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]

-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)

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Rubi [A]  time = 0.102601, antiderivative size = 55, normalized size of antiderivative = 1., 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 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]

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 \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*}

Mathematica [A]  time = 0.0739833, 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]

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

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Maple [A]  time = 0.157, size = 39, normalized size = 0.7 \begin{align*}{\frac{1}{d{a}^{2}} \left ({\frac{2}{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{1}{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{1}{6\, \left ( \sin \left ( dx+c \right ) \right ) ^{6}}} \right ) } \end{align*}

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

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Maxima [A]  time = 1.10102, size = 49, normalized size = 0.89 \begin{align*} -\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}} \end{align*}

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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Fricas [A]  time = 1.03921, size = 177, normalized size = 3.22 \begin{align*} -\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 )}} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.2168, size = 49, normalized size = 0.89 \begin{align*} -\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}} \end{align*}

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)