Integrand size = 32, antiderivative size = 129 \[ \int \frac {\sin ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {19 A x}{2 a^3}-\frac {4 A \cos (c+d x)}{a^3 d}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {41 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}-\frac {199 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))} \] Output:
-19/2*A*x/a^3-4*A*cos(d*x+c)/a^3/d+1/2*A*cos(d*x+c)*sin(d*x+c)/a^3/d-2/5*A *cos(d*x+c)/a^3/d/(1+sin(d*x+c))^3+41/15*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c)) ^2-199/15*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 6.18 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.64 \[ \int \frac {\sin ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {A \sec (c+d x) \sqrt {1-\sin (c+d x)} \left (140 \sqrt {2} \sqrt {a} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (1+\sin (c+d x))-360 \arcsin \left (\frac {\sqrt {a (1+\sin (c+d x))}}{\sqrt {2} \sqrt {a}}\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \sqrt {a (1+\sin (c+d x))}+\sqrt {a} \sqrt {1-\sin (c+d x)} \left (-308-639 \sin (c+d x)-433 \sin ^2(c+d x)-75 \sin ^3(c+d x)+15 \sin ^4(c+d x)\right )\right )}{30 a^{7/2} d (1+\sin (c+d x))^2} \] Input:
Integrate[(Sin[c + d*x]^4*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]
Output:
(A*Sec[c + d*x]*Sqrt[1 - Sin[c + d*x]]*(140*Sqrt[2]*Sqrt[a]*Hypergeometric 2F1[-3/2, -3/2, -1/2, (1 + Sin[c + d*x])/2]*(1 + Sin[c + d*x]) - 360*ArcSi n[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Sqrt[a*(1 + Sin[c + d*x])] + Sqrt[a]*Sqrt[1 - Sin[c + d*x]]* (-308 - 639*Sin[c + d*x] - 433*Sin[c + d*x]^2 - 75*Sin[c + d*x]^3 + 15*Sin [c + d*x]^4)))/(30*a^(7/2)*d*(1 + Sin[c + d*x])^2)
Time = 0.44 (sec) , antiderivative size = 129, 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.
\(\displaystyle \int \frac {\sin ^4(c+d x) (A-A \sin (c+d x))}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^4 (A-A \sin (c+d x))}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3445 |
\(\displaystyle \int \left (-\frac {A \sin ^2(c+d x)}{a^3}+\frac {4 A \sin (c+d x)}{a^3}+\frac {16 A}{a^3 (\sin (c+d x)+1)}-\frac {9 A}{a^3 (\sin (c+d x)+1)^2}+\frac {2 A}{a^3 (\sin (c+d x)+1)^3}-\frac {9 A}{a^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 A \cos (c+d x)}{a^3 d}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {199 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}+\frac {41 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}-\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac {19 A x}{2 a^3}\) |
Input:
Int[(Sin[c + d*x]^4*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]
Output:
(-19*A*x)/(2*a^3) - (4*A*Cos[c + d*x])/(a^3*d) + (A*Cos[c + d*x]*Sin[c + d *x])/(2*a^3*d) - (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) + (41*A *Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x])^2) - (199*A*Cos[c + d*x])/(15* 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 = 154, normalized size of antiderivative = 1.19
method | result | size |
derivativedivides | \(\frac {32 A \left (-\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+4}{16 \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {19 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32}-\frac {1}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {9}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\right )}{d \,a^{3}}\) | \(154\) |
default | \(\frac {32 A \left (-\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+4}{16 \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {19 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32}-\frac {1}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {9}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\right )}{d \,a^{3}}\) | \(154\) |
risch | \(-\frac {19 A x}{2 a^{3}}-\frac {i A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,a^{3}}-\frac {2 A \,{\mathrm e}^{i \left (d x +c \right )}}{a^{3} d}-\frac {2 A \,{\mathrm e}^{-i \left (d x +c \right )}}{a^{3} d}+\frac {i A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,a^{3}}-\frac {2 \left (825 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+240 A \,{\mathrm e}^{4 i \left (d x +c \right )}-755 i A \,{\mathrm e}^{i \left (d x +c \right )}-1165 A \,{\mathrm e}^{2 i \left (d x +c \right )}+199 A \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5}}\) | \(159\) |
parallelrisch | \(\frac {A \left (\left (76 d x -120\right ) \sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+\left (380 d x -128\right ) \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+\left (76 d x +\frac {3484}{15}\right ) \cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+\left (-760 d x -304\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )-380 d x \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-760 d x \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\cos \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )-\frac {2068 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )}{3}+11 \sin \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )-\sin \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )-\frac {2456 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{3}+11 \cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )\right )}{8 d \,a^{3} \left (-\sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+5 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-5 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\) | \(234\) |
norman | \(\frac {-\frac {95 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {285 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}-\frac {665 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}-\frac {1235 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a}-\frac {1919 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 a}-\frac {2565 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}-\frac {2945 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 a}-\frac {2945 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}-\frac {2565 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 a}-\frac {1919 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 a}-\frac {1235 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 a}-\frac {665 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{2 a}-\frac {285 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{2 a}-\frac {95 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{2 a}-\frac {19 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{2 a}-\frac {19 A x}{2 a}-\frac {391 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 a d}-\frac {353 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a d}-\frac {2300 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}-\frac {1308 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a d}-\frac {5599 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}-\frac {6979 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 a d}-\frac {7192 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}-\frac {2232 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{a d}-\frac {5117 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 a d}-\frac {5751 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{5 a d}-\frac {1900 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 a d}-\frac {836 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3 a d}-\frac {95 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{a d}-\frac {19 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{a d}-\frac {448 A}{15 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(596\) |
Input:
int(sin(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBO SE)
Output:
32/d*A/a^3*(-1/16*(1/2*tan(1/2*d*x+1/2*c)^3+4*tan(1/2*d*x+1/2*c)^2-1/2*tan (1/2*d*x+1/2*c)+4)/(1+tan(1/2*d*x+1/2*c)^2)^2-19/32*arctan(tan(1/2*d*x+1/2 *c))-1/10/(tan(1/2*d*x+1/2*c)+1)^5+1/4/(tan(1/2*d*x+1/2*c)+1)^4+1/24/(tan( 1/2*d*x+1/2*c)+1)^3-5/16/(tan(1/2*d*x+1/2*c)+1)^2-9/16/(tan(1/2*d*x+1/2*c) +1))
Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (119) = 238\).
Time = 0.08 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.92 \[ \int \frac {\sin ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {15 \, A \cos \left (d x + c\right )^{5} + 90 \, A \cos \left (d x + c\right )^{4} + {\left (285 \, A d x + 683 \, A\right )} \cos \left (d x + c\right )^{3} - 1140 \, A d x + {\left (855 \, A d x - 526 \, A\right )} \cos \left (d x + c\right )^{2} - 6 \, {\left (95 \, A d x + 191 \, A\right )} \cos \left (d x + c\right ) - {\left (15 \, A \cos \left (d x + c\right )^{4} - 75 \, A \cos \left (d x + c\right )^{3} + 1140 \, A d x - 19 \, {\left (15 \, A d x - 32 \, A\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (95 \, A d x + 189 \, A\right )} \cos \left (d x + c\right ) - 12 \, A\right )} \sin \left (d x + c\right ) - 12 \, A}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="f ricas")
Output:
-1/30*(15*A*cos(d*x + c)^5 + 90*A*cos(d*x + c)^4 + (285*A*d*x + 683*A)*cos (d*x + c)^3 - 1140*A*d*x + (855*A*d*x - 526*A)*cos(d*x + c)^2 - 6*(95*A*d* x + 191*A)*cos(d*x + c) - (15*A*cos(d*x + c)^4 - 75*A*cos(d*x + c)^3 + 114 0*A*d*x - 19*(15*A*d*x - 32*A)*cos(d*x + c)^2 + 6*(95*A*d*x + 189*A)*cos(d *x + c) - 12*A)*sin(d*x + c) - 12*A)/(a^3*d*cos(d*x + c)^3 + 3*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)^2 - 2*a^3 *d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))
Leaf count of result is larger than twice the leaf count of optimal. 3614 vs. \(2 (126) = 252\).
Time = 25.14 (sec) , antiderivative size = 3614, normalized size of antiderivative = 28.02 \[ \int \frac {\sin ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)**4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((-285*A*d*x*tan(c/2 + d*x/2)**9/(30*a**3*d*tan(c/2 + d*x/2)**9 + 150*a**3*d*tan(c/2 + d*x/2)**8 + 360*a**3*d*tan(c/2 + d*x/2)**7 + 600*a** 3*d*tan(c/2 + d*x/2)**6 + 780*a**3*d*tan(c/2 + d*x/2)**5 + 780*a**3*d*tan( c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**3 + 360*a**3*d*tan(c/2 + d* x/2)**2 + 150*a**3*d*tan(c/2 + d*x/2) + 30*a**3*d) - 1425*A*d*x*tan(c/2 + d*x/2)**8/(30*a**3*d*tan(c/2 + d*x/2)**9 + 150*a**3*d*tan(c/2 + d*x/2)**8 + 360*a**3*d*tan(c/2 + d*x/2)**7 + 600*a**3*d*tan(c/2 + d*x/2)**6 + 780*a* *3*d*tan(c/2 + d*x/2)**5 + 780*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan (c/2 + d*x/2)**3 + 360*a**3*d*tan(c/2 + d*x/2)**2 + 150*a**3*d*tan(c/2 + d *x/2) + 30*a**3*d) - 3420*A*d*x*tan(c/2 + d*x/2)**7/(30*a**3*d*tan(c/2 + d *x/2)**9 + 150*a**3*d*tan(c/2 + d*x/2)**8 + 360*a**3*d*tan(c/2 + d*x/2)**7 + 600*a**3*d*tan(c/2 + d*x/2)**6 + 780*a**3*d*tan(c/2 + d*x/2)**5 + 780*a **3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**3 + 360*a**3*d*ta n(c/2 + d*x/2)**2 + 150*a**3*d*tan(c/2 + d*x/2) + 30*a**3*d) - 5700*A*d*x* tan(c/2 + d*x/2)**6/(30*a**3*d*tan(c/2 + d*x/2)**9 + 150*a**3*d*tan(c/2 + d*x/2)**8 + 360*a**3*d*tan(c/2 + d*x/2)**7 + 600*a**3*d*tan(c/2 + d*x/2)** 6 + 780*a**3*d*tan(c/2 + d*x/2)**5 + 780*a**3*d*tan(c/2 + d*x/2)**4 + 600* a**3*d*tan(c/2 + d*x/2)**3 + 360*a**3*d*tan(c/2 + d*x/2)**2 + 150*a**3*d*t an(c/2 + d*x/2) + 30*a**3*d) - 7410*A*d*x*tan(c/2 + d*x/2)**5/(30*a**3*d*t an(c/2 + d*x/2)**9 + 150*a**3*d*tan(c/2 + d*x/2)**8 + 360*a**3*d*tan(c/...
Leaf count of result is larger than twice the leaf count of optimal. 715 vs. \(2 (119) = 238\).
Time = 0.13 (sec) , antiderivative size = 715, normalized size of antiderivative = 5.54 \[ \int \frac {\sin ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="m axima")
Output:
-1/15*(A*((1325*sin(d*x + c)/(cos(d*x + c) + 1) + 2673*sin(d*x + c)^2/(cos (d*x + c) + 1)^2 + 3805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 4329*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3575*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2275*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 975*sin(d*x + c)^7/(cos(d*x + c ) + 1)^7 + 195*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 304)/(a^3 + 5*a^3*sin (d*x + c)/(cos(d*x + c) + 1) + 12*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 20*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 26*a^3*sin(d*x + c)^4/(cos( d*x + c) + 1)^4 + 26*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 20*a^3*sin( d*x + c)^6/(cos(d*x + c) + 1)^6 + 12*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1) ^7 + 5*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a^3*sin(d*x + c)^9/(cos(d *x + c) + 1)^9) + 195*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) + 6*A*( (105*sin(d*x + c)/(cos(d*x + c) + 1) + 189*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 200*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 160*sin(d*x + c)^4/(cos(d *x + c) + 1)^4 + 75*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^ 6/(cos(d*x + c) + 1)^6 + 24)/(a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 11*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^3*sin(d*x + c)^3/(cos( d*x + c) + 1)^3 + 15*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 11*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*arctan(sin(d*x + c)/(cos (d*x + c) + 1))/a^3))/d
Time = 0.29 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.21 \[ \int \frac {\sin ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {285 \, {\left (d x + c\right )} A}{a^{3}} + \frac {30 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} + \frac {4 \, {\left (135 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 615 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1025 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 685 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 164 \, A\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{30 \, d} \] Input:
integrate(sin(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="g iac")
Output:
-1/30*(285*(d*x + c)*A/a^3 + 30*(A*tan(1/2*d*x + 1/2*c)^3 + 8*A*tan(1/2*d* x + 1/2*c)^2 - A*tan(1/2*d*x + 1/2*c) + 8*A)/((tan(1/2*d*x + 1/2*c)^2 + 1) ^2*a^3) + 4*(135*A*tan(1/2*d*x + 1/2*c)^4 + 615*A*tan(1/2*d*x + 1/2*c)^3 + 1025*A*tan(1/2*d*x + 1/2*c)^2 + 685*A*tan(1/2*d*x + 1/2*c) + 164*A)/(a^3* (tan(1/2*d*x + 1/2*c) + 1)^5))/d
Time = 38.84 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.53 \[ \int \frac {\sin ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {\left (\frac {95\,A\,\left (c+d\,x\right )}{2}-\frac {A\,\left (1425\,c+1425\,d\,x+570\right )}{30}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (114\,A\,\left (c+d\,x\right )-\frac {A\,\left (3420\,c+3420\,d\,x+2850\right )}{30}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (190\,A\,\left (c+d\,x\right )-\frac {A\,\left (5700\,c+5700\,d\,x+6650\right )}{30}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (247\,A\,\left (c+d\,x\right )-\frac {A\,\left (7410\,c+7410\,d\,x+10450\right )}{30}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (247\,A\,\left (c+d\,x\right )-\frac {A\,\left (7410\,c+7410\,d\,x+12846\right )}{30}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (190\,A\,\left (c+d\,x\right )-\frac {A\,\left (5700\,c+5700\,d\,x+11270\right )}{30}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (114\,A\,\left (c+d\,x\right )-\frac {A\,\left (3420\,c+3420\,d\,x+7902\right )}{30}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (\frac {95\,A\,\left (c+d\,x\right )}{2}-\frac {A\,\left (1425\,c+1425\,d\,x+3910\right )}{30}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {19\,A\,\left (c+d\,x\right )}{2}-\frac {A\,\left (285\,c+285\,d\,x+896\right )}{30}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}-\frac {19\,A\,x}{2\,a^3} \] Input:
int((sin(c + d*x)^4*(A - A*sin(c + d*x)))/(a + a*sin(c + d*x))^3,x)
Output:
(tan(c/2 + (d*x)/2)*((95*A*(c + d*x))/2 - (A*(1425*c + 1425*d*x + 3910))/3 0) + tan(c/2 + (d*x)/2)^8*((95*A*(c + d*x))/2 - (A*(1425*c + 1425*d*x + 57 0))/30) + tan(c/2 + (d*x)/2)^7*(114*A*(c + d*x) - (A*(3420*c + 3420*d*x + 2850))/30) + tan(c/2 + (d*x)/2)^2*(114*A*(c + d*x) - (A*(3420*c + 3420*d*x + 7902))/30) + tan(c/2 + (d*x)/2)^6*(190*A*(c + d*x) - (A*(5700*c + 5700* d*x + 6650))/30) + tan(c/2 + (d*x)/2)^3*(190*A*(c + d*x) - (A*(5700*c + 57 00*d*x + 11270))/30) + tan(c/2 + (d*x)/2)^5*(247*A*(c + d*x) - (A*(7410*c + 7410*d*x + 10450))/30) + tan(c/2 + (d*x)/2)^4*(247*A*(c + d*x) - (A*(741 0*c + 7410*d*x + 12846))/30) + (19*A*(c + d*x))/2 - (A*(285*c + 285*d*x + 896))/30)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)^5*(tan(c/2 + (d*x)/2)^2 + 1)^2) - (19*A*x)/(2*a^3)
Time = 0.17 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.15 \[ \int \frac {\sin ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+75 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-285 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x +379 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-570 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x +391 \cos \left (d x +c \right ) \sin \left (d x +c \right )-285 \cos \left (d x +c \right ) d x +114 \cos \left (d x +c \right )-15 \sin \left (d x +c \right )^{5}+90 \sin \left (d x +c \right )^{4}+285 \sin \left (d x +c \right )^{3} d x +972 \sin \left (d x +c \right )^{3}+855 \sin \left (d x +c \right )^{2} d x +1348 \sin \left (d x +c \right )^{2}+855 \sin \left (d x +c \right ) d x +391 \sin \left (d x +c \right )+285 d x -114}{30 a^{2} d \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )+\cos \left (d x +c \right )-\sin \left (d x +c \right )^{3}-3 \sin \left (d x +c \right )^{2}-3 \sin \left (d x +c \right )-1\right )} \] Input:
int(sin(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x)
Output:
( - 15*cos(c + d*x)*sin(c + d*x)**4 + 75*cos(c + d*x)*sin(c + d*x)**3 - 28 5*cos(c + d*x)*sin(c + d*x)**2*d*x + 379*cos(c + d*x)*sin(c + d*x)**2 - 57 0*cos(c + d*x)*sin(c + d*x)*d*x + 391*cos(c + d*x)*sin(c + d*x) - 285*cos( c + d*x)*d*x + 114*cos(c + d*x) - 15*sin(c + d*x)**5 + 90*sin(c + d*x)**4 + 285*sin(c + d*x)**3*d*x + 972*sin(c + d*x)**3 + 855*sin(c + d*x)**2*d*x + 1348*sin(c + d*x)**2 + 855*sin(c + d*x)*d*x + 391*sin(c + d*x) + 285*d*x - 114)/(30*a**2*d*(cos(c + d*x)*sin(c + d*x)**2 + 2*cos(c + d*x)*sin(c + d*x) + cos(c + d*x) - sin(c + d*x)**3 - 3*sin(c + d*x)**2 - 3*sin(c + d*x) - 1))