3.714 \(\int \frac{\cos ^3(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=150 \[ \frac{15 \cos (c+d x)}{8 a d}+\frac{5 \cot ^3(c+d x)}{6 a d}-\frac{5 \cot (c+d x)}{2 a d}-\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{\cos (c+d x) \cot ^4(c+d x)}{4 a d}+\frac{5 \cos (c+d x) \cot ^2(c+d x)}{8 a d}-\frac{15 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac{5 x}{2 a} \]

[Out]

(-5*x)/(2*a) - (15*ArcTanh[Cos[c + d*x]])/(8*a*d) + (15*Cos[c + d*x])/(8*a*d) - (5*Cot[c + d*x])/(2*a*d) + (5*
Cos[c + d*x]*Cot[c + d*x]^2)/(8*a*d) + (5*Cot[c + d*x]^3)/(6*a*d) - (Cos[c + d*x]^2*Cot[c + d*x]^3)/(2*a*d) -
(Cos[c + d*x]*Cot[c + d*x]^4)/(4*a*d)

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Rubi [A]  time = 0.176904, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2839, 2592, 288, 321, 206, 2591, 302, 203} \[ \frac{15 \cos (c+d x)}{8 a d}+\frac{5 \cot ^3(c+d x)}{6 a d}-\frac{5 \cot (c+d x)}{2 a d}-\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{\cos (c+d x) \cot ^4(c+d x)}{4 a d}+\frac{5 \cos (c+d x) \cot ^2(c+d x)}{8 a d}-\frac{15 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac{5 x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*Cot[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

(-5*x)/(2*a) - (15*ArcTanh[Cos[c + d*x]])/(8*a*d) + (15*Cos[c + d*x])/(8*a*d) - (5*Cot[c + d*x])/(2*a*d) + (5*
Cos[c + d*x]*Cot[c + d*x]^2)/(8*a*d) + (5*Cot[c + d*x]^3)/(6*a*d) - (Cos[c + d*x]^2*Cot[c + d*x]^3)/(2*a*d) -
(Cos[c + d*x]*Cot[c + d*x]^4)/(4*a*d)

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac{\int \cos ^2(c+d x) \cot ^4(c+d x) \, dx}{a}+\frac{\int \cos (c+d x) \cot ^5(c+d x) \, dx}{a}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^3} \, dx,x,\cos (c+d x)\right )}{a d}+\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{a d}\\ &=-\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{\cos (c+d x) \cot ^4(c+d x)}{4 a d}+\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{4 a d}+\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 a d}\\ &=\frac{5 \cos (c+d x) \cot ^2(c+d x)}{8 a d}-\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{\cos (c+d x) \cot ^4(c+d x)}{4 a d}-\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{8 a d}+\frac{5 \operatorname{Subst}\left (\int \left (-1+x^2+\frac{1}{1+x^2}\right ) \, dx,x,\cot (c+d x)\right )}{2 a d}\\ &=\frac{15 \cos (c+d x)}{8 a d}-\frac{5 \cot (c+d x)}{2 a d}+\frac{5 \cos (c+d x) \cot ^2(c+d x)}{8 a d}+\frac{5 \cot ^3(c+d x)}{6 a d}-\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{\cos (c+d x) \cot ^4(c+d x)}{4 a d}-\frac{15 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{8 a d}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 a d}\\ &=-\frac{5 x}{2 a}-\frac{15 \tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac{15 \cos (c+d x)}{8 a d}-\frac{5 \cot (c+d x)}{2 a d}+\frac{5 \cos (c+d x) \cot ^2(c+d x)}{8 a d}+\frac{5 \cot ^3(c+d x)}{6 a d}-\frac{\cos ^2(c+d x) \cot ^3(c+d x)}{2 a d}-\frac{\cos (c+d x) \cot ^4(c+d x)}{4 a d}\\ \end{align*}

Mathematica [A]  time = 0.733414, size = 252, normalized size = 1.68 \[ -\frac{\csc ^4(c+d x) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (95 \sin (2 (c+d x))-68 \sin (4 (c+d x))+3 \sin (6 (c+d x))+60 c \cos (4 (c+d x))-30 \cos (c+d x)+90 \cos (3 (c+d x))+60 d x \cos (4 (c+d x))-12 \cos (5 (c+d x))-135 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+45 \cos (4 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+135 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-60 \cos (2 (c+d x)) \left (-3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 c+4 d x\right )-45 \cos (4 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+180 c+180 d x\right )}{192 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

-(Csc[c + d*x]^4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(180*c + 180*d*x - 30*Cos[c + d*x] + 90*Cos[3*(c + d*
x)] + 60*c*Cos[4*(c + d*x)] + 60*d*x*Cos[4*(c + d*x)] - 12*Cos[5*(c + d*x)] + 135*Log[Cos[(c + d*x)/2]] + 45*C
os[4*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 60*Cos[2*(c + d*x)]*(4*c + 4*d*x + 3*Log[Cos[(c + d*x)/2]] - 3*Log[Sin
[(c + d*x)/2]]) - 135*Log[Sin[(c + d*x)/2]] - 45*Cos[4*(c + d*x)]*Log[Sin[(c + d*x)/2]] + 95*Sin[2*(c + d*x)]
- 68*Sin[4*(c + d*x)] + 3*Sin[6*(c + d*x)]))/(192*a*d*(1 + Sin[c + d*x]))

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Maple [B]  time = 0.148, size = 310, normalized size = 2.1 \begin{align*}{\frac{1}{64\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}}-{\frac{1}{24\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{4\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{9}{8\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{1}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-5\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}}-{\frac{1}{64\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}+{\frac{1}{24\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{1}{4\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{9}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{15}{8\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c)),x)

[Out]

1/64/d/a*tan(1/2*d*x+1/2*c)^4-1/24/d/a*tan(1/2*d*x+1/2*c)^3-1/4/d/a*tan(1/2*d*x+1/2*c)^2+9/8/d/a*tan(1/2*d*x+1
/2*c)+1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c
)^2-1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2-5/a/d*arctan(tan(1/2*
d*x+1/2*c))-1/64/d/a/tan(1/2*d*x+1/2*c)^4+1/24/d/a/tan(1/2*d*x+1/2*c)^3+1/4/d/a/tan(1/2*d*x+1/2*c)^2-9/8/d/a/t
an(1/2*d*x+1/2*c)+15/8/d/a*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.59504, size = 459, normalized size = 3.06 \begin{align*} \frac{\frac{\frac{216 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a} + \frac{\frac{8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{42 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{200 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{477 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{616 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{432 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{24 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 3}{\frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac{960 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{360 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/192*((216*sin(d*x + c)/(cos(d*x + c) + 1) - 48*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 8*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a + (8*sin(d*x + c)/(cos(d*x + c) + 1) + 42*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 - 200*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 477*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- 616*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 432*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 24*sin(d*x + c)^7/(cos(d
*x + c) + 1)^7 - 3)/(a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 2*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d
*x + c)^8/(cos(d*x + c) + 1)^8) - 960*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 360*log(sin(d*x + c)/(cos(d*
x + c) + 1))/a)/d

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Fricas [A]  time = 1.17538, size = 536, normalized size = 3.57 \begin{align*} -\frac{120 \, d x \cos \left (d x + c\right )^{4} - 48 \, \cos \left (d x + c\right )^{5} - 240 \, d x \cos \left (d x + c\right )^{2} + 150 \, \cos \left (d x + c\right )^{3} + 120 \, d x + 45 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 45 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 8 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 20 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 90 \, \cos \left (d x + c\right )}{48 \,{\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/48*(120*d*x*cos(d*x + c)^4 - 48*cos(d*x + c)^5 - 240*d*x*cos(d*x + c)^2 + 150*cos(d*x + c)^3 + 120*d*x + 45
*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) - 45*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 +
1)*log(-1/2*cos(d*x + c) + 1/2) + 8*(3*cos(d*x + c)^5 - 20*cos(d*x + c)^3 + 15*cos(d*x + c))*sin(d*x + c) - 9
0*cos(d*x + c))/(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*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)**8*csc(d*x+c)**5/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.51842, size = 302, normalized size = 2.01 \begin{align*} -\frac{\frac{480 \,{\left (d x + c\right )}}{a} - \frac{360 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} - \frac{192 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a} - \frac{3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 48 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 216 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{4}} + \frac{750 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 216 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 48 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(480*(d*x + c)/a - 360*log(abs(tan(1/2*d*x + 1/2*c)))/a - 192*(tan(1/2*d*x + 1/2*c)^3 + 2*tan(1/2*d*x +
1/2*c)^2 - tan(1/2*d*x + 1/2*c) + 2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a) - (3*a^3*tan(1/2*d*x + 1/2*c)^4 - 8*a
^3*tan(1/2*d*x + 1/2*c)^3 - 48*a^3*tan(1/2*d*x + 1/2*c)^2 + 216*a^3*tan(1/2*d*x + 1/2*c))/a^4 + (750*tan(1/2*d
*x + 1/2*c)^4 + 216*tan(1/2*d*x + 1/2*c)^3 - 48*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 3)/(a*tan(1/
2*d*x + 1/2*c)^4))/d