Integrand size = 32, antiderivative size = 105 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A \text {arctanh}(\cos (c+d x))}{4 d}-\frac {2 a^3 A \cot ^3(c+d x)}{3 d}-\frac {a^3 A \cot ^5(c+d x)}{5 d}+\frac {a^3 A \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d} \] Output:
1/4*a^3*A*arctanh(cos(d*x+c))/d-2/3*a^3*A*cot(d*x+c)^3/d-1/5*a^3*A*cot(d*x +c)^5/d+1/4*a^3*A*cot(d*x+c)*csc(d*x+c)/d-1/2*a^3*A*cot(d*x+c)*csc(d*x+c)^ 3/d
Leaf count is larger than twice the leaf count of optimal. \(268\) vs. \(2(105)=210\).
Time = 0.47 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.55 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=a^3 A \left (\frac {7 \cot \left (\frac {1}{2} (c+d x)\right )}{30 d}+\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{16 d}-\frac {19 \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{480 d}-\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^4\left (\frac {1}{2} (c+d x)\right )}{160 d}+\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{16 d}+\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {7 \tan \left (\frac {1}{2} (c+d x)\right )}{30 d}+\frac {19 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{480 d}+\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{160 d}\right ) \] Input:
Integrate[Csc[c + d*x]^6*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]
Output:
a^3*A*((7*Cot[(c + d*x)/2])/(30*d) + Csc[(c + d*x)/2]^2/(16*d) - (19*Cot[( c + d*x)/2]*Csc[(c + d*x)/2]^2)/(480*d) - Csc[(c + d*x)/2]^4/(32*d) - (Cot [(c + d*x)/2]*Csc[(c + d*x)/2]^4)/(160*d) + Log[Cos[(c + d*x)/2]]/(4*d) - Log[Sin[(c + d*x)/2]]/(4*d) - Sec[(c + d*x)/2]^2/(16*d) + Sec[(c + d*x)/2] ^4/(32*d) - (7*Tan[(c + d*x)/2])/(30*d) + (19*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(480*d) + (Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])/(160*d))
Time = 0.51 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.02, 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.
| \(\displaystyle \int \csc ^6(c+d x) (a \sin (c+d x)+a)^3 (A-A \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^3 (A-A \sin (c+d x))}{\sin (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3429 |
| \(\displaystyle a^3 A^3 \int \frac {\cot ^6(c+d x)}{(A-A \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 A^3 \int \frac {1}{(A-A \sin (c+d x))^2 \tan (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3188 |
| \(\displaystyle \frac {a^3 \int \left (A^4 \csc ^6(c+d x)+2 A^4 \csc ^5(c+d x)-2 A^4 \csc ^3(c+d x)-A^4 \csc ^2(c+d x)\right )dx}{A^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 \left (\frac {A^4 \text {arctanh}(\cos (c+d x))}{4 d}-\frac {A^4 \cot ^5(c+d x)}{5 d}-\frac {2 A^4 \cot ^3(c+d x)}{3 d}-\frac {A^4 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {A^4 \cot (c+d x) \csc (c+d x)}{4 d}\right )}{A^3}\) |
Input:
Int[Csc[c + d*x]^6*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]
Output:
(a^3*((A^4*ArcTanh[Cos[c + d*x]])/(4*d) - (2*A^4*Cot[c + d*x]^3)/(3*d) - ( A^4*Cot[c + d*x]^5)/(5*d) + (A^4*Cot[c + d*x]*Csc[c + d*x])/(4*d) - (A^4*C ot[c + d*x]*Csc[c + d*x]^3)/(2*d)))/A^3
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 = 1.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.15
| method | result | size |
| parallelrisch | \(-\frac {A \left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+5 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {25 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-30 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+40 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) a^{3}}{160 d}\) | \(121\) |
| derivativedivides | \(\frac {a^{3} A \cot \left (d x +c \right )-2 a^{3} A \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a^{3} A \left (\left (-\frac {\csc \left (d x +c \right )^{3}}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a^{3} A \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(140\) |
| default | \(\frac {a^{3} A \cot \left (d x +c \right )-2 a^{3} A \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a^{3} A \left (\left (-\frac {\csc \left (d x +c \right )^{3}}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a^{3} A \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(140\) |
| risch | \(-\frac {a^{3} A \left (-60 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}+240 i {\mathrm e}^{6 i \left (d x +c \right )}+90 \,{\mathrm e}^{7 i \left (d x +c \right )}-40 i {\mathrm e}^{4 i \left (d x +c \right )}+80 i {\mathrm e}^{2 i \left (d x +c \right )}-90 \,{\mathrm e}^{3 i \left (d x +c \right )}-28 i-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {a^{3} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d}+\frac {a^{3} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d}\) | \(161\) |
Input:
int(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x,method=_RETURNVERBO SE)
Output:
-1/160*A*(cot(1/2*d*x+1/2*c)^5-tan(1/2*d*x+1/2*c)^5+5*cot(1/2*d*x+1/2*c)^4 -5*tan(1/2*d*x+1/2*c)^4+25/3*cot(1/2*d*x+1/2*c)^3-25/3*tan(1/2*d*x+1/2*c)^ 3-30*cot(1/2*d*x+1/2*c)+30*tan(1/2*d*x+1/2*c)+40*ln(tan(1/2*d*x+1/2*c)))*a ^3/d
Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (95) = 190\).
Time = 0.09 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.91 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {56 \, A a^{3} \cos \left (d x + c\right )^{5} - 80 \, A a^{3} \cos \left (d x + c\right )^{3} + 15 \, {\left (A a^{3} \cos \left (d x + c\right )^{4} - 2 \, A a^{3} \cos \left (d x + c\right )^{2} + A a^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (A a^{3} \cos \left (d x + c\right )^{4} - 2 \, A a^{3} \cos \left (d x + c\right )^{2} + A a^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (A a^{3} \cos \left (d x + c\right )^{3} + A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:
integrate(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="f ricas")
Output:
1/120*(56*A*a^3*cos(d*x + c)^5 - 80*A*a^3*cos(d*x + c)^3 + 15*(A*a^3*cos(d *x + c)^4 - 2*A*a^3*cos(d*x + c)^2 + A*a^3)*log(1/2*cos(d*x + c) + 1/2)*si n(d*x + c) - 15*(A*a^3*cos(d*x + c)^4 - 2*A*a^3*cos(d*x + c)^2 + A*a^3)*lo g(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 30*(A*a^3*cos(d*x + c)^3 + A*a^3 *cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)* sin(d*x + c))
Timed out. \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)**6*(a+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.67 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {15 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {120 \, A a^{3}}{\tan \left (d x + c\right )} - \frac {8 \, {\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} A a^{3}}{\tan \left (d x + c\right )^{5}}}{120 \, d} \] Input:
integrate(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="m axima")
Output:
1/120*(15*A*a^3*(2*(3*cos(d*x + c)^3 - 5*cos(d*x + c))/(cos(d*x + c)^4 - 2 *cos(d*x + c)^2 + 1) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) - 60*A*a^3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) + 120*A*a^3/tan(d*x + c) - 8*(15*tan(d*x + c)^4 + 1 0*tan(d*x + c)^2 + 3)*A*a^3/tan(d*x + c)^5)/d
Time = 0.27 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.66 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 90 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {274 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 90 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 25 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \] Input:
integrate(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="g iac")
Output:
1/480*(3*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*A*a^3*tan(1/2*d*x + 1/2*c)^4 + 25*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 120*A*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 90*A*a^3*tan(1/2*d*x + 1/2*c) + (274*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 90* A*a^3*tan(1/2*d*x + 1/2*c)^4 - 25*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 15*A*a^3* tan(1/2*d*x + 1/2*c) - 3*A*a^3)/tan(1/2*d*x + 1/2*c)^5)/d
Time = 34.91 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.32 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {A\,a^3\,\left (3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-15\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-25\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+90\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-90\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+25\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+120\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}{480\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:
int(((A - A*sin(c + d*x))*(a + a*sin(c + d*x))^3)/sin(c + d*x)^6,x)
Output:
-(A*a^3*(3*cos(c/2 + (d*x)/2)^10 - 3*sin(c/2 + (d*x)/2)^10 - 15*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^9 + 15*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2) - 25*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 + 90*cos(c/2 + (d*x)/2)^4* sin(c/2 + (d*x)/2)^6 - 90*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 25*c os(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 120*log(sin(c/2 + (d*x)/2)/cos( c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5))/(480*d*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)
Time = 0.18 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.02 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^{4} \left (28 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )-12 \cos \left (d x +c \right )-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}\right )}{60 \sin \left (d x +c \right )^{5} d} \] Input:
int(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x)
Output:
(a**4*(28*cos(c + d*x)*sin(c + d*x)**4 + 15*cos(c + d*x)*sin(c + d*x)**3 - 16*cos(c + d*x)*sin(c + d*x)**2 - 30*cos(c + d*x)*sin(c + d*x) - 12*cos(c + d*x) - 15*log(tan((c + d*x)/2))*sin(c + d*x)**5))/(60*sin(c + d*x)**5*d )