3.1112 \(\int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=178 \[ -\frac{b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{24 a^2 d}-\frac{3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}+\frac{17 a b \cot (c+d x)}{12 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{8 d}+2 a b x \]

[Out]

2*a*b*x - (3*(a^2 - 4*b^2)*ArcTanh[Cos[c + d*x]])/(8*d) - (b^2*(39*a^2 + 2*b^2)*Cos[c + d*x])/(24*a^2*d) + (17
*a*b*Cot[c + d*x])/(12*d) + (5*Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^2)/(8*d) + (b*Cot[c + d*x]*Csc[c
+ d*x]^2*(a + b*Sin[c + d*x])^3)/(12*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(4*a*d)

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Rubi [A]  time = 0.462778, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2893, 3047, 3031, 3023, 2735, 3770} \[ -\frac{b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{24 a^2 d}-\frac{3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}+\frac{17 a b \cot (c+d x)}{12 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{8 d}+2 a b x \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a*b*x - (3*(a^2 - 4*b^2)*ArcTanh[Cos[c + d*x]])/(8*d) - (b^2*(39*a^2 + 2*b^2)*Cos[c + d*x])/(24*a^2*d) + (17
*a*b*Cot[c + d*x])/(12*d) + (5*Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^2)/(8*d) + (b*Cot[c + d*x]*Csc[c
+ d*x]^2*(a + b*Sin[c + d*x])^3)/(12*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(4*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 3023

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

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 (c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}-\frac{\int \csc ^3(c+d x) (a+b \sin (c+d x))^2 \left (15 a^2+2 a b \sin (c+d x)-\left (12 a^2+b^2\right ) \sin ^2(c+d x)\right ) \, dx}{12 a^2}\\ &=\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac{b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}-\frac{\int \csc ^2(c+d x) (a+b \sin (c+d x)) \left (34 a^2 b-a \left (9 a^2-2 b^2\right ) \sin (c+d x)-b \left (39 a^2+2 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{24 a^2}\\ &=\frac{17 a b \cot (c+d x)}{12 d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac{b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}+\frac{\int \csc (c+d x) \left (9 a^2 \left (a^2-4 b^2\right )+48 a^3 b \sin (c+d x)+b^2 \left (39 a^2+2 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{24 a^2}\\ &=-\frac{b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{24 a^2 d}+\frac{17 a b \cot (c+d x)}{12 d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac{b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}+\frac{\int \csc (c+d x) \left (9 a^2 \left (a^2-4 b^2\right )+48 a^3 b \sin (c+d x)\right ) \, dx}{24 a^2}\\ &=2 a b x-\frac{b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{24 a^2 d}+\frac{17 a b \cot (c+d x)}{12 d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac{b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}+\frac{1}{8} \left (3 \left (a^2-4 b^2\right )\right ) \int \csc (c+d x) \, dx\\ &=2 a b x-\frac{3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{24 a^2 d}+\frac{17 a b \cot (c+d x)}{12 d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac{b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}\\ \end{align*}

Mathematica [A]  time = 2.71773, size = 270, normalized size = 1.52 \[ \frac{-3 a^2 \csc ^4\left (\frac{1}{2} (c+d x)\right )+30 a^2 \csc ^2\left (\frac{1}{2} (c+d x)\right )+3 a^2 \sec ^4\left (\frac{1}{2} (c+d x)\right )-30 a^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )+72 a^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-72 a^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-256 a b \tan \left (\frac{1}{2} (c+d x)\right )+256 a b \cot \left (\frac{1}{2} (c+d x)\right )+128 a b \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-8 a b \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )+384 a b c+384 a b d x-192 b^2 \cos (c+d x)-24 b^2 \csc ^2\left (\frac{1}{2} (c+d x)\right )+24 b^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )-288 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+288 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(384*a*b*c + 384*a*b*d*x - 192*b^2*Cos[c + d*x] + 256*a*b*Cot[(c + d*x)/2] + 30*a^2*Csc[(c + d*x)/2]^2 - 24*b^
2*Csc[(c + d*x)/2]^2 - 3*a^2*Csc[(c + d*x)/2]^4 - 72*a^2*Log[Cos[(c + d*x)/2]] + 288*b^2*Log[Cos[(c + d*x)/2]]
+ 72*a^2*Log[Sin[(c + d*x)/2]] - 288*b^2*Log[Sin[(c + d*x)/2]] - 30*a^2*Sec[(c + d*x)/2]^2 + 24*b^2*Sec[(c +
d*x)/2]^2 + 3*a^2*Sec[(c + d*x)/2]^4 + 128*a*b*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 8*a*b*Csc[(c + d*x)/2]^4*Si
n[c + d*x] - 256*a*b*Tan[(c + d*x)/2])/(192*d)

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Maple [A]  time = 0.088, size = 223, normalized size = 1.3 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8\,d}}+{\frac{3\,{a}^{2}\cos \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}}-{\frac{2\,ab \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+2\,{\frac{ab\cot \left ( dx+c \right ) }{d}}+2\,abx+2\,{\frac{abc}{d}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d}}-{\frac{3\,{b}^{2}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{3\,{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^2,x)

[Out]

-1/4/d*a^2/sin(d*x+c)^4*cos(d*x+c)^5+1/8/d*a^2/sin(d*x+c)^2*cos(d*x+c)^5+1/8*a^2*cos(d*x+c)^3/d+3/8*a^2*cos(d*
x+c)/d+3/8/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-2/3*a*b*cot(d*x+c)^3/d+2*a*b*cot(d*x+c)/d+2*a*b*x+2/d*a*b*c-1/2/d*b
^2/sin(d*x+c)^2*cos(d*x+c)^5-1/2*b^2*cos(d*x+c)^3/d-3/2*b^2*cos(d*x+c)/d-3/2/d*b^2*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 1.61168, size = 224, normalized size = 1.26 \begin{align*} \frac{32 \,{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a b - 3 \, a^{2}{\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 )} + 12 \, b^{2}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/48*(32*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a*b - 3*a^2*(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)) + 12*b^2*(2*c
os(d*x + c)/(cos(d*x + c)^2 - 1) - 4*cos(d*x + c) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)))/d

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Fricas [A]  time = 1.83193, size = 666, normalized size = 3.74 \begin{align*} \frac{96 \, a b d x \cos \left (d x + c\right )^{4} - 48 \, b^{2} \cos \left (d x + c\right )^{5} - 192 \, a b d x \cos \left (d x + c\right )^{2} + 96 \, a b d x - 30 \,{\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 18 \,{\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right ) - 9 \,{\left ({\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} - 4 \, b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 9 \,{\left ({\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} - 4 \, b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 32 \,{\left (4 \, a b \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, 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)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/48*(96*a*b*d*x*cos(d*x + c)^4 - 48*b^2*cos(d*x + c)^5 - 192*a*b*d*x*cos(d*x + c)^2 + 96*a*b*d*x - 30*(a^2 -
4*b^2)*cos(d*x + c)^3 + 18*(a^2 - 4*b^2)*cos(d*x + c) - 9*((a^2 - 4*b^2)*cos(d*x + c)^4 - 2*(a^2 - 4*b^2)*cos(
d*x + c)^2 + a^2 - 4*b^2)*log(1/2*cos(d*x + c) + 1/2) + 9*((a^2 - 4*b^2)*cos(d*x + c)^4 - 2*(a^2 - 4*b^2)*cos(
d*x + c)^2 + a^2 - 4*b^2)*log(-1/2*cos(d*x + c) + 1/2) - 32*(4*a*b*cos(d*x + c)^3 - 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)

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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)**5*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.37084, size = 329, normalized size = 1.85 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 16 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 384 \,{\left (d x + c\right )} a b - 240 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 72 \,{\left (a^{2} - 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{384 \, b^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} - \frac{150 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 600 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 240 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 16 \, 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 )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 + 16*a*b*tan(1/2*d*x + 1/2*c)^3 - 24*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*t
an(1/2*d*x + 1/2*c)^2 + 384*(d*x + c)*a*b - 240*a*b*tan(1/2*d*x + 1/2*c) + 72*(a^2 - 4*b^2)*log(abs(tan(1/2*d*
x + 1/2*c))) - 384*b^2/(tan(1/2*d*x + 1/2*c)^2 + 1) - (150*a^2*tan(1/2*d*x + 1/2*c)^4 - 600*b^2*tan(1/2*d*x +
1/2*c)^4 - 240*a*b*tan(1/2*d*x + 1/2*c)^3 - 24*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*tan(1/2*d*x + 1/2*c)^2 + 16
*a*b*tan(1/2*d*x + 1/2*c) + 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d