3.1067 \(\int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=148 \[ \frac {\left (2 a^2+5 b^2\right ) \cot (c+d x)}{15 d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 d}+\frac {a b \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac {a b \cot (c+d x) \csc ^3(c+d x)}{10 d}+\frac {a b \cot (c+d x) \csc (c+d x)}{4 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2}{5 d} \]

[Out]

1/4*a*b*arctanh(cos(d*x+c))/d+1/15*(2*a^2+5*b^2)*cot(d*x+c)/d+1/4*a*b*cot(d*x+c)*csc(d*x+c)/d+1/15*(a^2-2*b^2)
*cot(d*x+c)*csc(d*x+c)^2/d-1/10*a*b*cot(d*x+c)*csc(d*x+c)^3/d-1/5*cot(d*x+c)*csc(d*x+c)^4*(a+b*sin(d*x+c))^2/d

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Rubi [A]  time = 0.39, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2889, 3048, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {\left (2 a^2+5 b^2\right ) \cot (c+d x)}{15 d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 d}+\frac {a b \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac {a b \cot (c+d x) \csc ^3(c+d x)}{10 d}+\frac {a b \cot (c+d x) \csc (c+d x)}{4 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2889

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

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 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 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(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, 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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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

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Mathematica [A]  time = 0.82, size = 236, normalized size = 1.59 \[ \frac {\csc ^5(c+d x) \left (-40 \left (4 a^2+b^2\right ) \cos (c+d x)+20 \left (b^2-2 a^2\right ) \cos (3 (c+d x))+8 a^2 \cos (5 (c+d x))-180 a b \sin (2 (c+d x))-30 a b \sin (4 (c+d x))-150 a b \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+75 a b \sin (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-15 a b \sin (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+150 a b \sin (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-75 a b \sin (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+15 a b \sin (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+20 b^2 \cos (5 (c+d x))\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^5*(-40*(4*a^2 + b^2)*Cos[c + d*x] + 20*(-2*a^2 + b^2)*Cos[3*(c + d*x)] + 8*a^2*Cos[5*(c + d*x)]
+ 20*b^2*Cos[5*(c + d*x)] + 150*a*b*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] - 150*a*b*Log[Sin[(c + d*x)/2]]*Sin[c +
d*x] - 180*a*b*Sin[2*(c + d*x)] - 75*a*b*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d*x)] + 75*a*b*Log[Sin[(c + d*x)/2]
]*Sin[3*(c + d*x)] - 30*a*b*Sin[4*(c + d*x)] + 15*a*b*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] - 15*a*b*Log[Sin[
(c + d*x)/2]]*Sin[5*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.69, size = 195, normalized size = 1.32 \[ \frac {8 \, {\left (2 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 40 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(8*(2*a^2 + 5*b^2)*cos(d*x + c)^5 - 40*(a^2 + b^2)*cos(d*x + c)^3 + 15*(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) - 15*(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) - 30*(a*b*cos(d*x + c)^3 + a*b*cos(d*x + c))*sin(d*x + c))/((d*co
s(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))

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giac [A]  time = 0.22, size = 222, normalized size = 1.50 \[ \frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {274 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 20 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*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 + 5*a^2*tan(1/2*d*x + 1/2*c)^3 + 20*b^2*ta
n(1/2*d*x + 1/2*c)^3 - 120*a*b*log(abs(tan(1/2*d*x + 1/2*c))) - 30*a^2*tan(1/2*d*x + 1/2*c) - 60*b^2*tan(1/2*d
*x + 1/2*c) + (274*a*b*tan(1/2*d*x + 1/2*c)^5 + 30*a^2*tan(1/2*d*x + 1/2*c)^4 + 60*b^2*tan(1/2*d*x + 1/2*c)^4
- 5*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

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maple [A]  time = 0.49, size = 156, normalized size = 1.05 \[ -\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}}-\frac {2 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{15 d \sin \left (d x +c \right )^{3}}-\frac {a b \left (\cos ^{3}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{4}}-\frac {a b \left (\cos ^{3}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{2}}-\frac {a b \cos \left (d x +c \right )}{4 d}-\frac {a b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{4 d}-\frac {b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.38, size = 108, normalized size = 0.73 \[ -\frac {15 \, a b {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {40 \, b^{2}}{\tan \left (d x + c\right )^{3}} + \frac {8 \, {\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{2}}{\tan \left (d x + c\right )^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*a*b*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - log(cos(d*x + c) +
1) + log(cos(d*x + c) - 1)) + 40*b^2/tan(d*x + c)^3 + 8*(5*tan(d*x + c)^2 + 3)*a^2/tan(d*x + c)^5)/d

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mupad [B]  time = 9.36, size = 187, normalized size = 1.26 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{3}+\frac {4\,b^2}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2+4\,b^2\right )+\frac {a^2}{5}+a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{32\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2}{16}+\frac {b^2}{8}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a^2}{96}+\frac {b^2}{24}\right )}{d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}-\frac {a\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*tan(c/2 + (d*x)/2)^5)/(160*d) - (cot(c/2 + (d*x)/2)^5*(tan(c/2 + (d*x)/2)^2*(a^2/3 + (4*b^2)/3) - tan(c/2
+ (d*x)/2)^4*(2*a^2 + 4*b^2) + a^2/5 + a*b*tan(c/2 + (d*x)/2)))/(32*d) - (tan(c/2 + (d*x)/2)*(a^2/16 + b^2/8)
)/d + (tan(c/2 + (d*x)/2)^3*(a^2/96 + b^2/24))/d + (a*b*tan(c/2 + (d*x)/2)^4)/(32*d) - (a*b*log(tan(c/2 + (d*x
)/2)))/(4*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**6*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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