3.1.56 \(\int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx\) [56]

Optimal. Leaf size=57 \[ \frac {x}{a^2}-\frac {5 \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {\sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]

[Out]

x/a^2-5/3*sin(d*x+c)/a^2/d/(1+cos(d*x+c))+1/3*sin(d*x+c)/d/(a+a*cos(d*x+c))^2

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Rubi [A]
time = 0.06, antiderivative size = 57, normalized size of antiderivative = 1.00, 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 = {2837, 2814, 2727} \begin {gather*} -\frac {5 \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {x}{a^2}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2814

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

Rule 2837

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*Cos[e + f*x]*((
a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(a*m - b
*(2*m + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=\frac {\sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {-2 a+3 a \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=\frac {x}{a^2}+\frac {\sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {5 \int \frac {1}{a+a \cos (c+d x)} \, dx}{3 a}\\ &=\frac {x}{a^2}+\frac {\sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {5 \sin (c+d x)}{3 d \left (a^2+a^2 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 105, normalized size = 1.84 \begin {gather*} \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (6 d x \cos ^3\left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-10 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (1+\cos (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*(6*d*x*Cos[(c + d*x)/2]^3 + Sec[c/2]*Sin[(d*x)/2] - 10*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*
x)/2] + Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[c + d*x])^2)

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Maple [A]
time = 0.08, size = 46, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(46\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(46\)
risch	\(\frac {x}{a^{2}}-\frac {2 i \left (6 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(53\)
norman	\(\frac {\frac {x}{a}+\frac {x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {17 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a d}+\frac {2 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 72, normalized size = 1.26 \begin {gather*} -\frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c)/(
cos(d*x + c) + 1))/a^2)/d

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Fricas [A]
time = 0.43, size = 80, normalized size = 1.40 \begin {gather*} \frac {3 \, d x \cos \left (d x + c\right )^{2} + 6 \, d x \cos \left (d x + c\right ) + 3 \, d x - {\left (5 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*d*x*cos(d*x + c)^2 + 6*d*x*cos(d*x + c) + 3*d*x - (5*cos(d*x + c) + 4)*sin(d*x + c))/(a^2*d*cos(d*x + c
)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [A]
time = 0.67, size = 56, normalized size = 0.98 \begin {gather*} \begin {cases} \frac {x}{a^{2}} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} - \frac {3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 50, normalized size = 0.88 \begin {gather*} \frac {\frac {6 \, {\left (d x + c\right )}}{a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(d*x + c)/a^2 + (a^4*tan(1/2*d*x + 1/2*c)^3 - 9*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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Mupad [B]
time = 0.36, size = 35, normalized size = 0.61 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,d\,x}{6\,a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3 - 9*tan(c/2 + (d*x)/2) + 6*d*x)/(6*a^2*d)

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