3.215 \(\int \cot (c+d x) \csc ^6(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=73 \[ -\frac{a^3 \csc ^6(c+d x)}{6 d}-\frac{3 a^3 \csc ^5(c+d x)}{5 d}-\frac{3 a^3 \csc ^4(c+d x)}{4 d}-\frac{a^3 \csc ^3(c+d x)}{3 d} \]

[Out]

-(a^3*Csc[c + d*x]^3)/(3*d) - (3*a^3*Csc[c + d*x]^4)/(4*d) - (3*a^3*Csc[c + d*x]^5)/(5*d) - (a^3*Csc[c + d*x]^
6)/(6*d)

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Rubi [A]  time = 0.0719408, antiderivative size = 73, normalized size of antiderivative = 1., 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} \[ -\frac{a^3 \csc ^6(c+d x)}{6 d}-\frac{3 a^3 \csc ^5(c+d x)}{5 d}-\frac{3 a^3 \csc ^4(c+d x)}{4 d}-\frac{a^3 \csc ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]*Csc[c + d*x]^6*(a + a*Sin[c + d*x])^3,x]

[Out]

-(a^3*Csc[c + d*x]^3)/(3*d) - (3*a^3*Csc[c + d*x]^4)/(4*d) - (3*a^3*Csc[c + d*x]^5)/(5*d) - (a^3*Csc[c + d*x]^
6)/(6*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 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])

Rubi steps

\begin{align*} \int \cot (c+d x) \csc ^6(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^7 (a+x)^3}{x^7} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac{a^6 \operatorname{Subst}\left (\int \frac{(a+x)^3}{x^7} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^6 \operatorname{Subst}\left (\int \left (\frac{a^3}{x^7}+\frac{3 a^2}{x^6}+\frac{3 a}{x^5}+\frac{1}{x^4}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{a^3 \csc ^3(c+d x)}{3 d}-\frac{3 a^3 \csc ^4(c+d x)}{4 d}-\frac{3 a^3 \csc ^5(c+d x)}{5 d}-\frac{a^3 \csc ^6(c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.0303022, size = 73, normalized size = 1. \[ -\frac{a^3 \csc ^6(c+d x)}{6 d}-\frac{3 a^3 \csc ^5(c+d x)}{5 d}-\frac{3 a^3 \csc ^4(c+d x)}{4 d}-\frac{a^3 \csc ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]*Csc[c + d*x]^6*(a + a*Sin[c + d*x])^3,x]

[Out]

-(a^3*Csc[c + d*x]^3)/(3*d) - (3*a^3*Csc[c + d*x]^4)/(4*d) - (3*a^3*Csc[c + d*x]^5)/(5*d) - (a^3*Csc[c + d*x]^
6)/(6*d)

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Maple [A]  time = 0.041, size = 49, normalized size = 0.7 \begin{align*}{\frac{{a}^{3}}{d} \left ( -{\frac{3}{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{3}{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{1}{6\, \left ( \sin \left ( dx+c \right ) \right ) ^{6}}}-{\frac{1}{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*csc(d*x+c)^7*(a+a*sin(d*x+c))^3,x)

[Out]

1/d*a^3*(-3/5/sin(d*x+c)^5-3/4/sin(d*x+c)^4-1/6/sin(d*x+c)^6-1/3/sin(d*x+c)^3)

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Maxima [A]  time = 1.23501, size = 76, normalized size = 1.04 \begin{align*} -\frac{20 \, a^{3} \sin \left (d x + c\right )^{3} + 45 \, a^{3} \sin \left (d x + c\right )^{2} + 36 \, a^{3} \sin \left (d x + c\right ) + 10 \, a^{3}}{60 \, d \sin \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)^7*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(20*a^3*sin(d*x + c)^3 + 45*a^3*sin(d*x + c)^2 + 36*a^3*sin(d*x + c) + 10*a^3)/(d*sin(d*x + c)^6)

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Fricas [A]  time = 1.77253, size = 208, normalized size = 2.85 \begin{align*} -\frac{45 \, a^{3} \cos \left (d x + c\right )^{2} - 55 \, a^{3} + 4 \,{\left (5 \, a^{3} \cos \left (d x + c\right )^{2} - 14 \, a^{3}\right )} \sin \left (d x + c\right )}{60 \,{\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)^7*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(45*a^3*cos(d*x + c)^2 - 55*a^3 + 4*(5*a^3*cos(d*x + c)^2 - 14*a^3)*sin(d*x + c))/(d*cos(d*x + c)^6 - 3*
d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)

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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)**7*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.26843, size = 76, normalized size = 1.04 \begin{align*} -\frac{20 \, a^{3} \sin \left (d x + c\right )^{3} + 45 \, a^{3} \sin \left (d x + c\right )^{2} + 36 \, a^{3} \sin \left (d x + c\right ) + 10 \, a^{3}}{60 \, d \sin \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)^7*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(20*a^3*sin(d*x + c)^3 + 45*a^3*sin(d*x + c)^2 + 36*a^3*sin(d*x + c) + 10*a^3)/(d*sin(d*x + c)^6)