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

Optimal. Leaf size=127 \[ -\frac {43 \sin (c+d x)}{21 a^4 d (\cos (c+d x)+1)}+\frac {11 \sin (c+d x)}{21 a^4 d (\cos (c+d x)+1)^2}+\frac {x}{a^4}-\frac {\sin (c+d x) \cos ^3(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {2 \sin (c+d x) \cos ^2(c+d x)}{7 a d (a \cos (c+d x)+a)^3} \]

[Out]

x/a^4+11/21*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2-43/21*sin(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*cos(d*x+c)^3*sin(d*x+c
)/d/(a+a*cos(d*x+c))^4-2/7*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3

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Rubi [A]  time = 0.28, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2765, 2977, 2968, 3019, 2735, 2648} \[ -\frac {43 \sin (c+d x)}{21 a^4 d (\cos (c+d x)+1)}+\frac {11 \sin (c+d x)}{21 a^4 d (\cos (c+d x)+1)^2}+\frac {x}{a^4}-\frac {\sin (c+d x) \cos ^3(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {2 \sin (c+d x) \cos ^2(c+d x)}{7 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/a^4 + (11*Sin[c + d*x])/(21*a^4*d*(1 + Cos[c + d*x])^2) - (43*Sin[c + d*x])/(21*a^4*d*(1 + Cos[c + d*x])) -
(Cos[c + d*x]^3*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - (2*Cos[c + d*x]^2*Sin[c + d*x])/(7*a*d*(a + a*Cos
[c + d*x])^3)

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]

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 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 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 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 3019

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

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\frac {\cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {\int \frac {\cos ^2(c+d x) (3 a-7 a \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {\cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \cos ^2(c+d x) \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3}-\frac {\int \frac {\cos (c+d x) \left (20 a^2-35 a^2 \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {\cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \cos ^2(c+d x) \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3}-\frac {\int \frac {20 a^2 \cos (c+d x)-35 a^2 \cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=\frac {11 \sin (c+d x)}{21 a^4 d (1+\cos (c+d x))^2}-\frac {\cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \cos ^2(c+d x) \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {-110 a^3+105 a^3 \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=\frac {x}{a^4}+\frac {11 \sin (c+d x)}{21 a^4 d (1+\cos (c+d x))^2}-\frac {\cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \cos ^2(c+d x) \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3}-\frac {43 \int \frac {1}{a+a \cos (c+d x)} \, dx}{21 a^3}\\ &=\frac {x}{a^4}+\frac {11 \sin (c+d x)}{21 a^4 d (1+\cos (c+d x))^2}-\frac {\cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \cos ^2(c+d x) \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3}-\frac {43 \sin (c+d x)}{21 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 224, normalized size = 1.76 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (1652 \sin \left (c+\frac {d x}{2}\right )-1428 \sin \left (c+\frac {3 d x}{2}\right )+756 \sin \left (2 c+\frac {3 d x}{2}\right )-560 \sin \left (2 c+\frac {5 d x}{2}\right )+168 \sin \left (3 c+\frac {5 d x}{2}\right )-104 \sin \left (3 c+\frac {7 d x}{2}\right )+735 d x \cos \left (c+\frac {d x}{2}\right )+441 d x \cos \left (c+\frac {3 d x}{2}\right )+441 d x \cos \left (2 c+\frac {3 d x}{2}\right )+147 d x \cos \left (2 c+\frac {5 d x}{2}\right )+147 d x \cos \left (3 c+\frac {5 d x}{2}\right )+21 d x \cos \left (3 c+\frac {7 d x}{2}\right )+21 d x \cos \left (4 c+\frac {7 d x}{2}\right )-1988 \sin \left (\frac {d x}{2}\right )+735 d x \cos \left (\frac {d x}{2}\right )\right )}{2688 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(735*d*x*Cos[(d*x)/2] + 735*d*x*Cos[c + (d*x)/2] + 441*d*x*Cos[c + (3*d*x)/2] + 4
41*d*x*Cos[2*c + (3*d*x)/2] + 147*d*x*Cos[2*c + (5*d*x)/2] + 147*d*x*Cos[3*c + (5*d*x)/2] + 21*d*x*Cos[3*c + (
7*d*x)/2] + 21*d*x*Cos[4*c + (7*d*x)/2] - 1988*Sin[(d*x)/2] + 1652*Sin[c + (d*x)/2] - 1428*Sin[c + (3*d*x)/2]
+ 756*Sin[2*c + (3*d*x)/2] - 560*Sin[2*c + (5*d*x)/2] + 168*Sin[3*c + (5*d*x)/2] - 104*Sin[3*c + (7*d*x)/2]))/
(2688*a^4*d)

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fricas [A]  time = 0.82, size = 152, normalized size = 1.20 \[ \frac {21 \, d x \cos \left (d x + c\right )^{4} + 84 \, d x \cos \left (d x + c\right )^{3} + 126 \, d x \cos \left (d x + c\right )^{2} + 84 \, d x \cos \left (d x + c\right ) + 21 \, d x - {\left (52 \, \cos \left (d x + c\right )^{3} + 124 \, \cos \left (d x + c\right )^{2} + 107 \, \cos \left (d x + c\right ) + 32\right )} \sin \left (d x + c\right )}{21 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(21*d*x*cos(d*x + c)^4 + 84*d*x*cos(d*x + c)^3 + 126*d*x*cos(d*x + c)^2 + 84*d*x*cos(d*x + c) + 21*d*x -
(52*cos(d*x + c)^3 + 124*cos(d*x + c)^2 + 107*cos(d*x + c) + 32)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d
*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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giac [A]  time = 0.52, size = 83, normalized size = 0.65 \[ \frac {\frac {168 \, {\left (d x + c\right )}}{a^{4}} + \frac {3 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 77 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 315 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{168 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*(168*(d*x + c)/a^4 + (3*a^24*tan(1/2*d*x + 1/2*c)^7 - 21*a^24*tan(1/2*d*x + 1/2*c)^5 + 77*a^24*tan(1/2*d
*x + 1/2*c)^3 - 315*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 0.05, size = 94, normalized size = 0.74 \[ \frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 d \,a^{4}}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/56/d/a^4*tan(1/2*d*x+1/2*c)^7-1/8/d/a^4*tan(1/2*d*x+1/2*c)^5+11/24/d/a^4*tan(1/2*d*x+1/2*c)^3-15/8/d/a^4*tan
(1/2*d*x+1/2*c)+2/d/a^4*arctan(tan(1/2*d*x+1/2*c))

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maxima [A]  time = 1.26, size = 112, normalized size = 0.88 \[ -\frac {\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}}{168 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/168*((315*sin(d*x + c)/(cos(d*x + c) + 1) - 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5 - 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 336*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^
4)/d

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mupad [B]  time = 0.43, size = 102, normalized size = 0.80 \[ \frac {x}{a^4}+\frac {-\frac {52\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{21}+\frac {16\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{21}-\frac {5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{28}+\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/a^4 + (sin(c/2 + (d*x)/2)/56 - (5*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2))/28 + (16*cos(c/2 + (d*x)/2)^4*sin
(c/2 + (d*x)/2))/21 - (52*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2))/21)/(a^4*d*cos(c/2 + (d*x)/2)^7)

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sympy [A]  time = 12.00, size = 95, normalized size = 0.75 \[ \begin {cases} \frac {x}{a^{4}} + \frac {\tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {11 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} - \frac {15 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{4}{\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/a**4 + tan(c/2 + d*x/2)**7/(56*a**4*d) - tan(c/2 + d*x/2)**5/(8*a**4*d) + 11*tan(c/2 + d*x/2)**3/
(24*a**4*d) - 15*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*cos(c)**4/(a*cos(c) + a)**4, True))

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