Integrand size = 32, antiderivative size = 79 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {1}{2} a^3 A x-\frac {2 a^3 A \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^3 A \cos (c+d x)}{d}-\frac {a^3 A \cot (c+d x)}{d}+\frac {a^3 A \cos (c+d x) \sin (c+d x)}{2 d} \] Output:
-1/2*a^3*A*x-2*a^3*A*arctanh(cos(d*x+c))/d+2*a^3*A*cos(d*x+c)/d-a^3*A*cot( d*x+c)/d+1/2*a^3*A*cos(d*x+c)*sin(d*x+c)/d
Time = 0.45 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {a^3 A (2 c+2 d x+8 \text {arctanh}(\cos (c+d x))-8 \cos (c) \cos (d x)+4 \cot (c+d x)+8 \sin (c) \sin (d x)-\sin (2 (c+d x)))}{4 d} \] Input:
Integrate[Csc[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]
Output:
-1/4*(a^3*A*(2*c + 2*d*x + 8*ArcTanh[Cos[c + d*x]] - 8*Cos[c]*Cos[d*x] + 4 *Cot[c + d*x] + 8*Sin[c]*Sin[d*x] - Sin[2*(c + d*x)]))/d
Time = 0.44 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3042, 3429, 3042, 3188, 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.
| \(\displaystyle \int \csc ^2(c+d x) (a \sin (c+d x)+a)^3 (A-A \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^3 (A-A \sin (c+d x))}{\sin (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3429 |
| \(\displaystyle a A \int \cot ^2(c+d x) (\sin (c+d x) a+a)^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a A \int \frac {(\sin (c+d x) a+a)^2}{\tan (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3188 |
| \(\displaystyle \frac {A \int \left (\csc ^2(c+d x) a^4-\sin ^2(c+d x) a^4+2 \csc (c+d x) a^4-2 \sin (c+d x) a^4\right )dx}{a}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {A \left (-\frac {2 a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^4 \cos (c+d x)}{d}-\frac {a^4 \cot (c+d x)}{d}+\frac {a^4 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {a^4 x}{2}\right )}{a}\) |
Input:
Int[Csc[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]
Output:
(A*(-1/2*(a^4*x) - (2*a^4*ArcTanh[Cos[c + d*x]])/d + (2*a^4*Cos[c + d*x])/ d - (a^4*Cot[c + d*x])/d + (a^4*Cos[c + d*x]*Sin[c + d*x])/(2*d)))/a
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_ ), x_Symbol] :> Simp[a^p Int[ExpandIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])
Int[sin[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^n*c^n Int[Tan[e + f*x]^p*(a + b*Sin[e + f*x])^(m - n), x], x] /; FreeQ[{a, b, c , d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[p + 2*n, 0] && IntegerQ[n]
Time = 0.60 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.01
| method | result | size |
| derivativedivides | \(\frac {-a^{3} A \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \cos \left (d x +c \right ) a^{3}+2 a^{3} A \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-a^{3} A \cot \left (d x +c \right )}{d}\) | \(80\) |
| default | \(\frac {-a^{3} A \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \cos \left (d x +c \right ) a^{3}+2 a^{3} A \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-a^{3} A \cot \left (d x +c \right )}{d}\) | \(80\) |
| parallelrisch | \(-\frac {A \,a^{3} \left (-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\cos \left (d x +c \right )-\frac {\cos \left (2 d x +2 c \right )}{2}+4 \sin \left (d x +c \right )-\frac {5}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+d x \right )}{2 d}\) | \(84\) |
| risch | \(-\frac {a^{3} A x}{2}-\frac {i a^{3} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a^{3} A \,{\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a^{3} A \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i a^{3} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{3} A}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 a^{3} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {2 a^{3} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\) | \(146\) |
| norman | \(\frac {\frac {4 a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a^{3} A}{2 d}+\frac {4 a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {12 a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {12 a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d}+\frac {a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 d}+\frac {a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 d}-\frac {a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-2 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-3 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-2 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-\frac {a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {2 a^{3} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(284\) |
Input:
int(csc(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x,method=_RETURNVERBO SE)
Output:
1/d*(-a^3*A*(-1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c)+2*A*cos(d*x+c)*a^3+ 2*a^3*A*ln(csc(d*x+c)-cot(d*x+c))-a^3*A*cot(d*x+c))
Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.41 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {A a^{3} \cos \left (d x + c\right )^{3} + 2 \, A a^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, A a^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + A a^{3} \cos \left (d x + c\right ) + {\left (A a^{3} d x - 4 \, A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \sin \left (d x + c\right )} \] Input:
integrate(csc(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="f ricas")
Output:
-1/2*(A*a^3*cos(d*x + c)^3 + 2*A*a^3*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 2*A*a^3*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + A*a^3*cos(d*x + c) + (A*a^3*d*x - 4*A*a^3*cos(d*x + c))*sin(d*x + c))/(d*sin(d*x + c))
\[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=- A a^{3} \left (\int \left (- 2 \sin {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\right )\, dx + \int 2 \sin ^{3}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \left (- \csc ^{2}{\left (c + d x \right )}\right )\, dx\right ) \] Input:
integrate(csc(d*x+c)**2*(a+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)
Output:
-A*a**3*(Integral(-2*sin(c + d*x)*csc(c + d*x)**2, x) + Integral(2*sin(c + d*x)**3*csc(c + d*x)**2, x) + Integral(sin(c + d*x)**4*csc(c + d*x)**2, x ) + Integral(-csc(c + d*x)**2, x))
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 4 \, A a^{3} {\left (\log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 8 \, A a^{3} \cos \left (d x + c\right ) + \frac {4 \, A a^{3}}{\tan \left (d x + c\right )}}{4 \, d} \] Input:
integrate(csc(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="m axima")
Output:
-1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*A*a^3 + 4*A*a^3*(log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) - 8*A*a^3*cos(d*x + c) + 4*A*a^3/tan(d*x + c) )/d
Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (75) = 150\).
Time = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.94 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {{\left (d x + c\right )} A a^{3} - 4 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:
integrate(csc(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="g iac")
Output:
-1/2*((d*x + c)*A*a^3 - 4*A*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - A*a^3*tan (1/2*d*x + 1/2*c) + (4*A*a^3*tan(1/2*d*x + 1/2*c) + A*a^3)/tan(1/2*d*x + 1 /2*c) + 2*(A*a^3*tan(1/2*d*x + 1/2*c)^3 - 4*A*a^3*tan(1/2*d*x + 1/2*c)^2 - A*a^3*tan(1/2*d*x + 1/2*c) - 4*A*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 40.43 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.86 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {A\,a^3\,\mathrm {atan}\left (\frac {A^2\,a^6}{4\,A^2\,a^6+A^2\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {4\,A^2\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,A^2\,a^6+A^2\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {3\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-8\,A\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+A\,a^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {2\,A\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {A\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \] Input:
int(((A - A*sin(c + d*x))*(a + a*sin(c + d*x))^3)/sin(c + d*x)^2,x)
Output:
(A*a^3*atan((A^2*a^6)/(4*A^2*a^6 + A^2*a^6*tan(c/2 + (d*x)/2)) - (4*A^2*a^ 6*tan(c/2 + (d*x)/2))/(4*A^2*a^6 + A^2*a^6*tan(c/2 + (d*x)/2))))/d - (A*a^ 3 - 8*A*a^3*tan(c/2 + (d*x)/2) - 8*A*a^3*tan(c/2 + (d*x)/2)^3 + 3*A*a^3*ta n(c/2 + (d*x)/2)^4)/(d*(2*tan(c/2 + (d*x)/2) + 4*tan(c/2 + (d*x)/2)^3 + 2* tan(c/2 + (d*x)/2)^5)) + (2*A*a^3*log(tan(c/2 + (d*x)/2)))/d + (A*a^3*tan( c/2 + (d*x)/2))/(2*d)
Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^{4} \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+4 \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \cos \left (d x +c \right )+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )-\sin \left (d x +c \right ) d x -4 \sin \left (d x +c \right )\right )}{2 \sin \left (d x +c \right ) d} \] Input:
int(csc(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x)
Output:
(a**4*(cos(c + d*x)*sin(c + d*x)**2 + 4*cos(c + d*x)*sin(c + d*x) - 2*cos( c + d*x) + 4*log(tan((c + d*x)/2))*sin(c + d*x) - sin(c + d*x)*d*x - 4*sin (c + d*x)))/(2*sin(c + d*x)*d)