Integrand size = 29, antiderivative size = 54 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin ^3(c+d x) \, dx=a A x-\frac {2 a A \cos (c+d x)}{d}+\frac {a A \cos ^3(c+d x)}{3 d}-\frac {a A \cos (c+d x) \sin (c+d x)}{d} \] Output:
a*A*x-2*a*A*cos(d*x+c)/d+1/3*a*A*cos(d*x+c)^3/d-a*A*cos(d*x+c)*sin(d*x+c)/ d
Time = 0.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.80 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin ^3(c+d x) \, dx=\frac {a A (12 c+12 d x-21 \cos (c+d x)+\cos (3 (c+d x))-6 \sin (2 (c+d x)))}{12 d} \] Input:
Integrate[(a + a*Csc[c + d*x])*(A + A*Csc[c + d*x])*Sin[c + d*x]^3,x]
Output:
(a*A*(12*c + 12*d*x - 21*Cos[c + d*x] + Cos[3*(c + d*x)] - 6*Sin[2*(c + d* x)]))/(12*d)
Time = 0.47 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.26, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {2011, 3042, 4275, 3042, 3115, 24, 4532, 3042, 3492, 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.
\(\displaystyle \int \sin ^3(c+d x) (a \csc (c+d x)+a) (A \csc (c+d x)+A) \, dx\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle \frac {A \int (\csc (c+d x) a+a)^2 \sin ^3(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \int \frac {(\csc (c+d x) a+a)^2}{\csc (c+d x)^3}dx}{a}\) |
\(\Big \downarrow \) 4275 |
\(\displaystyle \frac {A \left (2 a^2 \int \sin ^2(c+d x)dx+\int \left (\csc ^2(c+d x) a^2+a^2\right ) \sin ^3(c+d x)dx\right )}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \left (2 a^2 \int \sin (c+d x)^2dx+\int \frac {\csc (c+d x)^2 a^2+a^2}{\csc (c+d x)^3}dx\right )}{a}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {A \left (\int \frac {\csc (c+d x)^2 a^2+a^2}{\csc (c+d x)^3}dx+2 a^2 \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )}{a}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {A \left (\int \frac {\csc (c+d x)^2 a^2+a^2}{\csc (c+d x)^3}dx+2 a^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )}{a}\) |
\(\Big \downarrow \) 4532 |
\(\displaystyle \frac {A \left (\int \sin (c+d x) \left (\sin ^2(c+d x) a^2+a^2\right )dx+2 a^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \left (\int \sin (c+d x) \left (\sin (c+d x)^2 a^2+a^2\right )dx+2 a^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )}{a}\) |
\(\Big \downarrow \) 3492 |
\(\displaystyle \frac {A \left (2 a^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\int \left (2 a^2-a^2 \cos ^2(c+d x)\right )d\cos (c+d x)}{d}\right )}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {A \left (2 a^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {2 a^2 \cos (c+d x)-\frac {1}{3} a^2 \cos ^3(c+d x)}{d}\right )}{a}\) |
Input:
Int[(a + a*Csc[c + d*x])*(A + A*Csc[c + d*x])*Sin[c + d*x]^3,x]
Output:
(A*(-((2*a^2*Cos[c + d*x] - (a^2*Cos[c + d*x]^3)/3)/d) + 2*a^2*(x/2 - (Cos [c + d*x]*Sin[c + d*x])/(2*d))))/a
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2 ), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[2*a*(b/d) Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[ {e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]
Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78
method | result | size |
parallelrisch | \(\frac {A a \left (12 d x -21 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )-6 \sin \left (2 d x +2 c \right )-20\right )}{12 d}\) | \(42\) |
risch | \(a A x -\frac {7 a A \cos \left (d x +c \right )}{4 d}+\frac {A a \cos \left (3 d x +3 c \right )}{12 d}-\frac {A a \sin \left (2 d x +2 c \right )}{2 d}\) | \(51\) |
derivativedivides | \(\frac {-A \cos \left (d x +c \right ) a +2 A a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {A a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}}{d}\) | \(62\) |
default | \(\frac {-A \cos \left (d x +c \right ) a +2 A a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {A a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}}{d}\) | \(62\) |
parts | \(-\frac {A a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}-\frac {a A \cos \left (d x +c \right )}{d}+\frac {2 A a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(67\) |
norman | \(\frac {a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-\frac {2 A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {8 A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {10 A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d}-\frac {2 A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+3 a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+3 a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(177\) |
Input:
int((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c)^3,x,method=_RETURNVERBOSE )
Output:
1/12*A*a*(12*d*x-21*cos(d*x+c)+cos(3*d*x+3*c)-6*sin(2*d*x+2*c)-20)/d
Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin ^3(c+d x) \, dx=\frac {A a \cos \left (d x + c\right )^{3} + 3 \, A a d x - 3 \, A a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, A a \cos \left (d x + c\right )}{3 \, d} \] Input:
integrate((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c)^3,x, algorithm="fri cas")
Output:
1/3*(A*a*cos(d*x + c)^3 + 3*A*a*d*x - 3*A*a*cos(d*x + c)*sin(d*x + c) - 6* A*a*cos(d*x + c))/d
\[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin ^3(c+d x) \, dx=A a \left (\int 2 \sin ^{3}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c)**3,x)
Output:
A*a*(Integral(2*sin(c + d*x)**3*csc(c + d*x), x) + Integral(sin(c + d*x)** 3*csc(c + d*x)**2, x) + Integral(sin(c + d*x)**3, x))
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.11 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin ^3(c+d x) \, dx=\frac {2 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} A a + 3 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 6 \, A a \cos \left (d x + c\right )}{6 \, d} \] Input:
integrate((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c)^3,x, algorithm="max ima")
Output:
1/6*(2*(cos(d*x + c)^3 - 3*cos(d*x + c))*A*a + 3*(2*d*x + 2*c - sin(2*d*x + 2*c))*A*a - 6*A*a*cos(d*x + c))/d
Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.76 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin ^3(c+d x) \, dx=\frac {3 \, {\left (d x + c\right )} A a + \frac {2 \, {\left (3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, A a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \] Input:
integrate((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c)^3,x, algorithm="gia c")
Output:
1/3*(3*(d*x + c)*A*a + 2*(3*A*a*tan(1/2*d*x + 1/2*c)^5 - 3*A*a*tan(1/2*d*x + 1/2*c)^4 - 12*A*a*tan(1/2*d*x + 1/2*c)^2 - 3*A*a*tan(1/2*d*x + 1/2*c) - 5*A*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 16.79 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.83 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin ^3(c+d x) \, dx=A\,a\,x-\frac {-2\,A\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (A\,a\,\left (3\,c+3\,d\,x\right )-\frac {A\,a\,\left (9\,c+9\,d\,x-6\right )}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (A\,a\,\left (3\,c+3\,d\,x\right )-\frac {A\,a\,\left (9\,c+9\,d\,x-24\right )}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,A\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {A\,a\,\left (3\,c+3\,d\,x\right )}{3}-\frac {A\,a\,\left (3\,c+3\,d\,x-10\right )}{3}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \] Input:
int(sin(c + d*x)^3*(A + A/sin(c + d*x))*(a + a/sin(c + d*x)),x)
Output:
A*a*x - (tan(c/2 + (d*x)/2)^4*(A*a*(3*c + 3*d*x) - (A*a*(9*c + 9*d*x - 6)) /3) + tan(c/2 + (d*x)/2)^2*(A*a*(3*c + 3*d*x) - (A*a*(9*c + 9*d*x - 24))/3 ) - 2*A*a*tan(c/2 + (d*x)/2)^5 + (A*a*(3*c + 3*d*x))/3 - (A*a*(3*c + 3*d*x - 10))/3 + 2*A*a*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^2 + 1)^3)
Time = 0.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin ^3(c+d x) \, dx=\frac {a^{2} \left (-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right )-5 \cos \left (d x +c \right )+3 d x +5\right )}{3 d} \] Input:
int((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c)^3,x)
Output:
(a**2*( - cos(c + d*x)*sin(c + d*x)**2 - 3*cos(c + d*x)*sin(c + d*x) - 5*c os(c + d*x) + 3*d*x + 5))/(3*d)