3.1368 \(\int \frac{\csc ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=250 \[ -\frac{b^7 \log (a+b \sin (c+d x))}{a^2 d \left (a^2-b^2\right )^3}-\frac{\left (15 a^2+37 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac{\left (15 a^2-37 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac{b \log (\sin (c+d x))}{a^2 d}+\frac{7 a+9 b}{16 d (a+b)^2 (1-\sin (c+d x))}-\frac{7 a-9 b}{16 d (a-b)^2 (\sin (c+d x)+1)}+\frac{1}{16 d (a+b) (1-\sin (c+d x))^2}-\frac{1}{16 d (a-b) (\sin (c+d x)+1)^2}-\frac{\csc (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.404545, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2837, 12, 894} \[ -\frac{b^7 \log (a+b \sin (c+d x))}{a^2 d \left (a^2-b^2\right )^3}-\frac{\left (15 a^2+37 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac{\left (15 a^2-37 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac{b \log (\sin (c+d x))}{a^2 d}+\frac{7 a+9 b}{16 d (a+b)^2 (1-\sin (c+d x))}-\frac{7 a-9 b}{16 d (a-b)^2 (\sin (c+d x)+1)}+\frac{1}{16 d (a+b) (1-\sin (c+d x))^2}-\frac{1}{16 d (a-b) (\sin (c+d x)+1)^2}-\frac{\csc (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 6.20033, size = 257, normalized size = 1.03 \[ \frac{b^7 \left (-\frac{\left (15 a^2+37 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 b^7 (a+b)^3}-\frac{\log (\sin (c+d x))}{a^2 b^6}+\frac{\left (15 a^2-37 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{16 b^7 (a-b)^3}-\frac{\log (a+b \sin (c+d x))}{a^2 (a-b)^3 (a+b)^3}-\frac{7 a-9 b}{16 b^6 (a-b)^2 (b \sin (c+d x)+b)}+\frac{7 a+9 b}{16 b^6 (a+b)^2 (b-b \sin (c+d x))}+\frac{1}{16 b^5 (a+b) (b-b \sin (c+d x))^2}-\frac{1}{16 b^5 (a-b) (b \sin (c+d x)+b)^2}-\frac{\csc (c+d x)}{a b^7}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^7*(-(Csc[c + d*x]/(a*b^7)) - ((15*a^2 + 37*a*b + 24*b^2)*Log[1 - Sin[c + d*x]])/(16*b^7*(a + b)^3) - Log[Si
n[c + d*x]]/(a^2*b^6) + ((15*a^2 - 37*a*b + 24*b^2)*Log[1 + Sin[c + d*x]])/(16*(a - b)^3*b^7) - Log[a + b*Sin[
c + d*x]]/(a^2*(a - b)^3*(a + b)^3) + 1/(16*b^5*(a + b)*(b - b*Sin[c + d*x])^2) + (7*a + 9*b)/(16*b^6*(a + b)^
2*(b - b*Sin[c + d*x])) - 1/(16*(a - b)*b^5*(b + b*Sin[c + d*x])^2) - (7*a - 9*b)/(16*(a - b)^2*b^6*(b + b*Sin
[c + d*x]))))/d

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Maple [A]  time = 0.102, size = 340, normalized size = 1.4 \begin{align*} -{\frac{{b}^{7}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}{a}^{2}}}+{\frac{1}{2\,d \left ( 8\,a+8\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{7\,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{15\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ){a}^{2}}{16\,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{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ){b}^{2}}{2\,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{7\,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{15\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ){a}^{2}}{16\,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{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ){b}^{2}}{2\,d \left ( a-b \right ) ^{3}}}-{\frac{1}{da\sin \left ( dx+c \right ) }}-{\frac{b\ln \left ( \sin \left ( dx+c \right ) \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^5/(a+b*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.02319, size = 487, normalized size = 1.95 \begin{align*} -\frac{\frac{16 \, b^{7} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}} - \frac{{\left (15 \, a^{2} - 37 \, a b + 24 \, 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 (15 \, a^{2} + 37 \, a b + 24 \, 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 (15 \, a^{4} - 27 \, a^{2} b^{2} + 8 \, b^{4}\right )} \sin \left (d x + c\right )^{4} + 8 \, a^{4} - 16 \, a^{2} b^{2} + 8 \, b^{4} - 4 \,{\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )^{3} -{\left (25 \, a^{4} - 45 \, a^{2} b^{2} + 16 \, b^{4}\right )} \sin \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{3} b - 5 \, a b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{5} - 2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{3} +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )} + \frac{16 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(16*b^7*log(b*sin(d*x + c) + a)/(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6) - (15*a^2 - 37*a*b + 24*b^2)*log
(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (15*a^2 + 37*a*b + 24*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3
*a^2*b + 3*a*b^2 + b^3) + 2*((15*a^4 - 27*a^2*b^2 + 8*b^4)*sin(d*x + c)^4 + 8*a^4 - 16*a^2*b^2 + 8*b^4 - 4*(a^
3*b - 2*a*b^3)*sin(d*x + c)^3 - (25*a^4 - 45*a^2*b^2 + 16*b^4)*sin(d*x + c)^2 + 2*(3*a^3*b - 5*a*b^3)*sin(d*x
+ c))/((a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c)^5 - 2*(a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c)^3 + (a^5 - 2*a^3*b^
2 + a*b^4)*sin(d*x + c)) + 16*b*log(sin(d*x + c))/a^2)/d

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Fricas [A]  time = 14.2377, size = 980, normalized size = 3.92 \begin{align*} -\frac{16 \, b^{7} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) - 4 \, a^{7} + 8 \, a^{5} b^{2} - 4 \, a^{3} b^{4} + 16 \,{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) -{\left (15 \, a^{7} + 8 \, a^{6} b - 42 \, a^{5} b^{2} - 24 \, a^{4} b^{3} + 35 \, a^{3} b^{4} + 24 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) +{\left (15 \, a^{7} - 8 \, a^{6} b - 42 \, a^{5} b^{2} + 24 \, a^{4} b^{3} + 35 \, a^{3} b^{4} - 24 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 2 \,{\left (15 \, a^{7} - 42 \, a^{5} b^{2} + 35 \, a^{3} b^{4} - 8 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (5 \, a^{7} - 14 \, a^{5} b^{2} + 9 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5} + 2 \,{\left (a^{6} b - 3 \, a^{4} b^{3} + 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \,{\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(16*b^7*cos(d*x + c)^4*log(b*sin(d*x + c) + a)*sin(d*x + c) - 4*a^7 + 8*a^5*b^2 - 4*a^3*b^4 + 16*(a^6*b
- 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)^4*log(1/2*sin(d*x + c))*sin(d*x + c) - (15*a^7 + 8*a^6*b - 42*a^5*
b^2 - 24*a^4*b^3 + 35*a^3*b^4 + 24*a^2*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1)*sin(d*x + c) + (15*a^7 - 8*a^
6*b - 42*a^5*b^2 + 24*a^4*b^3 + 35*a^3*b^4 - 24*a^2*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)*sin(d*x + c) +
2*(15*a^7 - 42*a^5*b^2 + 35*a^3*b^4 - 8*a*b^6)*cos(d*x + c)^4 - 2*(5*a^7 - 14*a^5*b^2 + 9*a^3*b^4)*cos(d*x + c
)^2 + 4*(a^6*b - 2*a^4*b^3 + a^2*b^5 + 2*(a^6*b - 3*a^4*b^3 + 2*a^2*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^8 -
3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^4*sin(d*x + c))

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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(csc(d*x+c)**2*sec(d*x+c)**5/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.29249, size = 564, normalized size = 2.26 \begin{align*} -\frac{\frac{16 \, b^{8} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}} - \frac{{\left (15 \, a^{2} - 37 \, a b + 24 \, 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 (15 \, a^{2} + 37 \, a b + 24 \, 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 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac{2 \,{\left (6 \, a^{4} b \sin \left (d x + c\right )^{4} - 18 \, a^{2} b^{3} \sin \left (d x + c\right )^{4} + 18 \, b^{5} \sin \left (d x + c\right )^{4} + 7 \, a^{5} \sin \left (d x + c\right )^{3} - 18 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 11 \, a b^{4} \sin \left (d x + c\right )^{3} - 16 \, a^{4} b \sin \left (d x + c\right )^{2} + 48 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} - 44 \, b^{5} \sin \left (d x + c\right )^{2} - 9 \, a^{5} \sin \left (d x + c\right ) + 22 \, a^{3} b^{2} \sin \left (d x + c\right ) - 13 \, a b^{4} \sin \left (d x + c\right ) + 12 \, a^{4} b - 34 \, a^{2} b^{3} + 28 \, b^{5}\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}} - \frac{16 \,{\left (b \sin \left (d x + c\right ) - a\right )}}{a^{2} \sin \left (d x + c\right )}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/16*(16*b^8*log(abs(b*sin(d*x + c) + a))/(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7) - (15*a^2 - 37*a*b + 24*b
^2)*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (15*a^2 + 37*a*b + 24*b^2)*log(abs(sin(d*x +
c) - 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 16*b*log(abs(sin(d*x + c)))/a^2 + 2*(6*a^4*b*sin(d*x + c)^4 - 18*a^
2*b^3*sin(d*x + c)^4 + 18*b^5*sin(d*x + c)^4 + 7*a^5*sin(d*x + c)^3 - 18*a^3*b^2*sin(d*x + c)^3 + 11*a*b^4*sin
(d*x + c)^3 - 16*a^4*b*sin(d*x + c)^2 + 48*a^2*b^3*sin(d*x + c)^2 - 44*b^5*sin(d*x + c)^2 - 9*a^5*sin(d*x + c)
+ 22*a^3*b^2*sin(d*x + c) - 13*a*b^4*sin(d*x + c) + 12*a^4*b - 34*a^2*b^3 + 28*b^5)/((a^6 - 3*a^4*b^2 + 3*a^2
*b^4 - b^6)*(sin(d*x + c)^2 - 1)^2) - 16*(b*sin(d*x + c) - a)/(a^2*sin(d*x + c)))/d