3.1253 \(\int \cot ^6(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=151 \[ -\frac{\left (a^2+b^2\right ) \cot ^7(c+d x)}{7 d}-\frac{a^2 \cot ^9(c+d x)}{9 d}+\frac{5 a b \tanh ^{-1}(\cos (c+d x))}{64 d}-\frac{a b \cot ^5(c+d x) \csc ^3(c+d x)}{4 d}+\frac{5 a b \cot ^3(c+d x) \csc ^3(c+d x)}{24 d}-\frac{5 a b \cot (c+d x) \csc ^3(c+d x)}{32 d}+\frac{5 a b \cot (c+d x) \csc (c+d x)}{64 d} \]

[Out]

(5*a*b*ArcTanh[Cos[c + d*x]])/(64*d) - ((a^2 + b^2)*Cot[c + d*x]^7)/(7*d) - (a^2*Cot[c + d*x]^9)/(9*d) + (5*a*
b*Cot[c + d*x]*Csc[c + d*x])/(64*d) - (5*a*b*Cot[c + d*x]*Csc[c + d*x]^3)/(32*d) + (5*a*b*Cot[c + d*x]^3*Csc[c
+ d*x]^3)/(24*d) - (a*b*Cot[c + d*x]^5*Csc[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.405466, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2911, 2611, 3768, 3770, 14} \[ -\frac{\left (a^2+b^2\right ) \cot ^7(c+d x)}{7 d}-\frac{a^2 \cot ^9(c+d x)}{9 d}+\frac{5 a b \tanh ^{-1}(\cos (c+d x))}{64 d}-\frac{a b \cot ^5(c+d x) \csc ^3(c+d x)}{4 d}+\frac{5 a b \cot ^3(c+d x) \csc ^3(c+d x)}{24 d}-\frac{5 a b \cot (c+d x) \csc ^3(c+d x)}{32 d}+\frac{5 a b \cot (c+d x) \csc (c+d x)}{64 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]

[Out]

(5*a*b*ArcTanh[Cos[c + d*x]])/(64*d) - ((a^2 + b^2)*Cot[c + d*x]^7)/(7*d) - (a^2*Cot[c + d*x]^9)/(9*d) + (5*a*
b*Cot[c + d*x]*Csc[c + d*x])/(64*d) - (5*a*b*Cot[c + d*x]*Csc[c + d*x]^3)/(32*d) + (5*a*b*Cot[c + d*x]^3*Csc[c
+ d*x]^3)/(24*d) - (a*b*Cot[c + d*x]^5*Csc[c + d*x]^3)/(4*d)

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

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 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]

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]]

Rubi steps

\begin{align*} \int \cot ^6(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \cot ^6(c+d x) \csc ^3(c+d x) \, dx+\int \cot ^6(c+d x) \csc ^4(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac{a b \cot ^5(c+d x) \csc ^3(c+d x)}{4 d}-\frac{1}{4} (5 a b) \int \cot ^4(c+d x) \csc ^3(c+d x) \, dx+\frac{\operatorname{Subst}\left (\int \frac{a^2+\left (a^2+b^2\right ) x^2}{x^{10}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{5 a b \cot ^3(c+d x) \csc ^3(c+d x)}{24 d}-\frac{a b \cot ^5(c+d x) \csc ^3(c+d x)}{4 d}+\frac{1}{8} (5 a b) \int \cot ^2(c+d x) \csc ^3(c+d x) \, dx+\frac{\operatorname{Subst}\left (\int \left (\frac{a^2}{x^{10}}+\frac{a^2+b^2}{x^8}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\left (a^2+b^2\right ) \cot ^7(c+d x)}{7 d}-\frac{a^2 \cot ^9(c+d x)}{9 d}-\frac{5 a b \cot (c+d x) \csc ^3(c+d x)}{32 d}+\frac{5 a b \cot ^3(c+d x) \csc ^3(c+d x)}{24 d}-\frac{a b \cot ^5(c+d x) \csc ^3(c+d x)}{4 d}-\frac{1}{32} (5 a b) \int \csc ^3(c+d x) \, dx\\ &=-\frac{\left (a^2+b^2\right ) \cot ^7(c+d x)}{7 d}-\frac{a^2 \cot ^9(c+d x)}{9 d}+\frac{5 a b \cot (c+d x) \csc (c+d x)}{64 d}-\frac{5 a b \cot (c+d x) \csc ^3(c+d x)}{32 d}+\frac{5 a b \cot ^3(c+d x) \csc ^3(c+d x)}{24 d}-\frac{a b \cot ^5(c+d x) \csc ^3(c+d x)}{4 d}-\frac{1}{64} (5 a b) \int \csc (c+d x) \, dx\\ &=\frac{5 a b \tanh ^{-1}(\cos (c+d x))}{64 d}-\frac{\left (a^2+b^2\right ) \cot ^7(c+d x)}{7 d}-\frac{a^2 \cot ^9(c+d x)}{9 d}+\frac{5 a b \cot (c+d x) \csc (c+d x)}{64 d}-\frac{5 a b \cot (c+d x) \csc ^3(c+d x)}{32 d}+\frac{5 a b \cot ^3(c+d x) \csc ^3(c+d x)}{24 d}-\frac{a b \cot ^5(c+d x) \csc ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.1199, size = 204, normalized size = 1.35 \[ -\frac{\csc ^9(c+d x) \left (4032 \left (8 a^2+b^2\right ) \cos (c+d x)+18816 a^2 \cos (3 (c+d x))+5760 a^2 \cos (5 (c+d x))+576 a^2 \cos (7 (c+d x))-64 a^2 \cos (9 (c+d x))+18270 a b \sin (2 (c+d x))+10458 a b \sin (4 (c+d x))+8022 a b \sin (6 (c+d x))+315 a b \sin (8 (c+d x))-2304 b^2 \cos (5 (c+d x))-1440 b^2 \cos (7 (c+d x))-288 b^2 \cos (9 (c+d x))\right )+40320 a b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-40320 a b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{516096 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]

[Out]

-(-40320*a*b*Log[Cos[(c + d*x)/2]] + 40320*a*b*Log[Sin[(c + d*x)/2]] + Csc[c + d*x]^9*(4032*(8*a^2 + b^2)*Cos[
c + d*x] + 18816*a^2*Cos[3*(c + d*x)] + 5760*a^2*Cos[5*(c + d*x)] - 2304*b^2*Cos[5*(c + d*x)] + 576*a^2*Cos[7*
(c + d*x)] - 1440*b^2*Cos[7*(c + d*x)] - 64*a^2*Cos[9*(c + d*x)] - 288*b^2*Cos[9*(c + d*x)] + 18270*a*b*Sin[2*
(c + d*x)] + 10458*a*b*Sin[4*(c + d*x)] + 8022*a*b*Sin[6*(c + d*x)] + 315*a*b*Sin[8*(c + d*x)]))/(516096*d)

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Maple [A]  time = 0.096, size = 232, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{9\,d \left ( \sin \left ( dx+c \right ) \right ) ^{9}}}-{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{63\,d \left ( \sin \left ( dx+c \right ) \right ) ^{7}}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{8}}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{24\,d \left ( \sin \left ( dx+c \right ) \right ) ^{6}}}+{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{96\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{64\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{64\,d}}-{\frac{5\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{192\,d}}-{\frac{5\,ab\cos \left ( dx+c \right ) }{64\,d}}-{\frac{5\,ab\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{64\,d}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{7\,d \left ( \sin \left ( dx+c \right ) \right ) ^{7}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^10*(a+b*sin(d*x+c))^2,x)

[Out]

-1/9/d*a^2/sin(d*x+c)^9*cos(d*x+c)^7-2/63/d*a^2/sin(d*x+c)^7*cos(d*x+c)^7-1/4/d*a*b/sin(d*x+c)^8*cos(d*x+c)^7-
1/24/d*a*b/sin(d*x+c)^6*cos(d*x+c)^7+1/96/d*a*b/sin(d*x+c)^4*cos(d*x+c)^7-1/64/d*a*b/sin(d*x+c)^2*cos(d*x+c)^7
-1/64*a*b*cos(d*x+c)^5/d-5/192*a*b*cos(d*x+c)^3/d-5/64*a*b*cos(d*x+c)/d-5/64/d*a*b*ln(csc(d*x+c)-cot(d*x+c))-1
/7/d*b^2/sin(d*x+c)^7*cos(d*x+c)^7

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Maxima [A]  time = 0.990969, size = 208, normalized size = 1.38 \begin{align*} -\frac{21 \, a b{\left (\frac{2 \,{\left (15 \, \cos \left (d x + c\right )^{7} + 73 \, \cos \left (d x + c\right )^{5} - 55 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{8} - 4 \, \cos \left (d x + c\right )^{6} + 6 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + 1} - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac{1152 \, b^{2}}{\tan \left (d x + c\right )^{7}} + \frac{128 \,{\left (9 \, \tan \left (d x + c\right )^{2} + 7\right )} a^{2}}{\tan \left (d x + c\right )^{9}}}{8064 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^10*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/8064*(21*a*b*(2*(15*cos(d*x + c)^7 + 73*cos(d*x + c)^5 - 55*cos(d*x + c)^3 + 15*cos(d*x + c))/(cos(d*x + c)
^8 - 4*cos(d*x + c)^6 + 6*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + 1) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x +
c) - 1)) + 1152*b^2/tan(d*x + c)^7 + 128*(9*tan(d*x + c)^2 + 7)*a^2/tan(d*x + c)^9)/d

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Fricas [B]  time = 1.89782, size = 782, normalized size = 5.18 \begin{align*} \frac{128 \,{\left (2 \, a^{2} + 9 \, b^{2}\right )} \cos \left (d x + c\right )^{9} - 1152 \,{\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{7} + 315 \,{\left (a b \cos \left (d x + c\right )^{8} - 4 \, a b \cos \left (d x + c\right )^{6} + 6 \, a b \cos \left (d x + c\right )^{4} - 4 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 315 \,{\left (a b \cos \left (d x + c\right )^{8} - 4 \, a b \cos \left (d x + c\right )^{6} + 6 \, a b \cos \left (d x + c\right )^{4} - 4 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 42 \,{\left (15 \, a b \cos \left (d x + c\right )^{7} + 73 \, a b \cos \left (d x + c\right )^{5} - 55 \, a b \cos \left (d x + c\right )^{3} + 15 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8064 \,{\left (d \cos \left (d x + c\right )^{8} - 4 \, d \cos \left (d x + c\right )^{6} + 6 \, d \cos \left (d x + c\right )^{4} - 4 \, 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)^10*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/8064*(128*(2*a^2 + 9*b^2)*cos(d*x + c)^9 - 1152*(a^2 + b^2)*cos(d*x + c)^7 + 315*(a*b*cos(d*x + c)^8 - 4*a*b
*cos(d*x + c)^6 + 6*a*b*cos(d*x + c)^4 - 4*a*b*cos(d*x + c)^2 + a*b)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c)
- 315*(a*b*cos(d*x + c)^8 - 4*a*b*cos(d*x + c)^6 + 6*a*b*cos(d*x + c)^4 - 4*a*b*cos(d*x + c)^2 + a*b)*log(-1/2
*cos(d*x + c) + 1/2)*sin(d*x + c) - 42*(15*a*b*cos(d*x + c)^7 + 73*a*b*cos(d*x + c)^5 - 55*a*b*cos(d*x + c)^3
+ 15*a*b*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^8 - 4*d*cos(d*x + c)^6 + 6*d*cos(d*x + c)^4 - 4*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)**10*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.20013, size = 551, normalized size = 3.65 \begin{align*} \frac{14 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 63 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 54 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 72 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 336 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 504 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 504 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 336 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1512 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1008 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 5040 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 756 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2520 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{14258 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 756 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 2520 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 1008 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 336 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 1512 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 504 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 504 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 336 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 54 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 72 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 63 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 14 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9}}}{64512 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^10*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/64512*(14*a^2*tan(1/2*d*x + 1/2*c)^9 + 63*a*b*tan(1/2*d*x + 1/2*c)^8 - 54*a^2*tan(1/2*d*x + 1/2*c)^7 + 72*b^
2*tan(1/2*d*x + 1/2*c)^7 - 336*a*b*tan(1/2*d*x + 1/2*c)^6 - 504*b^2*tan(1/2*d*x + 1/2*c)^5 + 504*a*b*tan(1/2*d
*x + 1/2*c)^4 + 336*a^2*tan(1/2*d*x + 1/2*c)^3 + 1512*b^2*tan(1/2*d*x + 1/2*c)^3 + 1008*a*b*tan(1/2*d*x + 1/2*
c)^2 - 5040*a*b*log(abs(tan(1/2*d*x + 1/2*c))) - 756*a^2*tan(1/2*d*x + 1/2*c) - 2520*b^2*tan(1/2*d*x + 1/2*c)
+ (14258*a*b*tan(1/2*d*x + 1/2*c)^9 + 756*a^2*tan(1/2*d*x + 1/2*c)^8 + 2520*b^2*tan(1/2*d*x + 1/2*c)^8 - 1008*
a*b*tan(1/2*d*x + 1/2*c)^7 - 336*a^2*tan(1/2*d*x + 1/2*c)^6 - 1512*b^2*tan(1/2*d*x + 1/2*c)^6 - 504*a*b*tan(1/
2*d*x + 1/2*c)^5 + 504*b^2*tan(1/2*d*x + 1/2*c)^4 + 336*a*b*tan(1/2*d*x + 1/2*c)^3 + 54*a^2*tan(1/2*d*x + 1/2*
c)^2 - 72*b^2*tan(1/2*d*x + 1/2*c)^2 - 63*a*b*tan(1/2*d*x + 1/2*c) - 14*a^2)/tan(1/2*d*x + 1/2*c)^9)/d