\(\int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx\) [1112]

Optimal result

Integrand size = 27, antiderivative size = 178 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=2 a b x-\frac {3 \left (a^2-4 b^2\right ) \text {arctanh}(\cos (c+d x))}{8 d}-\frac {b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{24 a^2 d}+\frac {17 a b \cot (c+d x)}{12 d}+\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d} \]

[Out]

2*a*b*x-3/8*(a^2-4*b^2)*arctanh(cos(d*x+c))/d-1/24*b^2*(39*a^2+2*b^2)*cos(d*x+c)/a^2/d+17/12*a*b*cot(d*x+c)/d+
5/8*cot(d*x+c)*csc(d*x+c)*(a+b*sin(d*x+c))^2/d+1/12*b*cot(d*x+c)*csc(d*x+c)^2*(a+b*sin(d*x+c))^3/a^2/d-1/4*cot
(d*x+c)*csc(d*x+c)^3*(a+b*sin(d*x+c))^3/a/d

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2972, 3126, 3110, 3102, 2814, 3855} \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {3 \left (a^2-4 b^2\right ) \text {arctanh}(\cos (c+d x))}{8 d}-\frac {b^2 \left (39 a^2+2 b^2\right ) \cos (c+d x)}{24 a^2 d}+\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{12 a^2 d}+\frac {17 a b \cot (c+d x)}{12 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 a d}+\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{8 d}+2 a b x \]

[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 2814

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 2972

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 3102

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 3110

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] &&
NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3126

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {arctanh}(\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] (verified)

Time = 5.18 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.52 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {384 a b c+384 a b d x-192 b^2 \cos (c+d x)+256 a b \cot \left (\frac {1}{2} (c+d x)\right )+30 a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )-24 b^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )-3 a^2 \csc ^4\left (\frac {1}{2} (c+d x)\right )-72 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+288 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+72 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-288 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-30 a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+24 b^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+3 a^2 \sec ^4\left (\frac {1}{2} (c+d x)\right )+128 a b \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-8 a b \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)-256 a b \tan \left (\frac {1}{2} (c+d x)\right )}{192 d} \]

[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)

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {\cos ^{5}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos ^{5}\left (d x +c \right )}{8 \sin \left (d x +c \right )^{2}}+\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+2 a b \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+b^{2} \left (-\frac {\cos ^{5}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(167\)
default	\(\frac {a^{2} \left (-\frac {\cos ^{5}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos ^{5}\left (d x +c \right )}{8 \sin \left (d x +c \right )^{2}}+\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+2 a b \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+b^{2} \left (-\frac {\cos ^{5}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(167\)
parallelrisch	\(\frac {576 \left (a^{2}-4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 \left (\cos \left (d x +c \right )-\frac {2 \cos \left (2 d x +2 c \right )}{3}+\frac {5 \cos \left (3 d x +3 c \right )}{3}+\frac {\cos \left (4 d x +4 c \right )}{6}+\frac {1}{2}\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} \left (\csc ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-128 a b \left (\sec ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (3 d x +3 c \right )-288 b^{2} \left (\cos \left (d x +c \right )+\frac {3 \cos \left (2 d x +2 c \right )}{4}-\frac {\cos \left (3 d x +3 c \right )}{3}-\frac {3}{4}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3072 a b x d}{1536 d}\)	\(194\)
risch	\(2 a b x -\frac {b^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 d}-\frac {-96 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+15 a^{2} {\mathrm e}^{7 i \left (d x +c \right )}-12 b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+192 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+9 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-160 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+9 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+64 i a b +15 a^{2} {\mathrm e}^{i \left (d x +c \right )}-12 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{2 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{2 d}\)	\(299\)
norman	\(\frac {-\frac {a^{2}}{64 d}+\frac {a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {\left (3 a^{2}-4 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}-\frac {\left (3 a^{2}-4 b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}+\frac {\left (15 a^{2}-80 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}+\frac {\left (15 a^{2}-80 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d}+\frac {13 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {7 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {7 a b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {13 a b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+2 a b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 a b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a b x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {3 \left (a^{2}-4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}\)	\(343\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.46 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {96 \, a b d x \cos \left (d x + c\right )^{4} - 48 \, b^{2} \cos \left (d x + c\right )^{5} - 192 \, a b d x \cos \left (d x + c\right )^{2} + 96 \, a b d x - 30 \, {\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 18 \, {\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right ) - 9 \, {\left ({\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} - 4 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 9 \, {\left ({\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} - 4 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 32 \, {\left (4 \, a b \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.93 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {32 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a b - 3 \, a^{2} {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 12 \, b^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.37 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 384 \, {\left (d x + c\right )} a b - 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, {\left (a^{2} - 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {384 \, b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - \frac {150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 600 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 12.77 (sec) , antiderivative size = 825, normalized size of antiderivative = 4.63 \[ \int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Too large to display} \]

[In]

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

[Out]

-(3*a^2*cos(c/2 + (d*x)/2)^10 - 3*a^2*sin(c/2 + (d*x)/2)^10 + 21*a^2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8
+ 24*a^2*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 - 24*a^2*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 - 21*a^
2*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 - 24*b^2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 - 24*b^2*cos(c/
2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 408*b^2*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 24*b^2*cos(c/2 + (d*
x)/2)^8*sin(c/2 + (d*x)/2)^2 - 16*a*b*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^9 + 16*a*b*cos(c/2 + (d*x)/2)^9*si
n(c/2 + (d*x)/2) - 72*a^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6
- 72*a^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 288*b^2*log(s
in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 288*b^2*log(sin(c/2 + (d*x)/
2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 224*a*b*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x
)/2)^7 - 224*a*b*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^3 + 768*a*b*atan((3*a^2*sin(c/2 + (d*x)/2) - 12*b^2*s
in(c/2 + (d*x)/2) + 16*a*b*cos(c/2 + (d*x)/2))/(12*b^2*cos(c/2 + (d*x)/2) - 3*a^2*cos(c/2 + (d*x)/2) + 16*a*b*
sin(c/2 + (d*x)/2)))*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 768*a*b*atan((3*a^2*sin(c/2 + (d*x)/2) - 12*b
^2*sin(c/2 + (d*x)/2) + 16*a*b*cos(c/2 + (d*x)/2))/(12*b^2*cos(c/2 + (d*x)/2) - 3*a^2*cos(c/2 + (d*x)/2) + 16*
a*b*sin(c/2 + (d*x)/2)))*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4)/(192*d*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x
)/2)^4*(cos(c/2 + (d*x)/2)^2 + sin(c/2 + (d*x)/2)^2))