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

Optimal. Leaf size=192 \[ -\frac{b \left (4 a^2+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac{a \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3} \]

[Out]

-((b*(4*a^2 + b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d)) + (a*S
in[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) + ((2*a^2 + 3*b^2)*Sin[c + d*x])/(6*(a^2 - b^2)^2*d*(a +
b*Cos[c + d*x])^2) + (a*(2*a^2 + 13*b^2)*Sin[c + d*x])/(6*(a^2 - b^2)^3*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.225992, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2754, 12, 2659, 205} \[ -\frac{b \left (4 a^2+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac{a \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(4*a^2 + b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a + b)^(7/2)*d)) + (a*S
in[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) + ((2*a^2 + 3*b^2)*Sin[c + d*x])/(6*(a^2 - b^2)^2*d*(a +
b*Cos[c + d*x])^2) + (a*(2*a^2 + 13*b^2)*Sin[c + d*x])/(6*(a^2 - b^2)^3*d*(a + b*Cos[c + d*x]))

Rule 2754

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

Rule 12

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x)}{(a+b \cos (c+d x))^4} \, dx &=\frac{a \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{\int \frac{3 b-2 a \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac{a \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{\int \frac{-10 a b+\left (2 a^2+3 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 \left (a^2-b^2\right )^2}\\ &=\frac{a \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac{\int \frac{3 b \left (4 a^2+b^2\right )}{a+b \cos (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3}\\ &=\frac{a \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac{\left (b \left (4 a^2+b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac{a \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac{\left (b \left (4 a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac{b \left (4 a^2+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac{a \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{a \left (2 a^2+13 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.04692, size = 164, normalized size = 0.85 \[ \frac{\frac{\sin (c+d x) \left (a b^2 \left (2 a^2+13 b^2\right ) \cos ^2(c+d x)-3 b \left (-9 a^2 b^2-2 a^4+b^4\right ) \cos (c+d x)+10 a^3 b^2+6 a^5-a b^4\right )}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))^3}-\frac{6 b \left (4 a^2+b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{7/2}}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*b*(4*a^2 + b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(7/2) + ((6*a^5 + 10*a
^3*b^2 - a*b^4 - 3*b*(-2*a^4 - 9*a^2*b^2 + b^4)*Cos[c + d*x] + a*b^2*(2*a^2 + 13*b^2)*Cos[c + d*x]^2)*Sin[c +
d*x])/((a - b)^3*(a + b)^3*(a + b*Cos[c + d*x])^3))/(6*d)

________________________________________________________________________________________

Maple [B]  time = 0.089, size = 931, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*
c)^5+2/d*a^2*b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d
*x+1/2*c)^5+6/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/
2*d*x+1/2*c)^5*b^2+1/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*t
an(1/2*d*x+1/2*c)^5*b^3+4/d*a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a^2-2*a*b+b^2)/(a^2+2*a
*b+b^2)*tan(1/2*d*x+1/2*c)^3+28/3/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a/(a^2-2*a*b+b^2)/(a
^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*b^2+2/d*a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^
3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)-2/d*a^2*b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b
)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)+6/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a/(a+
b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*b^2-1/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/
(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*b^3-4/d*a^2*b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(
1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-1/d*b^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^
(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.23402, size = 1960, normalized size = 10.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(4*a^5*b + a^3*b^3 + (4*a^2*b^4 + b^6)*cos(d*x + c)^3 + 3*(4*a^3*b^3 + a*b^5)*cos(d*x + c)^2 + 3*(4*a
^4*b^2 + a^2*b^4)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sq
rt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2
)) + 2*(6*a^7 + 4*a^5*b^2 - 11*a^3*b^4 + a*b^6 + (2*a^5*b^2 + 11*a^3*b^4 - 13*a*b^6)*cos(d*x + c)^2 + 3*(2*a^6
*b + 7*a^4*b^3 - 10*a^2*b^5 + b^7)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 +
b^11)*d*cos(d*x + c)^3 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 + 3*(a^10*
b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c) + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 +
a^3*b^8)*d), -1/6*(3*(4*a^5*b + a^3*b^3 + (4*a^2*b^4 + b^6)*cos(d*x + c)^3 + 3*(4*a^3*b^3 + a*b^5)*cos(d*x + c
)^2 + 3*(4*a^4*b^2 + a^2*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(
d*x + c))) - (6*a^7 + 4*a^5*b^2 - 11*a^3*b^4 + a*b^6 + (2*a^5*b^2 + 11*a^3*b^4 - 13*a*b^6)*cos(d*x + c)^2 + 3*
(2*a^6*b + 7*a^4*b^3 - 10*a^2*b^5 + b^7)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2
*b^9 + b^11)*d*cos(d*x + c)^3 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 + 3*
(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c) + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*
b^6 + a^3*b^8)*d)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.35768, size = 576, normalized size = 3. \begin{align*} \frac{\frac{3 \,{\left (4 \, a^{2} b + b^{3}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} + \frac{6 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 27 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 16 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 28 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(4*a^2*b + b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b
*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + (6*a^5*tan(1/
2*d*x + 1/2*c)^5 - 6*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 12*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 27*a^2*b^3*tan(1/2*d*x
+ 1/2*c)^5 + 12*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*b^5*tan(1/2*d*x + 1/2*c)^5 + 12*a^5*tan(1/2*d*x + 1/2*c)^3 +
16*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 28*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 6*a^5*tan(1/2*d*x + 1/2*c) + 6*a^4*b*ta
n(1/2*d*x + 1/2*c) + 12*a^3*b^2*tan(1/2*d*x + 1/2*c) + 27*a^2*b^3*tan(1/2*d*x + 1/2*c) + 12*a*b^4*tan(1/2*d*x
+ 1/2*c) - 3*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(
1/2*d*x + 1/2*c)^2 + a + b)^3))/d