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

Optimal. Leaf size=237 \[ -\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac {(A b-a B) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac {\left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \sin (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))} \]

[Out]

(2*A*a^3+3*A*a*b^2-4*B*a^2*b-B*b^3)*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*b-B*a)*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^3-1/6*(5*A*a*b-2*B*a^2-3*B*b^2)*sin(d*x+c)/(a^2-b^2)^
2/d/(a+b*cos(d*x+c))^2-1/6*(11*A*a^2*b+4*A*b^3-2*B*a^3-13*B*a*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 = 0.48, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2754, 12, 2659, 205} \[ \frac {\left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {\left (11 a^2 A b-2 a^3 B-13 a b^2 B+4 A b^3\right ) \sin (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac {(A b-a B) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^3*A + 3*a*A*b^2 - 4*a^2*b*B - b^3*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*
(a + b)^(7/2)*d) - ((A*b - a*B)*Sin[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) - ((5*a*A*b - 2*a^2*B -
3*b^2*B)*Sin[c + d*x])/(6*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])^2) - ((11*a^2*A*b + 4*A*b^3 - 2*a^3*B - 13*a*b
^2*B)*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 {A+B \cos (c+d x)}{(a+b \cos (c+d x))^4} \, dx &=-\frac {(A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {\int \frac {-3 (a A-b B)+2 (A b-a B) \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {(A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {\left (5 a A b-2 a^2 B-3 b^2 B\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\int \frac {2 \left (3 a^2 A+2 A b^2-5 a b B\right )-\left (5 a A b-2 a^2 B-3 b^2 B\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 \left (a^2-b^2\right )^2}\\ &=-\frac {(A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {\left (5 a A b-2 a^2 B-3 b^2 B\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {\left (11 a^2 A b+4 A b^3-2 a^3 B-13 a b^2 B\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac {\int -\frac {3 \left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right )}{a+b \cos (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3}\\ &=-\frac {(A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {\left (5 a A b-2 a^2 B-3 b^2 B\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {\left (11 a^2 A b+4 A b^3-2 a^3 B-13 a b^2 B\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac {(A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {\left (5 a A b-2 a^2 B-3 b^2 B\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {\left (11 a^2 A b+4 A b^3-2 a^3 B-13 a b^2 B\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\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 {\left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\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 b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {\left (5 a A b-2 a^2 B-3 b^2 B\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {\left (11 a^2 A b+4 A b^3-2 a^3 B-13 a b^2 B\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 = 2.37, size = 227, normalized size = 0.96 \[ \frac {\frac {\left (2 a^2 B-5 a A b+3 b^2 B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}+\frac {\left (2 a^3 B-11 a^2 A b+13 a b^2 B-4 A b^3\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))}+\frac {6 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{7/2}}+\frac {2 (a B-A b) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^3}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*(2*a^3*A + 3*a*A*b^2 - 4*a^2*b*B - b^3*B)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^
2)^(7/2) + (2*(-(A*b) + a*B)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^3) + ((-5*a*A*b + 2*a^2*B + 3
*b^2*B)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])^2) + ((-11*a^2*A*b - 4*A*b^3 + 2*a^3*B + 13*a*
b^2*B)*Sin[c + d*x])/((a - b)^3*(a + b)^3*(a + b*Cos[c + d*x])))/(6*d)

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fricas [B]  time = 1.13, size = 1228, normalized size = 5.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*(2*A*a^6 - 4*B*a^5*b + 3*A*a^4*b^2 - B*a^3*b^3 + (2*A*a^3*b^3 - 4*B*a^2*b^4 + 3*A*a*b^5 - B*b^6)*cos
(d*x + c)^3 + 3*(2*A*a^4*b^2 - 4*B*a^3*b^3 + 3*A*a^2*b^4 - B*a*b^5)*cos(d*x + c)^2 + 3*(2*A*a^5*b - 4*B*a^4*b^
2 + 3*A*a^3*b^3 - B*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*sqrt(-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*B*a^7 - 18*A*a^6*b + 4*B*a^5*b^2 + 23*A*a^4*b^3 - 11*B*a^3*b^4 - 7*A*a^2*b^5 + B*a*b^6 +
2*A*b^7 + (2*B*a^5*b^2 - 11*A*a^4*b^3 + 11*B*a^3*b^4 + 7*A*a^2*b^5 - 13*B*a*b^6 + 4*A*b^7)*cos(d*x + c)^2 + 3*
(2*B*a^6*b - 9*A*a^5*b^2 + 7*B*a^4*b^3 + 8*A*a^3*b^4 - 10*B*a^2*b^5 + A*a*b^6 + B*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*(2*A*a^6 - 4*B*a^5*b + 3*A*a^4*b^2 - B
*a^3*b^3 + (2*A*a^3*b^3 - 4*B*a^2*b^4 + 3*A*a*b^5 - B*b^6)*cos(d*x + c)^3 + 3*(2*A*a^4*b^2 - 4*B*a^3*b^3 + 3*A
*a^2*b^4 - B*a*b^5)*cos(d*x + c)^2 + 3*(2*A*a^5*b - 4*B*a^4*b^2 + 3*A*a^3*b^3 - B*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*B*a^7 - 18*A*a^6*b + 4*B*a^5*b^2
+ 23*A*a^4*b^3 - 11*B*a^3*b^4 - 7*A*a^2*b^5 + B*a*b^6 + 2*A*b^7 + (2*B*a^5*b^2 - 11*A*a^4*b^3 + 11*B*a^3*b^4 +
7*A*a^2*b^5 - 13*B*a*b^6 + 4*A*b^7)*cos(d*x + c)^2 + 3*(2*B*a^6*b - 9*A*a^5*b^2 + 7*B*a^4*b^3 + 8*A*a^3*b^4 -
10*B*a^2*b^5 + A*a*b^6 + B*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 = 1.22, size = 691, normalized size = 2.92 \[ -\frac {\frac {3 \, {\left (2 \, A a^{3} - 4 \, B a^{2} b + 3 \, A a b^{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 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 27 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 32 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B 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((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,x, algorithm="giac")

[Out]

-1/3*(3*(2*A*a^3 - 4*B*a^2*b + 3*A*a*b^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*B*a^5*tan(1/2*d*x + 1/2*c)^5 - 18*A*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 6*B*a^4*b*tan(1/2*d*x + 1/
2*c)^5 + 27*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^2*b^3*tan(1/2*d*x +
1/2*c)^5 - 27*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 12*B*a*b^4*tan(1/2*d*x +
1/2*c)^5 - 6*A*b^5*tan(1/2*d*x + 1/2*c)^5 + 3*B*b^5*tan(1/2*d*x + 1/2*c)^5 + 12*B*a^5*tan(1/2*d*x + 1/2*c)^3 -
36*A*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 16*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 32*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^3
- 28*B*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*A*b^5*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^5*tan(1/2*d*x + 1/2*c) - 18*A*a^
4*b*tan(1/2*d*x + 1/2*c) + 6*B*a^4*b*tan(1/2*d*x + 1/2*c) - 27*A*a^3*b^2*tan(1/2*d*x + 1/2*c) + 12*B*a^3*b^2*t
an(1/2*d*x + 1/2*c) - 6*A*a^2*b^3*tan(1/2*d*x + 1/2*c) + 27*B*a^2*b^3*tan(1/2*d*x + 1/2*c) - 3*A*a*b^4*tan(1/2
*d*x + 1/2*c) + 12*B*a*b^4*tan(1/2*d*x + 1/2*c) - 6*A*b^5*tan(1/2*d*x + 1/2*c) - 3*B*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 = 0.08, size = 1727, normalized size = 7.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6/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*A-3/d*b^2/(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*A-2/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*A*b^3+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*B+2/d*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^2/(a-b)/(a^3+3*
a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*B+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-b)/(a^3
+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*B*a*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)^5*b^3*B-12/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^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-4/3/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^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A*b^3+4/d*a^3/(a*tan(1/2*d*x+1/2*c)^2-t
an(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*B+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^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B*a*b^2-6/d*a^2*b/(a*t
an(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)*A+3/d*b^2
/(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)*A-
2/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)*A
*b^3+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)*B-2/d*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^2/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(
1/2*d*x+1/2*c)*B+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+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan
(1/2*d*x+1/2*c)*B*a*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*B+2/d*a^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))*A+3/d*a*b^2/(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))*A-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))*B-1/d/(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))*b^3*B

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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((A+B*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 = 4.00, size = 440, normalized size = 1.86 \[ \frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,A\,a^3-4\,B\,a^2\,b+3\,A\,a\,b^2-B\,b^3\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-3\,B\,a^3+9\,A\,a^2\,b-7\,B\,a\,b^2+A\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,B\,a^3-2\,A\,b^3+B\,b^3-3\,A\,a\,b^2-6\,A\,a^2\,b+6\,B\,a\,b^2+2\,B\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b^3-2\,B\,a^3+B\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2+2\,B\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((tan(c/2 + (d*x)/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)))*(2*A*a
^3 - B*b^3 + 3*A*a*b^2 - 4*B*a^2*b))/(d*(a + b)^(7/2)*(a - b)^(7/2)) - ((4*tan(c/2 + (d*x)/2)^3*(A*b^3 - 3*B*a
^3 + 9*A*a^2*b - 7*B*a*b^2))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) - (tan(c/2 + (d*x)/2)^5*(2*B*a^3 - 2*A*b^3 + B*
b^3 - 3*A*a*b^2 - 6*A*a^2*b + 6*B*a*b^2 + 2*B*a^2*b))/((a + b)^3*(a - b)) + (tan(c/2 + (d*x)/2)*(2*A*b^3 - 2*B
*a^3 + B*b^3 - 3*A*a*b^2 + 6*A*a^2*b - 6*B*a*b^2 + 2*B*a^2*b))/((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)))

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

[Out]

Timed out

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