3.12 \(\int \frac{1}{(a+a \csc (c+d x))^3} \, dx\)

Optimal. Leaf size=88 \[ \frac{22 \cot (c+d x)}{15 d \left (a^3 \csc (c+d x)+a^3\right )}+\frac{x}{a^3}+\frac{7 \cot (c+d x)}{15 a d (a \csc (c+d x)+a)^2}+\frac{\cot (c+d x)}{5 d (a \csc (c+d x)+a)^3} \]

[Out]

x/a^3 + Cot[c + d*x]/(5*d*(a + a*Csc[c + d*x])^3) + (7*Cot[c + d*x])/(15*a*d*(a + a*Csc[c + d*x])^2) + (22*Cot
[c + d*x])/(15*d*(a^3 + a^3*Csc[c + d*x]))

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Rubi [A]  time = 0.109679, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3777, 3922, 3919, 3794} \[ \frac{22 \cot (c+d x)}{15 d \left (a^3 \csc (c+d x)+a^3\right )}+\frac{x}{a^3}+\frac{7 \cot (c+d x)}{15 a d (a \csc (c+d x)+a)^2}+\frac{\cot (c+d x)}{5 d (a \csc (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Csc[c + d*x])^(-3),x]

[Out]

x/a^3 + Cot[c + d*x]/(5*d*(a + a*Csc[c + d*x])^3) + (7*Cot[c + d*x])/(15*a*d*(a + a*Csc[c + d*x])^2) + (22*Cot
[c + d*x])/(15*d*(a^3 + a^3*Csc[c + d*x]))

Rule 3777

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(Cot[c + d*x]*(a + b*Csc[c + d*x])^n)/(d*(
2*n + 1)), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3922

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[((b
*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(b*f*(2*m + 1)), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e
+ f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f},
x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \csc (c+d x))^3} \, dx &=\frac{\cot (c+d x)}{5 d (a+a \csc (c+d x))^3}-\frac{\int \frac{-5 a+2 a \csc (c+d x)}{(a+a \csc (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{\cot (c+d x)}{5 d (a+a \csc (c+d x))^3}+\frac{7 \cot (c+d x)}{15 a d (a+a \csc (c+d x))^2}+\frac{\int \frac{15 a^2-7 a^2 \csc (c+d x)}{a+a \csc (c+d x)} \, dx}{15 a^4}\\ &=\frac{x}{a^3}+\frac{\cot (c+d x)}{5 d (a+a \csc (c+d x))^3}+\frac{7 \cot (c+d x)}{15 a d (a+a \csc (c+d x))^2}-\frac{22 \int \frac{\csc (c+d x)}{a+a \csc (c+d x)} \, dx}{15 a^2}\\ &=\frac{x}{a^3}+\frac{\cot (c+d x)}{5 d (a+a \csc (c+d x))^3}+\frac{7 \cot (c+d x)}{15 a d (a+a \csc (c+d x))^2}+\frac{22 \cot (c+d x)}{15 d \left (a^3+a^3 \csc (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.94444, size = 123, normalized size = 1.4 \[ \frac{\frac{2 \sin \left (\frac{1}{2} (c+d x)\right ) (-51 \sin (c+d x)+16 \cos (2 (c+d x))-38)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5}-\frac{13}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{3}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}+15 c+15 d x}{15 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Csc[c + d*x])^(-3),x]

[Out]

(15*c + 15*d*x + 3/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 13/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (2*S
in[(c + d*x)/2]*(-38 + 16*Cos[2*(c + d*x)] - 51*Sin[c + d*x]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)/(15*a^
3*d)

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Maple [A]  time = 0.079, size = 125, normalized size = 1.4 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-4\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}+{\frac{8}{5\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}+{\frac{4}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+2\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+2\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*csc(d*x+c))^3,x)

[Out]

2/d/a^3*arctan(tan(1/2*d*x+1/2*c))-4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^4+8/5/d/a^3/(tan(1/2*d*x+1/2*c)+1)^5+4/3/d/a
^3/(tan(1/2*d*x+1/2*c)+1)^3+2/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2+2/d/a^3/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.50908, size = 308, normalized size = 3.5 \begin{align*} \frac{2 \,{\left (\frac{\frac{95 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{145 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{75 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 22}{a^{3} + \frac{5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac{15 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*csc(d*x+c))^3,x, algorithm="maxima")

[Out]

2/15*((95*sin(d*x + c)/(cos(d*x + c) + 1) + 145*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 75*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 22)/(a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 1
0*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*a^3*sin(d*x + c)^4/
(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a
^3)/d

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Fricas [B]  time = 0.486009, size = 475, normalized size = 5.4 \begin{align*} \frac{{\left (15 \, d x + 32\right )} \cos \left (d x + c\right )^{3} +{\left (45 \, d x - 19\right )} \cos \left (d x + c\right )^{2} - 60 \, d x - 6 \,{\left (5 \, d x + 9\right )} \cos \left (d x + c\right ) +{\left ({\left (15 \, d x - 32\right )} \cos \left (d x + c\right )^{2} - 60 \, d x - 3 \,{\left (10 \, d x + 17\right )} \cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 3}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d +{\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*csc(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*((15*d*x + 32)*cos(d*x + c)^3 + (45*d*x - 19)*cos(d*x + c)^2 - 60*d*x - 6*(5*d*x + 9)*cos(d*x + c) + ((15
*d*x - 32)*cos(d*x + c)^2 - 60*d*x - 3*(10*d*x + 17)*cos(d*x + c) + 3)*sin(d*x + c) - 3)/(a^3*d*cos(d*x + c)^3
+ 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*
a^3*d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\csc ^{3}{\left (c + d x \right )} + 3 \csc ^{2}{\left (c + d x \right )} + 3 \csc{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*csc(d*x+c))**3,x)

[Out]

Integral(1/(csc(c + d*x)**3 + 3*csc(c + d*x)**2 + 3*csc(c + d*x) + 1), x)/a**3

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Giac [A]  time = 1.25117, size = 116, normalized size = 1.32 \begin{align*} \frac{\frac{15 \,{\left (d x + c\right )}}{a^{3}} + \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 75 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 145 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 95 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 22\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*csc(d*x+c))^3,x, algorithm="giac")

[Out]

1/15*(15*(d*x + c)/a^3 + 2*(15*tan(1/2*d*x + 1/2*c)^4 + 75*tan(1/2*d*x + 1/2*c)^3 + 145*tan(1/2*d*x + 1/2*c)^2
+ 95*tan(1/2*d*x + 1/2*c) + 22)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5))/d