3.1360 \(\int \frac{\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=221 \[ \frac{a^7 \log (a+b \sin (c+d x))}{b^2 d \left (a^2-b^2\right )^3}-\frac{\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac{\left (24 a^2-37 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac{\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}-\frac{\sin (c+d x)}{b d} \]

[Out]

-((24*a^2 + 37*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) - ((24*a^2 - 37*a*b + 15*b^2)*Log[1 + Sin
[c + d*x]])/(16*(a - b)^3*d) + (a^7*Log[a + b*Sin[c + d*x]])/(b^2*(a^2 - b^2)^3*d) - Sin[c + d*x]/(b*d) + (Sec
[c + d*x]^4*(a - b*Sin[c + d*x]))/(4*(a^2 - b^2)*d) - (Sec[c + d*x]^2*(4*a*(3*a^2 - 2*b^2) - b*(13*a^2 - 9*b^2
)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d)

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Rubi [A]  time = 0.539119, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2837, 12, 1647, 1629} \[ \frac{a^7 \log (a+b \sin (c+d x))}{b^2 d \left (a^2-b^2\right )^3}-\frac{\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac{\left (24 a^2-37 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac{\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}-\frac{\sin (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^2*Tan[c + d*x]^5)/(a + b*Sin[c + d*x]),x]

[Out]

-((24*a^2 + 37*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) - ((24*a^2 - 37*a*b + 15*b^2)*Log[1 + Sin
[c + d*x]])/(16*(a - b)^3*d) + (a^7*Log[a + b*Sin[c + d*x]])/(b^2*(a^2 - b^2)^3*d) - Sin[c + d*x]/(b*d) + (Sec
[c + d*x]^4*(a - b*Sin[c + d*x]))/(4*(a^2 - b^2)*d) - (Sec[c + d*x]^2*(4*a*(3*a^2 - 2*b^2) - b*(13*a^2 - 9*b^2
)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d)

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 1629

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

Rubi steps

\begin{align*} \int \frac{\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x^7}{b^7 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^7}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{b^2 d}\\ &=\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}+\frac{\operatorname{Subst}\left (\int \frac{\frac{a b^8}{a^2-b^2}-\frac{b^6 \left (4 a^2-b^2\right ) x}{a^2-b^2}-4 b^4 x^3-4 b^2 x^5}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^4 d}\\ &=\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac{\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{a b^8 \left (11 a^2-7 b^2\right )}{\left (a^2-b^2\right )^2}+\frac{b^6 \left (16 a^4-19 a^2 b^2+7 b^4\right ) x}{\left (a^2-b^2\right )^2}+8 b^4 x^3}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^6 d}\\ &=\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac{\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac{\operatorname{Subst}\left (\int \left (-8 b^4+\frac{b^6 \left (24 a^2+37 a b+15 b^2\right )}{2 (a+b)^3 (b-x)}+\frac{8 a^7 b^4}{(a-b)^3 (a+b)^3 (a+x)}+\frac{b^6 \left (24 a^2-37 a b+15 b^2\right )}{2 (-a+b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^6 d}\\ &=-\frac{\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac{\left (24 a^2-37 a b+15 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac{a^7 \log (a+b \sin (c+d x))}{b^2 \left (a^2-b^2\right )^3 d}-\frac{\sin (c+d x)}{b d}+\frac{\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac{\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ \end{align*}

Mathematica [A]  time = 2.42689, size = 198, normalized size = 0.9 \[ \frac{\frac{16 a^7 \log (a+b \sin (c+d x))}{b^2 (a-b)^3 (a+b)^3}-\frac{\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}-\frac{\left (24 a^2-37 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{(a-b)^3}+\frac{11 a+9 b}{(a+b)^2 (\sin (c+d x)-1)}+\frac{9 b-11 a}{(a-b)^2 (\sin (c+d x)+1)}+\frac{1}{(a+b) (\sin (c+d x)-1)^2}+\frac{1}{(a-b) (\sin (c+d x)+1)^2}-\frac{16 \sin (c+d x)}{b}}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^5)/(a + b*Sin[c + d*x]),x]

[Out]

(-(((24*a^2 + 37*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(a + b)^3) - ((24*a^2 - 37*a*b + 15*b^2)*Log[1 + Sin[c +
d*x]])/(a - b)^3 + (16*a^7*Log[a + b*Sin[c + d*x]])/((a - b)^3*b^2*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x]
)^2) + (11*a + 9*b)/((a + b)^2*(-1 + Sin[c + d*x])) - (16*Sin[c + d*x])/b + 1/((a - b)*(1 + Sin[c + d*x])^2) +
(-11*a + 9*b)/((a - b)^2*(1 + Sin[c + d*x])))/(16*d)

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Maple [A]  time = 0.095, size = 321, normalized size = 1.5 \begin{align*} -{\frac{\sin \left ( dx+c \right ) }{bd}}+{\frac{{a}^{7}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{2} \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}+{\frac{1}{2\,d \left ( 8\,a+8\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{11\,a}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{9\,b}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ){a}^{2}}{2\,d \left ( a+b \right ) ^{3}}}-{\frac{37\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) ab}{16\,d \left ( a+b \right ) ^{3}}}-{\frac{15\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ){b}^{2}}{16\,d \left ( a+b \right ) ^{3}}}+{\frac{1}{2\,d \left ( 8\,a-8\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{11\,a}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{9\,b}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ){a}^{2}}{2\,d \left ( a-b \right ) ^{3}}}+{\frac{37\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) ab}{16\,d \left ( a-b \right ) ^{3}}}-{\frac{15\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ){b}^{2}}{16\,d \left ( a-b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x)

[Out]

-sin(d*x+c)/b/d+1/d/b^2*a^7/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))+1/2/d/(8*a+8*b)/(sin(d*x+c)-1)^2+11/16/d/(a+b)^
2/(sin(d*x+c)-1)*a+9/16/d/(a+b)^2/(sin(d*x+c)-1)*b-3/2/d/(a+b)^3*ln(sin(d*x+c)-1)*a^2-37/16/d/(a+b)^3*ln(sin(d
*x+c)-1)*a*b-15/16/d/(a+b)^3*ln(sin(d*x+c)-1)*b^2+1/2/d/(8*a-8*b)/(1+sin(d*x+c))^2-11/16/d/(a-b)^2/(1+sin(d*x+
c))*a+9/16/d/(a-b)^2/(1+sin(d*x+c))*b-3/2/d/(a-b)^3*ln(1+sin(d*x+c))*a^2+37/16/d/(a-b)^3*ln(1+sin(d*x+c))*a*b-
15/16/d/(a-b)^3*ln(1+sin(d*x+c))*b^2

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Maxima [A]  time = 1.03794, size = 409, normalized size = 1.85 \begin{align*} \frac{\frac{16 \, a^{7} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}} - \frac{{\left (24 \, a^{2} - 37 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{{\left (24 \, a^{2} + 37 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{2 \,{\left ({\left (13 \, a^{2} b - 9 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 10 \, a^{3} - 6 \, a b^{2} - 4 \,{\left (3 \, a^{3} - 2 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} -{\left (11 \, a^{2} b - 7 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}} - \frac{16 \, \sin \left (d x + c\right )}{b}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*(16*a^7*log(b*sin(d*x + c) + a)/(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8) - (24*a^2 - 37*a*b + 15*b^2)*log(
sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (24*a^2 + 37*a*b + 15*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*
a^2*b + 3*a*b^2 + b^3) - 2*((13*a^2*b - 9*b^3)*sin(d*x + c)^3 + 10*a^3 - 6*a*b^2 - 4*(3*a^3 - 2*a*b^2)*sin(d*x
+ c)^2 - (11*a^2*b - 7*b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2
*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2) - 16*sin(d*x + c)/b)/d

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Fricas [A]  time = 3.47397, size = 780, normalized size = 3.53 \begin{align*} \frac{16 \, a^{7} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{5} b^{2} - 8 \, a^{3} b^{4} + 4 \, a b^{6} -{\left (24 \, a^{5} b^{2} + 35 \, a^{4} b^{3} - 24 \, a^{3} b^{4} - 42 \, a^{2} b^{5} + 8 \, a b^{6} + 15 \, b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (24 \, a^{5} b^{2} - 35 \, a^{4} b^{3} - 24 \, a^{3} b^{4} + 42 \, a^{2} b^{5} + 8 \, a b^{6} - 15 \, b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \,{\left (3 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (2 \, a^{4} b^{3} - 4 \, a^{2} b^{5} + 2 \, b^{7} + 8 \,{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} -{\left (13 \, a^{4} b^{3} - 22 \, a^{2} b^{5} + 9 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \,{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(16*a^7*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + 4*a^5*b^2 - 8*a^3*b^4 + 4*a*b^6 - (24*a^5*b^2 + 35*a^4*b
^3 - 24*a^3*b^4 - 42*a^2*b^5 + 8*a*b^6 + 15*b^7)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (24*a^5*b^2 - 35*a^4*b
^3 - 24*a^3*b^4 + 42*a^2*b^5 + 8*a*b^6 - 15*b^7)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 8*(3*a^5*b^2 - 5*a^3*
b^4 + 2*a*b^6)*cos(d*x + c)^2 - 2*(2*a^4*b^3 - 4*a^2*b^5 + 2*b^7 + 8*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos
(d*x + c)^4 - (13*a^4*b^3 - 22*a^2*b^5 + 9*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^
6 - b^8)*d*cos(d*x + c)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**7/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.26077, size = 518, normalized size = 2.34 \begin{align*} \frac{\frac{16 \, a^{7} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}} - \frac{{\left (24 \, a^{2} - 37 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{{\left (24 \, a^{2} + 37 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{16 \, \sin \left (d x + c\right )}{b} + \frac{2 \,{\left (18 \, a^{5} \sin \left (d x + c\right )^{4} - 18 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} + 6 \, a b^{4} \sin \left (d x + c\right )^{4} - 13 \, a^{4} b \sin \left (d x + c\right )^{3} + 22 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} - 9 \, b^{5} \sin \left (d x + c\right )^{3} - 24 \, a^{5} \sin \left (d x + c\right )^{2} + 16 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} - 4 \, a b^{4} \sin \left (d x + c\right )^{2} + 11 \, a^{4} b \sin \left (d x + c\right ) - 18 \, a^{2} b^{3} \sin \left (d x + c\right ) + 7 \, b^{5} \sin \left (d x + c\right ) + 8 \, a^{5} - 2 \, a^{3} b^{2}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*(16*a^7*log(abs(b*sin(d*x + c) + a))/(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8) - (24*a^2 - 37*a*b + 15*b^2)
*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (24*a^2 + 37*a*b + 15*b^2)*log(abs(sin(d*x + c)
- 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 16*sin(d*x + c)/b + 2*(18*a^5*sin(d*x + c)^4 - 18*a^3*b^2*sin(d*x + c)
^4 + 6*a*b^4*sin(d*x + c)^4 - 13*a^4*b*sin(d*x + c)^3 + 22*a^2*b^3*sin(d*x + c)^3 - 9*b^5*sin(d*x + c)^3 - 24*
a^5*sin(d*x + c)^2 + 16*a^3*b^2*sin(d*x + c)^2 - 4*a*b^4*sin(d*x + c)^2 + 11*a^4*b*sin(d*x + c) - 18*a^2*b^3*s
in(d*x + c) + 7*b^5*sin(d*x + c) + 8*a^5 - 2*a^3*b^2)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(sin(d*x + c)^2 - 1
)^2))/d