3.1108 \(\int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=116 \[ \frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos (c+d x)}{d}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a b \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac {3 a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac {3 a b x}{4}-\frac {b^2 \cos ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.45, antiderivative size = 190, normalized size of antiderivative = 1.64, 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 = {2895, 3049, 3033, 3023, 2735, 3770} \[ -\frac {\left (-14 a^2 b^2+a^4+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}-\frac {a \left (2 a^2-27 b^2\right ) \sin (c+d x) \cos (c+d x)}{60 b d}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{5 b d}+\frac {3 a b x}{4} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*Cot[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

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

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 2895

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d*f*(m
+ n + 3)*(m + n + 4)), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e
+ f*x])^(m + 1))/(b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[m + n + 4, 0]

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 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x))^2 \left (-20 b^2+2 a b \sin (c+d x)-2 \left (a^2-12 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{20 b^2}\\ &=-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x)) \left (-60 a b^2+2 b \left (a^2-6 b^2\right ) \sin (c+d x)-2 a \left (2 a^2-27 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{60 b^2}\\ &=-\frac {a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac {\int \csc (c+d x) \left (-120 a^2 b^2-90 a b^3 \sin (c+d x)-8 \left (a^4-14 a^2 b^2+3 b^4\right ) \sin ^2(c+d x)\right ) \, dx}{120 b^2}\\ &=-\frac {\left (a^4-14 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac {a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac {\int \csc (c+d x) \left (-120 a^2 b^2-90 a b^3 \sin (c+d x)\right ) \, dx}{120 b^2}\\ &=\frac {3 a b x}{4}-\frac {\left (a^4-14 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac {a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}+a^2 \int \csc (c+d x) \, dx\\ &=\frac {3 a b x}{4}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {\left (a^4-14 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac {a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 125, normalized size = 1.08 \[ \frac {30 \left (10 a^2-b^2\right ) \cos (c+d x)+5 \left (4 a^2-3 b^2\right ) \cos (3 (c+d x))+15 a \left (4 \left (4 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 b (c+d x)\right )+8 b \sin (2 (c+d x))+b \sin (4 (c+d x))\right )-3 b^2 \cos (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*Cot[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(30*(10*a^2 - b^2)*Cos[c + d*x] + 5*(4*a^2 - 3*b^2)*Cos[3*(c + d*x)] - 3*b^2*Cos[5*(c + d*x)] + 15*a*(4*(3*b*(
c + d*x) - 4*a*Log[Cos[(c + d*x)/2]] + 4*a*Log[Sin[(c + d*x)/2]]) + 8*b*Sin[2*(c + d*x)] + b*Sin[4*(c + d*x)])
)/(240*d)

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fricas [A]  time = 0.64, size = 112, normalized size = 0.97 \[ -\frac {12 \, b^{2} \cos \left (d x + c\right )^{5} - 20 \, a^{2} \cos \left (d x + c\right )^{3} - 45 \, a b d x - 60 \, a^{2} \cos \left (d x + c\right ) + 30 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 30 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (2 \, a b \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*b^2*cos(d*x + c)^5 - 20*a^2*cos(d*x + c)^3 - 45*a*b*d*x - 60*a^2*cos(d*x + c) + 30*a^2*log(1/2*cos(d
*x + c) + 1/2) - 30*a^2*log(-1/2*cos(d*x + c) + 1/2) - 15*(2*a*b*cos(d*x + c)^3 + 3*a*b*cos(d*x + c))*sin(d*x
+ c))/d

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giac [B]  time = 0.21, size = 213, normalized size = 1.84 \[ \frac {45 \, {\left (d x + c\right )} a b + 60 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 440 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 280 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 80 \, a^{2} + 12 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(45*(d*x + c)*a*b + 60*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(75*a*b*tan(1/2*d*x + 1/2*c)^9 - 120*a^2*ta
n(1/2*d*x + 1/2*c)^8 + 60*b^2*tan(1/2*d*x + 1/2*c)^8 + 30*a*b*tan(1/2*d*x + 1/2*c)^7 - 360*a^2*tan(1/2*d*x + 1
/2*c)^6 - 440*a^2*tan(1/2*d*x + 1/2*c)^4 + 120*b^2*tan(1/2*d*x + 1/2*c)^4 - 30*a*b*tan(1/2*d*x + 1/2*c)^3 - 28
0*a^2*tan(1/2*d*x + 1/2*c)^2 - 75*a*b*tan(1/2*d*x + 1/2*c) - 80*a^2 + 12*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/
d

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maple [A]  time = 0.54, size = 123, normalized size = 1.06 \[ \frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} \cos \left (d x +c \right )}{d}+\frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {a b \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{4 d}+\frac {3 a b x}{4}+\frac {3 a b c}{4 d}-\frac {b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.39, size = 97, normalized size = 0.84 \[ -\frac {48 \, b^{2} \cos \left (d x + c\right )^{5} - 40 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(48*b^2*cos(d*x + c)^5 - 40*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*
x + c) - 1))*a^2 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a*b)/d

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mupad [B]  time = 11.12, size = 319, normalized size = 2.75 \[ \frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (4\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {44\,a^2}{3}-4\,b^2\right )+\frac {28\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {8\,a^2}{3}-\frac {2\,b^2}{5}+a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {5\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+\frac {5\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,b\,\mathrm {atan}\left (\frac {9\,a^2\,b^2}{4\,\left (3\,a^3\,b-\frac {9\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )}+\frac {3\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,a^3\,b-\frac {9\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*log(tan(c/2 + (d*x)/2)))/d + (tan(c/2 + (d*x)/2)^8*(4*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*((44*a^2)/3 - 4
*b^2) + (28*a^2*tan(c/2 + (d*x)/2)^2)/3 + 12*a^2*tan(c/2 + (d*x)/2)^6 + (8*a^2)/3 - (2*b^2)/5 + a*b*tan(c/2 +
(d*x)/2)^3 - a*b*tan(c/2 + (d*x)/2)^7 - (5*a*b*tan(c/2 + (d*x)/2)^9)/2 + (5*a*b*tan(c/2 + (d*x)/2))/2)/(d*(5*t
an(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (
d*x)/2)^10 + 1)) + (3*a*b*atan((9*a^2*b^2)/(4*(3*a^3*b - (9*a^2*b^2*tan(c/2 + (d*x)/2))/4)) + (3*a^3*b*tan(c/2
+ (d*x)/2))/(3*a^3*b - (9*a^2*b^2*tan(c/2 + (d*x)/2))/4)))/(2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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