3.1083 \(\int \frac{\cot ^2(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=157 \[ \frac{2 b \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 d \sqrt{a^2-b^2}}+\frac{\left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac{3 b \cot (c+d x)}{a^3 d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))} \]

[Out]

(2*b*(2*a^2 - 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*Sqrt[a^2 - b^2]*d) + ((a^2 - 6*b^2
)*ArcTanh[Cos[c + d*x]])/(2*a^4*d) + (3*b*Cot[c + d*x])/(a^3*d) - (3*Cot[c + d*x]*Csc[c + d*x])/(2*a^2*d) + (C
ot[c + d*x]*Csc[c + d*x])/(a*d*(a + b*Sin[c + d*x]))

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Rubi [A]  time = 0.767619, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2889, 3056, 3055, 3001, 3770, 2660, 618, 204} \[ \frac{2 b \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 d \sqrt{a^2-b^2}}+\frac{\left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac{3 b \cot (c+d x)}{a^3 d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(2*a^2 - 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*Sqrt[a^2 - b^2]*d) + ((a^2 - 6*b^2
)*ArcTanh[Cos[c + d*x]])/(2*a^4*d) + (3*b*Cot[c + d*x])/(a^3*d) - (3*Cot[c + d*x]*Csc[c + d*x])/(2*a^2*d) + (C
ot[c + d*x]*Csc[c + d*x])/(a*d*(a + b*Sin[c + d*x]))

Rule 2889

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

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cot ^2(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \frac{\csc ^3(c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx\\ &=\frac{\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}+\frac{\int \frac{\csc ^3(c+d x) \left (3 \left (a^2-b^2\right )-2 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}+\frac{\int \frac{\csc ^2(c+d x) \left (-6 b \left (a^2-b^2\right )-a \left (a^2-b^2\right ) \sin (c+d x)+3 b \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac{3 b \cot (c+d x)}{a^3 d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}+\frac{\int \frac{\csc (c+d x) \left (-a^4+7 a^2 b^2-6 b^4+3 a b \left (a^2-b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{3 b \cot (c+d x)}{a^3 d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}-\frac{\left (a^2-6 b^2\right ) \int \csc (c+d x) \, dx}{2 a^4}+\frac{\left (b \left (2 a^2-3 b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^4}\\ &=\frac{\left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac{3 b \cot (c+d x)}{a^3 d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}+\frac{\left (2 b \left (2 a^2-3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 d}\\ &=\frac{\left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac{3 b \cot (c+d x)}{a^3 d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}-\frac{\left (4 b \left (2 a^2-3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 d}\\ &=\frac{2 b \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^4 \sqrt{a^2-b^2} d}+\frac{\left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac{3 b \cot (c+d x)}{a^3 d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.09739, size = 196, normalized size = 1.25 \[ \frac{-\frac{16 b \left (3 b^2-2 a^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-4 \left (a^2-6 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 \left (a^2-6 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-a^2 \csc ^2\left (\frac{1}{2} (c+d x)\right )+a^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )+\frac{8 a b^2 \cos (c+d x)}{a+b \sin (c+d x)}-8 a b \tan \left (\frac{1}{2} (c+d x)\right )+8 a b \cot \left (\frac{1}{2} (c+d x)\right )}{8 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-16*b*(-2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 8*a*b*Cot[(c + d*
x)/2] - a^2*Csc[(c + d*x)/2]^2 + 4*(a^2 - 6*b^2)*Log[Cos[(c + d*x)/2]] - 4*(a^2 - 6*b^2)*Log[Sin[(c + d*x)/2]]
+ a^2*Sec[(c + d*x)/2]^2 + (8*a*b^2*Cos[c + d*x])/(a + b*Sin[c + d*x]) - 8*a*b*Tan[(c + d*x)/2])/(8*a^4*d)

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Maple [B]  time = 0.19, size = 307, normalized size = 2. \begin{align*}{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{b}{d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{{b}^{3}\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{4} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+2\,{\frac{{b}^{2}}{d{a}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+4\,{\frac{b}{d{a}^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-6\,{\frac{{b}^{3}}{d{a}^{4}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{1}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){b}^{2}}{d{a}^{4}}}+{\frac{b}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^3/(a+b*sin(d*x+c))^2,x)

[Out]

1/8/d/a^2*tan(1/2*d*x+1/2*c)^2-1/d/a^3*tan(1/2*d*x+1/2*c)*b+2/d*b^3/a^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+
1/2*c)*b+a)*tan(1/2*d*x+1/2*c)+2/d*b^2/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)+4/d/a^2*b/(a^2-b^
2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-6/d*b^3/a^4/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*
tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-1/8/d/a^2/tan(1/2*d*x+1/2*c)^2-1/2/d/a^2*ln(tan(1/2*d*x+1/2*c))+3/d/a
^4*ln(tan(1/2*d*x+1/2*c))*b^2+1/d*b/a^3/tan(1/2*d*x+1/2*c)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.37467, size = 2500, normalized size = 15.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/4*(12*(a^3*b^2 - a*b^4)*cos(d*x + c)^3 - 6*(a^4*b - a^2*b^3)*cos(d*x + c)*sin(d*x + c) - 2*(2*a^3*b - 3*a*b
^3 - (2*a^3*b - 3*a*b^3)*cos(d*x + c)^2 + (2*a^2*b^2 - 3*b^4 - (2*a^2*b^2 - 3*b^4)*cos(d*x + c)^2)*sin(d*x + c
))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*si
n(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 2*(a^5
- 7*a^3*b^2 + 6*a*b^4)*cos(d*x + c) - (a^5 - 7*a^3*b^2 + 6*a*b^4 - (a^5 - 7*a^3*b^2 + 6*a*b^4)*cos(d*x + c)^2
+ (a^4*b - 7*a^2*b^3 + 6*b^5 - (a^4*b - 7*a^2*b^3 + 6*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c)
+ 1/2) + (a^5 - 7*a^3*b^2 + 6*a*b^4 - (a^5 - 7*a^3*b^2 + 6*a*b^4)*cos(d*x + c)^2 + (a^4*b - 7*a^2*b^3 + 6*b^5
- (a^4*b - 7*a^2*b^3 + 6*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^7 - a^5*b^2)*d*
cos(d*x + c)^2 - (a^7 - a^5*b^2)*d + ((a^6*b - a^4*b^3)*d*cos(d*x + c)^2 - (a^6*b - a^4*b^3)*d)*sin(d*x + c)),
1/4*(12*(a^3*b^2 - a*b^4)*cos(d*x + c)^3 - 6*(a^4*b - a^2*b^3)*cos(d*x + c)*sin(d*x + c) + 4*(2*a^3*b - 3*a*b
^3 - (2*a^3*b - 3*a*b^3)*cos(d*x + c)^2 + (2*a^2*b^2 - 3*b^4 - (2*a^2*b^2 - 3*b^4)*cos(d*x + c)^2)*sin(d*x + c
))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 2*(a^5 - 7*a^3*b^2 + 6*a*b^4
)*cos(d*x + c) - (a^5 - 7*a^3*b^2 + 6*a*b^4 - (a^5 - 7*a^3*b^2 + 6*a*b^4)*cos(d*x + c)^2 + (a^4*b - 7*a^2*b^3
+ 6*b^5 - (a^4*b - 7*a^2*b^3 + 6*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^5 - 7*a^3
*b^2 + 6*a*b^4 - (a^5 - 7*a^3*b^2 + 6*a*b^4)*cos(d*x + c)^2 + (a^4*b - 7*a^2*b^3 + 6*b^5 - (a^4*b - 7*a^2*b^3
+ 6*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^7 - a^5*b^2)*d*cos(d*x + c)^2 - (a^7
- a^5*b^2)*d + ((a^6*b - a^4*b^3)*d*cos(d*x + c)^2 - (a^6*b - a^4*b^3)*d)*sin(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{2}{\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**3/(a+b*sin(d*x+c))**2,x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)**3/(a + b*sin(c + d*x))**2, x)

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Giac [A]  time = 1.33662, size = 347, normalized size = 2.21 \begin{align*} -\frac{\frac{4 \,{\left (a^{2} - 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac{16 \,{\left (2 \, a^{2} b - 3 \, b^{3}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} a^{4}} - \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{4}} - \frac{16 \,{\left (b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a b^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )} a^{4}} - \frac{6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 36 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2}}{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/8*(4*(a^2 - 6*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 - 16*(2*a^2*b - 3*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1
/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^4) - (a^2*tan(1/2*d*x +
1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c))/a^4 - 16*(b^3*tan(1/2*d*x + 1/2*c) + a*b^2)/((a*tan(1/2*d*x + 1/2*c)^2
+ 2*b*tan(1/2*d*x + 1/2*c) + a)*a^4) - (6*a^2*tan(1/2*d*x + 1/2*c)^2 - 36*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*a*b*t
an(1/2*d*x + 1/2*c) - a^2)/(a^4*tan(1/2*d*x + 1/2*c)^2))/d