3.541 \(\int \frac{\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=55 \[ \frac{\cot ^4(c+d x)}{4 a d}-\frac{\csc ^5(c+d x)}{5 a d}+\frac{\csc ^3(c+d x)}{3 a d} \]

[Out]

Cot[c + d*x]^4/(4*a*d) + Csc[c + d*x]^3/(3*a*d) - Csc[c + d*x]^5/(5*a*d)

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Rubi [A]  time = 0.135552, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2835, 2606, 14, 2607, 30} \[ \frac{\cot ^4(c+d x)}{4 a d}-\frac{\csc ^5(c+d x)}{5 a d}+\frac{\csc ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Cot[c + d*x]^4/(4*a*d) + Csc[c + d*x]^3/(3*a*d) - Csc[c + d*x]^5/(5*a*d)

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rule 2606

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

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]]

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]

Rubi steps

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

Mathematica [A]  time = 0.110942, size = 48, normalized size = 0.87 \[ \frac{\csc ^2(c+d x) \left (-12 \csc ^3(c+d x)+15 \csc ^2(c+d x)+20 \csc (c+d x)-30\right )}{60 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^2*(-30 + 20*Csc[c + d*x] + 15*Csc[c + d*x]^2 - 12*Csc[c + d*x]^3))/(60*a*d)

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Maple [A]  time = 0.132, size = 49, normalized size = 0.9 \begin{align*}{\frac{1}{da} \left ( -{\frac{1}{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}+{\frac{1}{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{1}{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c)),x)

[Out]

1/d/a*(-1/5/sin(d*x+c)^5+1/4/sin(d*x+c)^4+1/3/sin(d*x+c)^3-1/2/sin(d*x+c)^2)

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Maxima [A]  time = 1.14586, size = 62, normalized size = 1.13 \begin{align*} -\frac{30 \, \sin \left (d x + c\right )^{3} - 20 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{60 \, a d \sin \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(30*sin(d*x + c)^3 - 20*sin(d*x + c)^2 - 15*sin(d*x + c) + 12)/(a*d*sin(d*x + c)^5)

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Fricas [A]  time = 1.05824, size = 185, normalized size = 3.36 \begin{align*} -\frac{20 \, \cos \left (d x + c\right )^{2} - 15 \,{\left (2 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 8}{60 \,{\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a 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)^5*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(20*cos(d*x + c)^2 - 15*(2*cos(d*x + c)^2 - 1)*sin(d*x + c) - 8)/((a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x +
c)^2 + a*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)**5*csc(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.29617, size = 62, normalized size = 1.13 \begin{align*} -\frac{30 \, \sin \left (d x + c\right )^{3} - 20 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{60 \, a d \sin \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(30*sin(d*x + c)^3 - 20*sin(d*x + c)^2 - 15*sin(d*x + c) + 12)/(a*d*sin(d*x + c)^5)