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]

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

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Rubi [A]  time = 0.06, antiderivative size = 76, 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 = {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 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]

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])

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*}

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Mathematica [A]  time = 0.23, 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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fricas [A]  time = 1.17, size = 57, normalized size = 0.75 \[ \frac {3 \, d x \cos \left (d x + c\right ) + 3 \, d x + {\left (\cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

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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giac [A]  time = 0.36, size = 73, normalized size = 0.96 \[ \frac {\frac {3 \, {\left (d x + c\right )}}{a} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \]

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

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maple [A]  time = 0.08, size = 103, normalized size = 1.36 \[ -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d*tan(1/2*d*x+1/2*c)-3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-1/a/d/(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 = 1.33, size = 133, normalized size = 1.75 \[ -\frac {\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

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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mupad [B]  time = 0.41, size = 89, normalized size = 1.17 \[ -\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (c+d\,x\right )}{2}+3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.32, size = 325, normalized size = 4.28 \[ \begin {cases} \frac {3 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} + \frac {6 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} + \frac {3 d x}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {2 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {10 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {4 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{3}{\relax (c )}}{a \cos {\relax (c )} + a} & \text {otherwise} \end {cases} \]

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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