\(\int \cot ^6(c+d x) (a+a \sin (c+d x)) \, dx\) [581]

Optimal result

Integrand size = 19, antiderivative size = 122 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x)) \, dx=-a x-\frac {15 a \text {arctanh}(\cos (c+d x))}{8 d}+\frac {15 a \cos (c+d x)}{8 d}-\frac {a \cot (c+d x)}{d}+\frac {5 a \cos (c+d x) \cot ^2(c+d x)}{8 d}+\frac {a \cot ^3(c+d x)}{3 d}-\frac {a \cos (c+d x) \cot ^4(c+d x)}{4 d}-\frac {a \cot ^5(c+d x)}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2789, 2672, 294, 327, 212, 3554, 8} \[ \int \cot ^6(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {15 a \text {arctanh}(\cos (c+d x))}{8 d}+\frac {15 a \cos (c+d x)}{8 d}-\frac {a \cot ^5(c+d x)}{5 d}+\frac {a \cot ^3(c+d x)}{3 d}-\frac {a \cot (c+d x)}{d}-\frac {a \cos (c+d x) \cot ^4(c+d x)}{4 d}+\frac {5 a \cos (c+d x) \cot ^2(c+d x)}{8 d}-a x \]

[In]

Int[Cot[c + d*x]^6*(a + a*Sin[c + d*x]),x]

[Out]

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

Rule 8

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

Rule 212

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

Rule 294

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 327

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 2672

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 2789

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.34 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \cos (c+d x)}{d}+\frac {9 a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {a \cot ^5(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2(c+d x)\right )}{5 d}-\frac {15 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {15 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {9 a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {a \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d} \]

[In]

Integrate[Cot[c + d*x]^6*(a + a*Sin[c + d*x]),x]

[Out]

(a*Cos[c + d*x])/d + (9*a*Csc[(c + d*x)/2]^2)/(32*d) - (a*Csc[(c + d*x)/2]^4)/(64*d) - (a*Cot[c + d*x]^5*Hyper
geometric2F1[-5/2, 1, -3/2, -Tan[c + d*x]^2])/(5*d) - (15*a*Log[Cos[(c + d*x)/2]])/(8*d) + (15*a*Log[Sin[(c +
d*x)/2]])/(8*d) - (9*a*Sec[(c + d*x)/2]^2)/(32*d) + (a*Sec[(c + d*x)/2]^4)/(64*d)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cos ^{7}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}+\frac {3 \left (\cos ^{7}\left (d x +c \right )\right )}{8 \sin \left (d x +c \right )^{2}}+\frac {3 \left (\cos ^{5}\left (d x +c \right )\right )}{8}+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {15 \cos \left (d x +c \right )}{8}+\frac {15 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a \left (-\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(129\)
default	\(\frac {a \left (-\frac {\cos ^{7}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}+\frac {3 \left (\cos ^{7}\left (d x +c \right )\right )}{8 \sin \left (d x +c \right )^{2}}+\frac {3 \left (\cos ^{5}\left (d x +c \right )\right )}{8}+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {15 \cos \left (d x +c \right )}{8}+\frac {15 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a \left (-\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(129\)
risch	\(-a x +\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a \left (360 i {\mathrm e}^{8 i \left (d x +c \right )}+135 \,{\mathrm e}^{9 i \left (d x +c \right )}-720 i {\mathrm e}^{6 i \left (d x +c \right )}-150 \,{\mathrm e}^{7 i \left (d x +c \right )}+1120 i {\mathrm e}^{4 i \left (d x +c \right )}-560 i {\mathrm e}^{2 i \left (d x +c \right )}+150 \,{\mathrm e}^{3 i \left (d x +c \right )}+184 i-135 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {15 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}+\frac {15 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}\)	\(186\)
parallelrisch	\(\frac {a \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (-1125 \left (\sin \left (3 d x +3 c \right )-\frac {\sin \left (5 d x +5 c \right )}{5}-2 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+600 d x \sin \left (3 d x +3 c \right )-120 d x \sin \left (5 d x +5 c \right )-1200 d x \sin \left (d x +c \right )+750 \sin \left (3 d x +3 c \right )-510 \sin \left (4 d x +4 c \right )-150 \sin \left (5 d x +5 c \right )+60 \sin \left (6 d x +6 c \right )-400 \cos \left (d x +c \right )+200 \cos \left (3 d x +3 c \right )-184 \cos \left (5 d x +5 c \right )-1500 \sin \left (d x +c \right )+600 \sin \left (2 d x +2 c \right )\right )}{61440 d}\)	\(200\)
norman	\(\frac {-\frac {a}{160 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d}+\frac {a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {15 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {59 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}+\frac {59 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {15 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}-a x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {5 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {15 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}\)	\(250\)

[In]

int(cos(d*x+c)^6*csc(d*x+c)^6*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (110) = 220\).

Time = 0.30 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.82 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {368 \, a \cos \left (d x + c\right )^{5} - 560 \, a \cos \left (d x + c\right )^{3} + 225 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 225 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 240 \, a \cos \left (d x + c\right ) + 30 \, {\left (8 \, a d x \cos \left (d x + c\right )^{4} - 8 \, a \cos \left (d x + c\right )^{5} - 16 \, a d x \cos \left (d x + c\right )^{2} + 25 \, a \cos \left (d x + c\right )^{3} + 8 \, a d x - 15 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

-1/240*(368*a*cos(d*x + c)^5 - 560*a*cos(d*x + c)^3 + 225*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(1/2*
cos(d*x + c) + 1/2)*sin(d*x + c) - 225*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(-1/2*cos(d*x + c) + 1/2
)*sin(d*x + c) + 240*a*cos(d*x + c) + 30*(8*a*d*x*cos(d*x + c)^4 - 8*a*cos(d*x + c)^5 - 16*a*d*x*cos(d*x + c)^
2 + 25*a*cos(d*x + c)^3 + 8*a*d*x - 15*a*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 +
d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \cot ^6(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)**6*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {16 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a + 15 \, a {\left (\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

[In]

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

[Out]

-1/240*(16*(15*d*x + 15*c + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/tan(d*x + c)^5)*a + 15*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(cos(d*x + c) + 1)
- 15*log(cos(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.63 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x)) \, dx=\frac {6 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 70 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 960 \, {\left (d x + c\right )} a + 1800 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 660 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {1920 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - \frac {4110 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 660 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 70 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

[In]

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

[Out]

1/960*(6*a*tan(1/2*d*x + 1/2*c)^5 + 15*a*tan(1/2*d*x + 1/2*c)^4 - 70*a*tan(1/2*d*x + 1/2*c)^3 - 240*a*tan(1/2*
d*x + 1/2*c)^2 - 960*(d*x + c)*a + 1800*a*log(abs(tan(1/2*d*x + 1/2*c))) + 660*a*tan(1/2*d*x + 1/2*c) + 1920*a
/(tan(1/2*d*x + 1/2*c)^2 + 1) - (4110*a*tan(1/2*d*x + 1/2*c)^5 + 660*a*tan(1/2*d*x + 1/2*c)^4 - 240*a*tan(1/2*
d*x + 1/2*c)^3 - 70*a*tan(1/2*d*x + 1/2*c)^2 + 15*a*tan(1/2*d*x + 1/2*c) + 6*a)/tan(1/2*d*x + 1/2*c)^5)/d

Mupad [B] (verification not implemented)

Time = 10.60 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.39 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x)) \, dx=\frac {11\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {22\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-72\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {59\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {32\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {15\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}+\frac {2\,a\,\mathrm {atan}\left (\frac {4\,a^2}{\frac {15\,a^2}{2}+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {15\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {15\,a^2}{2}+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )}{d} \]

[In]

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

[Out]

(11*a*tan(c/2 + (d*x)/2))/(16*d) - (a/5 + (a*tan(c/2 + (d*x)/2))/2 - (32*a*tan(c/2 + (d*x)/2)^2)/15 - (15*a*ta
n(c/2 + (d*x)/2)^3)/2 + (59*a*tan(c/2 + (d*x)/2)^4)/3 - 72*a*tan(c/2 + (d*x)/2)^5 + 22*a*tan(c/2 + (d*x)/2)^6)
/(d*(32*tan(c/2 + (d*x)/2)^5 + 32*tan(c/2 + (d*x)/2)^7)) - (a*tan(c/2 + (d*x)/2)^2)/(4*d) - (7*a*tan(c/2 + (d*
x)/2)^3)/(96*d) + (a*tan(c/2 + (d*x)/2)^4)/(64*d) + (a*tan(c/2 + (d*x)/2)^5)/(160*d) + (15*a*log(tan(c/2 + (d*
x)/2)))/(8*d) + (2*a*atan((4*a^2)/((15*a^2)/2 + 4*a^2*tan(c/2 + (d*x)/2)) - (15*a^2*tan(c/2 + (d*x)/2))/(2*((1
5*a^2)/2 + 4*a^2*tan(c/2 + (d*x)/2)))))/d