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

Optimal. Leaf size=51 \[ -\frac {\cot ^4(c+d x)}{4 a d}+\frac {\csc ^3(c+d x)}{3 a d}-\frac {\csc (c+d x)}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2706, 2607, 30, 2606} \[ -\frac {\cot ^4(c+d x)}{4 a d}+\frac {\csc ^3(c+d x)}{3 a d}-\frac {\csc (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 30, normalized size = 0.59 \[ -\frac {(\csc (c+d x)-1)^3 (3 \csc (c+d x)+5)}{12 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*((-1 + Csc[c + d*x])^3*(5 + 3*Csc[c + d*x]))/(a*d)

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 63, normalized size = 1.24 \[ -\frac {6 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) - 3}{12 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(6*cos(d*x + c)^2 - 4*(3*cos(d*x + c)^2 - 2)*sin(d*x + c) - 3)/(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^
2 + a*d)

________________________________________________________________________________________

giac [A]  time = 0.27, size = 46, normalized size = 0.90 \[ -\frac {12 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 3}{12 \, a d \sin \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*sin(d*x + c)^3 - 6*sin(d*x + c)^2 - 4*sin(d*x + c) + 3)/(a*d*sin(d*x + c)^4)

________________________________________________________________________________________

maple [A]  time = 0.23, size = 49, normalized size = 0.96 \[ \frac {-\frac {1}{\sin \left (d x +c \right )}+\frac {1}{2 \sin \left (d x +c \right )^{2}}-\frac {1}{4 \sin \left (d x +c \right )^{4}}+\frac {1}{3 \sin \left (d x +c \right )^{3}}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 46, normalized size = 0.90 \[ -\frac {12 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 3}{12 \, a d \sin \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*sin(d*x + c)^3 - 6*sin(d*x + c)^2 - 4*sin(d*x + c) + 3)/(a*d*sin(d*x + c)^4)

________________________________________________________________________________________

mupad [B]  time = 6.56, size = 45, normalized size = 0.88 \[ \frac {-{\sin \left (c+d\,x\right )}^3+\frac {{\sin \left (c+d\,x\right )}^2}{2}+\frac {\sin \left (c+d\,x\right )}{3}-\frac {1}{4}}{a\,d\,{\sin \left (c+d\,x\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cot ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5/(a+a*sin(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**5/(sin(c + d*x) + 1), x)/a

________________________________________________________________________________________