3.64 \(\int \frac{\cos ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=119 \[ \frac{9 \sin (c+d x)}{5 a^3 d}+\frac{3 \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{3 x}{a^3}-\frac{\sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{3 \sin (c+d x) \cos ^2(c+d x)}{5 a d (a \cos (c+d x)+a)^2} \]

[Out]

(-3*x)/a^3 + (9*Sin[c + d*x])/(5*a^3*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - (3*Cos[
c + d*x]^2*Sin[c + d*x])/(5*a*d*(a + a*Cos[c + d*x])^2) + (3*Sin[c + d*x])/(d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.27257, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2765, 2977, 2968, 3023, 12, 2735, 2648} \[ \frac{9 \sin (c+d x)}{5 a^3 d}+\frac{3 \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{3 x}{a^3}-\frac{\sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{3 \sin (c+d x) \cos ^2(c+d x)}{5 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^3,x]

[Out]

(-3*x)/a^3 + (9*Sin[c + d*x])/(5*a^3*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - (3*Cos[
c + d*x]^2*Sin[c + d*x])/(5*a*d*(a + a*Cos[c + d*x])^2) + (3*Sin[c + d*x])/(d*(a^3 + a^3*Cos[c + d*x]))

Rule 2765

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

Rule 2977

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

Rule 2968

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

Rule 3023

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2735

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 2648

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]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{\int \frac{\cos ^2(c+d x) (3 a-6 a \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{3 \cos ^2(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac{\int \frac{\cos (c+d x) \left (18 a^2-27 a^2 \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{3 \cos ^2(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac{\int \frac{18 a^2 \cos (c+d x)-27 a^2 \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac{9 \sin (c+d x)}{5 a^3 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{3 \cos ^2(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac{\int \frac{45 a^3 \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^5}\\ &=\frac{9 \sin (c+d x)}{5 a^3 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{3 \cos ^2(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac{3 \int \frac{\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=-\frac{3 x}{a^3}+\frac{9 \sin (c+d x)}{5 a^3 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{3 \cos ^2(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac{3 \int \frac{1}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=-\frac{3 x}{a^3}+\frac{9 \sin (c+d x)}{5 a^3 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{3 \cos ^2(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac{3 \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.545771, size = 161, normalized size = 1.35 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left (20 (\sin (c+d x)-3 d x) \cos ^5\left (\frac{1}{2} (c+d x)\right )-12 \tan \left (\frac{c}{2}\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right )+\tan \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+96 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right )-12 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )\right )}{5 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^3,x]

[Out]

(2*Cos[(c + d*x)/2]*(Sec[c/2]*Sin[(d*x)/2] - 12*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + 96*Cos[(c + d*x)/2]
^4*Sec[c/2]*Sin[(d*x)/2] + 20*Cos[(c + d*x)/2]^5*(-3*d*x + Sin[c + d*x]) + Cos[(c + d*x)/2]*Tan[c/2] - 12*Cos[
(c + d*x)/2]^3*Tan[c/2]))/(5*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.048, size = 107, normalized size = 0.9 \begin{align*}{\frac{1}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{17}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+cos(d*x+c)*a)^3,x)

[Out]

1/20/d/a^3*tan(1/2*d*x+1/2*c)^5-1/2/d/a^3*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*tan(1/2*d*x+1/2*c)+2/d/a^3*tan(1/2*d
*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-6/d/a^3*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.66513, size = 185, normalized size = 1.55 \begin{align*} \frac{\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac{a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} + \frac{\frac{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/20*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/
(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 120*a
rctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 1.62797, size = 325, normalized size = 2.73 \begin{align*} -\frac{15 \, d x \cos \left (d x + c\right )^{3} + 45 \, d x \cos \left (d x + c\right )^{2} + 45 \, d x \cos \left (d x + c\right ) + 15 \, d x -{\left (5 \, \cos \left (d x + c\right )^{3} + 39 \, \cos \left (d x + c\right )^{2} + 57 \, \cos \left (d x + c\right ) + 24\right )} \sin \left (d x + c\right )}{5 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/5*(15*d*x*cos(d*x + c)^3 + 45*d*x*cos(d*x + c)^2 + 45*d*x*cos(d*x + c) + 15*d*x - (5*cos(d*x + c)^3 + 39*co
s(d*x + c)^2 + 57*cos(d*x + c) + 24)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*co
s(d*x + c) + a^3*d)

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Sympy [A]  time = 18.245, size = 240, normalized size = 2.02 \begin{align*} \begin{cases} - \frac{60 d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 20 a^{3} d} - \frac{60 d x}{20 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 20 a^{3} d} + \frac{\tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 20 a^{3} d} - \frac{9 \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 20 a^{3} d} + \frac{75 \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 20 a^{3} d} + \frac{125 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 20 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \cos ^{4}{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\right )^{3}} & \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*cos(d*x+c))**3,x)

[Out]

Piecewise((-60*d*x*tan(c/2 + d*x/2)**2/(20*a**3*d*tan(c/2 + d*x/2)**2 + 20*a**3*d) - 60*d*x/(20*a**3*d*tan(c/2
+ d*x/2)**2 + 20*a**3*d) + tan(c/2 + d*x/2)**7/(20*a**3*d*tan(c/2 + d*x/2)**2 + 20*a**3*d) - 9*tan(c/2 + d*x/
2)**5/(20*a**3*d*tan(c/2 + d*x/2)**2 + 20*a**3*d) + 75*tan(c/2 + d*x/2)**3/(20*a**3*d*tan(c/2 + d*x/2)**2 + 20
*a**3*d) + 125*tan(c/2 + d*x/2)/(20*a**3*d*tan(c/2 + d*x/2)**2 + 20*a**3*d), Ne(d, 0)), (x*cos(c)**4/(a*cos(c)
+ a)**3, True))

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Giac [A]  time = 1.23387, size = 130, normalized size = 1.09 \begin{align*} -\frac{\frac{60 \,{\left (d x + c\right )}}{a^{3}} - \frac{40 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac{a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 85 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-1/20*(60*(d*x + c)/a^3 - 40*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) - (a^12*tan(1/2*d*x + 1/2
*c)^5 - 10*a^12*tan(1/2*d*x + 1/2*c)^3 + 85*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d