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]

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

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Rubi [A]  time = 0.07, antiderivative size = 73, 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} \[ -\frac {a^3 \csc ^6(c+d x)}{6 d}-\frac {3 a^3 \csc ^5(c+d x)}{5 d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^3(c+d x)}{3 d} \]

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 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 ^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*}

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Mathematica [A]  time = 0.03, size = 73, normalized size = 1.00 \[ -\frac {a^3 \csc ^6(c+d x)}{6 d}-\frac {3 a^3 \csc ^5(c+d x)}{5 d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a^3*Csc[c + d*x]^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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fricas [A]  time = 0.45, size = 86, normalized size = 1.18 \[ -\frac {45 \, a^{3} \cos \left (d x + c\right )^{2} - 55 \, a^{3} + 4 \, {\left (5 \, a^{3} \cos \left (d x + c\right )^{2} - 14 \, a^{3}\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \]

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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giac [A]  time = 0.20, size = 56, normalized size = 0.77 \[ -\frac {20 \, a^{3} \sin \left (d x + c\right )^{3} + 45 \, a^{3} \sin \left (d x + c\right )^{2} + 36 \, a^{3} \sin \left (d x + c\right ) + 10 \, a^{3}}{60 \, d \sin \left (d x + c\right )^{6}} \]

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)

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maple [A]  time = 0.16, size = 49, normalized size = 0.67 \[ \frac {a^{3} \left (-\frac {1}{6 \sin \left (d x +c \right )^{6}}-\frac {3}{5 \sin \left (d x +c \right )^{5}}-\frac {3}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 56, normalized size = 0.77 \[ -\frac {20 \, a^{3} \sin \left (d x + c\right )^{3} + 45 \, a^{3} \sin \left (d x + c\right )^{2} + 36 \, a^{3} \sin \left (d x + c\right ) + 10 \, a^{3}}{60 \, d \sin \left (d x + c\right )^{6}} \]

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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mupad [B]  time = 8.60, size = 56, normalized size = 0.77 \[ -\frac {a^3\,\left (-20\,{\sin \left (c+d\,x\right )}^6+20\,{\sin \left (c+d\,x\right )}^3+45\,{\sin \left (c+d\,x\right )}^2+36\,\sin \left (c+d\,x\right )+10\right )}{60\,d\,{\sin \left (c+d\,x\right )}^6} \]

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)^7,x)

[Out]

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

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

[Out]

Timed out

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