3.593 \(\int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=158 \[ \frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {7 a^2 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {25 a^2 x}{16} \]

[Out]

-25/16*a^2*x-2*a^2*arctanh(cos(d*x+c))/d+2*a^2*cos(d*x+c)/d+2/3*a^2*cos(d*x+c)^3/d+2/5*a^2*cos(d*x+c)^5/d-a^2*
cot(d*x+c)/d-7/16*a^2*cos(d*x+c)*sin(d*x+c)/d-7/24*a^2*cos(d*x+c)*sin(d*x+c)^3/d+1/6*a^2*cos(d*x+c)*sin(d*x+c)
^5/d

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Rubi [A]  time = 0.23, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2872, 3770, 3767, 8, 2638, 2633, 2635} \[ \frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {7 a^2 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {25 a^2 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*a^2*x)/16 - (2*a^2*ArcTanh[Cos[c + d*x]])/d + (2*a^2*Cos[c + d*x])/d + (2*a^2*Cos[c + d*x]^3)/(3*d) + (2*
a^2*Cos[c + d*x]^5)/(5*d) - (a^2*Cot[c + d*x])/d - (7*a^2*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (7*a^2*Cos[c + d
*x]*Sin[c + d*x]^3)/(24*d) + (a^2*Cos[c + d*x]*Sin[c + d*x]^5)/(6*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 2638

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

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 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 ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {\int \left (-2 a^8+2 a^8 \csc (c+d x)+a^8 \csc ^2(c+d x)-6 a^8 \sin (c+d x)+6 a^8 \sin ^3(c+d x)+2 a^8 \sin ^4(c+d x)-2 a^8 \sin ^5(c+d x)-a^8 \sin ^6(c+d x)\right ) \, dx}{a^6}\\ &=-2 a^2 x+a^2 \int \csc ^2(c+d x) \, dx-a^2 \int \sin ^6(c+d x) \, dx+\left (2 a^2\right ) \int \csc (c+d x) \, dx+\left (2 a^2\right ) \int \sin ^4(c+d x) \, dx-\left (2 a^2\right ) \int \sin ^5(c+d x) \, dx-\left (6 a^2\right ) \int \sin (c+d x) \, dx+\left (6 a^2\right ) \int \sin ^3(c+d x) \, dx\\ &=-2 a^2 x-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {6 a^2 \cos (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin ^3(c+d x)}{2 d}+\frac {a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac {1}{6} \left (5 a^2\right ) \int \sin ^4(c+d x) \, dx+\frac {1}{2} \left (3 a^2\right ) \int \sin ^2(c+d x) \, dx-\frac {a^2 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (6 a^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-2 a^2 x-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cot (c+d x)}{d}-\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}-\frac {7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac {1}{8} \left (5 a^2\right ) \int \sin ^2(c+d x) \, dx+\frac {1}{4} \left (3 a^2\right ) \int 1 \, dx\\ &=-\frac {5 a^2 x}{4}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cot (c+d x)}{d}-\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac {1}{16} \left (5 a^2\right ) \int 1 \, dx\\ &=-\frac {25 a^2 x}{16}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cot (c+d x)}{d}-\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 110, normalized size = 0.70 \[ \frac {a^2 \left (-255 \sin (2 (c+d x))+15 \sin (4 (c+d x))+5 \sin (6 (c+d x))+2640 \cos (c+d x)+280 \cos (3 (c+d x))+24 \cos (5 (c+d x))-960 \cot (c+d x)+1920 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-1920 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-1500 c-1500 d x\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-1500*c - 1500*d*x + 2640*Cos[c + d*x] + 280*Cos[3*(c + d*x)] + 24*Cos[5*(c + d*x)] - 960*Cot[c + d*x] -
1920*Log[Cos[(c + d*x)/2]] + 1920*Log[Sin[(c + d*x)/2]] - 255*Sin[2*(c + d*x)] + 15*Sin[4*(c + d*x)] + 5*Sin[
6*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.80, size = 161, normalized size = 1.02 \[ -\frac {40 \, a^{2} \cos \left (d x + c\right )^{7} - 50 \, a^{2} \cos \left (d x + c\right )^{5} - 125 \, a^{2} \cos \left (d x + c\right )^{3} + 240 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 240 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 375 \, a^{2} \cos \left (d x + c\right ) - {\left (96 \, a^{2} \cos \left (d x + c\right )^{5} + 160 \, a^{2} \cos \left (d x + c\right )^{3} - 375 \, a^{2} d x + 480 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(40*a^2*cos(d*x + c)^7 - 50*a^2*cos(d*x + c)^5 - 125*a^2*cos(d*x + c)^3 + 240*a^2*log(1/2*cos(d*x + c)
+ 1/2)*sin(d*x + c) - 240*a^2*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 375*a^2*cos(d*x + c) - (96*a^2*cos(d
*x + c)^5 + 160*a^2*cos(d*x + c)^3 - 375*a^2*d*x + 480*a^2*cos(d*x + c))*sin(d*x + c))/(d*sin(d*x + c))

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giac [A]  time = 0.27, size = 274, normalized size = 1.73 \[ -\frac {375 \, {\left (d x + c\right )} a^{2} - 480 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 120 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {120 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1440 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 4320 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6720 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2976 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 736 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(375*(d*x + c)*a^2 - 480*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 120*a^2*tan(1/2*d*x + 1/2*c) + 120*(4*a^2
*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c) - 2*(105*a^2*tan(1/2*d*x + 1/2*c)^11 + 1440*a^2*tan(1/2*d*x
+ 1/2*c)^10 + 595*a^2*tan(1/2*d*x + 1/2*c)^9 + 4320*a^2*tan(1/2*d*x + 1/2*c)^8 - 150*a^2*tan(1/2*d*x + 1/2*c)^
7 + 7360*a^2*tan(1/2*d*x + 1/2*c)^6 + 150*a^2*tan(1/2*d*x + 1/2*c)^5 + 6720*a^2*tan(1/2*d*x + 1/2*c)^4 - 595*a
^2*tan(1/2*d*x + 1/2*c)^3 + 2976*a^2*tan(1/2*d*x + 1/2*c)^2 - 105*a^2*tan(1/2*d*x + 1/2*c) + 736*a^2)/(tan(1/2
*d*x + 1/2*c)^2 + 1)^6)/d

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maple [A]  time = 0.42, size = 175, normalized size = 1.11 \[ -\frac {5 a^{2} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6 d}-\frac {25 a^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{24 d}-\frac {25 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{16 d}-\frac {25 a^{2} x}{16}-\frac {25 a^{2} c}{16 d}+\frac {2 a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5 d}+\frac {2 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {2 a^{2} \cos \left (d x +c \right )}{d}+\frac {2 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/6*a^2*cos(d*x+c)^5*sin(d*x+c)/d-25/24*a^2*cos(d*x+c)^3*sin(d*x+c)/d-25/16*a^2*cos(d*x+c)*sin(d*x+c)/d-25/16
*a^2*x-25/16/d*a^2*c+2/5*a^2*cos(d*x+c)^5/d+2/3*a^2*cos(d*x+c)^3/d+2*a^2*cos(d*x+c)/d+2/d*a^2*ln(csc(d*x+c)-co
t(d*x+c))-1/d*a^2/sin(d*x+c)*cos(d*x+c)^7

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maxima [A]  time = 0.78, size = 173, normalized size = 1.09 \[ \frac {64 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 120 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(64*(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*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a^2 - 120
*(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 = 8.97, size = 401, normalized size = 2.54 \[ \frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {25\,a^2\,\mathrm {atan}\left (\frac {625\,a^4}{64\,\left (\frac {25\,a^4}{2}+\frac {625\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}-\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {25\,a^4}{2}+\frac {625\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}\right )}{8\,d}+\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{4}+24\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\frac {47\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{12}+72\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {35\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+\frac {368\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {35\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+112\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {299\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12}+\frac {248\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}-\frac {31\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {184\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}-a^2}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

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)^2,x)

[Out]

(2*a^2*log(tan(c/2 + (d*x)/2)))/d + (25*a^2*atan((625*a^4)/(64*((25*a^4)/2 + (625*a^4*tan(c/2 + (d*x)/2))/64))
- (25*a^4*tan(c/2 + (d*x)/2))/(2*((25*a^4)/2 + (625*a^4*tan(c/2 + (d*x)/2))/64))))/(8*d) + ((248*a^2*tan(c/2
+ (d*x)/2)^3)/5 - (31*a^2*tan(c/2 + (d*x)/2)^2)/4 - (299*a^2*tan(c/2 + (d*x)/2)^4)/12 + 112*a^2*tan(c/2 + (d*x
)/2)^5 - (35*a^2*tan(c/2 + (d*x)/2)^6)/2 + (368*a^2*tan(c/2 + (d*x)/2)^7)/3 - (35*a^2*tan(c/2 + (d*x)/2)^8)/2
+ 72*a^2*tan(c/2 + (d*x)/2)^9 + (47*a^2*tan(c/2 + (d*x)/2)^10)/12 + 24*a^2*tan(c/2 + (d*x)/2)^11 + (3*a^2*tan(
c/2 + (d*x)/2)^12)/4 - a^2 + (184*a^2*tan(c/2 + (d*x)/2))/15)/(d*(2*tan(c/2 + (d*x)/2) + 12*tan(c/2 + (d*x)/2)
^3 + 30*tan(c/2 + (d*x)/2)^5 + 40*tan(c/2 + (d*x)/2)^7 + 30*tan(c/2 + (d*x)/2)^9 + 12*tan(c/2 + (d*x)/2)^11 +
2*tan(c/2 + (d*x)/2)^13)) + (a^2*tan(c/2 + (d*x)/2))/(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)**2*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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