\(\int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [65]

Optimal result

Integrand size = 21, antiderivative size = 23 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {(a-a \sin (c+d x))^3}{3 a^5 d} \]

[Out]

-1/3*(a-a*sin(d*x+c))^3/a^5/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2746, 32} \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {(a-a \sin (c+d x))^3}{3 a^5 d} \]

[In]

Int[Cos[c + d*x]^5/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/3*(a - a*Sin[c + d*x])^3/(a^5*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a-x)^2 \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = -\frac {(a-a \sin (c+d x))^3}{3 a^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin (c+d x) \left (3-3 \sin (c+d x)+\sin ^2(c+d x)\right )}{3 a^2 d} \]

[In]

Integrate[Cos[c + d*x]^5/(a + a*Sin[c + d*x])^2,x]

[Out]

(Sin[c + d*x]*(3 - 3*Sin[c + d*x] + Sin[c + d*x]^2))/(3*a^2*d)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\left (\sin \left (d x +c \right )-1\right )^{3}}{3 a^{2} d}\)	\(19\)
default	\(\frac {\left (\sin \left (d x +c \right )-1\right )^{3}}{3 a^{2} d}\)	\(19\)
parallelrisch	\(\frac {-\sin \left (3 d x +3 c \right )+15 \sin \left (d x +c \right )-6+6 \cos \left (2 d x +2 c \right )}{12 a^{2} d}\)	\(41\)
risch	\(\frac {5 \sin \left (d x +c \right )}{4 a^{2} d}-\frac {\sin \left (3 d x +3 c \right )}{12 a^{2} d}+\frac {\cos \left (2 d x +2 c \right )}{2 a^{2} d}\)	\(50\)
norman	\(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {2 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {10 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {10 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {28 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {28 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {14 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {14 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {44 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {44 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(260\)

[In]

int(cos(d*x+c)^5/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/3/a^2/d*(sin(d*x+c)-1)^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right )}{3 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(3*cos(d*x + c)^2 - (cos(d*x + c)^2 - 4)*sin(d*x + c))/(a^2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (19) = 38\).

Time = 12.83 (sec) , antiderivative size = 394, normalized size of antiderivative = 17.13 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} \frac {6 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} - \frac {12 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} + \frac {20 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} - \frac {12 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} + \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{5}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**5/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((6*tan(c/2 + d*x/2)**5/(3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c
/2 + d*x/2)**2 + 3*a**2*d) - 12*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)*
*4 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 20*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*
d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 12*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 +
d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 6*tan(c/2 + d*x/2)/(3*a*
*2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d), Ne(d, 0)),
(x*cos(c)**5/(a*sin(c) + a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right )}{3 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3*(sin(d*x + c)^3 - 3*sin(d*x + c)^2 + 3*sin(d*x + c))/(a^2*d)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right )}{3 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(sin(d*x + c)^3 - 3*sin(d*x + c)^2 + 3*sin(d*x + c))/(a^2*d)

Mupad [B] (verification not implemented)

Time = 5.98 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin \left (c+d\,x\right )\,\left ({\sin \left (c+d\,x\right )}^2-3\,\sin \left (c+d\,x\right )+3\right )}{3\,a^2\,d} \]

[In]

int(cos(c + d*x)^5/(a + a*sin(c + d*x))^2,x)

[Out]

(sin(c + d*x)*(sin(c + d*x)^2 - 3*sin(c + d*x) + 3))/(3*a^2*d)