3.1323 \(\int \frac{\cos ^5(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=252 \[ -\frac{a \left (a^2-3 b^2\right ) \cos (c+d x)}{b^4 d}+\frac{2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a b^5 d}+\frac{\left (a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b^3 d}-\frac{x \left (a^2-3 b^2\right )}{2 b^3}-\frac{x \left (-3 a^2 b^2+a^4+3 b^4\right )}{b^5}+\frac{a \cos ^3(c+d x)}{3 b^2 d}-\frac{a \cos (c+d x)}{b^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}+\frac{\sin ^3(c+d x) \cos (c+d x)}{4 b d}+\frac{3 \sin (c+d x) \cos (c+d x)}{8 b d}-\frac{3 x}{8 b} \]

[Out]

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

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Rubi [A]  time = 0.285735, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2897, 3770, 2638, 2635, 8, 2633, 2660, 618, 204} \[ -\frac{a \left (a^2-3 b^2\right ) \cos (c+d x)}{b^4 d}+\frac{2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a b^5 d}+\frac{\left (a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b^3 d}-\frac{x \left (a^2-3 b^2\right )}{2 b^3}-\frac{x \left (-3 a^2 b^2+a^4+3 b^4\right )}{b^5}+\frac{a \cos ^3(c+d x)}{3 b^2 d}-\frac{a \cos (c+d x)}{b^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}+\frac{\sin ^3(c+d x) \cos (c+d x)}{4 b d}+\frac{3 \sin (c+d x) \cos (c+d x)}{8 b d}-\frac{3 x}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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)} \, dx &=\int \left (\frac{-a^4+3 a^2 b^2-3 b^4}{b^5}+\frac{\csc (c+d x)}{a}+\frac{a \left (a^2-3 b^2\right ) \sin (c+d x)}{b^4}+\frac{\left (-a^2+3 b^2\right ) \sin ^2(c+d x)}{b^3}+\frac{a \sin ^3(c+d x)}{b^2}-\frac{\sin ^4(c+d x)}{b}+\frac{\left (a^2-b^2\right )^3}{a b^5 (a+b \sin (c+d x))}\right ) \, dx\\ &=-\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) x}{b^5}+\frac{\int \csc (c+d x) \, dx}{a}+\frac{a \int \sin ^3(c+d x) \, dx}{b^2}-\frac{\int \sin ^4(c+d x) \, dx}{b}+\frac{\left (a \left (a^2-3 b^2\right )\right ) \int \sin (c+d x) \, dx}{b^4}-\frac{\left (a^2-3 b^2\right ) \int \sin ^2(c+d x) \, dx}{b^3}+\frac{\left (a^2-b^2\right )^3 \int \frac{1}{a+b \sin (c+d x)} \, dx}{a b^5}\\ &=-\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) x}{b^5}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{a \left (a^2-3 b^2\right ) \cos (c+d x)}{b^4 d}+\frac{\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac{\cos (c+d x) \sin ^3(c+d x)}{4 b d}-\frac{3 \int \sin ^2(c+d x) \, dx}{4 b}-\frac{\left (a^2-3 b^2\right ) \int 1 \, dx}{2 b^3}-\frac{a \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 )^3\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^5 d}\\ &=-\frac{\left (a^2-3 b^2\right ) x}{2 b^3}-\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) x}{b^5}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{a \cos (c+d x)}{b^2 d}-\frac{a \left (a^2-3 b^2\right ) \cos (c+d x)}{b^4 d}+\frac{a \cos ^3(c+d x)}{3 b^2 d}+\frac{3 \cos (c+d x) \sin (c+d x)}{8 b d}+\frac{\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac{\cos (c+d x) \sin ^3(c+d x)}{4 b d}-\frac{3 \int 1 \, dx}{8 b}-\frac{\left (4 \left (a^2-b^2\right )^3\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^5 d}\\ &=-\frac{3 x}{8 b}-\frac{\left (a^2-3 b^2\right ) x}{2 b^3}-\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) x}{b^5}+\frac{2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a b^5 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{a \cos (c+d x)}{b^2 d}-\frac{a \left (a^2-3 b^2\right ) \cos (c+d x)}{b^4 d}+\frac{a \cos ^3(c+d x)}{3 b^2 d}+\frac{3 \cos (c+d x) \sin (c+d x)}{8 b d}+\frac{\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac{\cos (c+d x) \sin ^3(c+d x)}{4 b d}\\ \end{align*}

Mathematica [A]  time = 0.542313, size = 220, normalized size = 0.87 \[ -\frac{-24 a^3 b^2 \sin (2 (c+d x))+24 a^2 b \left (4 a^2-9 b^2\right ) \cos (c+d x)-8 a^2 b^3 \cos (3 (c+d x))-192 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )-240 a^3 b^2 c-240 a^3 b^2 d x+96 a^5 c+96 a^5 d x+48 a b^4 \sin (2 (c+d x))+3 a b^4 \sin (4 (c+d x))+180 a b^4 c+180 a b^4 d x-96 b^5 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+96 b^5 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{96 a b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(96*a^5*c - 240*a^3*b^2*c + 180*a*b^4*c + 96*a^5*d*x - 240*a^3*b^2*d*x + 180*a*b^4*d*x - 192*(a^2 - b^2)^(5/2
)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 24*a^2*b*(4*a^2 - 9*b^2)*Cos[c + d*x] - 8*a^2*b^3*Cos[3*(
c + d*x)] + 96*b^5*Log[Cos[(c + d*x)/2]] - 96*b^5*Log[Sin[(c + d*x)/2]] - 24*a^3*b^2*Sin[2*(c + d*x)] + 48*a*b
^4*Sin[2*(c + d*x)] + 3*a*b^4*Sin[4*(c + d*x)])/(96*a*b^5*d)

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Maple [B]  time = 0.112, size = 827, normalized size = 3.3 \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)),x)

[Out]

-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*a^2+9/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/
2*c)^7-2/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^6*a^3+6/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*
d*x+1/2*c)^6*a-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*a^2+1/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*
tan(1/2*d*x+1/2*c)^5-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^4*a^3+14/d/b^2/(1+tan(1/2*d*x+1/2*c
)^2)^4*tan(1/2*d*x+1/2*c)^4*a+1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*a^2-1/4/d/b/(1+tan(1/2*d
*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2*a^3+38/3/d/b^2/(1+
tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2*a+1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*a^2-9/4/d
/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)-2/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*a^3+14/3/d/b^2/(1+tan(1/2*
d*x+1/2*c)^2)^4*a-2/d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^4+5/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a^2-15/4/d/b*arcta
n(tan(1/2*d*x+1/2*c))+2/d*a^5/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))-6/d
*a^3/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))+6/d/b*a/(a^2-b^2)^(1/2)*arct
an(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-2/d/a*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))+1/d/a*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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.80253, size = 1212, normalized size = 4.81 \begin{align*} \left [\frac{8 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} - 12 \, b^{5} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 12 \, b^{5} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \,{\left (8 \, a^{5} - 20 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x + 12 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 24 \,{\left (a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right ) - 3 \,{\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} -{\left (4 \, a^{3} b^{2} - 7 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, a b^{5} d}, \frac{8 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} - 12 \, b^{5} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 12 \, b^{5} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \,{\left (8 \, a^{5} - 20 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x - 24 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 24 \,{\left (a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right ) - 3 \,{\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} -{\left (4 \, a^{3} b^{2} - 7 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, a b^{5} d}\right ] \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)),x, algorithm="fricas")

[Out]

[1/24*(8*a^2*b^3*cos(d*x + c)^3 - 12*b^5*log(1/2*cos(d*x + c) + 1/2) + 12*b^5*log(-1/2*cos(d*x + c) + 1/2) - 3
*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*d*x + 12*(a^4 - 2*a^2*b^2 + b^4)*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)) - 24*(a^4*b - 2*a^2*b^3)*cos(d*x + c) - 3*(2*a*b^4*cos(
d*x + c)^3 - (4*a^3*b^2 - 7*a*b^4)*cos(d*x + c))*sin(d*x + c))/(a*b^5*d), 1/24*(8*a^2*b^3*cos(d*x + c)^3 - 12*
b^5*log(1/2*cos(d*x + c) + 1/2) + 12*b^5*log(-1/2*cos(d*x + c) + 1/2) - 3*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*d*x
- 24*(a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 24
*(a^4*b - 2*a^2*b^3)*cos(d*x + c) - 3*(2*a*b^4*cos(d*x + c)^3 - (4*a^3*b^2 - 7*a*b^4)*cos(d*x + c))*sin(d*x +
c))/(a*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)),x)

[Out]

Timed out

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Giac [A]  time = 1.27358, size = 537, normalized size = 2.13 \begin{align*} \frac{\frac{24 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} - \frac{3 \,{\left (8 \, a^{4} - 20 \, a^{2} b^{2} + 15 \, b^{4}\right )}{\left (d x + c\right )}}{b^{5}} + \frac{48 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, 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 b^{5}} - \frac{2 \,{\left (12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 27 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 72 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 72 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 168 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 72 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 152 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, a^{3} - 56 \, a b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4} b^{4}}}{24 \, 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)),x, algorithm="giac")

[Out]

1/24*(24*log(abs(tan(1/2*d*x + 1/2*c)))/a - 3*(8*a^4 - 20*a^2*b^2 + 15*b^4)*(d*x + c)/b^5 + 48*(a^6 - 3*a^4*b^
2 + 3*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)))/(sqrt(a^2 - b^2)*a*b^5) - 2*(12*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 27*b^3*tan(1/2*d*x + 1/2*c)^7 + 24*a^3*
tan(1/2*d*x + 1/2*c)^6 - 72*a*b^2*tan(1/2*d*x + 1/2*c)^6 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 3*b^3*tan(1/2*d*x
+ 1/2*c)^5 + 72*a^3*tan(1/2*d*x + 1/2*c)^4 - 168*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 12*a^2*b*tan(1/2*d*x + 1/2*c)
^3 + 3*b^3*tan(1/2*d*x + 1/2*c)^3 + 72*a^3*tan(1/2*d*x + 1/2*c)^2 - 152*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 12*a^2*
b*tan(1/2*d*x + 1/2*c) + 27*b^3*tan(1/2*d*x + 1/2*c) + 24*a^3 - 56*a*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*b^4)
)/d