3.630 \(\int \frac{\cos ^3(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=106 \[ -\frac{3 \cos (c+d x)}{2 a d}+\frac{3 \cot (c+d x)}{2 a d}-\frac{\cos ^2(c+d x) \cot (c+d x)}{2 a d}-\frac{\cos (c+d x) \cot ^2(c+d x)}{2 a d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{3 x}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.161383, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2839, 2592, 288, 321, 206, 2591, 203} \[ -\frac{3 \cos (c+d x)}{2 a d}+\frac{3 \cot (c+d x)}{2 a d}-\frac{\cos ^2(c+d x) \cot (c+d x)}{2 a d}-\frac{\cos (c+d x) \cot ^2(c+d x)}{2 a d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{3 x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

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

Rule 2839

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

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 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 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 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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.491853, size = 152, normalized size = 1.43 \[ -\frac{\left (\csc \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (-10 \sin (2 (c+d x))+\sin (4 (c+d x))+12 \cos (c+d x)-4 \cos (3 (c+d x))+12 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-12 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+12 \cos (2 (c+d x)) \left (-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )-12 c-12 d x\right )}{64 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

-((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^2*(-12*c - 12*d*x + 12*Cos[c + d*x] - 4*Cos[3*(c + d*x)] - 12*Log[Cos[
(c + d*x)/2]] + 12*Cos[2*(c + d*x)]*(c + d*x + Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]) + 12*Log[Sin[(c
+ d*x)/2]] - 10*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(64*a*d*(1 + Sin[c + d*x]))

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Maple [B]  time = 0.126, size = 234, normalized size = 2.2 \begin{align*}{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{2\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{1}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+3\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}}-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{3}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

1/8/d/a*tan(1/2*d*x+1/2*c)^2-1/2/d/a*tan(1/2*d*x+1/2*c)-1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-
2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^2+1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-2/d/
a/(1+tan(1/2*d*x+1/2*c)^2)^2+3/a/d*arctan(tan(1/2*d*x+1/2*c))-1/8/d/a/tan(1/2*d*x+1/2*c)^2+1/2/d/a/tan(1/2*d*x
+1/2*c)-3/2/d/a*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.53292, size = 352, normalized size = 3.32 \begin{align*} -\frac{\frac{\frac{4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a} - \frac{\frac{4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{18 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{17 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{4 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 1}{\frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{2 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac{24 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/8*((4*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a - (4*sin(d*x + c)/(cos(d*x +
c) + 1) - 18*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 17*sin(d*x + c)^4
/(cos(d*x + c) + 1)^4 - 4*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1)/(a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*
a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) - 24*arctan(sin(d*x + c)/(cos(d
*x + c) + 1))/a + 12*log(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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Fricas [A]  time = 1.16939, size = 347, normalized size = 3.27 \begin{align*} \frac{6 \, d x \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right )^{3} - 6 \, d x + 3 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right )}{4 \,{\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(6*d*x*cos(d*x + c)^2 - 4*cos(d*x + c)^3 - 6*d*x + 3*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2) - 3*
(cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2) + 2*(cos(d*x + c)^3 - 3*cos(d*x + c))*sin(d*x + c) + 6*cos(d
*x + c))/(a*d*cos(d*x + c)^2 - a*d)

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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)**3/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.37917, size = 225, normalized size = 2.12 \begin{align*} \frac{\frac{12 \,{\left (d x + c\right )}}{a} - \frac{12 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} + \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} + \frac{6 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 5 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 16 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{2} a}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*(12*(d*x + c)/a - 12*log(abs(tan(1/2*d*x + 1/2*c)))/a + (a*tan(1/2*d*x + 1/2*c)^2 - 4*a*tan(1/2*d*x + 1/2*
c))/a^2 + (6*tan(1/2*d*x + 1/2*c)^6 - 4*tan(1/2*d*x + 1/2*c)^5 - 5*tan(1/2*d*x + 1/2*c)^4 + 16*tan(1/2*d*x + 1
/2*c)^3 - 12*tan(1/2*d*x + 1/2*c)^2 + 4*tan(1/2*d*x + 1/2*c) - 1)/((tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2
*c))^2*a))/d