3.193 \(\int \frac{\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=232 \[ \frac{a^3}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{a^2 \left (a^2+3 b^2\right )}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{a \left (8 a^2 b^2+a^4+3 b^4\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}+\frac{\sec ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}+\frac{(2 a-b) \log (1-\sin (c+d x))}{4 d (a+b)^4}+\frac{(2 a+b) \log (\sin (c+d x)+1)}{4 d (a-b)^4} \]

[Out]

((2*a - b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^4*d) + ((2*a + b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^4*d) - (a*(a^
4 + 8*a^2*b^2 + 3*b^4)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^4*d) + a^3/(2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x]
)^2) + (a^2*(a^2 + 3*b^2))/((a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) + (Sec[c + d*x]^2*(a*(a^2 + 3*b^2) - b*(3*a^
2 + b^2)*Sin[c + d*x]))/(2*(a^2 - b^2)^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.482381, antiderivative size = 232, normalized size of antiderivative = 1., 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 = {2721, 1647, 1629} \[ \frac{a^3}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{a^2 \left (a^2+3 b^2\right )}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{a \left (8 a^2 b^2+a^4+3 b^4\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}+\frac{\sec ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}+\frac{(2 a-b) \log (1-\sin (c+d x))}{4 d (a+b)^4}+\frac{(2 a+b) \log (\sin (c+d x)+1)}{4 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/(a + b*Sin[c + d*x])^3,x]

[Out]

((2*a - b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^4*d) + ((2*a + b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^4*d) - (a*(a^
4 + 8*a^2*b^2 + 3*b^4)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^4*d) + a^3/(2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x]
)^2) + (a^2*(a^2 + 3*b^2))/((a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) + (Sec[c + d*x]^2*(a*(a^2 + 3*b^2) - b*(3*a^
2 + b^2)*Sin[c + d*x]))/(2*(a^2 - b^2)^3*d)

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(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] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

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{\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{(a+x)^3 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}+\frac{\operatorname{Subst}\left (\int \frac{\frac{a^3 b^4 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3}-\frac{a^2 b^2 \left (2 a^4-3 a^2 b^2-3 b^4\right ) x}{\left (a^2-b^2\right )^3}-\frac{a b^4 \left (7 a^2-3 b^2\right ) x^2}{\left (a^2-b^2\right )^3}-\frac{b^4 \left (3 a^2+b^2\right ) x^3}{\left (a^2-b^2\right )^3}}{(a+x)^3 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac{\sec ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}+\frac{\operatorname{Subst}\left (\int \left (\frac{b^2 (-2 a+b)}{2 (a+b)^4 (b-x)}-\frac{2 a^3 b^2}{\left (a^2-b^2\right )^2 (a+x)^3}-\frac{2 a^2 b^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3 (a+x)^2}-\frac{2 a b^2 \left (a^4+8 a^2 b^2+3 b^4\right )}{\left (a^2-b^2\right )^4 (a+x)}+\frac{b^2 (2 a+b)}{2 (a-b)^4 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac{(2 a-b) \log (1-\sin (c+d x))}{4 (a+b)^4 d}+\frac{(2 a+b) \log (1+\sin (c+d x))}{4 (a-b)^4 d}-\frac{a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}+\frac{a^3}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{a^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{\sec ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}\\ \end{align*}

Mathematica [A]  time = 2.189, size = 196, normalized size = 0.84 \[ \frac{\frac{2 a^3}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{4 a^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{4 a \left (8 a^2 b^2+a^4+3 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4}-\frac{1}{(a+b)^3 (\sin (c+d x)-1)}+\frac{1}{(a-b)^3 (\sin (c+d x)+1)}+\frac{(2 a-b) \log (1-\sin (c+d x))}{(a+b)^4}+\frac{(2 a+b) \log (\sin (c+d x)+1)}{(a-b)^4}}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/(a + b*Sin[c + d*x])^3,x]

[Out]

(((2*a - b)*Log[1 - Sin[c + d*x]])/(a + b)^4 + ((2*a + b)*Log[1 + Sin[c + d*x]])/(a - b)^4 - (4*a*(a^4 + 8*a^2
*b^2 + 3*b^4)*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^4 - 1/((a + b)^3*(-1 + Sin[c + d*x])) + 1/((a - b)^3*(1 + S
in[c + d*x])) + (2*a^3)/((a^2 - b^2)^2*(a + b*Sin[c + d*x])^2) + (4*a^2*(a^2 + 3*b^2))/((a^2 - b^2)^3*(a + b*S
in[c + d*x])))/(4*d)

________________________________________________________________________________________

Maple [A]  time = 0.109, size = 323, normalized size = 1.4 \begin{align*}{\frac{{a}^{3}}{2\,d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{5}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}-8\,{\frac{{a}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{2}}{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}-3\,{\frac{a\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{4}}{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}+{\frac{{a}^{4}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( a+b\sin \left ( dx+c \right ) \right ) }}+3\,{\frac{{a}^{2}{b}^{2}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-{\frac{1}{4\,d \left ( a+b \right ) ^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) a}{2\,d \left ( a+b \right ) ^{4}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) b}{4\,d \left ( a+b \right ) ^{4}}}+{\frac{1}{4\,d \left ( a-b \right ) ^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) a}{2\,d \left ( a-b \right ) ^{4}}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) b}{4\,d \left ( a-b \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3/(a+b*sin(d*x+c))^3,x)

[Out]

1/2/d*a^3/(a+b)^2/(a-b)^2/(a+b*sin(d*x+c))^2-1/d*a^5/(a+b)^4/(a-b)^4*ln(a+b*sin(d*x+c))-8/d*a^3/(a+b)^4/(a-b)^
4*ln(a+b*sin(d*x+c))*b^2-3/d*a/(a+b)^4/(a-b)^4*ln(a+b*sin(d*x+c))*b^4+1/d*a^4/(a+b)^3/(a-b)^3/(a+b*sin(d*x+c))
+3/d*a^2/(a+b)^3/(a-b)^3/(a+b*sin(d*x+c))*b^2-1/4/d/(a+b)^3/(sin(d*x+c)-1)+1/2/d/(a+b)^4*ln(sin(d*x+c)-1)*a-1/
4/d/(a+b)^4*ln(sin(d*x+c)-1)*b+1/4/d/(a-b)^3/(1+sin(d*x+c))+1/2/d/(a-b)^4*ln(1+sin(d*x+c))*a+1/4/d/(a-b)^4*ln(
1+sin(d*x+c))*b

________________________________________________________________________________________

Maxima [A]  time = 1.87626, size = 595, normalized size = 2.56 \begin{align*} -\frac{\frac{4 \,{\left (a^{5} + 8 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac{{\left (2 \, a + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac{{\left (2 \, a - b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac{2 \,{\left (4 \, a^{5} + 8 \, a^{3} b^{2} -{\left (2 \, a^{4} b + 9 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{3} -{\left (3 \, a^{5} + 10 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{2} +{\left (a^{4} b + 11 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )}}{a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6} -{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \sin \left (d x + c\right )^{4} - 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (d x + c\right )^{3} -{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sin \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (d x + c\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/4*(4*(a^5 + 8*a^3*b^2 + 3*a*b^4)*log(b*sin(d*x + c) + a)/(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) -
(2*a + b)*log(sin(d*x + c) + 1)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (2*a - b)*log(sin(d*x + c) - 1)/
(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 2*(4*a^5 + 8*a^3*b^2 - (2*a^4*b + 9*a^2*b^3 + b^5)*sin(d*x + c)^
3 - (3*a^5 + 10*a^3*b^2 - a*b^4)*sin(d*x + c)^2 + (a^4*b + 11*a^2*b^3)*sin(d*x + c))/(a^8 - 3*a^6*b^2 + 3*a^4*
b^4 - a^2*b^6 - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*sin(d*x + c)^4 - 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*
b^7)*sin(d*x + c)^3 - (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*sin(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 +
3*a^3*b^5 - a*b^7)*sin(d*x + c)))/d

________________________________________________________________________________________

Fricas [B]  time = 3.96383, size = 1727, normalized size = 7.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + 2*(3*a^7 + 7*a^5*b^2 - 11*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + 4*
((a^5*b^2 + 8*a^3*b^4 + 3*a*b^6)*cos(d*x + c)^4 - 2*(a^6*b + 8*a^4*b^3 + 3*a^2*b^5)*cos(d*x + c)^2*sin(d*x + c
) - (a^7 + 9*a^5*b^2 + 11*a^3*b^4 + 3*a*b^6)*cos(d*x + c)^2)*log(b*sin(d*x + c) + a) - ((2*a^5*b^2 + 9*a^4*b^3
+ 16*a^3*b^4 + 14*a^2*b^5 + 6*a*b^6 + b^7)*cos(d*x + c)^4 - 2*(2*a^6*b + 9*a^5*b^2 + 16*a^4*b^3 + 14*a^3*b^4
+ 6*a^2*b^5 + a*b^6)*cos(d*x + c)^2*sin(d*x + c) - (2*a^7 + 9*a^6*b + 18*a^5*b^2 + 23*a^4*b^3 + 22*a^3*b^4 + 1
5*a^2*b^5 + 6*a*b^6 + b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((2*a^5*b^2 - 9*a^4*b^3 + 16*a^3*b^4 - 14*a
^2*b^5 + 6*a*b^6 - b^7)*cos(d*x + c)^4 - 2*(2*a^6*b - 9*a^5*b^2 + 16*a^4*b^3 - 14*a^3*b^4 + 6*a^2*b^5 - a*b^6)
*cos(d*x + c)^2*sin(d*x + c) - (2*a^7 - 9*a^6*b + 18*a^5*b^2 - 23*a^4*b^3 + 22*a^3*b^4 - 15*a^2*b^5 + 6*a*b^6
- b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 - (2*a^6*b + 7*a^4*b^3
- 8*a^2*b^5 - b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d
*x + c)^4 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)^2*sin(d*x + c) - (a^10 - 3*a^
8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c)^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{3}{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**3/(a + b*sin(c + d*x))**3, x)

________________________________________________________________________________________

Giac [B]  time = 2.10026, size = 626, normalized size = 2.7 \begin{align*} -\frac{\frac{4 \,{\left (a^{5} b + 8 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}} - \frac{{\left (2 \, a + b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac{{\left (2 \, a - b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac{2 \,{\left (a^{5} \sin \left (d x + c\right )^{2} + 8 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 3 \, a b^{4} \sin \left (d x + c\right )^{2} - 3 \, a^{4} b \sin \left (d x + c\right ) + 2 \, a^{2} b^{3} \sin \left (d x + c\right ) + b^{5} \sin \left (d x + c\right ) - 6 \, a^{3} b^{2} - 6 \, a b^{4}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}} - \frac{2 \,{\left (3 \, a^{5} b^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{3} b^{4} \sin \left (d x + c\right )^{2} + 9 \, a b^{6} \sin \left (d x + c\right )^{2} + 8 \, a^{6} b \sin \left (d x + c\right ) + 52 \, a^{4} b^{3} \sin \left (d x + c\right ) + 12 \, a^{2} b^{5} \sin \left (d x + c\right ) + 6 \, a^{7} + 26 \, a^{5} b^{2} + 4 \, a^{3} b^{4}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/4*(4*(a^5*b + 8*a^3*b^3 + 3*a*b^5)*log(abs(b*sin(d*x + c) + a))/(a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7
+ b^9) - (2*a + b)*log(abs(sin(d*x + c) + 1))/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (2*a - b)*log(abs(
sin(d*x + c) - 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + 2*(a^5*sin(d*x + c)^2 + 8*a^3*b^2*sin(d*x + c
)^2 + 3*a*b^4*sin(d*x + c)^2 - 3*a^4*b*sin(d*x + c) + 2*a^2*b^3*sin(d*x + c) + b^5*sin(d*x + c) - 6*a^3*b^2 -
6*a*b^4)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(sin(d*x + c)^2 - 1)) - 2*(3*a^5*b^2*sin(d*x + c)^2
+ 24*a^3*b^4*sin(d*x + c)^2 + 9*a*b^6*sin(d*x + c)^2 + 8*a^6*b*sin(d*x + c) + 52*a^4*b^3*sin(d*x + c) + 12*a^2
*b^5*sin(d*x + c) + 6*a^7 + 26*a^5*b^2 + 4*a^3*b^4)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(b*sin(d*
x + c) + a)^2))/d