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]

(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)

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Rubi [A]  time = 0.0649166, antiderivative size = 61, normalized size of antiderivative = 1., 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 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]

Rule 12

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

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 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]

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*}

Mathematica [A]  time = 0.0295041, 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]

-(a^3*Csc[c + d*x]^2)/(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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Maple [A]  time = 0.041, size = 49, normalized size = 0.8 \begin{align*}{\frac{{a}^{3}}{d} \left ( -{\frac{1}{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{3}{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}- \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)*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.138, size = 76, normalized size = 1.25 \begin{align*} -\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}} \end{align*}

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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Fricas [A]  time = 1.60963, size = 197, normalized size = 3.23 \begin{align*} \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 )} \end{align*}

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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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)*csc(d*x+c)**6*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.23292, size = 76, normalized size = 1.25 \begin{align*} -\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}} \end{align*}

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)