3.1113 \(\int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=209 \[ -\frac{\left (-14 a^2 b^2+3 a^4+b^4\right ) \cot (c+d x)}{15 a^2 d}+\frac{b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{60 a d}+\frac{\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{30 a^2 d}+\frac{b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac{3 a b \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}+b^2 x \]

[Out]

b^2*x - (3*a*b*ArcTanh[Cos[c + d*x]])/(4*d) - ((3*a^4 - 14*a^2*b^2 + b^4)*Cot[c + d*x])/(15*a^2*d) + (b*(27*a^
2 - 2*b^2)*Cot[c + d*x]*Csc[c + d*x])/(60*a*d) + ((12*a^2 - b^2)*Cot[c + d*x]*Csc[c + d*x]^2*(a + b*Sin[c + d*
x])^2)/(30*a^2*d) + (b*Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(10*a^2*d) - (Cot[c + d*x]*Csc[c +
d*x]^4*(a + b*Sin[c + d*x])^3)/(5*a*d)

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Rubi [A]  time = 0.515992, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2893, 3047, 3031, 3021, 2735, 3770} \[ -\frac{\left (-14 a^2 b^2+3 a^4+b^4\right ) \cot (c+d x)}{15 a^2 d}+\frac{b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{60 a d}+\frac{\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{30 a^2 d}+\frac{b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac{3 a b \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}+b^2 x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*Csc[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]

[Out]

b^2*x - (3*a*b*ArcTanh[Cos[c + d*x]])/(4*d) - ((3*a^4 - 14*a^2*b^2 + b^4)*Cot[c + d*x])/(15*a^2*d) + (b*(27*a^
2 - 2*b^2)*Cot[c + d*x]*Csc[c + d*x])/(60*a*d) + ((12*a^2 - b^2)*Cot[c + d*x]*Csc[c + d*x]^2*(a + b*Sin[c + d*
x])^2)/(30*a^2*d) + (b*Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(10*a^2*d) - (Cot[c + d*x]*Csc[c +
d*x]^4*(a + b*Sin[c + d*x])^3)/(5*a*d)

Rule 2893

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*d*f*(n + 1)), x] +
(-Dist[1/(a^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3770

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

Rubi steps

\begin{align*} \int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}-\frac{\int \csc ^4(c+d x) (a+b \sin (c+d x))^2 \left (2 \left (12 a^2-b^2\right )+2 a b \sin (c+d x)-20 a^2 \sin ^2(c+d x)\right ) \, dx}{20 a^2}\\ &=\frac{\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{30 a^2 d}+\frac{b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}-\frac{\int \csc ^3(c+d x) (a+b \sin (c+d x)) \left (2 b \left (27 a^2-2 b^2\right )-2 a \left (6 a^2-b^2\right ) \sin (c+d x)-60 a^2 b \sin ^2(c+d x)\right ) \, dx}{60 a^2}\\ &=\frac{b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{60 a d}+\frac{\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{30 a^2 d}+\frac{b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}+\frac{\int \csc ^2(c+d x) \left (8 \left (3 a^4-14 a^2 b^2+b^4\right )+90 a^3 b \sin (c+d x)+120 a^2 b^2 \sin ^2(c+d x)\right ) \, dx}{120 a^2}\\ &=-\frac{\left (3 a^4-14 a^2 b^2+b^4\right ) \cot (c+d x)}{15 a^2 d}+\frac{b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{60 a d}+\frac{\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{30 a^2 d}+\frac{b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}+\frac{\int \csc (c+d x) \left (90 a^3 b+120 a^2 b^2 \sin (c+d x)\right ) \, dx}{120 a^2}\\ &=b^2 x-\frac{\left (3 a^4-14 a^2 b^2+b^4\right ) \cot (c+d x)}{15 a^2 d}+\frac{b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{60 a d}+\frac{\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{30 a^2 d}+\frac{b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}+\frac{1}{4} (3 a b) \int \csc (c+d x) \, dx\\ &=b^2 x-\frac{3 a b \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{\left (3 a^4-14 a^2 b^2+b^4\right ) \cot (c+d x)}{15 a^2 d}+\frac{b \left (27 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{60 a d}+\frac{\left (12 a^2-b^2\right ) \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{30 a^2 d}+\frac{b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac{\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 a d}\\ \end{align*}

Mathematica [A]  time = 1.45704, size = 285, normalized size = 1.36 \[ \frac{\left (640 b^2-96 a^2\right ) \cot \left (\frac{1}{2} (c+d x)\right )+\csc ^4\left (\frac{1}{2} (c+d x)\right ) \left (\left (21 a^2-20 b^2\right ) \sin (c+d x)-30 a b\right )+96 a^2 \tan \left (\frac{1}{2} (c+d x)\right )+192 a^2 \sin ^6\left (\frac{1}{2} (c+d x)\right ) \csc ^5(c+d x)-336 a^2 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-3 a^2 \sin (c+d x) \csc ^6\left (\frac{1}{2} (c+d x)\right )+300 a b \csc ^2\left (\frac{1}{2} (c+d x)\right )+30 a b \sec ^4\left (\frac{1}{2} (c+d x)\right )-300 a b \sec ^2\left (\frac{1}{2} (c+d x)\right )+720 a b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-720 a b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-640 b^2 \tan \left (\frac{1}{2} (c+d x)\right )+320 b^2 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)+960 b^2 c+960 b^2 d x}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*Csc[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]

[Out]

(960*b^2*c + 960*b^2*d*x + (-96*a^2 + 640*b^2)*Cot[(c + d*x)/2] + 300*a*b*Csc[(c + d*x)/2]^2 - 720*a*b*Log[Cos
[(c + d*x)/2]] + 720*a*b*Log[Sin[(c + d*x)/2]] - 300*a*b*Sec[(c + d*x)/2]^2 + 30*a*b*Sec[(c + d*x)/2]^4 - 336*
a^2*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 320*b^2*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 192*a^2*Csc[c + d*x]^5*Sin
[(c + d*x)/2]^6 - 3*a^2*Csc[(c + d*x)/2]^6*Sin[c + d*x] + Csc[(c + d*x)/2]^4*(-30*a*b + (21*a^2 - 20*b^2)*Sin[
c + d*x]) + 96*a^2*Tan[(c + d*x)/2] - 640*b^2*Tan[(c + d*x)/2])/(960*d)

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Maple [A]  time = 0.089, size = 165, normalized size = 0.8 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,ab\cos \left ( dx+c \right ) }{4\,d}}+{\frac{3\,ab\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{4\,d}}-{\frac{{b}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{b}^{2}\cot \left ( dx+c \right ) }{d}}+{b}^{2}x+{\frac{{b}^{2}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^6*(a+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.51504, size = 166, normalized size = 0.79 \begin{align*} \frac{40 \,{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} b^{2} - 15 \, a b{\left (\frac{2 \,{\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - \frac{24 \, a^{2}}{\tan \left (d x + c\right )^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/120*(40*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*b^2 - 15*a*b*(2*(5*cos(d*x + c)^3 - 3*cos(d*x
+ c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)) - 24*a^2/ta
n(d*x + c)^5)/d

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Fricas [A]  time = 1.82271, size = 637, normalized size = 3.05 \begin{align*} -\frac{8 \,{\left (3 \, a^{2} - 20 \, b^{2}\right )} \cos \left (d x + c\right )^{5} + 280 \, b^{2} \cos \left (d x + c\right )^{3} - 120 \, b^{2} \cos \left (d x + c\right ) + 45 \,{\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 45 \,{\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 30 \,{\left (4 \, b^{2} d x \cos \left (d x + c\right )^{4} - 8 \, b^{2} d x \cos \left (d x + c\right )^{2} - 5 \, a b \cos \left (d x + c\right )^{3} + 4 \, b^{2} d x + 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/120*(8*(3*a^2 - 20*b^2)*cos(d*x + c)^5 + 280*b^2*cos(d*x + c)^3 - 120*b^2*cos(d*x + c) + 45*(a*b*cos(d*x +
c)^4 - 2*a*b*cos(d*x + c)^2 + a*b)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 45*(a*b*cos(d*x + c)^4 - 2*a*b*c
os(d*x + c)^2 + a*b)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 30*(4*b^2*d*x*cos(d*x + c)^4 - 8*b^2*d*x*cos(
d*x + c)^2 - 5*a*b*cos(d*x + c)^3 + 4*b^2*d*x + 3*a*b*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos
(d*x + c)^2 + 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)**4*csc(d*x+c)**6*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.35887, size = 355, normalized size = 1.7 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 120 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 480 \,{\left (d x + c\right )} b^{2} + 360 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 30 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 300 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{822 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 30 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 300 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 120 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 20 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/480*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 15*a*b*tan(1/2*d*x + 1/2*c)^4 - 15*a^2*tan(1/2*d*x + 1/2*c)^3 + 20*b^2*t
an(1/2*d*x + 1/2*c)^3 - 120*a*b*tan(1/2*d*x + 1/2*c)^2 + 480*(d*x + c)*b^2 + 360*a*b*log(abs(tan(1/2*d*x + 1/2
*c))) + 30*a^2*tan(1/2*d*x + 1/2*c) - 300*b^2*tan(1/2*d*x + 1/2*c) - (822*a*b*tan(1/2*d*x + 1/2*c)^5 + 30*a^2*
tan(1/2*d*x + 1/2*c)^4 - 300*b^2*tan(1/2*d*x + 1/2*c)^4 - 120*a*b*tan(1/2*d*x + 1/2*c)^3 - 15*a^2*tan(1/2*d*x
+ 1/2*c)^2 + 20*b^2*tan(1/2*d*x + 1/2*c)^2 + 15*a*b*tan(1/2*d*x + 1/2*c) + 3*a^2)/tan(1/2*d*x + 1/2*c)^5)/d