3.580 \(\int \cos (c+d x) \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=134 \[ \frac {15 a \cos (c+d x)}{8 d}-\frac {5 a \cot ^3(c+d x)}{6 d}+\frac {5 a \cot (c+d x)}{2 d}+\frac {a \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}-\frac {a \cos (c+d x) \cot ^4(c+d x)}{4 d}+\frac {5 a \cos (c+d x) \cot ^2(c+d x)}{8 d}-\frac {15 a \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {5 a x}{2} \]

[Out]

5/2*a*x-15/8*a*arctanh(cos(d*x+c))/d+15/8*a*cos(d*x+c)/d+5/2*a*cot(d*x+c)/d+5/8*a*cos(d*x+c)*cot(d*x+c)^2/d-5/
6*a*cot(d*x+c)^3/d+1/2*a*cos(d*x+c)^2*cot(d*x+c)^3/d-1/4*a*cos(d*x+c)*cot(d*x+c)^4/d

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Rubi [A]  time = 0.13, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2838, 2592, 288, 321, 206, 2591, 302, 203} \[ \frac {15 a \cos (c+d x)}{8 d}-\frac {5 a \cot ^3(c+d x)}{6 d}+\frac {5 a \cot (c+d x)}{2 d}+\frac {a \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}-\frac {a \cos (c+d x) \cot ^4(c+d x)}{4 d}+\frac {5 a \cos (c+d x) \cot ^2(c+d x)}{8 d}-\frac {15 a \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {5 a x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Cot[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

(5*a*x)/2 - (15*a*ArcTanh[Cos[c + d*x]])/(8*d) + (15*a*Cos[c + d*x])/(8*d) + (5*a*Cot[c + d*x])/(2*d) + (5*a*C
os[c + d*x]*Cot[c + d*x]^2)/(8*d) - (5*a*Cot[c + d*x]^3)/(6*d) + (a*Cos[c + d*x]^2*Cot[c + d*x]^3)/(2*d) - (a*
Cos[c + d*x]*Cot[c + d*x]^4)/(4*d)

Rule 203

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

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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2591

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

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 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rubi steps

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

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Mathematica [A]  time = 1.43, size = 138, normalized size = 1.03 \[ \frac {a \left (192 \cos (c+d x)-64 \cot (c+d x) \left (\csc ^2(c+d x)-7\right )+3 \left (16 \sin (2 (c+d x))-\csc ^4\left (\frac {1}{2} (c+d x)\right )+18 \csc ^2\left (\frac {1}{2} (c+d x)\right )+\sec ^4\left (\frac {1}{2} (c+d x)\right )-18 \sec ^2\left (\frac {1}{2} (c+d x)\right )+40 \left (3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 c+4 d x\right )\right )\right )}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Cot[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

(a*(192*Cos[c + d*x] - 64*Cot[c + d*x]*(-7 + Csc[c + d*x]^2) + 3*(18*Csc[(c + d*x)/2]^2 - Csc[(c + d*x)/2]^4 +
40*(4*c + 4*d*x - 3*Log[Cos[(c + d*x)/2]] + 3*Log[Sin[(c + d*x)/2]]) - 18*Sec[(c + d*x)/2]^2 + Sec[(c + d*x)/
2]^4 + 16*Sin[2*(c + d*x)])))/(192*d)

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fricas [A]  time = 0.70, size = 202, normalized size = 1.51 \[ \frac {120 \, a d x \cos \left (d x + c\right )^{4} + 48 \, a \cos \left (d x + c\right )^{5} - 240 \, a d x \cos \left (d x + c\right )^{2} - 150 \, a \cos \left (d x + c\right )^{3} + 120 \, a d x + 90 \, a \cos \left (d x + c\right ) - 45 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 45 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 8 \, {\left (3 \, a \cos \left (d x + c\right )^{5} - 20 \, a \cos \left (d x + c\right )^{3} + 15 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(120*a*d*x*cos(d*x + c)^4 + 48*a*cos(d*x + c)^5 - 240*a*d*x*cos(d*x + c)^2 - 150*a*cos(d*x + c)^3 + 120*a
*d*x + 90*a*cos(d*x + c) - 45*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(1/2*cos(d*x + c) + 1/2) + 45*(a*
cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(-1/2*cos(d*x + c) + 1/2) + 8*(3*a*cos(d*x + c)^5 - 20*a*cos(d*x +
c)^3 + 15*a*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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giac [A]  time = 0.24, size = 213, normalized size = 1.59 \[ \frac {3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 8 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 48 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 480 \, {\left (d x + c\right )} a + 360 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 216 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {192 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {750 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 216 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 48 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/192*(3*a*tan(1/2*d*x + 1/2*c)^4 + 8*a*tan(1/2*d*x + 1/2*c)^3 - 48*a*tan(1/2*d*x + 1/2*c)^2 + 480*(d*x + c)*a
+ 360*a*log(abs(tan(1/2*d*x + 1/2*c))) - 216*a*tan(1/2*d*x + 1/2*c) - 192*(a*tan(1/2*d*x + 1/2*c)^3 - 2*a*tan
(1/2*d*x + 1/2*c)^2 - a*tan(1/2*d*x + 1/2*c) - 2*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - (750*a*tan(1/2*d*x + 1/2*
c)^4 - 216*a*tan(1/2*d*x + 1/2*c)^3 - 48*a*tan(1/2*d*x + 1/2*c)^2 + 8*a*tan(1/2*d*x + 1/2*c) + 3*a)/tan(1/2*d*
x + 1/2*c)^4)/d

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maple [A]  time = 0.33, size = 221, normalized size = 1.65 \[ -\frac {a \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}+\frac {4 a \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )}+\frac {4 a \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {5 a x}{2}+\frac {5 c a}{2 d}-\frac {a \left (\cos ^{7}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}+\frac {3 a \left (\cos ^{7}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}+\frac {3 a \left (\cos ^{5}\left (d x +c \right )\right )}{8 d}+\frac {5 a \left (\cos ^{3}\left (d x +c \right )\right )}{8 d}+\frac {15 a \cos \left (d x +c \right )}{8 d}+\frac {15 a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^5*(a+a*sin(d*x+c)),x)

[Out]

-1/3/d*a/sin(d*x+c)^3*cos(d*x+c)^7+4/3/d*a/sin(d*x+c)*cos(d*x+c)^7+4/3*a*cos(d*x+c)^5*sin(d*x+c)/d+5/3*a*cos(d
*x+c)^3*sin(d*x+c)/d+5/2*a*cos(d*x+c)*sin(d*x+c)/d+5/2*a*x+5/2/d*c*a-1/4/d*a/sin(d*x+c)^4*cos(d*x+c)^7+3/8/d*a
/sin(d*x+c)^2*cos(d*x+c)^7+3/8*a*cos(d*x+c)^5/d+5/8*a*cos(d*x+c)^3/d+15/8*a*cos(d*x+c)/d+15/8/d*a*ln(csc(d*x+c
)-cot(d*x+c))

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maxima [A]  time = 0.61, size = 136, normalized size = 1.01 \[ \frac {8 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a - 3 \, a {\left (\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(8*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 - 2)/(tan(d*x + c)^5 + tan(d*x + c)^3))*a - 3*
a*(2*(9*cos(d*x + c)^3 - 7*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 16*cos(d*x + c) + 15*log(co
s(d*x + c) + 1) - 15*log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 8.82, size = 300, normalized size = 2.24 \[ \frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {9\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {15\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}+\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+36\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {154\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {159\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+\frac {50\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {a}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {5\,a\,\mathrm {atan}\left (\frac {25\,a^2}{\frac {75\,a^2}{4}-25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {75\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {75\,a^2}{4}-25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^6*(a + a*sin(c + d*x)))/sin(c + d*x)^5,x)

[Out]

(a*tan(c/2 + (d*x)/2)^3)/(24*d) - (a*tan(c/2 + (d*x)/2)^2)/(4*d) - (9*a*tan(c/2 + (d*x)/2))/(8*d) + (a*tan(c/2
+ (d*x)/2)^4)/(64*d) + (15*a*log(tan(c/2 + (d*x)/2)))/(8*d) + ((7*a*tan(c/2 + (d*x)/2)^2)/2 - (2*a*tan(c/2 +
(d*x)/2))/3 - a/4 + (50*a*tan(c/2 + (d*x)/2)^3)/3 + (159*a*tan(c/2 + (d*x)/2)^4)/4 + (154*a*tan(c/2 + (d*x)/2)
^5)/3 + 36*a*tan(c/2 + (d*x)/2)^6 + 2*a*tan(c/2 + (d*x)/2)^7)/(d*(16*tan(c/2 + (d*x)/2)^4 + 32*tan(c/2 + (d*x)
/2)^6 + 16*tan(c/2 + (d*x)/2)^8)) + (5*a*atan((25*a^2)/((75*a^2)/4 - 25*a^2*tan(c/2 + (d*x)/2)) + (75*a^2*tan(
c/2 + (d*x)/2))/(4*((75*a^2)/4 - 25*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)**5*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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