3.601 \(\int \frac{\cos ^4(c+d x) (1-\cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=237 \[ \frac{a \left (15 a^2-2 b^2\right ) \sin (c+d x)}{3 b^5 d}+\frac{2 a^3 \left (5 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^6 d \sqrt{a-b} \sqrt{a+b}}-\frac{\left (20 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}-\frac{x \left (-12 a^2 b^2+40 a^4-b^4\right )}{8 b^6}+\frac{5 a \sin (c+d x) \cos ^2(c+d x)}{3 b^3 d}+\frac{\sin (c+d x) \cos ^4(c+d x)}{b d (a+b \cos (c+d x))}-\frac{5 \sin (c+d x) \cos ^3(c+d x)}{4 b^2 d} \]

[Out]

-((40*a^4 - 12*a^2*b^2 - b^4)*x)/(8*b^6) + (2*a^3*(5*a^2 - 4*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a
+ b]])/(Sqrt[a - b]*b^6*Sqrt[a + b]*d) + (a*(15*a^2 - 2*b^2)*Sin[c + d*x])/(3*b^5*d) - ((20*a^2 - b^2)*Cos[c
+ d*x]*Sin[c + d*x])/(8*b^4*d) + (5*a*Cos[c + d*x]^2*Sin[c + d*x])/(3*b^3*d) - (5*Cos[c + d*x]^3*Sin[c + d*x])
/(4*b^2*d) + (Cos[c + d*x]^4*Sin[c + d*x])/(b*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.898475, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {3048, 3050, 3049, 3023, 2735, 2659, 205} \[ \frac{a \left (15 a^2-2 b^2\right ) \sin (c+d x)}{3 b^5 d}+\frac{2 a^3 \left (5 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^6 d \sqrt{a-b} \sqrt{a+b}}-\frac{\left (20 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}-\frac{x \left (-12 a^2 b^2+40 a^4-b^4\right )}{8 b^6}+\frac{5 a \sin (c+d x) \cos ^2(c+d x)}{3 b^3 d}+\frac{\sin (c+d x) \cos ^4(c+d x)}{b d (a+b \cos (c+d x))}-\frac{5 \sin (c+d x) \cos ^3(c+d x)}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*(1 - Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^2,x]

[Out]

-((40*a^4 - 12*a^2*b^2 - b^4)*x)/(8*b^6) + (2*a^3*(5*a^2 - 4*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a
+ b]])/(Sqrt[a - b]*b^6*Sqrt[a + b]*d) + (a*(15*a^2 - 2*b^2)*Sin[c + d*x])/(3*b^5*d) - ((20*a^2 - b^2)*Cos[c
+ d*x]*Sin[c + d*x])/(8*b^4*d) + (5*a*Cos[c + d*x]^2*Sin[c + d*x])/(3*b^3*d) - (5*Cos[c + d*x]^3*Sin[c + d*x])
/(4*b^2*d) + (Cos[c + d*x]^4*Sin[c + d*x])/(b*d*(a + b*Cos[c + d*x]))

Rule 3048

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

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n
*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0
] && NeQ[c, 0])))

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 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 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 ^4(c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx &=\frac{\cos ^4(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac{\int \frac{\cos ^3(c+d x) \left (-4 \left (a^2-b^2\right )+5 \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{5 \cos ^3(c+d x) \sin (c+d x)}{4 b^2 d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac{\int \frac{\cos ^2(c+d x) \left (15 a \left (a^2-b^2\right )-b \left (a^2-b^2\right ) \cos (c+d x)-20 a \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{4 b^2 \left (a^2-b^2\right )}\\ &=\frac{5 a \cos ^2(c+d x) \sin (c+d x)}{3 b^3 d}-\frac{5 \cos ^3(c+d x) \sin (c+d x)}{4 b^2 d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (-40 a^2 \left (a^2-b^2\right )+5 a b \left (a^2-b^2\right ) \cos (c+d x)+3 \left (a^2-b^2\right ) \left (20 a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{12 b^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (20 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{5 a \cos ^2(c+d x) \sin (c+d x)}{3 b^3 d}-\frac{5 \cos ^3(c+d x) \sin (c+d x)}{4 b^2 d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac{\int \frac{3 a \left (20 a^4-21 a^2 b^2+b^4\right )-b \left (a^2-b^2\right ) \left (20 a^2+3 b^2\right ) \cos (c+d x)-8 a \left (15 a^2-2 b^2\right ) \left (a^2-b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^4 \left (a^2-b^2\right )}\\ &=\frac{a \left (15 a^2-2 b^2\right ) \sin (c+d x)}{3 b^5 d}-\frac{\left (20 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{5 a \cos ^2(c+d x) \sin (c+d x)}{3 b^3 d}-\frac{5 \cos ^3(c+d x) \sin (c+d x)}{4 b^2 d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac{\int \frac{3 a b \left (20 a^4-21 a^2 b^2+b^4\right )+3 \left (40 a^6-52 a^4 b^2+11 a^2 b^4+b^6\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^5 \left (a^2-b^2\right )}\\ &=-\frac{\left (40 a^4-12 a^2 b^2-b^4\right ) x}{8 b^6}+\frac{a \left (15 a^2-2 b^2\right ) \sin (c+d x)}{3 b^5 d}-\frac{\left (20 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{5 a \cos ^2(c+d x) \sin (c+d x)}{3 b^3 d}-\frac{5 \cos ^3(c+d x) \sin (c+d x)}{4 b^2 d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}+\frac{\left (a^3 \left (5 a^2-4 b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^6}\\ &=-\frac{\left (40 a^4-12 a^2 b^2-b^4\right ) x}{8 b^6}+\frac{a \left (15 a^2-2 b^2\right ) \sin (c+d x)}{3 b^5 d}-\frac{\left (20 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{5 a \cos ^2(c+d x) \sin (c+d x)}{3 b^3 d}-\frac{5 \cos ^3(c+d x) \sin (c+d x)}{4 b^2 d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}+\frac{\left (2 a^3 \left (5 a^2-4 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 )}{b^6 d}\\ &=-\frac{\left (40 a^4-12 a^2 b^2-b^4\right ) x}{8 b^6}+\frac{2 a^3 \left (5 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^6 \sqrt{a+b} d}+\frac{a \left (15 a^2-2 b^2\right ) \sin (c+d x)}{3 b^5 d}-\frac{\left (20 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{5 a \cos ^2(c+d x) \sin (c+d x)}{3 b^3 d}-\frac{5 \cos ^3(c+d x) \sin (c+d x)}{4 b^2 d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.75048, size = 271, normalized size = 1.14 \[ \frac{\frac{240 a^3 b^2 \sin (2 (c+d x))+24 a^2 b \left (40 a^2-7 b^2\right ) \sin (c+d x)-40 a^2 b^3 \sin (3 (c+d x))+24 b \left (12 a^2 b^2-40 a^4+b^4\right ) (c+d x) \cos (c+d x)+288 a^3 b^2 c+288 a^3 b^2 d x-960 a^5 c-960 a^5 d x-32 a b^4 \sin (2 (c+d x))+10 a b^4 \sin (4 (c+d x))+24 a b^4 c+24 a b^4 d x-3 b^5 \sin (3 (c+d x))-3 b^5 \sin (5 (c+d x))}{a+b \cos (c+d x)}-\frac{384 a^3 \left (5 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}}{192 b^6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*(1 - Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^2,x]

[Out]

((-384*a^3*(5*a^2 - 4*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + (-960*a^5*
c + 288*a^3*b^2*c + 24*a*b^4*c - 960*a^5*d*x + 288*a^3*b^2*d*x + 24*a*b^4*d*x + 24*b*(-40*a^4 + 12*a^2*b^2 + b
^4)*(c + d*x)*Cos[c + d*x] + 24*a^2*b*(40*a^2 - 7*b^2)*Sin[c + d*x] + 240*a^3*b^2*Sin[2*(c + d*x)] - 32*a*b^4*
Sin[2*(c + d*x)] - 40*a^2*b^3*Sin[3*(c + d*x)] - 3*b^5*Sin[3*(c + d*x)] + 10*a*b^4*Sin[4*(c + d*x)] - 3*b^5*Si
n[5*(c + d*x)])/(a + b*Cos[c + d*x]))/(192*b^6*d)

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Maple [B]  time = 0.038, size = 708, normalized size = 3. \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)^4*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x)

[Out]

8/d/b^5/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^7*a^3+3/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2
*c)^7*a^2+1/4/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^7+24/d/b^5/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/
2*d*x+1/2*c)^5*a^3+3/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^5*a^2-16/3/d/b^3/(tan(1/2*d*x+1/2*c)^
2+1)^4*tan(1/2*d*x+1/2*c)^5*a-7/4/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^5+24/d/b^5/(tan(1/2*d*x+
1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^3*a^3-3/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^3*a^2+7/4/d/b^2/(
tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^3-16/3/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^3*a+8/
d/b^5/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)*a^3-3/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)*
a^2-1/4/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)-10/d/b^6*arctan(tan(1/2*d*x+1/2*c))*a^4+3/d/b^4*ar
ctan(tan(1/2*d*x+1/2*c))*a^2+1/4/d/b^2*arctan(tan(1/2*d*x+1/2*c))+2/d*a^4/b^5*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*
x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)+10/d*a^5/b^6/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)
*(a-b))^(1/2))-8/d*a^3/b^4/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))

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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)^4*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.02855, size = 1634, normalized size = 6.89 \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)^4*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/24*(3*(40*a^6*b - 52*a^4*b^3 + 11*a^2*b^5 + b^7)*d*x*cos(d*x + c) + 3*(40*a^7 - 52*a^5*b^2 + 11*a^3*b^4 +
a*b^6)*d*x - 12*(5*a^6 - 4*a^4*b^2 + (5*a^5*b - 4*a^3*b^3)*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)) - (120*a^6*b - 136*a^4*b^3 + 16*a^2*b^5 - 6*(a^2*b^5 - b^7)*cos(d*
x + c)^4 + 10*(a^3*b^4 - a*b^6)*cos(d*x + c)^3 - (20*a^4*b^3 - 23*a^2*b^5 + 3*b^7)*cos(d*x + c)^2 + (60*a^5*b^
2 - 73*a^3*b^4 + 13*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^2*b^7 - b^9)*d*cos(d*x + c) + (a^3*b^6 - a*b^8)*d),
-1/24*(3*(40*a^6*b - 52*a^4*b^3 + 11*a^2*b^5 + b^7)*d*x*cos(d*x + c) + 3*(40*a^7 - 52*a^5*b^2 + 11*a^3*b^4 +
a*b^6)*d*x - 24*(5*a^6 - 4*a^4*b^2 + (5*a^5*b - 4*a^3*b^3)*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))) - (120*a^6*b - 136*a^4*b^3 + 16*a^2*b^5 - 6*(a^2*b^5 - b^7)*cos(d*x +
c)^4 + 10*(a^3*b^4 - a*b^6)*cos(d*x + c)^3 - (20*a^4*b^3 - 23*a^2*b^5 + 3*b^7)*cos(d*x + c)^2 + (60*a^5*b^2 -
73*a^3*b^4 + 13*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^2*b^7 - b^9)*d*cos(d*x + c) + (a^3*b^6 - a*b^8)*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)**4*(1-cos(d*x+c)**2)/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.25701, size = 568, normalized size = 2.4 \begin{align*} \frac{\frac{48 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\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 )} b^{5}} - \frac{3 \,{\left (40 \, a^{4} - 12 \, a^{2} b^{2} - b^{4}\right )}{\left (d x + c\right )}}{b^{6}} - \frac{48 \,{\left (5 \, a^{5} - 4 \, a^{3} b^{2}\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 )}}{\sqrt{a^{2} - b^{2}} b^{6}} + \frac{2 \,{\left (96 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 36 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 3 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 288 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 64 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 21 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 288 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 64 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 96 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 36 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4} b^{5}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(48*a^4*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)*b^5) - 3*(40*
a^4 - 12*a^2*b^2 - b^4)*(d*x + c)/b^6 - 48*(5*a^5 - 4*a^3*b^2)*(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)))/(sqrt(a^2 - b^2)*b^6) + 2*(96
*a^3*tan(1/2*d*x + 1/2*c)^7 + 36*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 3*b^3*tan(1/2*d*x + 1/2*c)^7 + 288*a^3*tan(1/2
*d*x + 1/2*c)^5 + 36*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 64*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 21*b^3*tan(1/2*d*x + 1/2
*c)^5 + 288*a^3*tan(1/2*d*x + 1/2*c)^3 - 36*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 64*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 2
1*b^3*tan(1/2*d*x + 1/2*c)^3 + 96*a^3*tan(1/2*d*x + 1/2*c) - 36*a^2*b*tan(1/2*d*x + 1/2*c) - 3*b^3*tan(1/2*d*x
+ 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*b^5))/d