3.1246 \(\int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=180 \[ -\frac{\left (a^2-b^2\right ) \cos ^3(c+d x)}{3 d}-\frac{\left (2 a^2-b^2\right ) \cos (c+d x)}{d}+\frac{\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{15 a b \cot (c+d x)}{4 d}+\frac{a b \cos ^4(c+d x) \cot (c+d x)}{2 d}+\frac{5 a b \cos ^2(c+d x) \cot (c+d x)}{4 d}-\frac{15 a b x}{4}+\frac{b^2 \cos ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.304969, antiderivative size = 180, 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 = {2911, 2591, 288, 321, 203, 455, 1810, 206} \[ -\frac{\left (a^2-b^2\right ) \cos ^3(c+d x)}{3 d}-\frac{\left (2 a^2-b^2\right ) \cos (c+d x)}{d}+\frac{\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{15 a b \cot (c+d x)}{4 d}+\frac{a b \cos ^4(c+d x) \cot (c+d x)}{2 d}+\frac{5 a b \cos ^2(c+d x) \cot (c+d x)}{4 d}-\frac{15 a b x}{4}+\frac{b^2 \cos ^5(c+d x)}{5 d} \]

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]

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

Rule 2911

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

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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])

Rubi steps

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

Mathematica [A]  time = 6.1746, size = 250, normalized size = 1.39 \[ -\frac{\left (18 a^2-11 b^2\right ) \cos (c+d x)}{8 d}-\frac{\left (4 a^2-7 b^2\right ) \cos (3 (c+d x))}{48 d}+\frac{\left (2 b^2-5 a^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{\left (5 a^2-2 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}-\frac{a^2 \csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{a^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{15 a b (c+d x)}{4 d}-\frac{a b \sin (2 (c+d x))}{d}-\frac{a b \sin (4 (c+d x))}{16 d}+\frac{a b \tan \left (\frac{1}{2} (c+d x)\right )}{d}-\frac{a b \cot \left (\frac{1}{2} (c+d x)\right )}{d}+\frac{b^2 \cos (5 (c+d x))}{80 d} \]

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]

(-15*a*b*(c + d*x))/(4*d) - ((18*a^2 - 11*b^2)*Cos[c + d*x])/(8*d) - ((4*a^2 - 7*b^2)*Cos[3*(c + d*x)])/(48*d)
+ (b^2*Cos[5*(c + d*x)])/(80*d) - (a*b*Cot[(c + d*x)/2])/d - (a^2*Csc[(c + d*x)/2]^2)/(8*d) + ((5*a^2 - 2*b^2
)*Log[Cos[(c + d*x)/2]])/(2*d) + ((-5*a^2 + 2*b^2)*Log[Sin[(c + d*x)/2]])/(2*d) + (a^2*Sec[(c + d*x)/2]^2)/(8*
d) - (a*b*Sin[2*(c + d*x)])/d - (a*b*Sin[4*(c + d*x)])/(16*d) + (a*b*Tan[(c + d*x)/2])/d

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Maple [A]  time = 0.092, size = 261, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d}}-{\frac{5\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{6\,d}}-{\frac{5\,{a}^{2}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{5\,{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-2\,{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{d\sin \left ( dx+c \right ) }}-2\,{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }{d}}-{\frac{5\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{15\,ab\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{4\,d}}-{\frac{15\,abx}{4}}-{\frac{15\,abc}{4\,d}}+{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{b}^{2}\cos \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}} \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]

-1/2/d*a^2/sin(d*x+c)^2*cos(d*x+c)^7-1/2*a^2*cos(d*x+c)^5/d-5/6*a^2*cos(d*x+c)^3/d-5/2*a^2*cos(d*x+c)/d-5/2/d*
a^2*ln(csc(d*x+c)-cot(d*x+c))-2/d*a*b/sin(d*x+c)*cos(d*x+c)^7-2*a*b*cos(d*x+c)^5*sin(d*x+c)/d-5/2*a*b*cos(d*x+
c)^3*sin(d*x+c)/d-15/4*a*b*cos(d*x+c)*sin(d*x+c)/d-15/4*a*b*x-15/4/d*a*b*c+1/5*b^2*cos(d*x+c)^5/d+1/3*b^2*cos(
d*x+c)^3/d+b^2*cos(d*x+c)/d+1/d*b^2*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 1.51559, size = 257, normalized size = 1.43 \begin{align*} -\frac{5 \,{\left (4 \, \cos \left (d x + c\right )^{3} - \frac{6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} + 15 \,{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a b - 2 \,{\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} b^{2}}{60 \, 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="maxima")

[Out]

-1/60*(5*(4*cos(d*x + c)^3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos(d*x + c) - 15*log(cos(d*x + c) + 1)
+ 15*log(cos(d*x + c) - 1))*a^2 + 15*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(tan(d*x + c
)^5 + 2*tan(d*x + c)^3 + tan(d*x + c)))*a*b - 2*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*l
og(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1))*b^2)/d

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Fricas [A]  time = 1.88128, size = 610, normalized size = 3.39 \begin{align*} \frac{12 \, b^{2} \cos \left (d x + c\right )^{7} - 225 \, a b d x \cos \left (d x + c\right )^{2} - 4 \,{\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{5} + 225 \, a b d x - 20 \,{\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \,{\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right ) + 15 \,{\left ({\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 5 \, a^{2} + 2 \, b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 15 \,{\left ({\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 5 \, a^{2} + 2 \, b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 15 \,{\left (2 \, a b \cos \left (d x + c\right )^{5} + 5 \, a b \cos \left (d x + c\right )^{3} - 15 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \,{\left (d \cos \left (d x + c\right )^{2} - 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)^3*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/60*(12*b^2*cos(d*x + c)^7 - 225*a*b*d*x*cos(d*x + c)^2 - 4*(5*a^2 - 2*b^2)*cos(d*x + c)^5 + 225*a*b*d*x - 20
*(5*a^2 - 2*b^2)*cos(d*x + c)^3 + 30*(5*a^2 - 2*b^2)*cos(d*x + c) + 15*((5*a^2 - 2*b^2)*cos(d*x + c)^2 - 5*a^2
+ 2*b^2)*log(1/2*cos(d*x + c) + 1/2) - 15*((5*a^2 - 2*b^2)*cos(d*x + c)^2 - 5*a^2 + 2*b^2)*log(-1/2*cos(d*x +
c) + 1/2) - 15*(2*a*b*cos(d*x + c)^5 + 5*a*b*cos(d*x + c)^3 - 15*a*b*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x +
c)^2 - 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)**3*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.27225, size = 467, normalized size = 2.59 \begin{align*} \frac{15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 450 \,{\left (d x + c\right )} a b + 120 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 60 \,{\left (5 \, a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{15 \,{\left (30 \, 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} - 8 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} + \frac{4 \,{\left (135 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 180 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 180 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 150 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 600 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 360 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 800 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 560 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 150 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 520 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 280 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 135 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 140 \, a^{2} + 92 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, 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]

1/120*(15*a^2*tan(1/2*d*x + 1/2*c)^2 - 450*(d*x + c)*a*b + 120*a*b*tan(1/2*d*x + 1/2*c) - 60*(5*a^2 - 2*b^2)*l
og(abs(tan(1/2*d*x + 1/2*c))) + 15*(30*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(
1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2 + 4*(135*a*b*tan(1/2*d*x + 1/2*c)^9 - 180*a^2*tan(1/2*d*x + 1/2
*c)^8 + 180*b^2*tan(1/2*d*x + 1/2*c)^8 + 150*a*b*tan(1/2*d*x + 1/2*c)^7 - 600*a^2*tan(1/2*d*x + 1/2*c)^6 + 360
*b^2*tan(1/2*d*x + 1/2*c)^6 - 800*a^2*tan(1/2*d*x + 1/2*c)^4 + 560*b^2*tan(1/2*d*x + 1/2*c)^4 - 150*a*b*tan(1/
2*d*x + 1/2*c)^3 - 520*a^2*tan(1/2*d*x + 1/2*c)^2 + 280*b^2*tan(1/2*d*x + 1/2*c)^2 - 135*a*b*tan(1/2*d*x + 1/2
*c) - 140*a^2 + 92*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d