3.66 \(\int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=56 \[ \frac{3 \cos (c+d x)}{2 a^2 d}+\frac{\cos ^3(c+d x)}{2 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{3 x}{2 a^2} \]

[Out]

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

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Rubi [A]  time = 0.0847453, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2679, 2682, 8} \[ \frac{3 \cos (c+d x)}{2 a^2 d}+\frac{\cos ^3(c+d x)}{2 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{3 x}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 2679

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] + Dist[(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] && IntegersQ[2*m, 2*p]

Rule 2682

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] + Dist[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]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\cos ^3(c+d x)}{2 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{3 \int \frac{\cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx}{2 a}\\ &=\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 )}+\frac{3 \int 1 \, dx}{2 a^2}\\ &=\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 )}\\ \end{align*}

Mathematica [A]  time = 0.171297, size = 109, normalized size = 1.95 \[ -\frac{\left (\sqrt{\sin (c+d x)+1} \left (\sin ^2(c+d x)-5 \sin (c+d x)+4\right )-6 \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right ) \sqrt{1-\sin (c+d x)}\right ) \cos ^5(c+d x)}{2 a^2 d (\sin (c+d x)-1)^3 (\sin (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^2,x]

[Out]

-(Cos[c + d*x]^5*(-6*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*x]]*(4
- 5*Sin[c + d*x] + Sin[c + d*x]^2)))/(2*a^2*d*(-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^(5/2))

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Maple [B]  time = 0.072, size = 142, normalized size = 2.5 \begin{align*}{\frac{1}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{1}{d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+4\,{\frac{1}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+3\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x)

[Out]

1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3+4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^
2-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)+4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2+3/d/a^2*arctan(tan(
1/2*d*x+1/2*c))

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Maxima [B]  time = 1.43535, size = 189, normalized size = 3.38 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.986, size = 89, normalized size = 1.59 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 169.455, size = 461, normalized size = 8.23 \begin{align*} \begin{cases} \frac{33 d x \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{22 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 44 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 22 a^{2} d} + \frac{66 d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{22 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 44 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 22 a^{2} d} + \frac{33 d x}{22 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 44 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 22 a^{2} d} - \frac{42 \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{22 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 44 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 22 a^{2} d} + \frac{22 \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{22 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 44 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 22 a^{2} d} + \frac{4 \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{22 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 44 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 22 a^{2} d} - \frac{22 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{22 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 44 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 22 a^{2} d} + \frac{46}{22 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 44 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 22 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((33*d*x*tan(c/2 + d*x/2)**4/(22*a**2*d*tan(c/2 + d*x/2)**4 + 44*a**2*d*tan(c/2 + d*x/2)**2 + 22*a**2
*d) + 66*d*x*tan(c/2 + d*x/2)**2/(22*a**2*d*tan(c/2 + d*x/2)**4 + 44*a**2*d*tan(c/2 + d*x/2)**2 + 22*a**2*d) +
33*d*x/(22*a**2*d*tan(c/2 + d*x/2)**4 + 44*a**2*d*tan(c/2 + d*x/2)**2 + 22*a**2*d) - 42*tan(c/2 + d*x/2)**4/(
22*a**2*d*tan(c/2 + d*x/2)**4 + 44*a**2*d*tan(c/2 + d*x/2)**2 + 22*a**2*d) + 22*tan(c/2 + d*x/2)**3/(22*a**2*d
*tan(c/2 + d*x/2)**4 + 44*a**2*d*tan(c/2 + d*x/2)**2 + 22*a**2*d) + 4*tan(c/2 + d*x/2)**2/(22*a**2*d*tan(c/2 +
d*x/2)**4 + 44*a**2*d*tan(c/2 + d*x/2)**2 + 22*a**2*d) - 22*tan(c/2 + d*x/2)/(22*a**2*d*tan(c/2 + d*x/2)**4 +
44*a**2*d*tan(c/2 + d*x/2)**2 + 22*a**2*d) + 46/(22*a**2*d*tan(c/2 + d*x/2)**4 + 44*a**2*d*tan(c/2 + d*x/2)**
2 + 22*a**2*d), Ne(d, 0)), (x*cos(c)**4/(a*sin(c) + a)**2, True))

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Giac [A]  time = 1.12535, size = 99, normalized size = 1.77 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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