Integrand size = 21, antiderivative size = 49 \[ \int \frac {\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {x}{2 a}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d} \] Output:
1/2*x/a+1/3*cos(d*x+c)^3/a/d+1/2*cos(d*x+c)*sin(d*x+c)/a/d
Leaf count is larger than twice the leaf count of optimal. \(119\) vs. \(2(49)=98\).
Time = 0.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.43 \[ \int \frac {\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cos ^5(c+d x) \left (-6 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (2+\sin (c+d x)-5 \sin ^2(c+d x)+2 \sin ^3(c+d x)\right )\right )}{6 a d (-1+\sin (c+d x))^3 (1+\sin (c+d x))^{5/2}} \] Input:
Integrate[Cos[c + d*x]^4/(a + a*Sin[c + d*x]),x]
Output:
-1/6*(Cos[c + d*x]^5*(-6*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - S in[c + d*x]] + Sqrt[1 + Sin[c + d*x]]*(2 + Sin[c + d*x] - 5*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3)))/(a*d*(-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^(5/2) )
Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3161, 3042, 3115, 24}
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.
\(\displaystyle \int \frac {\cos ^4(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^4}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3161 |
\(\displaystyle \frac {\int \cos ^2(c+d x)dx}{a}+\frac {\cos ^3(c+d x)}{3 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx}{a}+\frac {\cos ^3(c+d x)}{3 a d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}}{a}+\frac {\cos ^3(c+d x)}{3 a d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\cos ^3(c+d x)}{3 a d}+\frac {\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}}{a}\) |
Input:
Int[Cos[c + d*x]^4/(a + a*Sin[c + d*x]),x]
Output:
Cos[c + d*x]^3/(3*a*d) + (x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d))/a
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si mp[g^2/a Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x ] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
Time = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {6 d x +3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )+3 \sin \left (2 d x +2 c \right )+4}{12 d a}\) | \(43\) |
risch | \(\frac {x}{2 a}+\frac {\cos \left (d x +c \right )}{4 a d}+\frac {\cos \left (3 d x +3 c \right )}{12 a d}+\frac {\sin \left (2 d x +2 c \right )}{4 d a}\) | \(56\) |
derivativedivides | \(\frac {\frac {2 \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {1}{3}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(75\) |
default | \(\frac {\frac {2 \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {1}{3}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(75\) |
norman | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d a}+\frac {x}{2 a}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+\frac {2 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {2 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a}+\frac {2 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {2 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 a}+\frac {2}{3 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 a d}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 a d}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(343\) |
Input:
int(cos(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/12*(6*d*x+3*cos(d*x+c)+cos(3*d*x+3*c)+3*sin(2*d*x+2*c)+4)/d/a
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 \, \cos \left (d x + c\right )^{3} + 3 \, d x + 3 \, \cos \left (d x + c\right ) \sin \left (d x + c\right )}{6 \, a d} \] Input:
integrate(cos(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/6*(2*cos(d*x + c)^3 + 3*d*x + 3*cos(d*x + c)*sin(d*x + c))/(a*d)
Leaf count of result is larger than twice the leaf count of optimal. 558 vs. \(2 (36) = 72\).
Time = 4.77 (sec) , antiderivative size = 558, normalized size of antiderivative = 11.39 \[ \int \frac {\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)**4/(a+a*sin(d*x+c)),x)
Output:
Piecewise((3*d*x*tan(c/2 + d*x/2)**6/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*t an(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 9*d*x*tan(c/2 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a* d*tan(c/2 + d*x/2)**2 + 6*a*d) + 9*d*x*tan(c/2 + d*x/2)**2/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6* a*d) + 3*d*x/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18* a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 6*tan(c/2 + d*x/2)**5/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a* d) + 12*tan(c/2 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 6*tan(c/2 + d*x/2)/(6*a* d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/ 2)**2 + 6*a*d) + 4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d), Ne(d, 0)), (x*cos(c)**4/(a*sin(c) + a), True))
Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (43) = 86\).
Time = 0.11 (sec) , antiderivative size = 156, normalized size of antiderivative = 3.18 \[ \int \frac {\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 2}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{3 \, d} \] Input:
integrate(cos(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/3*((3*sin(d*x + c)/(cos(d*x + c) + 1) + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2)/(a + 3*a*sin(d*x + c)^2 /(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin(d* x + c)^6/(cos(d*x + c) + 1)^6) + 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) /a)/d
Time = 0.13 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.53 \[ \int \frac {\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {3 \, {\left (d x + c\right )}}{a} - \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \] Input:
integrate(cos(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/6*(3*(d*x + c)/a - 2*(3*tan(1/2*d*x + 1/2*c)^5 - 6*tan(1/2*d*x + 1/2*c)^ 4 - 3*tan(1/2*d*x + 1/2*c) - 2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a))/d
Time = 27.70 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.35 \[ \int \frac {\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {x}{2\,a}+\frac {-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {2}{3}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \] Input:
int(cos(c + d*x)^4/(a + a*sin(c + d*x)),x)
Output:
x/(2*a) + (tan(c/2 + (d*x)/2) + 2*tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2 )^5 + 2/3)/(a*d*(tan(c/2 + (d*x)/2)^2 + 1)^3)
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \frac {\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 \cos \left (d x +c \right )+3 d x -2}{6 a d} \] Input:
int(cos(d*x+c)^4/(a+a*sin(d*x+c)),x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)**2 + 3*cos(c + d*x)*sin(c + d*x) + 2*cos(c + d*x) + 3*d*x - 2)/(6*a*d)