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

Optimal. Leaf size=23 \[ -\frac {(a-a \sin (c+d x))^3}{3 a^5 d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2667, 32} \[ -\frac {(a-a \sin (c+d x))^3}{3 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a - a*Sin[c + d*x])^3/(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 2667

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*} \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\operatorname {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*}

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Mathematica [A]  time = 0.06, size = 34, normalized size = 1.48 \[ \frac {\sin (c+d x) \left (\sin ^2(c+d x)-3 \sin (c+d x)+3\right )}{3 a^2 d} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.69, size = 37, normalized size = 1.61 \[ \frac {3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right )}{3 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 1.97, size = 35, normalized size = 1.52 \[ \frac {\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right )}{3 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.17, size = 19, normalized size = 0.83 \[ \frac {\left (\sin \left (d x +c \right )-1\right )^{3}}{3 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 35, normalized size = 1.52 \[ \frac {\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right )}{3 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.61, size = 32, normalized size = 1.39 \[ \frac {\sin \left (c+d\,x\right )\,\left ({\sin \left (c+d\,x\right )}^2-3\,\sin \left (c+d\,x\right )+3\right )}{3\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 37.35, size = 394, normalized size = 17.13 \[ \begin {cases} \frac {6 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} - \frac {12 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} + \frac {20 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} - \frac {12 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} + \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{5}{\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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