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

Optimal. Leaf size=139 \[ \frac{2 a^2 \cos (c+d x)}{d}-\frac{a^2 \cot ^5(c+d x)}{5 d}+\frac{a^2 \cot (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac{15 a^2 \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac{9 a^2 \cot (c+d x) \csc (c+d x)}{4 d}+\frac{3 a^2 x}{2} \]

[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)

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Rubi [A]  time = 0.249414, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2709, 3770, 3768, 3767, 2638, 2635, 8} \[ \frac{2 a^2 \cos (c+d x)}{d}-\frac{a^2 \cot ^5(c+d x)}{5 d}+\frac{a^2 \cot (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac{15 a^2 \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac{9 a^2 \cot (c+d x) \csc (c+d x)}{4 d}+\frac{3 a^2 x}{2} \]

Antiderivative was successfully verified.

[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 2709

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 3770

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

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 3767

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 2638

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

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 8

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

Rubi steps

\begin{align*} \int \cot ^6(c+d x) (a+a \sin (c+d x))^2 \, dx &=\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 \tanh ^{-1}(\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 \operatorname{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 ) \operatorname{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 \tanh ^{-1}(\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 \tanh ^{-1}(\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]  time = 1.32833, size = 264, normalized size = 1.9 \[ \frac{a^2 (\sin (c+d x)+1)^2 \left (240 (c+d x)+40 \sin (2 (c+d x))+320 \cos (c+d x)-64 \tan \left (\frac{1}{2} (c+d x)\right )+64 \cot \left (\frac{1}{2} (c+d x)\right )-5 \csc ^4\left (\frac{1}{2} (c+d x)\right )+90 \csc ^2\left (\frac{1}{2} (c+d x)\right )+5 \sec ^4\left (\frac{1}{2} (c+d x)\right )-90 \sec ^2\left (\frac{1}{2} (c+d x)\right )+600 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-600 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-56 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-\frac{1}{2} \sin (c+d x) \csc ^6\left (\frac{1}{2} (c+d x)\right )+\frac{7}{2} \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )+\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right )\right )}{160 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[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)

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Maple [B]  time = 0.086, size = 293, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{4\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{3\,d\sin \left ( dx+c \right ) }}+{\frac{4\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{5\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{5\,{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}x}{2}}+{\frac{3\,c{a}^{2}}{2\,d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{4\,d}}+{\frac{5\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{15\,{a}^{2}\cos \left ( dx+c \right ) }{4\,d}}+{\frac{15\,{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{4\,d}}-{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.64059, size = 248, normalized size = 1.78 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [B]  time = 1.23628, size = 686, normalized size = 4.94 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.29388, size = 367, normalized size = 2.64 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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