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

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

[Out]

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

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Rubi [A]  time = 0.111392, antiderivative size = 73, 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, 75} \[ -\frac{\csc ^6(c+d x)}{6 a d}+\frac{\csc ^5(c+d x)}{5 a d}+\frac{\csc ^4(c+d x)}{4 a d}-\frac{\csc ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Csc[c + d*x]^3/(3*a*d) + Csc[c + d*x]^4/(4*a*d) + Csc[c + d*x]^5/(5*a*d) - Csc[c + d*x]^6/(6*a*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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 0.10296, size = 48, normalized size = 0.66 \[ \frac{\csc ^3(c+d x) \left (-10 \csc ^3(c+d x)+12 \csc ^2(c+d x)+15 \csc (c+d x)-20\right )}{60 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^3*(-20 + 15*Csc[c + d*x] + 12*Csc[c + d*x]^2 - 10*Csc[c + d*x]^3))/(60*a*d)

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Maple [A]  time = 0.142, size = 49, normalized size = 0.7 \begin{align*}{\frac{1}{da} \left ({\frac{1}{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}}}-{\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)^5*csc(d*x+c)^7/(a+a*sin(d*x+c)),x)

[Out]

1/d/a*(1/5/sin(d*x+c)^5+1/4/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 = 1.12297, size = 62, normalized size = 0.85 \begin{align*} -\frac{20 \, \sin \left (d x + c\right )^{3} - 15 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right ) + 10}{60 \, a 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)),x, algorithm="maxima")

[Out]

-1/60*(20*sin(d*x + c)^3 - 15*sin(d*x + c)^2 - 12*sin(d*x + c) + 10)/(a*d*sin(d*x + c)^6)

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Fricas [A]  time = 1.01741, size = 193, normalized size = 2.64 \begin{align*} \frac{15 \, \cos \left (d x + c\right )^{2} - 4 \,{\left (5 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) - 5}{60 \,{\left (a d \cos \left (d x + c\right )^{6} - 3 \, a d \cos \left (d x + c\right )^{4} + 3 \, a d \cos \left (d x + c\right )^{2} - a 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)),x, algorithm="fricas")

[Out]

1/60*(15*cos(d*x + c)^2 - 4*(5*cos(d*x + c)^2 - 2)*sin(d*x + c) - 5)/(a*d*cos(d*x + c)^6 - 3*a*d*cos(d*x + c)^
4 + 3*a*d*cos(d*x + c)^2 - a*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)),x)

[Out]

Timed out

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Giac [A]  time = 1.32414, size = 62, normalized size = 0.85 \begin{align*} -\frac{20 \, \sin \left (d x + c\right )^{3} - 15 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right ) + 10}{60 \, a 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)),x, algorithm="giac")

[Out]

-1/60*(20*sin(d*x + c)^3 - 15*sin(d*x + c)^2 - 12*sin(d*x + c) + 10)/(a*d*sin(d*x + c)^6)