3.727 \(\int \frac {\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=119 \[ -\frac {\cos ^5(c+d x)}{5 a^2 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {\cos (c+d x)}{a^2 d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 a^2 d}-\frac {3 \sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {3 x}{4 a^2} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2875, 2873, 2635, 8, 2592, 302, 206, 2565, 30} \[ -\frac {\cos ^5(c+d x)}{5 a^2 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {\cos (c+d x)}{a^2 d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 a^2 d}-\frac {3 \sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {3 x}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^7*Cot[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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 2873

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

Rule 2875

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

Rubi steps

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

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Mathematica [A]  time = 0.71, size = 93, normalized size = 0.78 \[ \frac {270 \cos (c+d x)+5 \cos (3 (c+d x))-3 \left (40 \sin (2 (c+d x))+5 \sin (4 (c+d x))+\cos (5 (c+d x))-80 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+80 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+60 c+60 d x\right )}{240 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^7*Cot[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

(270*Cos[c + d*x] + 5*Cos[3*(c + d*x)] - 3*(60*c + 60*d*x + Cos[5*(c + d*x)] + 80*Log[Cos[(c + d*x)/2]] - 80*L
og[Sin[(c + d*x)/2]] + 40*Sin[2*(c + d*x)] + 5*Sin[4*(c + d*x)]))/(240*a^2*d)

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fricas [A]  time = 0.50, size = 94, normalized size = 0.79 \[ -\frac {12 \, \cos \left (d x + c\right )^{5} - 20 \, \cos \left (d x + c\right )^{3} + 45 \, d x + 15 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 60 \, \cos \left (d x + c\right ) + 30 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 30 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{60 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 156, normalized size = 1.31 \[ -\frac {\frac {45 \, {\left (d x + c\right )}}{a^{2}} - \frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {2 \, {\left (75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 360 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 320 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 280 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 68\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a^{2}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/60*(45*(d*x + c)/a^2 - 60*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 2*(75*tan(1/2*d*x + 1/2*c)^9 + 60*tan(1/2*d*
x + 1/2*c)^8 + 30*tan(1/2*d*x + 1/2*c)^7 + 360*tan(1/2*d*x + 1/2*c)^6 + 320*tan(1/2*d*x + 1/2*c)^4 - 30*tan(1/
2*d*x + 1/2*c)^3 + 280*tan(1/2*d*x + 1/2*c)^2 - 75*tan(1/2*d*x + 1/2*c) + 68)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*
a^2))/d

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maple [B]  time = 0.54, size = 329, normalized size = 2.76 \[ \frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {2 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {12 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {32 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {28 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {34}{15 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*csc(d*x+c)/(a+a*sin(d*x+c))^2,x)

[Out]

5/2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^9+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c
)^8+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^7+12/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/
2*c)^6+32/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^4-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d
*x+1/2*c)^3+28/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^2-5/2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*ta
n(1/2*d*x+1/2*c)+34/15/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5-3/2/d/a^2*arctan(tan(1/2*d*x+1/2*c))+1/d/a^2*ln(tan(1/
2*d*x+1/2*c))

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maxima [B]  time = 0.43, size = 333, normalized size = 2.80 \[ -\frac {\frac {\frac {75 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {280 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {320 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {360 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {30 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {60 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {75 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 68}{a^{2} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {45 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*((75*sin(d*x + c)/(cos(d*x + c) + 1) - 280*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 30*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 320*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 360*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 30*sin(d
*x + c)^7/(cos(d*x + c) + 1)^7 - 60*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 75*sin(d*x + c)^9/(cos(d*x + c) + 1)
^9 - 68)/(a^2 + 5*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*a^
2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a^2*sin(d*x + c)^10/(cos(d
*x + c) + 1)^10) + 45*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - 30*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^
2)/d

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mupad [B]  time = 10.76, size = 262, normalized size = 2.20 \[ \frac {3\,\mathrm {atan}\left (\frac {9}{4\,\left (\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+3\right )}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+3}\right )}{2\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {34}{15}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(sin(c + d*x)*(a + a*sin(c + d*x))^2),x)

[Out]

(3*atan(9/(4*((9*tan(c/2 + (d*x)/2))/4 + 3)) - (3*tan(c/2 + (d*x)/2))/((9*tan(c/2 + (d*x)/2))/4 + 3)))/(2*a^2*
d) + log(tan(c/2 + (d*x)/2))/(a^2*d) + ((28*tan(c/2 + (d*x)/2)^2)/3 - (5*tan(c/2 + (d*x)/2))/2 - tan(c/2 + (d*
x)/2)^3 + (32*tan(c/2 + (d*x)/2)^4)/3 + 12*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^7 + 2*tan(c/2 + (d*x)/2)^
8 + (5*tan(c/2 + (d*x)/2)^9)/2 + 34/15)/(d*(5*a^2*tan(c/2 + (d*x)/2)^2 + 10*a^2*tan(c/2 + (d*x)/2)^4 + 10*a^2*
tan(c/2 + (d*x)/2)^6 + 5*a^2*tan(c/2 + (d*x)/2)^8 + a^2*tan(c/2 + (d*x)/2)^10 + a^2))

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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)**8*csc(d*x+c)/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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