3.45 \(\int \frac{\cos ^3(c+d x)}{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=76 \[ -\frac{2 \sin (c+d x)}{a d}-\frac{\sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}+\frac{3 \sin (c+d x) \cos (c+d x)}{2 a d}+\frac{3 x}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0611472, antiderivative size = 76, normalized size of antiderivative = 1., 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 = {2767, 2734} \[ -\frac{2 \sin (c+d x)}{a d}-\frac{\sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}+\frac{3 \sin (c+d x) \cos (c+d x)}{2 a d}+\frac{3 x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + a*Cos[c + d*x]),x]

[Out]

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

Rule 2767

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(a + b*Sin[e + f*x])), x] - Dist[d/(a*b), Int[(c +
d*Sin[e + f*x])^(n - 2)*Simp[b*d*(n - 1) - a*c*n + (b*c*(n - 1) - a*d*n)*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && (IntegerQ[2
*n] || EqQ[c, 0])

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.234428, size = 117, normalized size = 1.54 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (-4 \sin \left (c+\frac{d x}{2}\right )-3 \sin \left (c+\frac{3 d x}{2}\right )-3 \sin \left (2 c+\frac{3 d x}{2}\right )+\sin \left (2 c+\frac{5 d x}{2}\right )+\sin \left (3 c+\frac{5 d x}{2}\right )+12 d x \cos \left (c+\frac{d x}{2}\right )-20 \sin \left (\frac{d x}{2}\right )+12 d x \cos \left (\frac{d x}{2}\right )\right )}{16 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + a*Cos[c + d*x]),x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]*(12*d*x*Cos[(d*x)/2] + 12*d*x*Cos[c + (d*x)/2] - 20*Sin[(d*x)/2] - 4*Sin[c + (d*x)/
2] - 3*Sin[c + (3*d*x)/2] - 3*Sin[2*c + (3*d*x)/2] + Sin[2*c + (5*d*x)/2] + Sin[3*c + (5*d*x)/2]))/(16*a*d)

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Maple [A]  time = 0.044, size = 103, normalized size = 1.4 \begin{align*} -{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-3\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{1}{da}\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}}+3\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+cos(d*x+c)*a),x)

[Out]

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

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Maxima [A]  time = 1.65559, size = 180, normalized size = 2.37 \begin{align*} -\frac{\frac{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \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} + \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.58943, size = 149, normalized size = 1.96 \begin{align*} \frac{3 \, d x \cos \left (d x + c\right ) + 3 \, d x +{\left (\cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right )}{2 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.69114, size = 325, normalized size = 4.28 \begin{align*} \begin{cases} \frac{3 d x \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d} + \frac{6 d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d} + \frac{3 d x}{2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d} - \frac{2 \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d} - \frac{10 \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d} - \frac{4 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d} & \text{for}\: d \neq 0 \\\frac{x \cos ^{3}{\left (c \right )}}{a \cos{\left (c \right )} + a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+a*cos(d*x+c)),x)

[Out]

Piecewise((3*d*x*tan(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 6*d*x*t
an(c/2 + d*x/2)**2/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 3*d*x/(2*a*d*tan(c/2 + d*
x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 2*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/
2 + d*x/2)**2 + 2*a*d) - 10*tan(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d
) - 4*tan(c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d), Ne(d, 0)), (x*cos(c)**
3/(a*cos(c) + a), True))

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Giac [A]  time = 1.37556, size = 99, normalized size = 1.3 \begin{align*} \frac{\frac{3 \,{\left (d x + c\right )}}{a} - \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} - \frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

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