3.214 \(\int \cot (c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=61 \[ \frac {\csc ^4(c+d x) (a \sin (c+d x)+a)^4}{20 a d}-\frac {\csc ^5(c+d x) (a \sin (c+d x)+a)^4}{5 a d} \]

[Out]

1/20*csc(d*x+c)^4*(a+a*sin(d*x+c))^4/a/d-1/5*csc(d*x+c)^5*(a+a*sin(d*x+c))^4/a/d

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Rubi [A]  time = 0.06, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2833, 12, 45, 37} \[ \frac {\csc ^4(c+d x) (a \sin (c+d x)+a)^4}{20 a d}-\frac {\csc ^5(c+d x) (a \sin (c+d x)+a)^4}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^4*(a + a*Sin[c + d*x])^4)/(20*a*d) - (Csc[c + d*x]^5*(a + a*Sin[c + d*x])^4)/(5*a*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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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))^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a^6 (a+x)^3}{x^6} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {a^5 \operatorname {Subst}\left (\int \frac {(a+x)^3}{x^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {\csc ^5(c+d x) (a+a \sin (c+d x))^4}{5 a d}-\frac {a^4 \operatorname {Subst}\left (\int \frac {(a+x)^3}{x^5} \, dx,x,a \sin (c+d x)\right )}{5 d}\\ &=\frac {\csc ^4(c+d x) (a+a \sin (c+d x))^4}{20 a d}-\frac {\csc ^5(c+d x) (a+a \sin (c+d x))^4}{5 a d}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 71, normalized size = 1.16 \[ -\frac {a^3 \csc ^5(c+d x)}{5 d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^3(c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 81, normalized size = 1.33 \[ \frac {20 \, a^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} + 5 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3}\right )} \sin \left (d x + c\right )}{20 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

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))^3,x, algorithm="fricas")

[Out]

1/20*(20*a^3*cos(d*x + c)^2 - 24*a^3 + 5*(2*a^3*cos(d*x + c)^2 - 5*a^3)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d
*cos(d*x + c)^2 + d)*sin(d*x + c))

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giac [A]  time = 0.18, size = 56, normalized size = 0.92 \[ -\frac {10 \, a^{3} \sin \left (d x + c\right )^{3} + 20 \, a^{3} \sin \left (d x + c\right )^{2} + 15 \, a^{3} \sin \left (d x + c\right ) + 4 \, a^{3}}{20 \, d \sin \left (d x + c\right )^{5}} \]

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))^3,x, algorithm="giac")

[Out]

-1/20*(10*a^3*sin(d*x + c)^3 + 20*a^3*sin(d*x + c)^2 + 15*a^3*sin(d*x + c) + 4*a^3)/(d*sin(d*x + c)^5)

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maple [A]  time = 0.14, size = 49, normalized size = 0.80 \[ \frac {a^{3} \left (-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}-\frac {3}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{\sin \left (d x +c \right )^{3}}\right )}{d} \]

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))^3,x)

[Out]

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

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maxima [A]  time = 0.51, size = 56, normalized size = 0.92 \[ -\frac {10 \, a^{3} \sin \left (d x + c\right )^{3} + 20 \, a^{3} \sin \left (d x + c\right )^{2} + 15 \, a^{3} \sin \left (d x + c\right ) + 4 \, a^{3}}{20 \, d \sin \left (d x + c\right )^{5}} \]

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))^3,x, algorithm="maxima")

[Out]

-1/20*(10*a^3*sin(d*x + c)^3 + 20*a^3*sin(d*x + c)^2 + 15*a^3*sin(d*x + c) + 4*a^3)/(d*sin(d*x + c)^5)

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mupad [B]  time = 8.60, size = 56, normalized size = 0.92 \[ -\frac {10\,a^3\,{\sin \left (c+d\,x\right )}^3+20\,a^3\,{\sin \left (c+d\,x\right )}^2+15\,a^3\,\sin \left (c+d\,x\right )+4\,a^3}{20\,d\,{\sin \left (c+d\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(a + a*sin(c + d*x))^3)/sin(c + d*x)^6,x)

[Out]

-(15*a^3*sin(c + d*x) + 4*a^3 + 20*a^3*sin(c + d*x)^2 + 10*a^3*sin(c + d*x)^3)/(20*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)*csc(d*x+c)**6*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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