3.200 \(\int \frac{\cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=202 \[ -\frac{\left (-9 a^2 b^2+2 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{3/2}}-\frac{\left (5 a^2-6 b^2\right ) \cot (c+d x)}{2 a^3 d \left (a^2-b^2\right )}+\frac{\left (2 a^2-3 b^2\right ) \cot (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cot (c+d x)}{2 a d (a+b \sin (c+d x))^2} \]

[Out]

-(((2*a^4 - 9*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*(a^2 - b^2)^(3/2)*d)) +
(3*b*ArcTanh[Cos[c + d*x]])/(a^4*d) - ((5*a^2 - 6*b^2)*Cot[c + d*x])/(2*a^3*(a^2 - b^2)*d) + Cot[c + d*x]/(2*a
*d*(a + b*Sin[c + d*x])^2) + ((2*a^2 - 3*b^2)*Cot[c + d*x])/(2*a^2*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

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Rubi [A]  time = 0.788453, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {2723, 3056, 3055, 3001, 3770, 2660, 618, 204} \[ -\frac{\left (-9 a^2 b^2+2 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{3/2}}-\frac{\left (5 a^2-6 b^2\right ) \cot (c+d x)}{2 a^3 d \left (a^2-b^2\right )}+\frac{\left (2 a^2-3 b^2\right ) \cot (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cot (c+d x)}{2 a d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2/(a + b*Sin[c + d*x])^3,x]

[Out]

-(((2*a^4 - 9*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*(a^2 - b^2)^(3/2)*d)) +
(3*b*ArcTanh[Cos[c + d*x]])/(a^4*d) - ((5*a^2 - 6*b^2)*Cot[c + d*x])/(2*a^3*(a^2 - b^2)*d) + Cot[c + d*x]/(2*a
*d*(a + b*Sin[c + d*x])^2) + ((2*a^2 - 3*b^2)*Cot[c + d*x])/(2*a^2*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

Rule 2723

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

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)}{(a+b \sin (c+d x))^3} \, dx &=\int \frac{\csc ^2(c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^3} \, dx\\ &=\frac{\cot (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac{\int \frac{\csc ^2(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))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{\cot (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac{\left (2 a^2-3 b^2\right ) \cot (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\int \frac{\csc ^2(c+d x) \left (5 a^4-11 a^2 b^2+6 b^4-a b \left (a^2-b^2\right ) \sin (c+d x)-\left (2 a^2-3 b^2\right ) \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 )^2}\\ &=-\frac{\left (5 a^2-6 b^2\right ) \cot (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\cot (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac{\left (2 a^2-3 b^2\right ) \cot (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\int \frac{\csc (c+d x) \left (-6 b \left (a^2-b^2\right )^2-a \left (2 a^4-5 a^2 b^2+3 b^4\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (5 a^2-6 b^2\right ) \cot (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\cot (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac{\left (2 a^2-3 b^2\right ) \cot (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{(3 b) \int \csc (c+d x) \, dx}{a^4}-\frac{\left (2 a^4-9 a^2 b^2+6 b^4\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )}\\ &=\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{\left (5 a^2-6 b^2\right ) \cot (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\cot (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac{\left (2 a^2-3 b^2\right ) \cot (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\left (2 a^4-9 a^2 b^2+6 b^4\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 \left (a^2-b^2\right ) d}\\ &=\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{\left (5 a^2-6 b^2\right ) \cot (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\cot (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac{\left (2 a^2-3 b^2\right ) \cot (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\left (2 \left (2 a^4-9 a^2 b^2+6 b^4\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 \left (a^2-b^2\right ) d}\\ &=-\frac{\left (2 a^4-9 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{3/2} d}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{\left (5 a^2-6 b^2\right ) \cot (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\cot (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac{\left (2 a^2-3 b^2\right ) \cot (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 5.78381, size = 195, normalized size = 0.97 \[ \frac{-\frac{2 \left (-9 a^2 b^2+2 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a b \left (4 b^2-3 a^2\right ) \cos (c+d x)}{(a-b) (a+b) (a+b \sin (c+d x))}-\frac{a^2 b \cos (c+d x)}{(a+b \sin (c+d x))^2}+a \tan \left (\frac{1}{2} (c+d x)\right )-a \cot \left (\frac{1}{2} (c+d x)\right )-6 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2/(a + b*Sin[c + d*x])^3,x]

[Out]

((-2*(2*a^4 - 9*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - a*Cot[(
c + d*x)/2] + 6*b*Log[Cos[(c + d*x)/2]] - 6*b*Log[Sin[(c + d*x)/2]] - (a^2*b*Cos[c + d*x])/(a + b*Sin[c + d*x]
)^2 + (a*b*(-3*a^2 + 4*b^2)*Cos[c + d*x])/((a - b)*(a + b)*(a + b*Sin[c + d*x])) + a*Tan[(c + d*x)/2])/(2*a^4*
d)

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Maple [B]  time = 0.116, size = 729, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/a^3*tan(1/2*d*x+1/2*c)-5/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a/(a^2-b^2)*tan(1/2*d*x+1
/2*c)^3*b^2+6/d/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b^4/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3-4/d
/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2-3/d/(tan(1/2*d*x+1/2*c)^
2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b^3/(a^2-b^2)/a^2*tan(1/2*d*x+1/2*c)^2+10/d/a^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(
1/2*d*x+1/2*c)*b+a)^2*b^5/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2-11/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a
)^2/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)*b^2+14/d/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b^4/(a^2-b
^2)*tan(1/2*d*x+1/2*c)-4/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*b+5/d/a^2/(tan(1/2*d*
x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b^3/(a^2-b^2)-2/d/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+
2*b)/(a^2-b^2)^(1/2))+9/d/a^2/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))*b^2-6/d
/a^4/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))*b^4-1/2/d/a^3/tan(1/2*d*x+1/2*c)
-3/d/a^4*b*ln(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(cot(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.54849, size = 3044, normalized size = 15.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*(5*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^3 - 2*(8*a^6*b - 17*a^4*b^3 + 9*a^2*b^5)*cos(d*x + c)
*sin(d*x + c) + (4*a^5*b - 18*a^3*b^3 + 12*a*b^5 - 2*(2*a^5*b - 9*a^3*b^3 + 6*a*b^5)*cos(d*x + c)^2 + (2*a^6 -
7*a^4*b^2 - 3*a^2*b^4 + 6*b^6 - (2*a^4*b^2 - 9*a^2*b^4 + 6*b^6)*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)*sin(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*(2*a^7 + a^5*b^2 - 9*a^
3*b^4 + 6*a*b^6)*cos(d*x + c) + 6*(2*a^5*b^2 - 4*a^3*b^4 + 2*a*b^6 - 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x +
c)^2 + (a^6*b - a^4*b^3 - a^2*b^5 + b^7 - (a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*c
os(d*x + c) + 1/2) - 6*(2*a^5*b^2 - 4*a^3*b^4 + 2*a*b^6 - 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^
6*b - a^4*b^3 - a^2*b^5 + b^7 - (a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c
) + 1/2))/(2*(a^9*b - 2*a^7*b^3 + a^5*b^5)*d*cos(d*x + c)^2 - 2*(a^9*b - 2*a^7*b^3 + a^5*b^5)*d + ((a^8*b^2 -
2*a^6*b^4 + a^4*b^6)*d*cos(d*x + c)^2 - (a^10 - a^8*b^2 - a^6*b^4 + a^4*b^6)*d)*sin(d*x + c)), -1/2*((5*a^5*b^
2 - 11*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^3 - (8*a^6*b - 17*a^4*b^3 + 9*a^2*b^5)*cos(d*x + c)*sin(d*x + c) + (4*a
^5*b - 18*a^3*b^3 + 12*a*b^5 - 2*(2*a^5*b - 9*a^3*b^3 + 6*a*b^5)*cos(d*x + c)^2 + (2*a^6 - 7*a^4*b^2 - 3*a^2*b
^4 + 6*b^6 - (2*a^4*b^2 - 9*a^2*b^4 + 6*b^6)*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^7 + a^5*b^2 - 9*a^3*b^4 + 6*a*b^6)*cos(d*x + c) + 3*(2*a^5*b^
2 - 4*a^3*b^4 + 2*a*b^6 - 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^6*b - a^4*b^3 - a^2*b^5 + b^7 -
(a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*(2*a^5*b^2 - 4*a^3*b
^4 + 2*a*b^6 - 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^6*b - a^4*b^3 - a^2*b^5 + b^7 - (a^4*b^3 -
2*a^2*b^5 + b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/(2*(a^9*b - 2*a^7*b^3 + a^5*b^5)*
d*cos(d*x + c)^2 - 2*(a^9*b - 2*a^7*b^3 + a^5*b^5)*d + ((a^8*b^2 - 2*a^6*b^4 + a^4*b^6)*d*cos(d*x + c)^2 - (a^
10 - a^8*b^2 - a^6*b^4 + a^4*b^6)*d)*sin(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{2}{\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(cot(d*x+c)**2/(a+b*sin(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.55189, size = 458, normalized size = 2.27 \begin{align*} -\frac{\frac{2 \,{\left (2 \, a^{4} - 9 \, a^{2} b^{2} + 6 \, b^{4}\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 )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{2 \,{\left (5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 10 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 11 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 14 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a^{4} b - 5 \, a^{2} b^{3}\right )}}{{\left (a^{6} - a^{4} 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 )}^{2}} + \frac{6 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}} - \frac{6 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a}{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(2*a^4 - 9*a^2*b^2 + 6*b^4)*(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)))/((a^6 - a^4*b^2)*sqrt(a^2 - b^2)) + 2*(5*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^4*tan(1
/2*d*x + 1/2*c)^3 + 4*a^4*b*tan(1/2*d*x + 1/2*c)^2 + 3*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 - 10*b^5*tan(1/2*d*x + 1
/2*c)^2 + 11*a^3*b^2*tan(1/2*d*x + 1/2*c) - 14*a*b^4*tan(1/2*d*x + 1/2*c) + 4*a^4*b - 5*a^2*b^3)/((a^6 - a^4*b
^2)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) + 6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 - ta
n(1/2*d*x + 1/2*c)/a^3 - (6*b*tan(1/2*d*x + 1/2*c) - a)/(a^4*tan(1/2*d*x + 1/2*c)))/d