Integrand size = 29, antiderivative size = 102 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3 x}{4 a^2}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\cos ^3(c+d x)}{a^2 d}-\frac {\cos ^5(c+d x)}{5 a^2 d}+\frac {3 \cos (c+d x) \sin (c+d x)}{4 a^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{2 a^2 d} \] Output:
-3/4*x/a^2-2*cos(d*x+c)/a^2/d+cos(d*x+c)^3/a^2/d-1/5*cos(d*x+c)^5/a^2/d+3/ 4*cos(d*x+c)*sin(d*x+c)/a^2/d+1/2*cos(d*x+c)*sin(d*x+c)^3/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(308\) vs. \(2(102)=204\).
Time = 1.11 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.02 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {5 (1+24 d x) \cos \left (\frac {c}{2}\right )+110 \cos \left (\frac {c}{2}+d x\right )+110 \cos \left (\frac {3 c}{2}+d x\right )-40 \cos \left (\frac {3 c}{2}+2 d x\right )+40 \cos \left (\frac {5 c}{2}+2 d x\right )-15 \cos \left (\frac {5 c}{2}+3 d x\right )-15 \cos \left (\frac {7 c}{2}+3 d x\right )+5 \cos \left (\frac {7 c}{2}+4 d x\right )-5 \cos \left (\frac {9 c}{2}+4 d x\right )+\cos \left (\frac {9 c}{2}+5 d x\right )+\cos \left (\frac {11 c}{2}+5 d x\right )-5 \sin \left (\frac {c}{2}\right )+120 d x \sin \left (\frac {c}{2}\right )-110 \sin \left (\frac {c}{2}+d x\right )+110 \sin \left (\frac {3 c}{2}+d x\right )-40 \sin \left (\frac {3 c}{2}+2 d x\right )-40 \sin \left (\frac {5 c}{2}+2 d x\right )+15 \sin \left (\frac {5 c}{2}+3 d x\right )-15 \sin \left (\frac {7 c}{2}+3 d x\right )+5 \sin \left (\frac {7 c}{2}+4 d x\right )+5 \sin \left (\frac {9 c}{2}+4 d x\right )-\sin \left (\frac {9 c}{2}+5 d x\right )+\sin \left (\frac {11 c}{2}+5 d x\right )}{160 a^2 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \] Input:
Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]
Output:
-1/160*(5*(1 + 24*d*x)*Cos[c/2] + 110*Cos[c/2 + d*x] + 110*Cos[(3*c)/2 + d *x] - 40*Cos[(3*c)/2 + 2*d*x] + 40*Cos[(5*c)/2 + 2*d*x] - 15*Cos[(5*c)/2 + 3*d*x] - 15*Cos[(7*c)/2 + 3*d*x] + 5*Cos[(7*c)/2 + 4*d*x] - 5*Cos[(9*c)/2 + 4*d*x] + Cos[(9*c)/2 + 5*d*x] + Cos[(11*c)/2 + 5*d*x] - 5*Sin[c/2] + 12 0*d*x*Sin[c/2] - 110*Sin[c/2 + d*x] + 110*Sin[(3*c)/2 + d*x] - 40*Sin[(3*c )/2 + 2*d*x] - 40*Sin[(5*c)/2 + 2*d*x] + 15*Sin[(5*c)/2 + 3*d*x] - 15*Sin[ (7*c)/2 + 3*d*x] + 5*Sin[(7*c)/2 + 4*d*x] + 5*Sin[(9*c)/2 + 4*d*x] - Sin[( 9*c)/2 + 5*d*x] + Sin[(11*c)/2 + 5*d*x])/(a^2*d*(Cos[c/2] + Sin[c/2]))
Time = 0.50 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3348, 3042, 3236, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^4(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^4}{(a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3348 |
| \(\displaystyle \frac {\int \sin ^3(c+d x) (a-a \sin (c+d x))^2dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sin (c+d x)^3 (a-a \sin (c+d x))^2dx}{a^4}\) |
| \(\Big \downarrow \) 3236 |
| \(\displaystyle \frac {\int \left (a^2 \sin ^5(c+d x)-2 a^2 \sin ^4(c+d x)+a^2 \sin ^3(c+d x)\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {a^2 \cos ^3(c+d x)}{d}-\frac {2 a^2 \cos (c+d x)}{d}+\frac {a^2 \sin ^3(c+d x) \cos (c+d x)}{2 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}-\frac {3 a^2 x}{4}}{a^4}\) |
Input:
Int[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]
Output:
((-3*a^2*x)/4 - (2*a^2*Cos[c + d*x])/d + (a^2*Cos[c + d*x]^3)/d - (a^2*Cos [c + d*x]^5)/(5*d) + (3*a^2*Cos[c + d*x]*Sin[c + d*x])/(4*d) + (a^2*Cos[c + d*x]*Sin[c + d*x]^3)/(2*d))/a^4
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt Q[m, 0] && RationalQ[n]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m) Int[(d* Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
Time = 0.71 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66
| method | result | size |
| parallelrisch | \(\frac {-60 d x +15 \cos \left (3 d x +3 c \right )-110 \cos \left (d x +c \right )-\cos \left (5 d x +5 c \right )-5 \sin \left (4 d x +4 c \right )+40 \sin \left (2 d x +2 c \right )-96}{80 d \,a^{2}}\) | \(67\) |
| risch | \(-\frac {3 x}{4 a^{2}}-\frac {11 \cos \left (d x +c \right )}{8 a^{2} d}-\frac {\cos \left (5 d x +5 c \right )}{80 d \,a^{2}}-\frac {\sin \left (4 d x +4 c \right )}{16 d \,a^{2}}+\frac {3 \cos \left (3 d x +3 c \right )}{16 d \,a^{2}}+\frac {\sin \left (2 d x +2 c \right )}{2 d \,a^{2}}\) | \(90\) |
| derivativedivides | \(\frac {\frac {16 \left (-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{32}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}-\frac {3}{20}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{d \,a^{2}}\) | \(129\) |
| default | \(\frac {\frac {16 \left (-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{32}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}-\frac {3}{20}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{d \,a^{2}}\) | \(129\) |
| norman | \(\frac {-\frac {12}{5 a d}-\frac {189 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 a}-\frac {105 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a}-\frac {3 x}{4 a}-\frac {383 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{10 d a}-\frac {361 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 d a}-\frac {221 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d a}-\frac {1387 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{10 d a}-\frac {653 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{10 d a}-\frac {1679 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 d a}-\frac {57 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{10 d a}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}}{2 d a}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{2 d a}-\frac {193 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 d a}-\frac {39 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d a}-\frac {311 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 d a}-\frac {269 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 d a}-\frac {115 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{2 d a}-\frac {63 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a}-\frac {105 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 a}-\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{2 d a}-\frac {71 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{2 d a}-\frac {9 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a}-\frac {63 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{2 a}-\frac {33 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{2 a}-\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{2 a}-\frac {9 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}}{4 a}-\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{17}}{4 a}-\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}-\frac {33 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}-\frac {147 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}-\frac {105 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{a}-\frac {189 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 a}-\frac {147 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 a}-\frac {105 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{2 a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\) | \(637\) |
Input:
int(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/80*(-60*d*x+15*cos(3*d*x+3*c)-110*cos(d*x+c)-cos(5*d*x+5*c)-5*sin(4*d*x+ 4*c)+40*sin(2*d*x+2*c)-96)/d/a^2
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {4 \, \cos \left (d x + c\right )^{5} - 20 \, \cos \left (d x + c\right )^{3} + 15 \, d x + 5 \, {\left (2 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 40 \, \cos \left (d x + c\right )}{20 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/20*(4*cos(d*x + c)^5 - 20*cos(d*x + c)^3 + 15*d*x + 5*(2*cos(d*x + c)^3 - 5*cos(d*x + c))*sin(d*x + c) + 40*cos(d*x + c))/(a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 1608 vs. \(2 (94) = 188\).
Time = 40.69 (sec) , antiderivative size = 1608, normalized size of antiderivative = 15.76 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**4*sin(d*x+c)**3/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((-15*d*x*tan(c/2 + d*x/2)**10/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2 *d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 75* d*x*tan(c/2 + d*x/2)**8/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c /2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x /2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 150*d*x*tan(c/2 + d *x/2)**6/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a* *2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 150*d*x*tan(c/2 + d*x/2)**4/(20*a* *2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*ta n(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 75*d*x*tan(c/2 + d*x/2)**2/(20*a**2*d*tan(c/2 + d *x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)** 6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a **2*d) - 15*d*x/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x /2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 30*tan(c/2 + d*x/2)**9/(20* a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d* tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 140*tan(c/2 + d*x/2)**7/(20*a**2*d*tan(c/2 +...
Leaf count of result is larger than twice the leaf count of optimal. 290 vs. \(2 (94) = 188\).
Time = 0.11 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.84 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {70 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {200 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {40 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {70 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {15 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 24}{a^{2} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{10 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/10*((15*sin(d*x + c)/(cos(d*x + c) + 1) - 120*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 70*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 200*sin(d*x + c)^4/(c os(d*x + c) + 1)^4 - 40*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 70*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 15*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 24)/ (a^2 + 5*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^2*sin(d*x + c)^4/( cos(d*x + c) + 1)^4 + 10*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a^2*s in(d*x + c)^8/(cos(d*x + c) + 1)^8 + a^2*sin(d*x + c)^10/(cos(d*x + c) + 1 )^10) - 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.17 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {15 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 70 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 200 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 70 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a^{2}}}{20 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-1/20*(15*(d*x + c)/a^2 + 2*(15*tan(1/2*d*x + 1/2*c)^9 + 70*tan(1/2*d*x + 1/2*c)^7 + 40*tan(1/2*d*x + 1/2*c)^6 + 200*tan(1/2*d*x + 1/2*c)^4 - 70*tan (1/2*d*x + 1/2*c)^3 + 120*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 24)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*a^2))/d
Time = 17.66 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3\,\cos \left (3\,c+3\,d\,x\right )}{16\,a^2\,d}-\frac {11\,\cos \left (c+d\,x\right )}{8\,a^2\,d}-\frac {3\,x}{4\,a^2}-\frac {\cos \left (5\,c+5\,d\,x\right )}{80\,a^2\,d}+\frac {\sin \left (2\,c+2\,d\,x\right )}{2\,a^2\,d}-\frac {\sin \left (4\,c+4\,d\,x\right )}{16\,a^2\,d} \] Input:
int((cos(c + d*x)^4*sin(c + d*x)^3)/(a + a*sin(c + d*x))^2,x)
Output:
(3*cos(3*c + 3*d*x))/(16*a^2*d) - (11*cos(c + d*x))/(8*a^2*d) - (3*x)/(4*a ^2) - cos(5*c + 5*d*x)/(80*a^2*d) + sin(2*c + 2*d*x)/(2*a^2*d) - sin(4*c + 4*d*x)/(16*a^2*d)
Time = 0.18 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+10 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right )-24 \cos \left (d x +c \right )-15 d x +24}{20 a^{2} d} \] Input:
int(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x)
Output:
( - 4*cos(c + d*x)*sin(c + d*x)**4 + 10*cos(c + d*x)*sin(c + d*x)**3 - 12* cos(c + d*x)*sin(c + d*x)**2 + 15*cos(c + d*x)*sin(c + d*x) - 24*cos(c + d *x) - 15*d*x + 24)/(20*a**2*d)