3.746 \(\int \frac{\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=97 \[ \frac{\cot ^3(c+d x)}{a^3 d}+\frac{\cot (c+d x)}{a^3 d}+\frac{13 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}-\frac{11 \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac{x}{a^3} \]

[Out]

x/a^3 + (13*ArcTanh[Cos[c + d*x]])/(8*a^3*d) + Cot[c + d*x]/(a^3*d) + Cot[c + d*x]^3/(a^3*d) - (11*Cot[c + d*x
]*Csc[c + d*x])/(8*a^3*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^3*d)

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Rubi [A]  time = 0.318165, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2875, 2873, 3473, 8, 2611, 3770, 2607, 30, 3768} \[ \frac{\cot ^3(c+d x)}{a^3 d}+\frac{\cot (c+d x)}{a^3 d}+\frac{13 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}-\frac{11 \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac{x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*Cot[c + d*x]^5)/(a + a*Sin[c + d*x])^3,x]

[Out]

x/a^3 + (13*ArcTanh[Cos[c + d*x]])/(8*a^3*d) + Cot[c + d*x]/(a^3*d) + Cot[c + d*x]^3/(a^3*d) - (11*Cot[c + d*x
]*Csc[c + d*x])/(8*a^3*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^3*d)

Rule 2875

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

Rule 2873

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

Rule 3473

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]

Rule 8

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

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 3770

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

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \cot ^2(c+d x) \csc ^3(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (-a^3 \cot ^2(c+d x)+3 a^3 \cot ^2(c+d x) \csc (c+d x)-3 a^3 \cot ^2(c+d x) \csc ^2(c+d x)+a^3 \cot ^2(c+d x) \csc ^3(c+d x)\right ) \, dx}{a^6}\\ &=-\frac{\int \cot ^2(c+d x) \, dx}{a^3}+\frac{\int \cot ^2(c+d x) \csc ^3(c+d x) \, dx}{a^3}+\frac{3 \int \cot ^2(c+d x) \csc (c+d x) \, dx}{a^3}-\frac{3 \int \cot ^2(c+d x) \csc ^2(c+d x) \, dx}{a^3}\\ &=\frac{\cot (c+d x)}{a^3 d}-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}-\frac{\int \csc ^3(c+d x) \, dx}{4 a^3}+\frac{\int 1 \, dx}{a^3}-\frac{3 \int \csc (c+d x) \, dx}{2 a^3}-\frac{3 \operatorname{Subst}\left (\int x^2 \, dx,x,-\cot (c+d x)\right )}{a^3 d}\\ &=\frac{x}{a^3}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{\cot (c+d x)}{a^3 d}+\frac{\cot ^3(c+d x)}{a^3 d}-\frac{11 \cot (c+d x) \csc (c+d x)}{8 a^3 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}-\frac{\int \csc (c+d x) \, dx}{8 a^3}\\ &=\frac{x}{a^3}+\frac{13 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac{\cot (c+d x)}{a^3 d}+\frac{\cot ^3(c+d x)}{a^3 d}-\frac{11 \cot (c+d x) \csc (c+d x)}{8 a^3 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}\\ \end{align*}

Mathematica [A]  time = 2.37377, size = 165, normalized size = 1.7 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6 \left (-22 \csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^4\left (\frac{1}{2} (c+d x)\right )+22 \sec ^2\left (\frac{1}{2} (c+d x)\right )+(4 \sin (c+d x)-1) \csc ^4\left (\frac{1}{2} (c+d x)\right )+8 \left (-13 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+13 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-8 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)+8 c+8 d x\right )\right )}{64 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^5)/(a + a*Sin[c + d*x])^3,x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(-22*Csc[(c + d*x)/2]^2 + 22*Sec[(c + d*x)/2]^2 + Sec[(c + d*x)/2]^4
+ 8*(8*c + 8*d*x + 13*Log[Cos[(c + d*x)/2]] - 13*Log[Sin[(c + d*x)/2]] - 8*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4)
+ Csc[(c + d*x)/2]^4*(-1 + 4*Sin[c + d*x])))/(64*a^3*d*(1 + Sin[c + d*x])^3)

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Maple [B]  time = 0.171, size = 188, normalized size = 1.9 \begin{align*}{\frac{1}{64\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}}-{\frac{1}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{3}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{8\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-{\frac{1}{64\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}+{\frac{1}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{\frac{3}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{1}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{13}{8\,d{a}^{3}}\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)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x)

[Out]

1/64/d/a^3*tan(1/2*d*x+1/2*c)^4-1/8/d/a^3*tan(1/2*d*x+1/2*c)^3+3/8/d/a^3*tan(1/2*d*x+1/2*c)^2-1/8/d/a^3*tan(1/
2*d*x+1/2*c)+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))-1/64/d/a^3/tan(1/2*d*x+1/2*c)^4+1/8/d/a^3/tan(1/2*d*x+1/2*c)^3
-3/8/d/a^3/tan(1/2*d*x+1/2*c)^2+1/8/d/a^3/tan(1/2*d*x+1/2*c)-13/8/d/a^3*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.55326, size = 294, normalized size = 3.03 \begin{align*} -\frac{\frac{\frac{8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{3}} - \frac{128 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac{104 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac{{\left (\frac{8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}{a^{3} \sin \left (d x + c\right )^{4}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/64*((8*sin(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a^3 - 128*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + 10
4*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - (8*sin(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 1)*(cos(d*x + c) + 1)^4/(a^3*sin(d*x + c)^4))/d

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Fricas [A]  time = 1.17891, size = 452, normalized size = 4.66 \begin{align*} \frac{16 \, d x \cos \left (d x + c\right )^{4} - 32 \, d x \cos \left (d x + c\right )^{2} + 22 \, \cos \left (d x + c\right )^{3} + 16 \, d x + 13 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 13 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 16 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 26 \, \cos \left (d x + c\right )}{16 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16*(16*d*x*cos(d*x + c)^4 - 32*d*x*cos(d*x + c)^2 + 22*cos(d*x + c)^3 + 16*d*x + 13*(cos(d*x + c)^4 - 2*cos(
d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) - 13*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c)
+ 1/2) + 16*cos(d*x + c)*sin(d*x + c) - 26*cos(d*x + c))/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3
*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)**8*csc(d*x+c)**5/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.3419, size = 224, normalized size = 2.31 \begin{align*} \frac{\frac{192 \,{\left (d x + c\right )}}{a^{3}} - \frac{312 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{650 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 72 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}} + \frac{3 \,{\left (a^{9} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8 \, a^{9} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, a^{9} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a^{9} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a^{12}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/192*(192*(d*x + c)/a^3 - 312*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + (650*tan(1/2*d*x + 1/2*c)^4 + 24*tan(1/2*d
*x + 1/2*c)^3 - 72*tan(1/2*d*x + 1/2*c)^2 + 24*tan(1/2*d*x + 1/2*c) - 3)/(a^3*tan(1/2*d*x + 1/2*c)^4) + 3*(a^9
*tan(1/2*d*x + 1/2*c)^4 - 8*a^9*tan(1/2*d*x + 1/2*c)^3 + 24*a^9*tan(1/2*d*x + 1/2*c)^2 - 8*a^9*tan(1/2*d*x + 1
/2*c))/a^12)/d