3.204 \(\int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=55 \[ -\frac {a^2 \csc ^5(c+d x)}{5 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {a^2 \csc ^3(c+d x)}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2833, 12, 43} \[ -\frac {a^2 \csc ^5(c+d x)}{5 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {a^2 \csc ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]*Csc[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

-(a^2*Csc[c + d*x]^3)/(3*d) - (a^2*Csc[c + d*x]^4)/(2*d) - (a^2*Csc[c + d*x]^5)/(5*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 55, normalized size = 1.00 \[ -\frac {a^2 \csc ^5(c+d x)}{5 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {a^2 \csc ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]*Csc[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 65, normalized size = 1.18 \[ \frac {10 \, a^{2} \cos \left (d x + c\right )^{2} - 15 \, a^{2} \sin \left (d x + c\right ) - 16 \, a^{2}}{30 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/30*(10*a^2*cos(d*x + c)^2 - 15*a^2*sin(d*x + c) - 16*a^2)/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d
*x + c))

________________________________________________________________________________________

giac [A]  time = 0.17, size = 43, normalized size = 0.78 \[ -\frac {10 \, a^{2} \sin \left (d x + c\right )^{2} + 15 \, a^{2} \sin \left (d x + c\right ) + 6 \, a^{2}}{30 \, d \sin \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/30*(10*a^2*sin(d*x + c)^2 + 15*a^2*sin(d*x + c) + 6*a^2)/(d*sin(d*x + c)^5)

________________________________________________________________________________________

maple [A]  time = 0.14, size = 39, normalized size = 0.71 \[ \frac {a^{2} \left (-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{2 \sin \left (d x +c \right )^{4}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 43, normalized size = 0.78 \[ -\frac {10 \, a^{2} \sin \left (d x + c\right )^{2} + 15 \, a^{2} \sin \left (d x + c\right ) + 6 \, a^{2}}{30 \, d \sin \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*(10*a^2*sin(d*x + c)^2 + 15*a^2*sin(d*x + c) + 6*a^2)/(d*sin(d*x + c)^5)

________________________________________________________________________________________

mupad [B]  time = 8.92, size = 43, normalized size = 0.78 \[ -\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^2}{3}+\frac {a^2\,\sin \left (c+d\,x\right )}{2}+\frac {a^2}{5}}{d\,{\sin \left (c+d\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(a + a*sin(c + d*x))^2)/sin(c + d*x)^6,x)

[Out]

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

________________________________________________________________________________________

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)*csc(d*x+c)**6*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________