3.1260 \(\int \frac{\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=254 \[ -\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a b^4 d}+\frac{4 \left (-3 a^4 b^2+2 a^6+b^6\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^4 d \sqrt{a^2-b^2}}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^2 b^3 d (a+b \sin (c+d x))}-\frac{3 x \left (a^2-b^2\right )}{b^4}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\cot (c+d x)}{a^2 d}-\frac{2 a \cos (c+d x)}{b^3 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 b^2 d}-\frac{x}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.341806, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.379, Rules used = {2897, 3770, 3767, 8, 2638, 2635, 2664, 12, 2660, 618, 204} \[ -\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a b^4 d}+\frac{4 \left (-3 a^4 b^2+2 a^6+b^6\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^4 d \sqrt{a^2-b^2}}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^2 b^3 d (a+b \sin (c+d x))}-\frac{3 x \left (a^2-b^2\right )}{b^4}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\cot (c+d x)}{a^2 d}-\frac{2 a \cos (c+d x)}{b^3 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 b^2 d}-\frac{x}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]

[Out]

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

Rule 2897

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

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2638

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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{\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \left (\frac{3 \left (-a^2+b^2\right )}{b^4}-\frac{2 b \csc (c+d x)}{a^3}+\frac{\csc ^2(c+d x)}{a^2}+\frac{2 a \sin (c+d x)}{b^3}-\frac{\sin ^2(c+d x)}{b^2}-\frac{\left (a^2-b^2\right )^3}{a^2 b^4 (a+b \sin (c+d x))^2}+\frac{2 \left (2 a^6-3 a^4 b^2+b^6\right )}{a^3 b^4 (a+b \sin (c+d x))}\right ) \, dx\\ &=-\frac{3 \left (a^2-b^2\right ) x}{b^4}+\frac{\int \csc ^2(c+d x) \, dx}{a^2}+\frac{(2 a) \int \sin (c+d x) \, dx}{b^3}-\frac{\int \sin ^2(c+d x) \, dx}{b^2}-\frac{(2 b) \int \csc (c+d x) \, dx}{a^3}-\frac{\left (a^2-b^2\right )^3 \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{a^2 b^4}+\frac{\left (2 \left (2 a^6-3 a^4 b^2+b^6\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^3 b^4}\\ &=-\frac{3 \left (a^2-b^2\right ) x}{b^4}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{2 a \cos (c+d x)}{b^3 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^2 b^3 d (a+b \sin (c+d x))}-\frac{\int 1 \, dx}{2 b^2}-\frac{\left (a^2-b^2\right )^2 \int \frac{a}{a+b \sin (c+d x)} \, dx}{a^2 b^4}-\frac{\operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}+\frac{\left (4 \left (2 a^6-3 a^4 b^2+b^6\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^3 b^4 d}\\ &=-\frac{x}{2 b^2}-\frac{3 \left (a^2-b^2\right ) x}{b^4}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{2 a \cos (c+d x)}{b^3 d}-\frac{\cot (c+d x)}{a^2 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^2 b^3 d (a+b \sin (c+d x))}-\frac{\left (a^2-b^2\right )^2 \int \frac{1}{a+b \sin (c+d x)} \, dx}{a b^4}-\frac{\left (8 \left (2 a^6-3 a^4 b^2+b^6\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^3 b^4 d}\\ &=-\frac{x}{2 b^2}-\frac{3 \left (a^2-b^2\right ) x}{b^4}+\frac{4 \left (2 a^6-3 a^4 b^2+b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 b^4 \sqrt{a^2-b^2} d}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{2 a \cos (c+d x)}{b^3 d}-\frac{\cot (c+d x)}{a^2 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^2 b^3 d (a+b \sin (c+d x))}-\frac{\left (2 \left (a^2-b^2\right )^2\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 b^4 d}\\ &=-\frac{x}{2 b^2}-\frac{3 \left (a^2-b^2\right ) x}{b^4}+\frac{4 \left (2 a^6-3 a^4 b^2+b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 b^4 \sqrt{a^2-b^2} d}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{2 a \cos (c+d x)}{b^3 d}-\frac{\cot (c+d x)}{a^2 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^2 b^3 d (a+b \sin (c+d x))}+\frac{\left (4 \left (a^2-b^2\right )^2\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 b^4 d}\\ &=-\frac{x}{2 b^2}-\frac{3 \left (a^2-b^2\right ) x}{b^4}-\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a b^4 d}+\frac{4 \left (2 a^6-3 a^4 b^2+b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 b^4 \sqrt{a^2-b^2} d}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{2 a \cos (c+d x)}{b^3 d}-\frac{\cot (c+d x)}{a^2 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^2 b^3 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.74542, size = 215, normalized size = 0.85 \[ \frac{\frac{2 \left (5 b^2-6 a^2\right ) (c+d x)}{b^4}+\frac{8 \left (3 a^2+2 b^2\right ) \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^4}-\frac{4 \left (a^2-b^2\right )^2 \cos (c+d x)}{a^2 b^3 (a+b \sin (c+d x))}-\frac{8 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}+\frac{8 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}+\frac{2 \tan \left (\frac{1}{2} (c+d x)\right )}{a^2}-\frac{2 \cot \left (\frac{1}{2} (c+d x)\right )}{a^2}-\frac{8 a \cos (c+d x)}{b^3}+\frac{\sin (2 (c+d x))}{b^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]

[Out]

((2*(-6*a^2 + 5*b^2)*(c + d*x))/b^4 + (8*(a^2 - b^2)^(3/2)*(3*a^2 + 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqr
t[a^2 - b^2]])/(a^3*b^4) - (8*a*Cos[c + d*x])/b^3 - (2*Cot[(c + d*x)/2])/a^2 + (8*b*Log[Cos[(c + d*x)/2]])/a^3
- (8*b*Log[Sin[(c + d*x)/2]])/a^3 - (4*(a^2 - b^2)^2*Cos[c + d*x])/(a^2*b^3*(a + b*Sin[c + d*x])) + Sin[2*(c
+ d*x)]/b^2 + (2*Tan[(c + d*x)/2])/a^2)/(4*d)

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Maple [B]  time = 0.172, size = 680, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x)

[Out]

1/2/d/a^2*tan(1/2*d*x+1/2*c)-1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-4/d/b^3/(1+tan(1/2*d*x+1/
2*c)^2)^2*tan(1/2*d*x+1/2*c)^2*a+1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-4/d/b^3/(1+tan(1/2*d*x+
1/2*c)^2)^2*a-6/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a^2+5/d/b^2*arctan(tan(1/2*d*x+1/2*c))-2/d/b^2/(tan(1/2*d*x+1
/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*a*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)
/a*tan(1/2*d*x+1/2*c)-2/d/a^3*b^2/(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^3
/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*a^2+4/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)
-2/d/a^2*b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)+6/d/b^4*a^3/(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))-8/d/b^2*a/(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))-2/d/a/(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))+4/d/a^3/(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))*b^2-1/2/d/a^2/tan(1/2*d*x+1/2*c)-2/d/a^3*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(cos(d*x+c)^6*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.67204, size = 2010, normalized size = 7.91 \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)^6*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/2*(3*a^4*b^2*cos(d*x + c)^3 + (6*a^5*b - 5*a^3*b^3)*d*x*cos(d*x + c)^2 - (6*a^5*b - 5*a^3*b^3)*d*x - (3*a^
4*b - a^2*b^3 - 2*b^5 - (3*a^4*b - a^2*b^3 - 2*b^5)*cos(d*x + c)^2 + (3*a^5 - a^3*b^2 - 2*a*b^4)*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)) - (3*a^4*b^
2 + 2*a^2*b^4)*cos(d*x + c) - 2*(b^6*cos(d*x + c)^2 - a*b^5*sin(d*x + c) - b^6)*log(1/2*cos(d*x + c) + 1/2) +
2*(b^6*cos(d*x + c)^2 - a*b^5*sin(d*x + c) - b^6)*log(-1/2*cos(d*x + c) + 1/2) - (a^3*b^3*cos(d*x + c)^3 + (6*
a^6 - 5*a^4*b^2)*d*x + (6*a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c))*sin(d*x + c))/(a^3*b^5*d*cos(d*x + c)^2 -
a^4*b^4*d*sin(d*x + c) - a^3*b^5*d), -1/2*(3*a^4*b^2*cos(d*x + c)^3 + (6*a^5*b - 5*a^3*b^3)*d*x*cos(d*x + c)^
2 - (6*a^5*b - 5*a^3*b^3)*d*x - 2*(3*a^4*b - a^2*b^3 - 2*b^5 - (3*a^4*b - a^2*b^3 - 2*b^5)*cos(d*x + c)^2 + (3
*a^5 - a^3*b^2 - 2*a*b^4)*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))) - (3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c) - 2*(b^6*cos(d*x + c)^2 - a*b^5*sin(d*x + c) - b^6)*log(1/2*cos(
d*x + c) + 1/2) + 2*(b^6*cos(d*x + c)^2 - a*b^5*sin(d*x + c) - b^6)*log(-1/2*cos(d*x + c) + 1/2) - (a^3*b^3*co
s(d*x + c)^3 + (6*a^6 - 5*a^4*b^2)*d*x + (6*a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c))*sin(d*x + c))/(a^3*b^5*
d*cos(d*x + c)^2 - a^4*b^4*d*sin(d*x + c) - a^3*b^5*d)]

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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(cos(d*x+c)**6*csc(d*x+c)**2/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.26405, size = 518, normalized size = 2.04 \begin{align*} -\frac{\frac{12 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} + \frac{3 \,{\left (6 \, a^{2} - 5 \, b^{2}\right )}{\left (d x + c\right )}}{b^{4}} + \frac{6 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}} - \frac{12 \,{\left (3 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} + 2 \, b^{6}\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^{3} b^{4}} - \frac{4 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 21 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, 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 ) - 3 \, a^{2} b^{3}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} a^{3} b^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(12*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 3*tan(1/2*d*x + 1/2*c)/a^2 + 3*(6*a^2 - 5*b^2)*(d*x + c)/b^4 +
6*(b*tan(1/2*d*x + 1/2*c)^3 + 4*a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c) + 4*a)/((tan(1/2*d*x + 1/2*
c)^2 + 1)^2*b^3) - 12*(3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*(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^3*b^4) - (4*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 1
2*a^4*b*tan(1/2*d*x + 1/2*c)^2 + 21*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 - 4*b^5*tan(1/2*d*x + 1/2*c)^2 - 12*a^5*tan
(1/2*d*x + 1/2*c) + 24*a^3*b^2*tan(1/2*d*x + 1/2*c) - 14*a*b^4*tan(1/2*d*x + 1/2*c) - 3*a^2*b^3)/((a*tan(1/2*d
*x + 1/2*c)^3 + 2*b*tan(1/2*d*x + 1/2*c)^2 + a*tan(1/2*d*x + 1/2*c))*a^3*b^3))/d