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

Optimal. Leaf size=140 \[ \frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cos (c+d x)}{d}-\frac {2 a^2 \cot (c+d x)}{d}+\frac {a^2 \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac {9 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {15 a^2 x}{4} \]

[Out]

-15/4*a^2*x+3/2*a^2*arctanh(cos(d*x+c))/d-a^2*cos(d*x+c)/d+1/5*a^2*cos(d*x+c)^5/d-2*a^2*cot(d*x+c)/d-1/2*a^2*c
ot(d*x+c)*csc(d*x+c)/d-9/4*a^2*cos(d*x+c)*sin(d*x+c)/d+1/2*a^2*cos(d*x+c)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.21, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2872, 3770, 3767, 8, 3768, 2635, 2633} \[ \frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cos (c+d x)}{d}-\frac {2 a^2 \cot (c+d x)}{d}+\frac {a^2 \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac {9 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {15 a^2 x}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*a^2*x)/4 + (3*a^2*ArcTanh[Cos[c + d*x]])/(2*d) - (a^2*Cos[c + d*x])/d + (a^2*Cos[c + d*x]^5)/(5*d) - (2*a
^2*Cot[c + d*x])/d - (a^2*Cot[c + d*x]*Csc[c + d*x])/(2*d) - (9*a^2*Cos[c + d*x]*Sin[c + d*x])/(4*d) + (a^2*Co
s[c + d*x]*Sin[c + d*x]^3)/(2*d)

Rule 8

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

Rule 2633

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

Rule 2635

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

Rule 2872

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

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 \cos ^3(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {\int \left (-6 a^8-2 a^8 \csc (c+d x)+2 a^8 \csc ^2(c+d x)+a^8 \csc ^3(c+d x)+6 a^8 \sin ^2(c+d x)+2 a^8 \sin ^3(c+d x)-2 a^8 \sin ^4(c+d x)-a^8 \sin ^5(c+d x)\right ) \, dx}{a^6}\\ &=-6 a^2 x+a^2 \int \csc ^3(c+d x) \, dx-a^2 \int \sin ^5(c+d x) \, dx-\left (2 a^2\right ) \int \csc (c+d x) \, dx+\left (2 a^2\right ) \int \csc ^2(c+d x) \, dx+\left (2 a^2\right ) \int \sin ^3(c+d x) \, dx-\left (2 a^2\right ) \int \sin ^4(c+d x) \, dx+\left (6 a^2\right ) \int \sin ^2(c+d x) \, dx\\ &=-6 a^2 x+\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{d}+\frac {a^2 \cos (c+d x) \sin ^3(c+d x)}{2 d}+\frac {1}{2} a^2 \int \csc (c+d x) \, dx-\frac {1}{2} \left (3 a^2\right ) \int \sin ^2(c+d x) \, dx+\left (3 a^2\right ) \int 1 \, dx+\frac {a^2 \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (2 a^2\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-3 a^2 x+\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {2 a^2 \cot (c+d x)}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {9 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 \cos (c+d x) \sin ^3(c+d x)}{2 d}-\frac {1}{4} \left (3 a^2\right ) \int 1 \, dx\\ &=-\frac {15 a^2 x}{4}+\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {2 a^2 \cot (c+d x)}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {9 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 \cos (c+d x) \sin ^3(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 5.53, size = 174, normalized size = 1.24 \[ \frac {(a \sin (c+d x)+a)^2 \left (-300 (c+d x)-80 \sin (2 (c+d x))-5 \sin (4 (c+d x))-70 \cos (c+d x)+5 \cos (3 (c+d x))+\cos (5 (c+d x))+80 \tan \left (\frac {1}{2} (c+d x)\right )-80 \cot \left (\frac {1}{2} (c+d x)\right )-10 \csc ^2\left (\frac {1}{2} (c+d x)\right )+10 \sec ^2\left (\frac {1}{2} (c+d x)\right )-120 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+120 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{80 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + a*Sin[c + d*x])^2*(-300*(c + d*x) - 70*Cos[c + d*x] + 5*Cos[3*(c + d*x)] + Cos[5*(c + d*x)] - 80*Cot[(c
+ d*x)/2] - 10*Csc[(c + d*x)/2]^2 + 120*Log[Cos[(c + d*x)/2]] - 120*Log[Sin[(c + d*x)/2]] + 10*Sec[(c + d*x)/2
]^2 - 80*Sin[2*(c + d*x)] - 5*Sin[4*(c + d*x)] + 80*Tan[(c + d*x)/2]))/(80*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)
/2])^4)

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fricas [A]  time = 0.93, size = 199, normalized size = 1.42 \[ \frac {4 \, a^{2} \cos \left (d x + c\right )^{7} - 4 \, a^{2} \cos \left (d x + c\right )^{5} - 75 \, a^{2} d x \cos \left (d x + c\right )^{2} - 20 \, a^{2} \cos \left (d x + c\right )^{3} + 75 \, a^{2} d x + 30 \, a^{2} \cos \left (d x + c\right ) + 15 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 5 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{5} + 5 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{20 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/20*(4*a^2*cos(d*x + c)^7 - 4*a^2*cos(d*x + c)^5 - 75*a^2*d*x*cos(d*x + c)^2 - 20*a^2*cos(d*x + c)^3 + 75*a^2
*d*x + 30*a^2*cos(d*x + c) + 15*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*cos(d*x + c) + 1/2) - 15*(a^2*cos(d*x + c)^
2 - a^2)*log(-1/2*cos(d*x + c) + 1/2) - 5*(2*a^2*cos(d*x + c)^5 + 5*a^2*cos(d*x + c)^3 - 15*a^2*cos(d*x + c))*
sin(d*x + c))/(d*cos(d*x + c)^2 - d)

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giac [A]  time = 0.26, size = 244, normalized size = 1.74 \[ \frac {5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 150 \, {\left (d x + c\right )} a^{2} - 60 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 40 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {5 \, {\left (18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {4 \, {\left (45 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 50 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 50 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 16 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{40 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/40*(5*a^2*tan(1/2*d*x + 1/2*c)^2 - 150*(d*x + c)*a^2 - 60*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 40*a^2*tan(1/
2*d*x + 1/2*c) + 5*(18*a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2 +
4*(45*a^2*tan(1/2*d*x + 1/2*c)^9 + 50*a^2*tan(1/2*d*x + 1/2*c)^7 - 80*a^2*tan(1/2*d*x + 1/2*c)^6 - 80*a^2*tan
(1/2*d*x + 1/2*c)^4 - 50*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*a^2*tan(1/2*d*x + 1/2*c)^2 - 45*a^2*tan(1/2*d*x + 1/2
*c) - 16*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

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maple [A]  time = 0.48, size = 199, normalized size = 1.42 \[ -\frac {3 a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{10 d}-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{2 d}-\frac {3 a^{2} \cos \left (d x +c \right )}{2 d}-\frac {3 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {2 a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {2 a^{2} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {5 a^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{2 d}-\frac {15 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{4 d}-\frac {15 a^{2} x}{4}-\frac {15 a^{2} c}{4 d}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

-3/10*a^2*cos(d*x+c)^5/d-1/2*a^2*cos(d*x+c)^3/d-3/2*a^2*cos(d*x+c)/d-3/2/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-2/d*a
^2/sin(d*x+c)*cos(d*x+c)^7-2*a^2*cos(d*x+c)^5*sin(d*x+c)/d-5/2*a^2*cos(d*x+c)^3*sin(d*x+c)/d-15/4*a^2*cos(d*x+
c)*sin(d*x+c)/d-15/4*a^2*x-15/4/d*a^2*c-1/2/d*a^2/sin(d*x+c)^2*cos(d*x+c)^7

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maxima [A]  time = 0.81, size = 191, normalized size = 1.36 \[ \frac {2 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 5 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 15 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/60*(2*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x +
c) - 1))*a^2 - 5*(4*cos(d*x + c)^3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos(d*x + c) - 15*log(cos(d*x +
c) + 1) + 15*log(cos(d*x + c) - 1))*a^2 - 15*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(tan
(d*x + c)^5 + 2*tan(d*x + c)^3 + tan(d*x + c)))*a^2)/d

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mupad [B]  time = 8.94, size = 377, normalized size = 2.69 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {3\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {15\,a^2\,\mathrm {atan}\left (\frac {225\,a^4}{4\,\left (\frac {45\,a^4}{2}-\frac {225\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )}+\frac {45\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {45\,a^4}{2}-\frac {225\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )}\right )}{2\,d}-\frac {-14\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{2}+\frac {69\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+40\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+37\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+60\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+37\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+38\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {89\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{10}+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^6*(a + a*sin(c + d*x))^2)/sin(c + d*x)^3,x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^2)/(8*d) - (3*a^2*log(tan(c/2 + (d*x)/2)))/(2*d) - (15*a^2*atan((225*a^4)/(4*((45*a^4)
/2 - (225*a^4*tan(c/2 + (d*x)/2))/4)) + (45*a^4*tan(c/2 + (d*x)/2))/(2*((45*a^4)/2 - (225*a^4*tan(c/2 + (d*x)/
2))/4))))/(2*d) - ((89*a^2*tan(c/2 + (d*x)/2)^2)/10 + 38*a^2*tan(c/2 + (d*x)/2)^3 + 37*a^2*tan(c/2 + (d*x)/2)^
4 + 60*a^2*tan(c/2 + (d*x)/2)^5 + 37*a^2*tan(c/2 + (d*x)/2)^6 + 40*a^2*tan(c/2 + (d*x)/2)^7 + (69*a^2*tan(c/2
+ (d*x)/2)^8)/2 + (a^2*tan(c/2 + (d*x)/2)^10)/2 - 14*a^2*tan(c/2 + (d*x)/2)^11 + a^2/2 + 4*a^2*tan(c/2 + (d*x)
/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 20*tan(c/2 + (d*x)/2)^4 + 40*tan(c/2 + (d*x)/2)^6 + 40*tan(c/2 + (d*x)/2)^8
+ 20*tan(c/2 + (d*x)/2)^10 + 4*tan(c/2 + (d*x)/2)^12)) + (a^2*tan(c/2 + (d*x)/2))/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)**6*csc(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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