3.553 \(\int \frac {\cot ^5(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=55 \[ -\frac {\csc ^6(c+d x)}{6 a^2 d}+\frac {2 \csc ^5(c+d x)}{5 a^2 d}-\frac {\csc ^4(c+d x)}{4 a^2 d} \]

[Out]

-1/4*csc(d*x+c)^4/a^2/d+2/5*csc(d*x+c)^5/a^2/d-1/6*csc(d*x+c)^6/a^2/d

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Rubi [A]  time = 0.10, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2836, 12, 43} \[ -\frac {\csc ^6(c+d x)}{6 a^2 d}+\frac {2 \csc ^5(c+d x)}{5 a^2 d}-\frac {\csc ^4(c+d x)}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Csc[c + d*x]^4/(4*a^2*d) + (2*Csc[c + d*x]^5)/(5*a^2*d) - Csc[c + d*x]^6/(6*a^2*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 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 38, normalized size = 0.69 \[ -\frac {\csc ^4(c+d x) \left (10 \csc ^2(c+d x)-24 \csc (c+d x)+15\right )}{60 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(Csc[c + d*x]^4*(15 - 24*Csc[c + d*x] + 10*Csc[c + d*x]^2))/(a^2*d)

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fricas [A]  time = 0.70, size = 72, normalized size = 1.31 \[ -\frac {15 \, \cos \left (d x + c\right )^{2} + 24 \, \sin \left (d x + c\right ) - 25}{60 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 3 \, a^{2} d \cos \left (d x + c\right )^{4} + 3 \, a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/60*(15*cos(d*x + c)^2 + 24*sin(d*x + c) - 25)/(a^2*d*cos(d*x + c)^6 - 3*a^2*d*cos(d*x + c)^4 + 3*a^2*d*cos(
d*x + c)^2 - a^2*d)

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giac [A]  time = 0.27, size = 36, normalized size = 0.65 \[ -\frac {15 \, \sin \left (d x + c\right )^{2} - 24 \, \sin \left (d x + c\right ) + 10}{60 \, a^{2} d \sin \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/60*(15*sin(d*x + c)^2 - 24*sin(d*x + c) + 10)/(a^2*d*sin(d*x + c)^6)

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maple [A]  time = 0.58, size = 39, normalized size = 0.71 \[ \frac {-\frac {1}{6 \sin \left (d x +c \right )^{6}}+\frac {2}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{4 \sin \left (d x +c \right )^{4}}}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x)

[Out]

1/d/a^2*(-1/6/sin(d*x+c)^6+2/5/sin(d*x+c)^5-1/4/sin(d*x+c)^4)

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maxima [A]  time = 0.96, size = 36, normalized size = 0.65 \[ -\frac {15 \, \sin \left (d x + c\right )^{2} - 24 \, \sin \left (d x + c\right ) + 10}{60 \, a^{2} d \sin \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/60*(15*sin(d*x + c)^2 - 24*sin(d*x + c) + 10)/(a^2*d*sin(d*x + c)^6)

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mupad [B]  time = 8.95, size = 36, normalized size = 0.65 \[ -\frac {\frac {{\sin \left (c+d\,x\right )}^2}{4}-\frac {2\,\sin \left (c+d\,x\right )}{5}+\frac {1}{6}}{a^2\,d\,{\sin \left (c+d\,x\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^7*(a + a*sin(c + d*x))^2),x)

[Out]

-(sin(c + d*x)^2/4 - (2*sin(c + d*x))/5 + 1/6)/(a^2*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)**5*csc(d*x+c)**7/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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