\(\int \frac {1}{(a+a \csc (c+d x))^3} \, dx\) [12]

Optimal result

Integrand size = 12, antiderivative size = 88 \[ \int \frac {1}{(a+a \csc (c+d x))^3} \, dx=\frac {x}{a^3}+\frac {\cot (c+d x)}{5 d (a+a \csc (c+d x))^3}+\frac {7 \cot (c+d x)}{15 a d (a+a \csc (c+d x))^2}+\frac {22 \cot (c+d x)}{15 d \left (a^3+a^3 \csc (c+d x)\right )} \]

[Out]

x/a^3+1/5*cot(d*x+c)/d/(a+a*csc(d*x+c))^3+7/15*cot(d*x+c)/a/d/(a+a*csc(d*x+c))^2+22/15*cot(d*x+c)/d/(a^3+a^3*c
sc(d*x+c))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3862, 4007, 4004, 3879} \[ \int \frac {1}{(a+a \csc (c+d x))^3} \, dx=\frac {22 \cot (c+d x)}{15 d \left (a^3 \csc (c+d x)+a^3\right )}+\frac {x}{a^3}+\frac {7 \cot (c+d x)}{15 a d (a \csc (c+d x)+a)^2}+\frac {\cot (c+d x)}{5 d (a \csc (c+d x)+a)^3} \]

[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 3862

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 3879

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]

Rule 4004

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 4007

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 1.05 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.40 \[ \int \frac {1}{(a+a \csc (c+d x))^3} \, dx=\frac {15 c+15 d x+\frac {3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}-\frac {13}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) (-38+16 \cos (2 (c+d x))-51 \sin (c+d x))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}}{15 a^3 d} \]

[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)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {x}{a^{3}}+\frac {-\frac {74 \,{\mathrm e}^{2 i \left (d x +c \right )}}{3}+18 i {\mathrm e}^{3 i \left (d x +c \right )}-\frac {46 i {\mathrm e}^{i \left (d x +c \right )}}{3}+6 \,{\mathrm e}^{4 i \left (d x +c \right )}+\frac {64}{15}}{d \,a^{3} \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )^{5}}\)	\(77\)
derivativedivides	\(\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8}}{d \,a^{3}}\)	\(97\)
default	\(\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8}}{d \,a^{3}}\)	\(97\)
parallelrisch	\(\frac {\left (15 d x -38\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (75 d x -160\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (150 d x -230\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (150 d x -90\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+75 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) x d +15 d x +6}{15 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(113\)
norman	\(\frac {\frac {x}{a}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {5 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}+\frac {10 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {10 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a}+\frac {5 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {44}{15 a d}+\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}+\frac {58 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d a}+\frac {38 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(188\)

[In]

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

[Out]

x/a^3+2/15*(-185*exp(2*I*(d*x+c))+135*I*exp(3*I*(d*x+c))-115*I*exp(I*(d*x+c))+45*exp(4*I*(d*x+c))+32)/d/a^3/(I
+exp(I*(d*x+c)))^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (82) = 164\).

Time = 0.25 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.06 \[ \int \frac {1}{(a+a \csc (c+d x))^3} \, dx=\frac {{\left (15 \, d x + 32\right )} \cos \left (d x + c\right )^{3} + {\left (45 \, d x - 19\right )} \cos \left (d x + c\right )^{2} - 60 \, d x - 6 \, {\left (5 \, d x + 9\right )} \cos \left (d x + c\right ) + {\left ({\left (15 \, d x - 32\right )} \cos \left (d x + c\right )^{2} - 60 \, d x - 3 \, {\left (10 \, d x + 17\right )} \cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 3}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

[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))

Sympy [F]

\[ \int \frac {1}{(a+a \csc (c+d x))^3} \, dx=\frac {\int \frac {1}{\csc ^{3}{\left (c + d x \right )} + 3 \csc ^{2}{\left (c + d x \right )} + 3 \csc {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (82) = 164\).

Time = 0.32 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.59 \[ \int \frac {1}{(a+a \csc (c+d x))^3} \, dx=\frac {2 \, {\left (\frac {\frac {95 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {145 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {75 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 22}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{15 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(a+a \csc (c+d x))^3} \, dx=\frac {\frac {15 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 145 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 95 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 22\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{15 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 19.65 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(a+a \csc (c+d x))^3} \, dx=\frac {x}{a^3}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {58\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {38\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {44}{15}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]

[In]

int(1/(a + a/sin(c + d*x))^3,x)

[Out]

x/a^3 + ((38*tan(c/2 + (d*x)/2))/3 + (58*tan(c/2 + (d*x)/2)^2)/3 + 10*tan(c/2 + (d*x)/2)^3 + 2*tan(c/2 + (d*x)
/2)^4 + 44/15)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)^5)