Integrand size = 21, antiderivative size = 56 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 x}{2 a^2}+\frac {3 \cos (c+d x)}{2 a^2 d}+\frac {\cos ^3(c+d x)}{2 d \left (a^2+a^2 \sin (c+d x)\right )} \] Output:
3/2*x/a^2+3/2*cos(d*x+c)/a^2/d+1/2*cos(d*x+c)^3/d/(a^2+a^2*sin(d*x+c))
Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.95 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\cos ^5(c+d x) \left (-6 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (4-5 \sin (c+d x)+\sin ^2(c+d x)\right )\right )}{2 a^2 d (-1+\sin (c+d x))^3 (1+\sin (c+d x))^{5/2}} \] Input:
Integrate[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^2,x]
Output:
-1/2*(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]]*(4 - 5*Sin[c + d*x] + Sin[c + d*x]^2 )))/(a^2*d*(-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^(5/2))
Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3158, 3042, 3161, 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.
\(\displaystyle \int \frac {\cos ^4(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^4}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3158 |
\(\displaystyle \frac {3 \int \frac {\cos ^2(c+d x)}{\sin (c+d x) a+a}dx}{2 a}+\frac {\cos ^3(c+d x)}{2 d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \frac {\cos (c+d x)^2}{\sin (c+d x) a+a}dx}{2 a}+\frac {\cos ^3(c+d x)}{2 d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3161 |
\(\displaystyle \frac {3 \left (\frac {\int 1dx}{a}+\frac {\cos (c+d x)}{a d}\right )}{2 a}+\frac {\cos ^3(c+d x)}{2 d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\cos ^3(c+d x)}{2 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {3 \left (\frac {\cos (c+d x)}{a d}+\frac {x}{a}\right )}{2 a}\) |
Input:
Int[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^2,x]
Output:
(3*(x/a + Cos[c + d*x]/(a*d)))/(2*a) + Cos[c + d*x]^3/(2*d*(a^2 + a^2*Sin[ c + d*x]))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(a*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p, 0] && In tegersQ[2*m, 2*p]
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 = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.61
method | result | size |
parallelrisch | \(\frac {6 d x +8 \cos \left (d x +c \right )-\sin \left (2 d x +2 c \right )-8}{4 d \,a^{2}}\) | \(34\) |
risch | \(\frac {3 x}{2 a^{2}}+\frac {2 \cos \left (d x +c \right )}{a^{2} d}-\frac {\sin \left (2 d x +2 c \right )}{4 d \,a^{2}}\) | \(39\) |
derivativedivides | \(\frac {\frac {2 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+2\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\) | \(77\) |
default | \(\frac {\frac {2 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+2\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\) | \(77\) |
norman | \(\frac {\frac {4}{a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d a}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}+\frac {36 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}+\frac {3 x}{2 a}+\frac {9 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+\frac {21 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}+\frac {39 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}+\frac {27 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {33 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a}+\frac {33 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {27 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a}+\frac {39 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}+\frac {21 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 a}+\frac {9 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 a}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d a}+\frac {21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a d}+\frac {28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}+\frac {44 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {42 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}+\frac {46 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\) | \(420\) |
Input:
int(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/4*(6*d*x+8*cos(d*x+c)-sin(2*d*x+2*c)-8)/d/a^2
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.62 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, d x - \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, \cos \left (d x + c\right )}{2 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
1/2*(3*d*x - cos(d*x + c)*sin(d*x + c) + 4*cos(d*x + c))/(a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 403 vs. \(2 (48) = 96\).
Time = 10.20 (sec) , antiderivative size = 403, normalized size of antiderivative = 7.20 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} \frac {3 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} + \frac {6 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} + \frac {3 d x}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} + \frac {2 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} + \frac {8 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} - \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} + \frac {8}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{4}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**4/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((3*d*x*tan(c/2 + d*x/2)**4/(2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a** 2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d) + 6*d*x*tan(c/2 + d*x/2)**2/(2*a**2*d* tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d) + 3*d*x/(2* a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d) + 2* tan(c/2 + d*x/2)**3/(2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x /2)**2 + 2*a**2*d) + 8*tan(c/2 + d*x/2)**2/(2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d) - 2*tan(c/2 + d*x/2)/(2*a**2*d*t an(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d) + 8/(2*a**2* d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d), Ne(d, 0) ), (x*cos(c)**4/(a*sin(c) + a)**2, True))
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (50) = 100\).
Time = 0.11 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.50 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 4}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{d} \] Input:
integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-((sin(d*x + c)/(cos(d*x + c) + 1) - 4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 4)/(a^2 + 2*a^2*sin(d*x + c)^2/(c os(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*arctan(s in(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.30 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \] Input:
integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/2*(3*(d*x + c)/a^2 + 2*(tan(1/2*d*x + 1/2*c)^3 + 4*tan(1/2*d*x + 1/2*c)^ 2 - tan(1/2*d*x + 1/2*c) + 4)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2))/d
Time = 25.68 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.57 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4\,\cos \left (c+d\,x\right )-\frac {\sin \left (2\,c+2\,d\,x\right )}{2}+3\,d\,x}{2\,a^2\,d} \] Input:
int(cos(c + d*x)^4/(a + a*sin(c + d*x))^2,x)
Output:
(4*cos(c + d*x) - sin(2*c + 2*d*x)/2 + 3*d*x)/(2*a^2*d)
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.64 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-\cos \left (d x +c \right ) \sin \left (d x +c \right )+4 \cos \left (d x +c \right )+3 d x -4}{2 a^{2} d} \] Input:
int(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x)
Output:
( - cos(c + d*x)*sin(c + d*x) + 4*cos(c + d*x) + 3*d*x - 4)/(2*a**2*d)