3.1261 \(\int \frac{\cos ^3(c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=251 \[ -\frac{6 \left (a^2+b^2\right ) \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 b^3 d}+\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^3 b^2 d (a+b \sin (c+d x))}+\frac{3 \left (a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{2 b \cot (c+d x)}{a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{2 a x}{b^3}+\frac{\cos (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.335801, antiderivative size = 251, 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, 3768, 2638, 2664, 12, 2660, 618, 204} \[ -\frac{6 \left (a^2+b^2\right ) \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 b^3 d}+\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^3 b^2 d (a+b \sin (c+d x))}+\frac{3 \left (a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{2 b \cot (c+d x)}{a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{2 a x}{b^3}+\frac{\cos (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*x)/b^3 + (2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*b^3*d) - (6*(a^2 - b
^2)^(3/2)*(a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*b^3*d) - ArcTanh[Cos[c + d*x]]/(2
*a^2*d) + (3*(a^2 - b^2)*ArcTanh[Cos[c + d*x]])/(a^4*d) + Cos[c + d*x]/(b^2*d) + (2*b*Cot[c + d*x])/(a^3*d) -
(Cot[c + d*x]*Csc[c + d*x])/(2*a^2*d) + ((a^2 - b^2)^2*Cos[c + d*x])/(a^3*b^2*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 3768

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

Rule 2638

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

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 ^3(c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \left (\frac{2 a}{b^3}-\frac{3 \left (a^2-b^2\right ) \csc (c+d x)}{a^4}-\frac{2 b \csc ^2(c+d x)}{a^3}+\frac{\csc ^3(c+d x)}{a^2}-\frac{\sin (c+d x)}{b^2}+\frac{\left (a^2-b^2\right )^3}{a^3 b^3 (a+b \sin (c+d x))^2}-\frac{3 \left (a^2-b^2\right )^2 \left (a^2+b^2\right )}{a^4 b^3 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac{2 a x}{b^3}+\frac{\int \csc ^3(c+d x) \, dx}{a^2}-\frac{\int \sin (c+d x) \, dx}{b^2}-\frac{(2 b) \int \csc ^2(c+d x) \, dx}{a^3}-\frac{\left (3 \left (a^2-b^2\right )\right ) \int \csc (c+d x) \, dx}{a^4}+\frac{\left (a^2-b^2\right )^3 \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{a^3 b^3}-\frac{\left (3 \left (a^2-b^2\right )^2 \left (a^2+b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^4 b^3}\\ &=\frac{2 a x}{b^3}+\frac{3 \left (a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^3 b^2 d (a+b \sin (c+d x))}+\frac{\int \csc (c+d x) \, dx}{2 a^2}+\frac{\left (a^2-b^2\right )^2 \int \frac{a}{a+b \sin (c+d x)} \, dx}{a^3 b^3}+\frac{(2 b) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}-\frac{\left (6 \left (a^2-b^2\right )^2 \left (a^2+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 b^3 d}\\ &=\frac{2 a x}{b^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac{3 \left (a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^2 d}+\frac{2 b \cot (c+d x)}{a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^3 b^2 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^2 b^3}+\frac{\left (12 \left (a^2-b^2\right )^2 \left (a^2+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 b^3 d}\\ &=\frac{2 a x}{b^3}-\frac{6 \left (a^2-b^2\right )^{3/2} \left (a^2+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 b^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac{3 \left (a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^2 d}+\frac{2 b \cot (c+d x)}{a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^3 b^2 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^2 b^3 d}\\ &=\frac{2 a x}{b^3}-\frac{6 \left (a^2-b^2\right )^{3/2} \left (a^2+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 b^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac{3 \left (a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^2 d}+\frac{2 b \cot (c+d x)}{a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^3 b^2 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^2 b^3 d}\\ &=\frac{2 a x}{b^3}+\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^2 b^3 d}-\frac{6 \left (a^2-b^2\right )^{3/2} \left (a^2+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 b^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac{3 \left (a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^2 d}+\frac{2 b \cot (c+d x)}{a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a^3 b^2 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.19018, size = 315, normalized size = 1.25 \[ \frac{\left (6 b^2-5 a^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac{\left (5 a^2-6 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac{-2 a^2 b^2 \cos (c+d x)+a^4 \cos (c+d x)+b^4 \cos (c+d x)}{a^3 b^2 d (a+b \sin (c+d x))}-\frac{2 \left (a^2-b^2\right )^{3/2} \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^4 b^3 d}-\frac{b \tan \left (\frac{1}{2} (c+d x)\right )}{a^3 d}+\frac{b \cot \left (\frac{1}{2} (c+d x)\right )}{a^3 d}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 a^2 d}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 a^2 d}+\frac{2 a (c+d x)}{b^3 d}+\frac{\cos (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.19, size = 618, normalized size = 2.5 \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)^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^2/(1+tan(1/2*d*x+1/2*c)^2)+4/d/b^3*arctan(ta
n(1/2*d*x+1/2*c))*a+2/d/b/(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)-4/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)*tan(1/2*d*x+1/2*c)+2/d/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)*b^3+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-4/d/a/(tan(1/2*
d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)+2/d/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*b^2-4/d*a^2
/b^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))+2/d/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))+8/d*b/a^2/(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)^2-5/2/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)^6*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 = 5.47255, size = 2700, normalized size = 10.76 \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)^3/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/4*(8*a^6*d*x*cos(d*x + c)^2 - 8*a^6*d*x + 4*(2*a^5*b - 2*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 + 2*(2*a^5 + a^3
*b^2 - 3*a*b^4 - (2*a^5 + a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 + (2*a^4*b + a^2*b^3 - 3*b^5 - (2*a^4*b + a^2*b^3
- 3*b^5)*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*(4*a^5*b - 5*a^3*b^3 + 6*a*b^5)*cos(d*x + c) - (5*a^3*b^3 - 6*a*b^5 - (5*a^3*b
^3 - 6*a*b^5)*cos(d*x + c)^2 + (5*a^2*b^4 - 6*b^6 - (5*a^2*b^4 - 6*b^6)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*
cos(d*x + c) + 1/2) + (5*a^3*b^3 - 6*a*b^5 - (5*a^3*b^3 - 6*a*b^5)*cos(d*x + c)^2 + (5*a^2*b^4 - 6*b^6 - (5*a^
2*b^4 - 6*b^6)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(4*a^5*b*d*x*cos(d*x + c)^2 + 2*
a^4*b^2*cos(d*x + c)^3 - 4*a^5*b*d*x - (2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/(a^5*b^3*d*cos(d*x
+ c)^2 - a^5*b^3*d + (a^4*b^4*d*cos(d*x + c)^2 - a^4*b^4*d)*sin(d*x + c)), 1/4*(8*a^6*d*x*cos(d*x + c)^2 - 8*a
^6*d*x + 4*(2*a^5*b - 2*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 - 4*(2*a^5 + a^3*b^2 - 3*a*b^4 - (2*a^5 + a^3*b^2 -
3*a*b^4)*cos(d*x + c)^2 + (2*a^4*b + a^2*b^3 - 3*b^5 - (2*a^4*b + a^2*b^3 - 3*b^5)*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*(4*a^5*b - 5*a^3*b^3 + 6*a
*b^5)*cos(d*x + c) - (5*a^3*b^3 - 6*a*b^5 - (5*a^3*b^3 - 6*a*b^5)*cos(d*x + c)^2 + (5*a^2*b^4 - 6*b^6 - (5*a^2
*b^4 - 6*b^6)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (5*a^3*b^3 - 6*a*b^5 - (5*a^3*b^3 -
6*a*b^5)*cos(d*x + c)^2 + (5*a^2*b^4 - 6*b^6 - (5*a^2*b^4 - 6*b^6)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(
d*x + c) + 1/2) + 2*(4*a^5*b*d*x*cos(d*x + c)^2 + 2*a^4*b^2*cos(d*x + c)^3 - 4*a^5*b*d*x - (2*a^4*b^2 + 3*a^2*
b^4)*cos(d*x + c))*sin(d*x + c))/(a^5*b^3*d*cos(d*x + c)^2 - a^5*b^3*d + (a^4*b^4*d*cos(d*x + c)^2 - a^4*b^4*d
)*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(cos(d*x+c)**6*csc(d*x+c)**3/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.27648, size = 625, normalized size = 2.49 \begin{align*} \frac{\frac{16 \,{\left (d x + c\right )} a}{b^{3}} - \frac{4 \,{\left (5 \, 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{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{30 \, 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}} - \frac{16 \,{\left (2 \, a^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, 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^{4} b^{3}} + \frac{16 \,{\left (a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, 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} b^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(16*(d*x + c)*a/b^3 - 4*(5*a^2 - 6*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/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 + (30*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*tan(
1/2*d*x + 1/2*c) - a^2)/(a^4*tan(1/2*d*x + 1/2*c)^2) - 16*(2*a^6 - a^4*b^2 - 4*a^2*b^4 + 3*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^4*b^3) +
16*(a^4*b*tan(1/2*d*x + 1/2*c)^3 - 2*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + b^5*tan(1/2*d*x + 1/2*c)^3 + 2*a^5*tan(
1/2*d*x + 1/2*c)^2 - 2*a^3*b^2*tan(1/2*d*x + 1/2*c)^2 + a*b^4*tan(1/2*d*x + 1/2*c)^2 + 3*a^4*b*tan(1/2*d*x + 1
/2*c) - 2*a^2*b^3*tan(1/2*d*x + 1/2*c) + b^5*tan(1/2*d*x + 1/2*c) + 2*a^5 - 2*a^3*b^2 + a*b^4)/((a*tan(1/2*d*x
+ 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*a^4*b^2)
)/d