + a)^m/(b^5*d*m^5 + 15*b^5*d*m^
4 + 85*b^5*d*m^3 + 225*b^5*d*m^2 + 274*b^5*d*m + 120*b^5*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+b*sin(d*x+c))**m,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1410 vs. \(2 (167) = 334\).
time = 4.55, size = 1410, normalized size = 8.44 \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*sin(d*x+c))^m,x, algorithm="giac")

[Out]

((b*sin(d*x + c) + a)^5*(b*sin(d*x + c) + a)^m*m^4 - 4*(b*sin(d*x + c) + a)^4*(b*sin(d*x + c) + a)^m*a*m^4 + 6
*(b*sin(d*x + c) + a)^3*(b*sin(d*x + c) + a)^m*a^2*m^4 - 4*(b*sin(d*x + c) + a)^2*(b*sin(d*x + c) + a)^m*a^3*m
^4 + (b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*a^4*m^4 - 2*(b*sin(d*x + c) + a)^3*(b*sin(d*x + c) + a)^m*b^2
*m^4 + 4*(b*sin(d*x + c) + a)^2*(b*sin(d*x + c) + a)^m*a*b^2*m^4 - 2*(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)
^m*a^2*b^2*m^4 + (b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*b^4*m^4 + 10*(b*sin(d*x + c) + a)^5*(b*sin(d*x +
c) + a)^m*m^3 - 44*(b*sin(d*x + c) + a)^4*(b*sin(d*x + c) + a)^m*a*m^3 + 72*(b*sin(d*x + c) + a)^3*(b*sin(d*x
+ c) + a)^m*a^2*m^3 - 52*(b*sin(d*x + c) + a)^2*(b*sin(d*x + c) + a)^m*a^3*m^3 + 14*(b*sin(d*x + c) + a)*(b*si
n(d*x + c) + a)^m*a^4*m^3 - 24*(b*sin(d*x + c) + a)^3*(b*sin(d*x + c) + a)^m*b^2*m^3 + 52*(b*sin(d*x + c) + a)
^2*(b*sin(d*x + c) + a)^m*a*b^2*m^3 - 28*(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*a^2*b^2*m^3 + 14*(b*sin(d
*x + c) + a)*(b*sin(d*x + c) + a)^m*b^4*m^3 + 35*(b*sin(d*x + c) + a)^5*(b*sin(d*x + c) + a)^m*m^2 - 164*(b*si
n(d*x + c) + a)^4*(b*sin(d*x + c) + a)^m*a*m^2 + 294*(b*sin(d*x + c) + a)^3*(b*sin(d*x + c) + a)^m*a^2*m^2 - 2
36*(b*sin(d*x + c) + a)^2*(b*sin(d*x + c) + a)^m*a^3*m^2 + 71*(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*a^4*
m^2 - 98*(b*sin(d*x + c) + a)^3*(b*sin(d*x + c) + a)^m*b^2*m^2 + 236*(b*sin(d*x + c) + a)^2*(b*sin(d*x + c) +
a)^m*a*b^2*m^2 - 142*(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*a^2*b^2*m^2 + 71*(b*sin(d*x + c) + a)*(b*sin(
d*x + c) + a)^m*b^4*m^2 + 50*(b*sin(d*x + c) + a)^5*(b*sin(d*x + c) + a)^m*m - 244*(b*sin(d*x + c) + a)^4*(b*s
in(d*x + c) + a)^m*a*m + 468*(b*sin(d*x + c) + a)^3*(b*sin(d*x + c) + a)^m*a^2*m - 428*(b*sin(d*x + c) + a)^2*
(b*sin(d*x + c) + a)^m*a^3*m + 154*(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*a^4*m - 156*(b*sin(d*x + c) + a
)^3*(b*sin(d*x + c) + a)^m*b^2*m + 428*(b*sin(d*x + c) + a)^2*(b*sin(d*x + c) + a)^m*a*b^2*m - 308*(b*sin(d*x
+ c) + a)*(b*sin(d*x + c) + a)^m*a^2*b^2*m + 154*(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*b^4*m + 24*(b*sin
(d*x + c) + a)^5*(b*sin(d*x + c) + a)^m - 120*(b*sin(d*x + c) + a)^4*(b*sin(d*x + c) + a)^m*a + 240*(b*sin(d*x
+ c) + a)^3*(b*sin(d*x + c) + a)^m*a^2 - 240*(b*sin(d*x + c) + a)^2*(b*sin(d*x + c) + a)^m*a^3 + 120*(b*sin(d
*x + c) + a)*(b*sin(d*x + c) + a)^m*a^4 - 80*(b*sin(d*x + c) + a)^3*(b*sin(d*x + c) + a)^m*b^2 + 240*(b*sin(d*
x + c) + a)^2*(b*sin(d*x + c) + a)^m*a*b^2 - 240*(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*a^2*b^2 + 120*(b*
sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*b^4)/((b^4*m^5 + 15*b^4*m^4 + 85*b^4*m^3 + 225*b^4*m^2 + 274*b^4*m +
120*b^4)*b*d)

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Mupad [B]
time = 11.62, size = 641, normalized size = 3.84 \begin {gather*} \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m\,\left (1920\,a\,b^4+1200\,b^5\,\sin \left (c+d\,x\right )+384\,a^5-1280\,a^3\,b^2+200\,b^5\,\sin \left (3\,c+3\,d\,x\right )+24\,b^5\,\sin \left (5\,c+5\,d\,x\right )-480\,a^3\,b^2\,m+738\,a\,b^4\,m^2+100\,a\,b^4\,m^3+6\,a\,b^4\,m^4+374\,b^5\,m\,\sin \left (3\,c+3\,d\,x\right )+50\,b^5\,m\,\sin \left (5\,c+5\,d\,x\right )+310\,b^5\,m^2\,\sin \left (c+d\,x\right )+36\,b^5\,m^3\,\sin \left (c+d\,x\right )+2\,b^5\,m^4\,\sin \left (c+d\,x\right )+32\,a^3\,b^2\,m^2+217\,b^5\,m^2\,\sin \left (3\,c+3\,d\,x\right )+46\,b^5\,m^3\,\sin \left (3\,c+3\,d\,x\right )+3\,b^5\,m^4\,\sin \left (3\,c+3\,d\,x\right )+35\,b^5\,m^2\,\sin \left (5\,c+5\,d\,x\right )+10\,b^5\,m^3\,\sin \left (5\,c+5\,d\,x\right )+b^5\,m^4\,\sin \left (5\,c+5\,d\,x\right )+2180\,a\,b^4\,m+1092\,b^5\,m\,\sin \left (c+d\,x\right )-96\,a^3\,b^2\,m\,\cos \left (2\,c+2\,d\,x\right )+376\,a\,b^4\,m^2\,\cos \left (2\,c+2\,d\,x\right )+112\,a\,b^4\,m^3\,\cos \left (2\,c+2\,d\,x\right )+8\,a\,b^4\,m^4\,\cos \left (2\,c+2\,d\,x\right )+22\,a\,b^4\,m^2\,\cos \left (4\,c+4\,d\,x\right )+12\,a\,b^4\,m^3\,\cos \left (4\,c+4\,d\,x\right )+2\,a\,b^4\,m^4\,\cos \left (4\,c+4\,d\,x\right )+32\,a^2\,b^3\,m\,\sin \left (3\,c+3\,d\,x\right )+432\,a^2\,b^3\,m^2\,\sin \left (c+d\,x\right )+16\,a^2\,b^3\,m^3\,\sin \left (c+d\,x\right )-384\,a^4\,b\,m\,\sin \left (c+d\,x\right )-96\,a^3\,b^2\,m^2\,\cos \left (2\,c+2\,d\,x\right )+48\,a^2\,b^3\,m^2\,\sin \left (3\,c+3\,d\,x\right )+16\,a^2\,b^3\,m^3\,\sin \left (3\,c+3\,d\,x\right )+272\,a\,b^4\,m\,\cos \left (2\,c+2\,d\,x\right )+12\,a\,b^4\,m\,\cos \left (4\,c+4\,d\,x\right )+1184\,a^2\,b^3\,m\,\sin \left (c+d\,x\right )\right )}{16\,b^5\,d\,\left (m^5+15\,m^4+85\,m^3+225\,m^2+274\,m+120\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + b*sin(c + d*x))^m,x)

[Out]

((a + b*sin(c + d*x))^m*(1920*a*b^4 + 1200*b^5*sin(c + d*x) + 384*a^5 - 1280*a^3*b^2 + 200*b^5*sin(3*c + 3*d*x
) + 24*b^5*sin(5*c + 5*d*x) - 480*a^3*b^2*m + 738*a*b^4*m^2 + 100*a*b^4*m^3 + 6*a*b^4*m^4 + 374*b^5*m*sin(3*c
+ 3*d*x) + 50*b^5*m*sin(5*c + 5*d*x) + 310*b^5*m^2*sin(c + d*x) + 36*b^5*m^3*sin(c + d*x) + 2*b^5*m^4*sin(c +
d*x) + 32*a^3*b^2*m^2 + 217*b^5*m^2*sin(3*c + 3*d*x) + 46*b^5*m^3*sin(3*c + 3*d*x) + 3*b^5*m^4*sin(3*c + 3*d*x
) + 35*b^5*m^2*sin(5*c + 5*d*x) + 10*b^5*m^3*sin(5*c + 5*d*x) + b^5*m^4*sin(5*c + 5*d*x) + 2180*a*b^4*m + 1092
*b^5*m*sin(c + d*x) - 96*a^3*b^2*m*cos(2*c + 2*d*x) + 376*a*b^4*m^2*cos(2*c + 2*d*x) + 112*a*b^4*m^3*cos(2*c +
2*d*x) + 8*a*b^4*m^4*cos(2*c + 2*d*x) + 22*a*b^4*m^2*cos(4*c + 4*d*x) + 12*a*b^4*m^3*cos(4*c + 4*d*x) + 2*a*b
^4*m^4*cos(4*c + 4*d*x) + 32*a^2*b^3*m*sin(3*c + 3*d*x) + 432*a^2*b^3*m^2*sin(c + d*x) + 16*a^2*b^3*m^3*sin(c
+ d*x) - 384*a^4*b*m*sin(c + d*x) - 96*a^3*b^2*m^2*cos(2*c + 2*d*x) + 48*a^2*b^3*m^2*sin(3*c + 3*d*x) + 16*a^2
*b^3*m^3*sin(3*c + 3*d*x) + 272*a*b^4*m*cos(2*c + 2*d*x) + 12*a*b^4*m*cos(4*c + 4*d*x) + 1184*a^2*b^3*m*sin(c
+ d*x)))/(16*b^5*d*(274*m + 225*m^2 + 85*m^3 + 15*m^4 + m^5 + 120))

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