\(\int \cot ^6(c+d x) (a+a \sin (c+d x))^2 \, dx\) [597]

Optimal result

Integrand size = 21, antiderivative size = 139 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 a^2 x}{2}-\frac {15 a^2 \text {arctanh}(\cos (c+d x))}{4 d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {9 a^2 \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

3/2*a^2*x-15/4*a^2*arctanh(cos(d*x+c))/d+2*a^2*cos(d*x+c)/d+a^2*cot(d*x+c)/d-1/5*a^2*cot(d*x+c)^5/d+9/4*a^2*co
t(d*x+c)*csc(d*x+c)/d-1/2*a^2*cot(d*x+c)*csc(d*x+c)^3/d+1/2*a^2*cos(d*x+c)*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2788, 3855, 3853, 3852, 2718, 2715, 8} \[ \int \cot ^6(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {15 a^2 \text {arctanh}(\cos (c+d x))}{4 d}+\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {9 a^2 \cot (c+d x) \csc (c+d x)}{4 d}+\frac {3 a^2 x}{2} \]

[In]

Int[Cot[c + d*x]^6*(a + a*Sin[c + d*x])^2,x]

[Out]

(3*a^2*x)/2 - (15*a^2*ArcTanh[Cos[c + d*x]])/(4*d) + (2*a^2*Cos[c + d*x])/d + (a^2*Cot[c + d*x])/d - (a^2*Cot[
c + d*x]^5)/(5*d) + (9*a^2*Cot[c + d*x]*Csc[c + d*x])/(4*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(2*d) + (a^2*C
os[c + d*x]*Sin[c + d*x])/(2*d)

Rule 8

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 2718

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

Rule 2788

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

Rule 3852

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 3853

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (2 a^8+6 a^8 \csc (c+d x)-6 a^8 \csc ^3(c+d x)-2 a^8 \csc ^4(c+d x)+2 a^8 \csc ^5(c+d x)+a^8 \csc ^6(c+d x)-2 a^8 \sin (c+d x)-a^8 \sin ^2(c+d x)\right ) \, dx}{a^6} \\ & = 2 a^2 x+a^2 \int \csc ^6(c+d x) \, dx-a^2 \int \sin ^2(c+d x) \, dx-\left (2 a^2\right ) \int \csc ^4(c+d x) \, dx+\left (2 a^2\right ) \int \csc ^5(c+d x) \, dx-\left (2 a^2\right ) \int \sin (c+d x) \, dx+\left (6 a^2\right ) \int \csc (c+d x) \, dx-\left (6 a^2\right ) \int \csc ^3(c+d x) \, dx \\ & = 2 a^2 x-\frac {6 a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} a^2 \int 1 \, dx+\frac {1}{2} \left (3 a^2\right ) \int \csc ^3(c+d x) \, dx-\left (3 a^2\right ) \int \csc (c+d x) \, dx-\frac {a^2 \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{d}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d} \\ & = \frac {3 a^2 x}{2}-\frac {3 a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {9 a^2 \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {1}{4} \left (3 a^2\right ) \int \csc (c+d x) \, dx \\ & = \frac {3 a^2 x}{2}-\frac {15 a^2 \text {arctanh}(\cos (c+d x))}{4 d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {9 a^2 \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.38 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.90 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 (1+\sin (c+d x))^2 \left (240 (c+d x)+320 \cos (c+d x)+64 \cot \left (\frac {1}{2} (c+d x)\right )+90 \csc ^2\left (\frac {1}{2} (c+d x)\right )-5 \csc ^4\left (\frac {1}{2} (c+d x)\right )-600 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+600 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-90 \sec ^2\left (\frac {1}{2} (c+d x)\right )+5 \sec ^4\left (\frac {1}{2} (c+d x)\right )-56 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+\frac {7}{2} \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)-\frac {1}{2} \csc ^6\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)+40 \sin (2 (c+d x))-64 \tan \left (\frac {1}{2} (c+d x)\right )+\sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{160 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

[In]

Integrate[Cot[c + d*x]^6*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(1 + Sin[c + d*x])^2*(240*(c + d*x) + 320*Cos[c + d*x] + 64*Cot[(c + d*x)/2] + 90*Csc[(c + d*x)/2]^2 - 5*
Csc[(c + d*x)/2]^4 - 600*Log[Cos[(c + d*x)/2]] + 600*Log[Sin[(c + d*x)/2]] - 90*Sec[(c + d*x)/2]^2 + 5*Sec[(c
+ d*x)/2]^4 - 56*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + (7*Csc[(c + d*x)/2]^4*Sin[c + d*x])/2 - (Csc[(c + d*x)/2]
^6*Sin[c + d*x])/2 + 40*Sin[2*(c + d*x)] - 64*Tan[(c + d*x)/2] + Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2]))/(160*d*
(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.53

method	result	size
parallelrisch	\(\frac {a^{2} \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (-3000 \left (\sin \left (3 d x +3 c \right )-\frac {\sin \left (5 d x +5 c \right )}{5}-2 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1200 d x \sin \left (3 d x +3 c \right )+240 d x \sin \left (5 d x +5 c \right )+2400 d x \sin \left (d x +c \right )+1600 \sin \left (2 d x +2 c \right )-2375 \sin \left (3 d x +3 c \right )-1360 \sin \left (4 d x +4 c \right )+475 \sin \left (5 d x +5 c \right )+160 \sin \left (6 d x +6 c \right )+100 \cos \left (d x +c \right )-820 \cos \left (3 d x +3 c \right )+228 \cos \left (5 d x +5 c \right )-20 \cos \left (7 d x +7 c \right )+4750 \sin \left (d x +c \right )\right )}{81920 d}\)	\(213\)
derivativedivides	\(\frac {a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {4 \left (\cos ^{7}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )}+\frac {4 \left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+2 a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}+\frac {3 \left (\cos ^{7}\left (d x +c \right )\right )}{8 \sin \left (d x +c \right )^{2}}+\frac {3 \left (\cos ^{5}\left (d x +c \right )\right )}{8}+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {15 \cos \left (d x +c \right )}{8}+\frac {15 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a^{2} \left (-\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(217\)
default	\(\frac {a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {4 \left (\cos ^{7}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )}+\frac {4 \left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+2 a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}+\frac {3 \left (\cos ^{7}\left (d x +c \right )\right )}{8 \sin \left (d x +c \right )^{2}}+\frac {3 \left (\cos ^{5}\left (d x +c \right )\right )}{8}+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {15 \cos \left (d x +c \right )}{8}+\frac {15 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a^{2} \left (-\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(217\)
risch	\(\frac {3 a^{2} x}{2}-\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {a^{2} \left (45 \,{\mathrm e}^{9 i \left (d x +c \right )}+80 i {\mathrm e}^{6 i \left (d x +c \right )}-50 \,{\mathrm e}^{7 i \left (d x +c \right )}-80 i {\mathrm e}^{4 i \left (d x +c \right )}+80 i {\mathrm e}^{2 i \left (d x +c \right )}+50 \,{\mathrm e}^{3 i \left (d x +c \right )}-16 i-45 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{10 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {15 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d}-\frac {15 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d}\)	\(220\)
norman	\(\frac {-\frac {a^{2}}{160 d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 d}+\frac {3 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}+\frac {7 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}+\frac {79 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}+\frac {47 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}-\frac {47 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}-\frac {79 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}-\frac {7 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {3 a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}+\frac {a^{2} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}+\frac {a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}+\frac {3 a^{2} x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+3 a^{2} x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {3 a^{2} x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {95 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}+\frac {95 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {15 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}\)	\(352\)

[In]

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

[Out]

1/81920*a^2*csc(1/2*d*x+1/2*c)^5*sec(1/2*d*x+1/2*c)^5*(-3000*(sin(3*d*x+3*c)-1/5*sin(5*d*x+5*c)-2*sin(d*x+c))*
ln(tan(1/2*d*x+1/2*c))-1200*d*x*sin(3*d*x+3*c)+240*d*x*sin(5*d*x+5*c)+2400*d*x*sin(d*x+c)+1600*sin(2*d*x+2*c)-
2375*sin(3*d*x+3*c)-1360*sin(4*d*x+4*c)+475*sin(5*d*x+5*c)+160*sin(6*d*x+6*c)+100*cos(d*x+c)-820*cos(3*d*x+3*c
)+228*cos(5*d*x+5*c)-20*cos(7*d*x+7*c)+4750*sin(d*x+c))/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (127) = 254\).

Time = 0.30 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.91 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {20 \, a^{2} \cos \left (d x + c\right )^{7} - 92 \, a^{2} \cos \left (d x + c\right )^{5} + 140 \, a^{2} \cos \left (d x + c\right )^{3} - 60 \, a^{2} \cos \left (d x + c\right ) + 75 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 75 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 10 \, {\left (6 \, a^{2} d x \cos \left (d x + c\right )^{4} + 8 \, a^{2} \cos \left (d x + c\right )^{5} - 12 \, a^{2} d x \cos \left (d x + c\right )^{2} - 25 \, a^{2} \cos \left (d x + c\right )^{3} + 6 \, a^{2} d x + 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

-1/40*(20*a^2*cos(d*x + c)^7 - 92*a^2*cos(d*x + c)^5 + 140*a^2*cos(d*x + c)^3 - 60*a^2*cos(d*x + c) + 75*(a^2*
cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 75*(a^2*cos(d*x + c)^4
- 2*a^2*cos(d*x + c)^2 + a^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 10*(6*a^2*d*x*cos(d*x + c)^4 + 8*a^
2*cos(d*x + c)^5 - 12*a^2*d*x*cos(d*x + c)^2 - 25*a^2*cos(d*x + c)^3 + 6*a^2*d*x + 15*a^2*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. \[ \int \cot ^6(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.32 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {20 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a^{2} - 8 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a^{2} - 15 \, a^{2} {\left (\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]

[In]

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

[Out]

1/120*(20*(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^2
- 8*(15*d*x + 15*c + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/tan(d*x + c)^5)*a^2 - 15*a^2*(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) - 1
5*log(cos(d*x + c) - 1)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (127) = 254\).

Time = 0.40 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.96 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 240 \, {\left (d x + c\right )} a^{2} + 600 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {160 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {1370 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{160 \, d} \]

[In]

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

[Out]

1/160*(a^2*tan(1/2*d*x + 1/2*c)^5 + 5*a^2*tan(1/2*d*x + 1/2*c)^4 - 5*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*a^2*tan(1
/2*d*x + 1/2*c)^2 + 240*(d*x + c)*a^2 + 600*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 70*a^2*tan(1/2*d*x + 1/2*c) -
160*(a^2*tan(1/2*d*x + 1/2*c)^3 - 4*a^2*tan(1/2*d*x + 1/2*c)^2 - a^2*tan(1/2*d*x + 1/2*c) - 4*a^2)/(tan(1/2*d
*x + 1/2*c)^2 + 1)^2 - (1370*a^2*tan(1/2*d*x + 1/2*c)^5 - 70*a^2*tan(1/2*d*x + 1/2*c)^4 - 80*a^2*tan(1/2*d*x +
1/2*c)^3 - 5*a^2*tan(1/2*d*x + 1/2*c)^2 + 5*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c)^5)/d

Mupad [B] (verification not implemented)

Time = 9.40 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.61 \[ \int \cot ^6(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {15\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}+\frac {-18\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+144\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+61\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+159\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {79\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+14\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^2}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {3\,a^2\,\mathrm {atan}\left (\frac {9\,a^4}{\frac {45\,a^4}{2}-9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {45\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {45\,a^4}{2}-9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )}{d}-\frac {7\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d} \]

[In]

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

[Out]

(a^2*tan(c/2 + (d*x)/2)^4)/(32*d) - (a^2*tan(c/2 + (d*x)/2)^3)/(32*d) - (a^2*tan(c/2 + (d*x)/2)^2)/(2*d) + (a^
2*tan(c/2 + (d*x)/2)^5)/(160*d) + (15*a^2*log(tan(c/2 + (d*x)/2)))/(4*d) + ((3*a^2*tan(c/2 + (d*x)/2)^2)/5 + 1
4*a^2*tan(c/2 + (d*x)/2)^3 + (79*a^2*tan(c/2 + (d*x)/2)^4)/5 + 159*a^2*tan(c/2 + (d*x)/2)^5 + 61*a^2*tan(c/2 +
(d*x)/2)^6 + 144*a^2*tan(c/2 + (d*x)/2)^7 - 18*a^2*tan(c/2 + (d*x)/2)^8 - a^2/5 - a^2*tan(c/2 + (d*x)/2))/(d*
(32*tan(c/2 + (d*x)/2)^5 + 64*tan(c/2 + (d*x)/2)^7 + 32*tan(c/2 + (d*x)/2)^9)) + (3*a^2*atan((9*a^4)/((45*a^4)
/2 - 9*a^4*tan(c/2 + (d*x)/2)) + (45*a^4*tan(c/2 + (d*x)/2))/(2*((45*a^4)/2 - 9*a^4*tan(c/2 + (d*x)/2)))))/d -
(7*a^2*tan(c/2 + (d*x)/2))/(16*d)