Integrand size = 32, antiderivative size = 153 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {18 A \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {10 A \cot (c+d x)}{a^3 d}-\frac {A \cot ^3(c+d x)}{3 a^3 d}+\frac {2 A \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac {13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {93 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))} \] Output:
18*A*arctanh(cos(d*x+c))/a^3/d-10*A*cot(d*x+c)/a^3/d-1/3*A*cot(d*x+c)^3/a^ 3/d+2*A*cot(d*x+c)*csc(d*x+c)/a^3/d-2/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^ 3-13/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^2-93/5*A*cos(d*x+c)/a^3/d/(1+sin( d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(348\) vs. \(2(153)=306\).
Time = 7.09 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.27 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {A \left (-\frac {29 \cot \left (\frac {1}{2} (c+d x)\right )}{6 d}+\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{2 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 d}+\frac {18 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {18 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{2 d}+\frac {4 \sin \left (\frac {1}{2} (c+d x)\right )}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}-\frac {2}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {26 \sin \left (\frac {1}{2} (c+d x)\right )}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {13}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {186 \sin \left (\frac {1}{2} (c+d x)\right )}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {29 \tan \left (\frac {1}{2} (c+d x)\right )}{6 d}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{24 d}\right )}{a^3} \] Input:
Integrate[(Csc[c + d*x]^4*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]
Output:
(A*((-29*Cot[(c + d*x)/2])/(6*d) + Csc[(c + d*x)/2]^2/(2*d) - (Cot[(c + d* x)/2]*Csc[(c + d*x)/2]^2)/(24*d) + (18*Log[Cos[(c + d*x)/2]])/d - (18*Log[ Sin[(c + d*x)/2]])/d - Sec[(c + d*x)/2]^2/(2*d) + (4*Sin[(c + d*x)/2])/(5* d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5) - 2/(5*d*(Cos[(c + d*x)/2] + Si n[(c + d*x)/2])^4) + (26*Sin[(c + d*x)/2])/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 13/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (186*S in[(c + d*x)/2])/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (29*Tan[(c + d*x)/2])/(6*d) + (Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(24*d)))/a^3
Time = 0.51 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3042, 3445, 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.
| \(\displaystyle \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A-A \sin (c+d x)}{\sin (c+d x)^4 (a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3445 |
| \(\displaystyle \int \left (\frac {16 A}{a^3 (\sin (c+d x)+1)}+\frac {7 A}{a^3 (\sin (c+d x)+1)^2}+\frac {2 A}{a^3 (\sin (c+d x)+1)^3}+\frac {A \csc ^4(c+d x)}{a^3}-\frac {4 A \csc ^3(c+d x)}{a^3}+\frac {9 A \csc ^2(c+d x)}{a^3}-\frac {16 A \csc (c+d x)}{a^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {18 A \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {A \cot ^3(c+d x)}{3 a^3 d}-\frac {10 A \cot (c+d x)}{a^3 d}-\frac {93 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}-\frac {13 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}-\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}+\frac {2 A \cot (c+d x) \csc (c+d x)}{a^3 d}\) |
Input:
Int[(Csc[c + d*x]^4*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]
Output:
(18*A*ArcTanh[Cos[c + d*x]])/(a^3*d) - (10*A*Cot[c + d*x])/(a^3*d) - (A*Co t[c + d*x]^3)/(3*a^3*d) + (2*A*Cot[c + d*x]*Csc[c + d*x])/(a^3*d) - (2*A*C os[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) - (13*A*Cos[c + d*x])/(5*a^3*d *(1 + Sin[c + d*x])^2) - (93*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x]))
Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[si n[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; FreeQ[{ a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ [m] && IntegerQ[n]
Time = 0.98 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {A \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+39 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {39}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-144 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {128}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {64}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {160}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {176}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {400}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}\right )}{8 d \,a^{3}}\) | \(174\) |
| default | \(\frac {A \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+39 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {39}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-144 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {128}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {64}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {160}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {176}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {400}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}\right )}{8 d \,a^{3}}\) | \(174\) |
| parallelrisch | \(-\frac {A \left (432 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-67 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+2637 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+8484 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-7 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+11384 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+67 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+7297 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {9934}{5}\right )}{24 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(174\) |
| risch | \(-\frac {4 \left (675 i A \,{\mathrm e}^{9 i \left (d x +c \right )}+135 \,{\mathrm e}^{10 i \left (d x +c \right )} A -3150 i A \,{\mathrm e}^{7 i \left (d x +c \right )}-1710 A \,{\mathrm e}^{8 i \left (d x +c \right )}+5180 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+4572 A \,{\mathrm e}^{6 i \left (d x +c \right )}-3590 i A \,{\mathrm e}^{3 i \left (d x +c \right )}-4906 A \,{\mathrm e}^{4 i \left (d x +c \right )}+925 i A \,{\mathrm e}^{i \left (d x +c \right )}+2081 A \,{\mathrm e}^{2 i \left (d x +c \right )}-212 A \right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} d \,a^{3}}+\frac {18 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}-\frac {18 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}\) | \(208\) |
| norman | \(\frac {-\frac {A}{24 a d}+\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 a d}-\frac {17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{6 a d}+\frac {17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 a d}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{24 a d}+\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{24 a d}-\frac {66469 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{120 a d}-\frac {8407 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{24 a d}-\frac {3479 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 a d}-\frac {3919 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{6 a d}-\frac {2443 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 a d}-\frac {845 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{8 a d}-\frac {411 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{5 a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {18 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}\) | \(312\) |
Input:
int(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBO SE)
Output:
1/8/d*A/a^3*(1/3*tan(1/2*d*x+1/2*c)^3-4*tan(1/2*d*x+1/2*c)^2+39*tan(1/2*d* x+1/2*c)-1/3/tan(1/2*d*x+1/2*c)^3+4/tan(1/2*d*x+1/2*c)^2-39/tan(1/2*d*x+1/ 2*c)-144*ln(tan(1/2*d*x+1/2*c))-128/5/(tan(1/2*d*x+1/2*c)+1)^5+64/(tan(1/2 *d*x+1/2*c)+1)^4-160/(tan(1/2*d*x+1/2*c)+1)^3+176/(tan(1/2*d*x+1/2*c)+1)^2 -400/(tan(1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 594 vs. \(2 (145) = 290\).
Time = 0.10 (sec) , antiderivative size = 594, normalized size of antiderivative = 3.88 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="f ricas")
Output:
1/15*(424*A*cos(d*x + c)^6 + 1002*A*cos(d*x + c)^5 - 944*A*cos(d*x + c)^4 - 2074*A*cos(d*x + c)^3 + 531*A*cos(d*x + c)^2 + 1077*A*cos(d*x + c) + 135 *(A*cos(d*x + c)^6 - 2*A*cos(d*x + c)^5 - 6*A*cos(d*x + c)^4 + 4*A*cos(d*x + c)^3 + 9*A*cos(d*x + c)^2 - 2*A*cos(d*x + c) - (A*cos(d*x + c)^5 + 3*A* cos(d*x + c)^4 - 3*A*cos(d*x + c)^3 - 7*A*cos(d*x + c)^2 + 2*A*cos(d*x + c ) + 4*A)*sin(d*x + c) - 4*A)*log(1/2*cos(d*x + c) + 1/2) - 135*(A*cos(d*x + c)^6 - 2*A*cos(d*x + c)^5 - 6*A*cos(d*x + c)^4 + 4*A*cos(d*x + c)^3 + 9* A*cos(d*x + c)^2 - 2*A*cos(d*x + c) - (A*cos(d*x + c)^5 + 3*A*cos(d*x + c) ^4 - 3*A*cos(d*x + c)^3 - 7*A*cos(d*x + c)^2 + 2*A*cos(d*x + c) + 4*A)*sin (d*x + c) - 4*A)*log(-1/2*cos(d*x + c) + 1/2) + (424*A*cos(d*x + c)^5 - 57 8*A*cos(d*x + c)^4 - 1522*A*cos(d*x + c)^3 + 552*A*cos(d*x + c)^2 + 1083*A *cos(d*x + c) + 6*A)*sin(d*x + c) - 6*A)/(a^3*d*cos(d*x + c)^6 - 2*a^3*d*c os(d*x + c)^5 - 6*a^3*d*cos(d*x + c)^4 + 4*a^3*d*cos(d*x + c)^3 + 9*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)^5 + 3*a^3*d*cos(d*x + c)^4 - 3*a^3*d*cos(d*x + c)^3 - 7*a^3*d*cos(d*x + c)^2 + 2*a^3*d*cos(d*x + c) + 4*a^3*d)*sin(d*x + c))
\[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=- \frac {A \left (\int \left (- \frac {\csc ^{4}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\right )\, dx + \int \frac {\sin {\left (c + d x \right )} \csc ^{4}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx\right )}{a^{3}} \] Input:
integrate(csc(d*x+c)**4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)
Output:
-A*(Integral(-csc(c + d*x)**4/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin (c + d*x) + 1), x) + Integral(sin(c + d*x)*csc(c + d*x)**4/(sin(c + d*x)** 3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x))/a**3
Leaf count of result is larger than twice the leaf count of optimal. 706 vs. \(2 (145) = 290\).
Time = 0.05 (sec) , antiderivative size = 706, normalized size of antiderivative = 4.61 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="m axima")
Output:
-1/120*(A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 2782*sin(d*x + c)^2/(cos (d*x + c) + 1)^2 + 9410*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 13645*sin(d* x + c)^4/(cos(d*x + c) + 1)^4 + 9285*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2580*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 15)/(a^3*sin(d*x + c)^2/(cos(d *x + c) + 1)^2 + 5*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 10*a^3*sin(d* x + c)^4/(cos(d*x + c) + 1)^4 + 10*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) - 15*(12*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(c os(d*x + c) + 1)^2)/a^3 + 780*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - A*((20*sin(d*x + c)/(cos(d*x + c) + 1) - 230*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4777*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 15785*sin(d*x + c)^4/( cos(d*x + c) + 1)^4 - 22390*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 14940*si n(d*x + c)^6/(cos(d*x + c) + 1)^6 - 4005*sin(d*x + c)^7/(cos(d*x + c) + 1) ^7 - 5)/(a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*a^3*sin(d*x + c)^4/(c os(d*x + c) + 1)^4 + 10*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 10*a^3*s in(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 5*(81*sin(d*x + c)/(cos( d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^3/(co s(d*x + c) + 1)^3)/a^3 - 1380*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3))/d
Time = 0.28 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.39 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {2160 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {5 \, {\left (792 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 117 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A\right )}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} + \frac {48 \, {\left (125 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 445 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 635 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 415 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 108 \, A\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}} - \frac {5 \, {\left (A a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 117 \, A a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{9}}}{120 \, d} \] Input:
integrate(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="g iac")
Output:
-1/120*(2160*A*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 5*(792*A*tan(1/2*d*x + 1/2*c)^3 - 117*A*tan(1/2*d*x + 1/2*c)^2 + 12*A*tan(1/2*d*x + 1/2*c) - A)/ (a^3*tan(1/2*d*x + 1/2*c)^3) + 48*(125*A*tan(1/2*d*x + 1/2*c)^4 + 445*A*ta n(1/2*d*x + 1/2*c)^3 + 635*A*tan(1/2*d*x + 1/2*c)^2 + 415*A*tan(1/2*d*x + 1/2*c) + 108*A)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5) - 5*(A*a^6*tan(1/2*d*x + 1/2*c)^3 - 12*A*a^6*tan(1/2*d*x + 1/2*c)^2 + 117*A*a^6*tan(1/2*d*x + 1/2 *c))/a^9)/d
Time = 37.06 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.05 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {A\,\left (335\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-35\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+7559\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+24610\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+33170\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+18670\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+1310\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-2375\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-335\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+2160\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+10800\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+21600\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+21600\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10800\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2160\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+5\right )}{120\,a^3\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \] Input:
int((A - A*sin(c + d*x))/(sin(c + d*x)^4*(a + a*sin(c + d*x))^3),x)
Output:
-(A*(335*tan(c/2 + (d*x)/2)^2 - 35*tan(c/2 + (d*x)/2) + 7559*tan(c/2 + (d* x)/2)^3 + 24610*tan(c/2 + (d*x)/2)^4 + 33170*tan(c/2 + (d*x)/2)^5 + 18670* tan(c/2 + (d*x)/2)^6 + 1310*tan(c/2 + (d*x)/2)^7 - 2375*tan(c/2 + (d*x)/2) ^8 - 335*tan(c/2 + (d*x)/2)^9 + 35*tan(c/2 + (d*x)/2)^10 - 5*tan(c/2 + (d* x)/2)^11 + 2160*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^3 + 10800*log(t an(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^4 + 21600*log(tan(c/2 + (d*x)/2))*ta n(c/2 + (d*x)/2)^5 + 21600*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^6 + 10800*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^7 + 2160*log(tan(c/2 + (d *x)/2))*tan(c/2 + (d*x)/2)^8 + 5))/(120*a^3*d*tan(c/2 + (d*x)/2)^3*(tan(c/ 2 + (d*x)/2) + 1)^5)
Time = 0.17 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.30 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {-2160 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-10800 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-21600 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-21600 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-10800 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2160 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}-35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+335 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+2637 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-16050 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-30550 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-23300 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-7297 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-335 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-5}{120 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \] Input:
int(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x)
Output:
( - 2160*log(tan((c + d*x)/2))*tan((c + d*x)/2)**8 - 10800*log(tan((c + d* x)/2))*tan((c + d*x)/2)**7 - 21600*log(tan((c + d*x)/2))*tan((c + d*x)/2)* *6 - 21600*log(tan((c + d*x)/2))*tan((c + d*x)/2)**5 - 10800*log(tan((c + d*x)/2))*tan((c + d*x)/2)**4 - 2160*log(tan((c + d*x)/2))*tan((c + d*x)/2) **3 + 5*tan((c + d*x)/2)**11 - 35*tan((c + d*x)/2)**10 + 335*tan((c + d*x) /2)**9 + 2637*tan((c + d*x)/2)**8 - 16050*tan((c + d*x)/2)**6 - 30550*tan( (c + d*x)/2)**5 - 23300*tan((c + d*x)/2)**4 - 7297*tan((c + d*x)/2)**3 - 3 35*tan((c + d*x)/2)**2 + 35*tan((c + d*x)/2) - 5)/(120*tan((c + d*x)/2)**3 *a**2*d*(tan((c + d*x)/2)**5 + 5*tan((c + d*x)/2)**4 + 10*tan((c + d*x)/2) **3 + 10*tan((c + d*x)/2)**2 + 5*tan((c + d*x)/2) + 1))