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((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/(a+b)^(7/2)/d+1/3*a*sin(d*x+c)/(
a^2-b^2)/d/(a+b*cos(d*x+c))^3+1/6*(2*a^2+3*b^2)*sin(d*x+c)/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2+1/6*a*(2*a^2+13*b^
2)*sin(d*x+c)/(a^2-b^2)^3/d/(a+b*cos(d*x+c))

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Rubi [A]  time = 0.23, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 12

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

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]

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 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]

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*}

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Mathematica [A]  time = 1.06, size = 164, normalized size = 0.85 \[ \frac {\frac {\sin (c+d x) \left (6 a^5+10 a^3 b^2+a b^2 \left (2 a^2+13 b^2\right ) \cos ^2(c+d x)-3 b \left (-2 a^4-9 a^2 b^2+b^4\right ) \cos (c+d x)-a b^4\right )}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))^3}-\frac {6 b \left (4 a^2+b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{7/2}}}{6 d} \]

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)

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fricas [B]  time = 0.70, size = 891, normalized size = 4.64 \[ \left [\frac {3 \, {\left (4 \, a^{5} b + a^{3} b^{3} + {\left (4 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 2 \, {\left (6 \, a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + a b^{6} + {\left (2 \, a^{5} b^{2} + 11 \, a^{3} b^{4} - 13 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, a^{6} b + 7 \, a^{4} b^{3} - 10 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d\right )}}, -\frac {3 \, {\left (4 \, a^{5} b + a^{3} b^{3} + {\left (4 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (6 \, a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + a b^{6} + {\left (2 \, a^{5} b^{2} + 11 \, a^{3} b^{4} - 13 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, a^{6} b + 7 \, a^{4} b^{3} - 10 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d\right )}}\right ] \]

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)]

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giac [B]  time = 0.95, size = 427, normalized size = 2.22 \[ \frac {\frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{3}}}{3 \, d} \]

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

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maple [B]  time = 0.05, size = 931, normalized size = 4.85 \[ \text {result too large to display} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 4.25, size = 382, normalized size = 1.99 \[ \frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^3+2\,a^2\,b+6\,a\,b^2+b^3\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^3+7\,a\,b^2\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^3-2\,a^2\,b+6\,a\,b^2-b^3\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^2+b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}\,\left (4\,a^2\,b+b^3\right )}\right )\,\left (4\,a^2+b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(c/2 + (d*x)/2)^5*(6*a*b^2 + 2*a^2*b + 2*a^3 + b^3))/((a + b)^3*(a - b)) + (4*tan(c/2 + (d*x)/2)^3*(7*a*b
^2 + 3*a^3))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/2 + (d*x)/2)*(6*a*b^2 - 2*a^2*b + 2*a^3 - b^3))/((a +
b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))/(d*(3*a*b^2 - tan(c/2 + (d*x)/2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3 - 3*b^3) -
tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) + 3*a^2*b + a^3 + b^3 + tan(c/2 + (d*x)/2)^6*(3*a*b^2
- 3*a^2*b + a^3 - b^3))) - (b*atan((b*tan(c/2 + (d*x)/2)*(4*a^2 + b^2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 -
b^3))/(2*(a + b)^(1/2)*(a - b)^(7/2)*(4*a^2*b + b^3)))*(4*a^2 + b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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