3.73 \(\int \frac{\cos ^5(c+d x)}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=150 \[ \frac{244 \sin (c+d x)}{105 a^4 d}-\frac{88 \sin (c+d x) \cos ^2(c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac{4 \sin (c+d x)}{a^4 d (\cos (c+d x)+1)}-\frac{4 x}{a^4}-\frac{\sin (c+d x) \cos ^4(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac{12 \sin (c+d x) \cos ^3(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

[Out]

(-4*x)/a^4 + (244*Sin[c + d*x])/(105*a^4*d) - (88*Cos[c + d*x]^2*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2
) + (4*Sin[c + d*x])/(a^4*d*(1 + Cos[c + d*x])) - (Cos[c + d*x]^4*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) -
(12*Cos[c + d*x]^3*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

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Rubi [A]  time = 0.372574, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{244 \sin (c+d x)}{105 a^4 d}-\frac{88 \sin (c+d x) \cos ^2(c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac{4 \sin (c+d x)}{a^4 d (\cos (c+d x)+1)}-\frac{4 x}{a^4}-\frac{\sin (c+d x) \cos ^4(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac{12 \sin (c+d x) \cos ^3(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + a*Cos[c + d*x])^4,x]

[Out]

(-4*x)/a^4 + (244*Sin[c + d*x])/(105*a^4*d) - (88*Cos[c + d*x]^2*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2
) + (4*Sin[c + d*x])/(a^4*d*(1 + Cos[c + d*x])) - (Cos[c + d*x]^4*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) -
(12*Cos[c + d*x]^3*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

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 ^5(c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\frac{\cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{\int \frac{\cos ^3(c+d x) (4 a-8 a \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{\cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{12 \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{\int \frac{\cos ^2(c+d x) \left (36 a^2-52 a^2 \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{88 \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{\cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{12 \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{\int \frac{\cos (c+d x) \left (176 a^3-244 a^3 \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=-\frac{88 \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{\cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{12 \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{\int \frac{176 a^3 \cos (c+d x)-244 a^3 \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=\frac{244 \sin (c+d x)}{105 a^4 d}-\frac{88 \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{\cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{12 \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{\int \frac{420 a^4 \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^7}\\ &=\frac{244 \sin (c+d x)}{105 a^4 d}-\frac{88 \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{\cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{12 \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{4 \int \frac{\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a^3}\\ &=-\frac{4 x}{a^4}+\frac{244 \sin (c+d x)}{105 a^4 d}-\frac{88 \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{\cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{12 \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{4 \int \frac{1}{a+a \cos (c+d x)} \, dx}{a^3}\\ &=-\frac{4 x}{a^4}+\frac{244 \sin (c+d x)}{105 a^4 d}-\frac{88 \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{\cos ^4(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{12 \cos ^3(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{4 \sin (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.399254, size = 263, normalized size = 1.75 \[ -\frac{\sec \left (\frac{c}{2}\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right ) \left (46130 \sin \left (c+\frac{d x}{2}\right )-46116 \sin \left (c+\frac{3 d x}{2}\right )+18060 \sin \left (2 c+\frac{3 d x}{2}\right )-19292 \sin \left (2 c+\frac{5 d x}{2}\right )+2100 \sin \left (3 c+\frac{5 d x}{2}\right )-3791 \sin \left (3 c+\frac{7 d x}{2}\right )-735 \sin \left (4 c+\frac{7 d x}{2}\right )-105 \sin \left (4 c+\frac{9 d x}{2}\right )-105 \sin \left (5 c+\frac{9 d x}{2}\right )+29400 d x \cos \left (c+\frac{d x}{2}\right )+17640 d x \cos \left (c+\frac{3 d x}{2}\right )+17640 d x \cos \left (2 c+\frac{3 d x}{2}\right )+5880 d x \cos \left (2 c+\frac{5 d x}{2}\right )+5880 d x \cos \left (3 c+\frac{5 d x}{2}\right )+840 d x \cos \left (3 c+\frac{7 d x}{2}\right )+840 d x \cos \left (4 c+\frac{7 d x}{2}\right )-60830 \sin \left (\frac{d x}{2}\right )+29400 d x \cos \left (\frac{d x}{2}\right )\right )}{26880 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + a*Cos[c + d*x])^4,x]

[Out]

-(Sec[c/2]*Sec[(c + d*x)/2]^7*(29400*d*x*Cos[(d*x)/2] + 29400*d*x*Cos[c + (d*x)/2] + 17640*d*x*Cos[c + (3*d*x)
/2] + 17640*d*x*Cos[2*c + (3*d*x)/2] + 5880*d*x*Cos[2*c + (5*d*x)/2] + 5880*d*x*Cos[3*c + (5*d*x)/2] + 840*d*x
*Cos[3*c + (7*d*x)/2] + 840*d*x*Cos[4*c + (7*d*x)/2] - 60830*Sin[(d*x)/2] + 46130*Sin[c + (d*x)/2] - 46116*Sin
[c + (3*d*x)/2] + 18060*Sin[2*c + (3*d*x)/2] - 19292*Sin[2*c + (5*d*x)/2] + 2100*Sin[3*c + (5*d*x)/2] - 3791*S
in[3*c + (7*d*x)/2] - 735*Sin[4*c + (7*d*x)/2] - 105*Sin[4*c + (9*d*x)/2] - 105*Sin[5*c + (9*d*x)/2]))/(26880*
a^4*d)

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Maple [A]  time = 0.046, size = 126, normalized size = 0.8 \begin{align*} -{\frac{1}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{7}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{23}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{49}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-8\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a+cos(d*x+c)*a)^4,x)

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7+7/40/d/a^4*tan(1/2*d*x+1/2*c)^5-23/24/d/a^4*tan(1/2*d*x+1/2*c)^3+49/8/d/a^4*t
an(1/2*d*x+1/2*c)+2/d/a^4*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-8/d/a^4*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.75742, size = 213, normalized size = 1.42 \begin{align*} \frac{\frac{1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac{a^{4} \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{5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{6720 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(1680*sin(d*x + c)/((a^4 + a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d*x
+ c)/(cos(d*x + c) + 1) - 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 -
15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 6720*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4)/d

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Fricas [A]  time = 1.71374, size = 444, normalized size = 2.96 \begin{align*} -\frac{420 \, d x \cos \left (d x + c\right )^{4} + 1680 \, d x \cos \left (d x + c\right )^{3} + 2520 \, d x \cos \left (d x + c\right )^{2} + 1680 \, d x \cos \left (d x + c\right ) + 420 \, d x -{\left (105 \, \cos \left (d x + c\right )^{4} + 1184 \, \cos \left (d x + c\right )^{3} + 2636 \, \cos \left (d x + c\right )^{2} + 2236 \, \cos \left (d x + c\right ) + 664\right )} \sin \left (d x + c\right )}{105 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/105*(420*d*x*cos(d*x + c)^4 + 1680*d*x*cos(d*x + c)^3 + 2520*d*x*cos(d*x + c)^2 + 1680*d*x*cos(d*x + c) + 4
20*d*x - (105*cos(d*x + c)^4 + 1184*cos(d*x + c)^3 + 2636*cos(d*x + c)^2 + 2236*cos(d*x + c) + 664)*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5/(a+a*cos(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.43507, size = 151, normalized size = 1.01 \begin{align*} -\frac{\frac{3360 \,{\left (d x + c\right )}}{a^{4}} - \frac{1680 \, \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^{4}} + \frac{15 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 147 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 805 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5145 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

-1/840*(3360*(d*x + c)/a^4 - 1680*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^4) + (15*a^24*tan(1/2*d
*x + 1/2*c)^7 - 147*a^24*tan(1/2*d*x + 1/2*c)^5 + 805*a^24*tan(1/2*d*x + 1/2*c)^3 - 5145*a^24*tan(1/2*d*x + 1/
2*c))/a^28)/d