3.1259 \(\int \frac{\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=266 \[ \frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{b^4 d}+\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^5 d}-\frac{2 \left (a^2-b^2\right )^{3/2} \left (5 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^5 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 d (a+b \sin (c+d x))}+\frac{2 a x \left (2 a^2-3 b^2\right )}{b^5}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{a \sin (c+d x) \cos (c+d x)}{b^3 d}+\frac{a x}{b^3}-\frac{\cos ^3(c+d x)}{3 b^2 d}+\frac{\cos (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.350575, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.407, Rules used = {2897, 3770, 2638, 2635, 8, 2633, 2664, 12, 2660, 618, 204} \[ \frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{b^4 d}+\frac{2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^5 d}-\frac{2 \left (a^2-b^2\right )^{3/2} \left (5 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^5 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 d (a+b \sin (c+d x))}+\frac{2 a x \left (2 a^2-3 b^2\right )}{b^5}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{a \sin (c+d x) \cos (c+d x)}{b^3 d}+\frac{a x}{b^3}-\frac{\cos ^3(c+d x)}{3 b^2 d}+\frac{\cos (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Rule 2633

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

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

Mathematica [A]  time = 1.68633, size = 207, normalized size = 0.78 \[ \frac{\frac{12 a \left (4 a^2-5 b^2\right ) (c+d x)}{b^5}+\frac{9 \left (4 a^2-3 b^2\right ) \cos (c+d x)}{b^4}-\frac{24 \left (4 a^2+b^2\right ) \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^5}+\frac{12 \left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 (a+b \sin (c+d x))}+\frac{12 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^2}-\frac{12 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^2}-\frac{6 a \sin (2 (c+d x))}{b^3}-\frac{\cos (3 (c+d x))}{b^2}}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*a*(4*a^2 - 5*b^2)*(c + d*x))/b^5 - (24*(a^2 - b^2)^(3/2)*(4*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sq
rt[a^2 - b^2]])/(a^2*b^5) + (9*(4*a^2 - 3*b^2)*Cos[c + d*x])/b^4 - Cos[3*(c + d*x)]/b^2 - (12*Log[Cos[(c + d*x
)/2]])/a^2 + (12*Log[Sin[(c + d*x)/2]])/a^2 + (12*(a^2 - b^2)^2*Cos[c + d*x])/(a*b^4*(a + b*Sin[c + d*x])) - (
6*a*Sin[2*(c + d*x)])/b^3)/(12*d)

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Maple [B]  time = 0.17, size = 778, normalized size = 2.9 \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)/(a+b*sin(d*x+c))^2,x)

[Out]

2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)^5+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c
)^4*a^2-6/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4+12/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*
x+1/2*c)^2*a^2-8/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2-2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a*ta
n(1/2*d*x+1/2*c)+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2-14/3/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3+8/d/b^5*arctan(t
an(1/2*d*x+1/2*c))*a^3-10/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a+2/d*a^2/b^3/(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/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^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^3/b^4/(tan(1/2*d*x+1/2*c)^2*
a+2*tan(1/2*d*x+1/2*c)*b+a)-4/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+2/d/a/(tan(1/2*d*x+1/2
*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)-8/d*a^4/b^5/(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))+14/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))-4/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))-2/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))+1/d/a^2*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)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.95416, size = 1598, normalized size = 6.01 \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)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.22426, size = 477, normalized size = 1.79 \begin{align*} \frac{\frac{3 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac{3 \,{\left (4 \, a^{3} - 5 \, a b^{2}\right )}{\left (d x + c\right )}}{b^{5}} - \frac{6 \,{\left (4 \, a^{6} - 7 \, a^{4} b^{2} + 2 \, a^{2} b^{4} + 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^{2} b^{5}} + \frac{2 \,{\left (3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 9 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a^{2} - 7 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} b^{4}} + \frac{6 \,{\left (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 ) + 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 )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )} a^{2} b^{4}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 3*(4*a^3 - 5*a*b^2)*(d*x + c)/b^5 - 6*(4*a^6 - 7*a^4*b^2 + 2*a^2*b
^4 + 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)))/(sq
rt(a^2 - b^2)*a^2*b^5) + 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*tan(1/2*d*x + 1/2*c)^4 - 9*b^2*tan(1/2*d*x +
1/2*c)^4 + 18*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 9*a^2
- 7*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^4) + 6*(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) + a^5 - 2*a^3*b^2 + a*b^4)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) +
a)*a^2*b^4))/d