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]

1/4*cot(d*x+c)^4/a/d+1/3*csc(d*x+c)^3/a/d-1/5*csc(d*x+c)^5/a/d

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Rubi [A]  time = 0.14, antiderivative size = 55, normalized size of antiderivative = 1.00, 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 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 30

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

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 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 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]))

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*}

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Mathematica [A]  time = 0.11, 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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fricas [A]  time = 0.66, size = 71, normalized size = 1.29 \[ -\frac {20 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (2 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 8}{60 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \]

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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giac [A]  time = 0.23, size = 46, normalized size = 0.84 \[ -\frac {30 \, \sin \left (d x + c\right )^{3} - 20 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{60 \, a d \sin \left (d x + c\right )^{5}} \]

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)

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maple [A]  time = 0.48, size = 49, normalized size = 0.89 \[ \frac {-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {1}{4 \sin \left (d x +c \right )^{4}}+\frac {1}{3 \sin \left (d x +c \right )^{3}}}{d a} \]

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

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maxima [A]  time = 0.77, size = 46, normalized size = 0.84 \[ -\frac {30 \, \sin \left (d x + c\right )^{3} - 20 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{60 \, a d \sin \left (d x + c\right )^{5}} \]

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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mupad [B]  time = 8.96, size = 46, normalized size = 0.84 \[ \frac {-30\,{\sin \left (c+d\,x\right )}^3+20\,{\sin \left (c+d\,x\right )}^2+15\,\sin \left (c+d\,x\right )-12}{60\,a\,d\,{\sin \left (c+d\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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