3.419 \(\int \frac {\cot ^4(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=124 \[ \frac {\cot ^5(c+d x)}{5 a d}+\frac {\cot ^3(c+d x)}{3 a d}+\frac {\tanh ^{-1}(\cos (c+d x))}{16 a d}-\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}+\frac {\cot (c+d x) \csc (c+d x)}{16 a d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2839, 2611, 3768, 3770, 2607, 14} \[ \frac {\cot ^5(c+d x)}{5 a d}+\frac {\cot ^3(c+d x)}{3 a d}+\frac {\tanh ^{-1}(\cos (c+d x))}{16 a d}-\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}+\frac {\cot (c+d x) \csc (c+d x)}{16 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^4*Csc[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

ArcTanh[Cos[c + d*x]]/(16*a*d) + Cot[c + d*x]^3/(3*a*d) + Cot[c + d*x]^5/(5*a*d) + (Cot[c + d*x]*Csc[c + d*x])
/(16*a*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(24*a*d) - (Cot[c + d*x]*Csc[c + d*x]^5)/(6*a*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^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 \frac {\cot ^4(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {\int \cot ^2(c+d x) \csc ^4(c+d x) \, dx}{a}+\frac {\int \cot ^2(c+d x) \csc ^5(c+d x) \, dx}{a}\\ &=-\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}-\frac {\int \csc ^5(c+d x) \, dx}{6 a}-\frac {\operatorname {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{a d}\\ &=\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}-\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}-\frac {\int \csc ^3(c+d x) \, dx}{8 a}-\frac {\operatorname {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{a d}\\ &=\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot ^5(c+d x)}{5 a d}+\frac {\cot (c+d x) \csc (c+d x)}{16 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}-\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}-\frac {\int \csc (c+d x) \, dx}{16 a}\\ &=\frac {\tanh ^{-1}(\cos (c+d x))}{16 a d}+\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot ^5(c+d x)}{5 a d}+\frac {\cot (c+d x) \csc (c+d x)}{16 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}-\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 229, normalized size = 1.85 \[ -\frac {\csc ^6(c+d x) \left (-480 \sin (2 (c+d x))-192 \sin (4 (c+d x))+32 \sin (6 (c+d x))+1140 \cos (c+d x)+170 \cos (3 (c+d x))-30 \cos (5 (c+d x))+150 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+225 \cos (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-90 \cos (4 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+15 \cos (6 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-150 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-225 \cos (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+90 \cos (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-15 \cos (6 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{7680 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^4*Csc[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

-1/7680*(Csc[c + d*x]^6*(1140*Cos[c + d*x] + 170*Cos[3*(c + d*x)] - 30*Cos[5*(c + d*x)] - 150*Log[Cos[(c + d*x
)/2]] + 225*Cos[2*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 90*Cos[4*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 15*Cos[6*(c +
d*x)]*Log[Cos[(c + d*x)/2]] + 150*Log[Sin[(c + d*x)/2]] - 225*Cos[2*(c + d*x)]*Log[Sin[(c + d*x)/2]] + 90*Cos
[4*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 15*Cos[6*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 480*Sin[2*(c + d*x)] - 192*S
in[4*(c + d*x)] + 32*Sin[6*(c + d*x)]))/(a*d)

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fricas [A]  time = 0.46, size = 188, normalized size = 1.52 \[ -\frac {30 \, \cos \left (d x + c\right )^{5} - 80 \, \cos \left (d x + c\right )^{3} - 15 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 32 \, {\left (2 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right ) - 30 \, \cos \left (d x + c\right )}{480 \, {\left (a d \cos \left (d x + c\right )^{6} - 3 \, a d \cos \left (d x + c\right )^{4} + 3 \, a d \cos \left (d x + c\right )^{2} - a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/480*(30*cos(d*x + c)^5 - 80*cos(d*x + c)^3 - 15*(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - 1)*
log(1/2*cos(d*x + c) + 1/2) + 15*(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x +
c) + 1/2) - 32*(2*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*sin(d*x + c) - 30*cos(d*x + c))/(a*d*cos(d*x + c)^6 - 3*
a*d*cos(d*x + c)^4 + 3*a*d*cos(d*x + c)^2 - a*d)

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giac [A]  time = 0.22, size = 216, normalized size = 1.74 \[ -\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {5 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 12 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 20 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}} - \frac {294 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6}}}{1920 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/1920*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a - (5*a^5*tan(1/2*d*x + 1/2*c)^6 - 12*a^5*tan(1/2*d*x + 1/2*c)^5
+ 15*a^5*tan(1/2*d*x + 1/2*c)^4 - 20*a^5*tan(1/2*d*x + 1/2*c)^3 - 15*a^5*tan(1/2*d*x + 1/2*c)^2 + 120*a^5*tan(
1/2*d*x + 1/2*c))/a^6 - (294*tan(1/2*d*x + 1/2*c)^6 - 120*tan(1/2*d*x + 1/2*c)^5 + 15*tan(1/2*d*x + 1/2*c)^4 +
20*tan(1/2*d*x + 1/2*c)^3 - 15*tan(1/2*d*x + 1/2*c)^2 + 12*tan(1/2*d*x + 1/2*c) - 5)/(a*tan(1/2*d*x + 1/2*c)^
6))/d

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maple [B]  time = 0.48, size = 246, normalized size = 1.98 \[ \frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{384 a d}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 a d}+\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{128 a d}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{96 a d}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{128 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}-\frac {1}{384 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}-\frac {1}{16 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a d}+\frac {1}{160 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{128 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{128 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {1}{96 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^7/(a+a*sin(d*x+c)),x)

[Out]

1/384/a/d*tan(1/2*d*x+1/2*c)^6-1/160/a/d*tan(1/2*d*x+1/2*c)^5+1/128/a/d*tan(1/2*d*x+1/2*c)^4-1/96/a/d*tan(1/2*
d*x+1/2*c)^3-1/128/a/d*tan(1/2*d*x+1/2*c)^2+1/16/a/d*tan(1/2*d*x+1/2*c)-1/384/a/d/tan(1/2*d*x+1/2*c)^6-1/16/a/
d/tan(1/2*d*x+1/2*c)-1/16/a/d*ln(tan(1/2*d*x+1/2*c))+1/160/a/d/tan(1/2*d*x+1/2*c)^5+1/128/a/d/tan(1/2*d*x+1/2*
c)^2-1/128/a/d/tan(1/2*d*x+1/2*c)^4+1/96/a/d/tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.36, size = 274, normalized size = 2.21 \[ \frac {\frac {\frac {120 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {12 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}{a} - \frac {120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {{\left (\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {120 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 5\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{6}}{a \sin \left (d x + c\right )^{6}}}{1920 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/1920*((120*sin(d*x + c)/(cos(d*x + c) + 1) - 15*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 20*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 12*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6)/a - 120*log(sin(d*x + c)/(cos(d*x + c) + 1))/a + (12*sin(d*x + c)/(cos(d*x + c)
+ 1) - 15*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(co
s(d*x + c) + 1)^4 - 120*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5)*(cos(d*x + c) + 1)^6/(a*sin(d*x + c)^6))/d

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mupad [B]  time = 9.68, size = 339, normalized size = 2.73 \[ -\frac {5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+12\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+20\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-20\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+120\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{1920\,a\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^7*(a + a*sin(c + d*x))),x)

[Out]

-(5*cos(c/2 + (d*x)/2)^12 - 5*sin(c/2 + (d*x)/2)^12 + 12*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^11 - 12*cos(c/2
+ (d*x)/2)^11*sin(c/2 + (d*x)/2) - 15*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^10 + 20*cos(c/2 + (d*x)/2)^3*si
n(c/2 + (d*x)/2)^9 + 15*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^8 - 120*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2
)^7 + 120*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^5 - 15*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4 - 20*cos(c/
2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^3 + 15*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^2 + 120*log(sin(c/2 + (d*x)/
2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^6)/(1920*a*d*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*
x)/2)^6)

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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)**4*csc(d*x+c)**7/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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