Integrand size = 14, antiderivative size = 74 \[ \int \left (a+b \csc ^2(c+d x)\right )^3 \, dx=a^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \cot (c+d x)}{d}-\frac {b^2 (3 a+2 b) \cot ^3(c+d x)}{3 d}-\frac {b^3 \cot ^5(c+d x)}{5 d} \] Output:
a^3*x-b*(3*a^2+3*a*b+b^2)*cot(d*x+c)/d-1/3*b^2*(3*a+2*b)*cot(d*x+c)^3/d-1/ 5*b^3*cot(d*x+c)^5/d
Time = 4.48 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.51 \[ \int \left (a+b \csc ^2(c+d x)\right )^3 \, dx=\frac {8 \left (a+b \csc ^2(c+d x)\right )^3 \left (-15 a^3 (c+d x)+b \cot (c+d x) \left (45 a^2+30 a b+8 b^2+b (15 a+4 b) \csc ^2(c+d x)+3 b^2 \csc ^4(c+d x)\right )\right ) \sin ^6(c+d x)}{15 d (-a-2 b+a \cos (2 (c+d x)))^3} \] Input:
Integrate[(a + b*Csc[c + d*x]^2)^3,x]
Output:
(8*(a + b*Csc[c + d*x]^2)^3*(-15*a^3*(c + d*x) + b*Cot[c + d*x]*(45*a^2 + 30*a*b + 8*b^2 + b*(15*a + 4*b)*Csc[c + d*x]^2 + 3*b^2*Csc[c + d*x]^4))*Si n[c + d*x]^6)/(15*d*(-a - 2*b + a*Cos[2*(c + d*x)])^3)
Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4616, 300, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (a+b \csc ^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \sec \left (c+d x+\frac {\pi }{2}\right )^2\right )^3dx\) |
\(\Big \downarrow \) 4616 |
\(\displaystyle -\frac {\int \frac {\left (b \cot ^2(c+d x)+a+b\right )^3}{\cot ^2(c+d x)+1}d\cot (c+d x)}{d}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle -\frac {\int \left (b^3 \cot ^4(c+d x)+b^2 (3 a+2 b) \cot ^2(c+d x)+b \left (3 a^2+3 b a+b^2\right )+\frac {a^3}{\cot ^2(c+d x)+1}\right )d\cot (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \arctan (\cot (c+d x))+b \left (3 a^2+3 a b+b^2\right ) \cot (c+d x)+\frac {1}{3} b^2 (3 a+2 b) \cot ^3(c+d x)+\frac {1}{5} b^3 \cot ^5(c+d x)}{d}\) |
Input:
Int[(a + b*Csc[c + d*x]^2)^3,x]
Output:
-((a^3*ArcTan[Cot[c + d*x]] + b*(3*a^2 + 3*a*b + b^2)*Cot[c + d*x] + (b^2* (3*a + 2*b)*Cot[c + d*x]^3)/3 + (b^3*Cot[c + d*x]^5)/5)/d)
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b + b*ff^2*x^2)^p /(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && NeQ[a + b, 0] && NeQ[p, -1]
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.12
method | result | size |
derivativedivides | \(\frac {a^{3} \left (d x +c \right )-3 a^{2} b \cot \left (d x +c \right )+3 a \,b^{2} \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )+b^{3} \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(83\) |
default | \(\frac {a^{3} \left (d x +c \right )-3 a^{2} b \cot \left (d x +c \right )+3 a \,b^{2} \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )+b^{3} \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(83\) |
parts | \(a^{3} x +\frac {b^{3} \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}+\frac {3 a \,b^{2} \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )}{d}-\frac {3 a^{2} b \cot \left (d x +c \right )}{d}\) | \(84\) |
parallelrisch | \(\frac {-3 b^{3} \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-60 a \,b^{2}-25 b^{3}\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-720 a^{2} b -540 a \,b^{2}-150 b^{3}\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+3 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (60 a \,b^{2}+25 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (720 a^{2} b +540 a \,b^{2}+150 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+480 a^{3} x d}{480 d}\) | \(150\) |
risch | \(a^{3} x -\frac {2 i b \left (45 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-180 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-90 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+270 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+210 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+80 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-180 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-150 a b \,{\mathrm e}^{2 i \left (d x +c \right )}-40 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+45 a^{2}+30 b a +8 b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}\) | \(165\) |
norman | \(\frac {a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-\frac {b^{3}}{160 d}+\frac {b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{160 d}-\frac {b \left (24 a^{2}+18 b a +5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16 d}+\frac {b \left (24 a^{2}+18 b a +5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{16 d}-\frac {b^{2} \left (12 a +5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{96 d}+\frac {b^{2} \left (12 a +5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{96 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}\) | \(173\) |
Input:
int((a+b*csc(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(a^3*(d*x+c)-3*a^2*b*cot(d*x+c)+3*a*b^2*(-2/3-1/3*csc(d*x+c)^2)*cot(d* x+c)+b^3*(-8/15-1/5*csc(d*x+c)^4-4/15*csc(d*x+c)^2)*cot(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (70) = 140\).
Time = 0.07 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.15 \[ \int \left (a+b \csc ^2(c+d x)\right )^3 \, dx=-\frac {{\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{5} - 5 \, {\left (18 \, a^{2} b + 15 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right ) - 15 \, {\left (a^{3} d x \cos \left (d x + c\right )^{4} - 2 \, a^{3} d x \cos \left (d x + c\right )^{2} + a^{3} d x\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:
integrate((a+b*csc(d*x+c)^2)^3,x, algorithm="fricas")
Output:
-1/15*((45*a^2*b + 30*a*b^2 + 8*b^3)*cos(d*x + c)^5 - 5*(18*a^2*b + 15*a*b ^2 + 4*b^3)*cos(d*x + c)^3 + 15*(3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c) - 1 5*(a^3*d*x*cos(d*x + c)^4 - 2*a^3*d*x*cos(d*x + c)^2 + a^3*d*x)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))
\[ \int \left (a+b \csc ^2(c+d x)\right )^3 \, dx=\int \left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{3}\, dx \] Input:
integrate((a+b*csc(d*x+c)**2)**3,x)
Output:
Integral((a + b*csc(c + d*x)**2)**3, x)
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.22 \[ \int \left (a+b \csc ^2(c+d x)\right )^3 \, dx=a^{3} x - \frac {3 \, a^{2} b}{d \tan \left (d x + c\right )} - \frac {{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a b^{2}}{d \tan \left (d x + c\right )^{3}} - \frac {{\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} b^{3}}{15 \, d \tan \left (d x + c\right )^{5}} \] Input:
integrate((a+b*csc(d*x+c)^2)^3,x, algorithm="maxima")
Output:
a^3*x - 3*a^2*b/(d*tan(d*x + c)) - (3*tan(d*x + c)^2 + 1)*a*b^2/(d*tan(d*x + c)^3) - 1/15*(15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 + 3)*b^3/(d*tan(d*x + c)^5)
Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (70) = 140\).
Time = 0.20 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.85 \[ \int \left (a+b \csc ^2(c+d x)\right )^3 \, dx=\frac {3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 60 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 25 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 480 \, {\left (d x + c\right )} a^{3} + 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 540 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 150 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 540 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 150 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 60 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 25 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, b^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \] Input:
integrate((a+b*csc(d*x+c)^2)^3,x, algorithm="giac")
Output:
1/480*(3*b^3*tan(1/2*d*x + 1/2*c)^5 + 60*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 25 *b^3*tan(1/2*d*x + 1/2*c)^3 + 480*(d*x + c)*a^3 + 720*a^2*b*tan(1/2*d*x + 1/2*c) + 540*a*b^2*tan(1/2*d*x + 1/2*c) + 150*b^3*tan(1/2*d*x + 1/2*c) - ( 720*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 540*a*b^2*tan(1/2*d*x + 1/2*c)^4 + 150* b^3*tan(1/2*d*x + 1/2*c)^4 + 60*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 25*b^3*tan( 1/2*d*x + 1/2*c)^2 + 3*b^3)/tan(1/2*d*x + 1/2*c)^5)/d
Time = 15.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95 \[ \int \left (a+b \csc ^2(c+d x)\right )^3 \, dx=a^3\,x-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {2\,b^3}{3}+a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )+\frac {b^3}{5}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5} \] Input:
int((a + b/sin(c + d*x)^2)^3,x)
Output:
a^3*x - (tan(c + d*x)^2*(a*b^2 + (2*b^3)/3) + tan(c + d*x)^4*(3*a*b^2 + 3* a^2*b + b^3) + b^3/5)/(d*tan(c + d*x)^5)
Time = 0.16 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.86 \[ \int \left (a+b \csc ^2(c+d x)\right )^3 \, dx=\frac {-45 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{2} b -30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{3}-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{3}-3 \cos \left (d x +c \right ) b^{3}+15 \sin \left (d x +c \right )^{5} a^{3} d x}{15 \sin \left (d x +c \right )^{5} d} \] Input:
int((a+b*csc(d*x+c)^2)^3,x)
Output:
( - 45*cos(c + d*x)*sin(c + d*x)**4*a**2*b - 30*cos(c + d*x)*sin(c + d*x)* *4*a*b**2 - 8*cos(c + d*x)*sin(c + d*x)**4*b**3 - 15*cos(c + d*x)*sin(c + d*x)**2*a*b**2 - 4*cos(c + d*x)*sin(c + d*x)**2*b**3 - 3*cos(c + d*x)*b**3 + 15*sin(c + d*x)**5*a**3*d*x)/(15*sin(c + d*x)**5*d)