3.610 \(\int \frac{\cos ^4(c+d x) (1-\cos ^2(c+d x))}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=326 \[ -\frac{\left (-59 a^2 b^2+60 a^4+2 b^4\right ) \sin (c+d x)}{6 b^5 d \left (a^2-b^2\right )}-\frac{a^2 \left (-33 a^2 b^2+20 a^4+12 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^6 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (5 a^2-4 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{\left (20 a^2-17 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{6 b^3 d \left (a^2-b^2\right )}+\frac{a \left (10 a^2-9 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b^4 d \left (a^2-b^2\right )}+\frac{a x \left (20 a^2-3 b^2\right )}{2 b^6}+\frac{\sin (c+d x) \cos ^4(c+d x)}{2 b d (a+b \cos (c+d x))^2} \]

[Out]

(a*(20*a^2 - 3*b^2)*x)/(2*b^6) - (a^2*(20*a^4 - 33*a^2*b^2 + 12*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqr
t[a + b]])/((a - b)^(3/2)*b^6*(a + b)^(3/2)*d) - ((60*a^4 - 59*a^2*b^2 + 2*b^4)*Sin[c + d*x])/(6*b^5*(a^2 - b^
2)*d) + (a*(10*a^2 - 9*b^2)*Cos[c + d*x]*Sin[c + d*x])/(2*b^4*(a^2 - b^2)*d) - ((20*a^2 - 17*b^2)*Cos[c + d*x]
^2*Sin[c + d*x])/(6*b^3*(a^2 - b^2)*d) + (Cos[c + d*x]^4*Sin[c + d*x])/(2*b*d*(a + b*Cos[c + d*x])^2) + ((5*a^
2 - 4*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(2*b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.04671, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3048, 3049, 3023, 2735, 2659, 205} \[ -\frac{\left (-59 a^2 b^2+60 a^4+2 b^4\right ) \sin (c+d x)}{6 b^5 d \left (a^2-b^2\right )}-\frac{a^2 \left (-33 a^2 b^2+20 a^4+12 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^6 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (5 a^2-4 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{\left (20 a^2-17 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{6 b^3 d \left (a^2-b^2\right )}+\frac{a \left (10 a^2-9 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b^4 d \left (a^2-b^2\right )}+\frac{a x \left (20 a^2-3 b^2\right )}{2 b^6}+\frac{\sin (c+d x) \cos ^4(c+d x)}{2 b d (a+b \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*(1 - Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^3,x]

[Out]

(a*(20*a^2 - 3*b^2)*x)/(2*b^6) - (a^2*(20*a^4 - 33*a^2*b^2 + 12*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqr
t[a + b]])/((a - b)^(3/2)*b^6*(a + b)^(3/2)*d) - ((60*a^4 - 59*a^2*b^2 + 2*b^4)*Sin[c + d*x])/(6*b^5*(a^2 - b^
2)*d) + (a*(10*a^2 - 9*b^2)*Cos[c + d*x]*Sin[c + d*x])/(2*b^4*(a^2 - b^2)*d) - ((20*a^2 - 17*b^2)*Cos[c + d*x]
^2*Sin[c + d*x])/(6*b^3*(a^2 - b^2)*d) + (Cos[c + d*x]^4*Sin[c + d*x])/(2*b*d*(a + b*Cos[c + d*x])^2) + ((5*a^
2 - 4*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(2*b^2*(a^2 - b^2)*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 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))^3} \, dx &=\frac{\cos ^4(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\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))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{\cos ^4(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (5 a^2-4 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\cos ^2(c+d x) \left (3 \left (5 a^4-9 a^2 b^2+4 b^4\right )-a b \left (a^2-b^2\right ) \cos (c+d x)-\left (20 a^2-17 b^2\right ) \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (20 a^2-17 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{6 b^3 \left (a^2-b^2\right ) d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (5 a^2-4 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\cos (c+d x) \left (-2 a \left (20 a^4-37 a^2 b^2+17 b^4\right )+b \left (5 a^4-7 a^2 b^2+2 b^4\right ) \cos (c+d x)+6 a \left (10 a^2-9 b^2\right ) \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{a \left (10 a^2-9 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right ) d}-\frac{\left (20 a^2-17 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{6 b^3 \left (a^2-b^2\right ) d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (5 a^2-4 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{6 a^2 \left (10 a^4-19 a^2 b^2+9 b^4\right )-2 a b \left (10 a^4-17 a^2 b^2+7 b^4\right ) \cos (c+d x)-2 \left (60 a^6-119 a^4 b^2+61 a^2 b^4-2 b^6\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{12 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (60 a^4-59 a^2 b^2+2 b^4\right ) \sin (c+d x)}{6 b^5 \left (a^2-b^2\right ) d}+\frac{a \left (10 a^2-9 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right ) d}-\frac{\left (20 a^2-17 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{6 b^3 \left (a^2-b^2\right ) d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (5 a^2-4 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{6 a^2 b \left (10 a^4-19 a^2 b^2+9 b^4\right )+6 a \left (20 a^2-3 b^2\right ) \left (a^2-b^2\right )^2 \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{12 b^5 \left (a^2-b^2\right )^2}\\ &=\frac{a \left (20 a^2-3 b^2\right ) x}{2 b^6}-\frac{\left (60 a^4-59 a^2 b^2+2 b^4\right ) \sin (c+d x)}{6 b^5 \left (a^2-b^2\right ) d}+\frac{a \left (10 a^2-9 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right ) d}-\frac{\left (20 a^2-17 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{6 b^3 \left (a^2-b^2\right ) d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (5 a^2-4 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (a^2 \left (20 a^4-33 a^2 b^2+12 b^4\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^6 \left (a^2-b^2\right )}\\ &=\frac{a \left (20 a^2-3 b^2\right ) x}{2 b^6}-\frac{\left (60 a^4-59 a^2 b^2+2 b^4\right ) \sin (c+d x)}{6 b^5 \left (a^2-b^2\right ) d}+\frac{a \left (10 a^2-9 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right ) d}-\frac{\left (20 a^2-17 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{6 b^3 \left (a^2-b^2\right ) d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (5 a^2-4 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (a^2 \left (20 a^4-33 a^2 b^2+12 b^4\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 \left (a^2-b^2\right ) d}\\ &=\frac{a \left (20 a^2-3 b^2\right ) x}{2 b^6}-\frac{a^2 \left (20 a^4-33 a^2 b^2+12 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} b^6 (a+b)^{3/2} d}-\frac{\left (60 a^4-59 a^2 b^2+2 b^4\right ) \sin (c+d x)}{6 b^5 \left (a^2-b^2\right ) d}+\frac{a \left (10 a^2-9 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right ) d}-\frac{\left (20 a^2-17 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{6 b^3 \left (a^2-b^2\right ) d}+\frac{\cos ^4(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (5 a^2-4 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [B]  time = 7.20272, size = 979, normalized size = 3. \[ \frac{-\frac{12 \left (-48 a (c+d x)-\frac{6 \left (16 a^6-40 b^2 a^4+30 b^4 a^2-5 b^6\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 )^{5/2}}+16 b \sin (c+d x)+\frac{a b \left (40 a^4-72 b^2 a^2+29 b^4\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}-\frac{b \left (8 a^4-8 b^2 a^2+b^4\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}\right )}{b^4}+12 \left (\frac{b \left (-4 a^2-3 b \cos (c+d x) a+b^2\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}-\frac{2 \left (2 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 )^{5/2}}\right )+\frac{6 \left (\frac{\left (a \left (2 a^2-5 b^2\right ) \cos (c+d x)-b \left (2 a^2+b^2\right )\right ) \sin (c+d x)}{(a+b \cos (c+d x))^2}-\frac{6 b^2 \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}}\right )}{(a-b)^2 (a+b)^2}+\frac{\frac{12 \left (640 a^8-1792 b^2 a^6+1680 b^4 a^4-560 b^6 a^2+35 b^8\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 )^{5/2}}+\frac{3840 c a^9+3840 d x a^9-3840 b \sin (c+d x) a^8-6912 b^2 c a^7-6912 b^2 d x a^7-2880 b^2 \sin (2 (c+d x)) a^7+7872 b^3 \sin (c+d x) a^6-320 b^3 \sin (3 (c+d x)) a^6+1728 b^4 c a^5+1728 b^4 d x a^5+6304 b^4 \sin (2 (c+d x)) a^5+40 b^4 \sin (4 (c+d x)) a^5-4256 b^5 \sin (c+d x) a^4+696 b^5 \sin (3 (c+d x)) a^4-8 b^5 \sin (5 (c+d x)) a^4+1920 b^6 c a^3+1920 b^6 d x a^3-4022 b^6 \sin (2 (c+d x)) a^3-80 b^6 \sin (4 (c+d x)) a^3+172 b^7 \sin (c+d x) a^2-432 b^7 \sin (3 (c+d x)) a^2+16 b^7 \sin (5 (c+d x)) a^2-576 b^8 c a-576 b^8 d x a+192 b^2 \left (10 a^2-3 b^2\right ) \left (a^2-b^2\right )^2 (c+d x) \cos (2 (c+d x)) a+607 b^8 \sin (2 (c+d x)) a+40 b^8 \sin (4 (c+d x)) a+768 b \left (10 a^2-3 b^2\right ) \left (a^3-a b^2\right )^2 (c+d x) \cos (c+d x)+70 b^9 \sin (c+d x)+56 b^9 \sin (3 (c+d x))-8 b^9 \sin (5 (c+d x))}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}}{b^6}}{384 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*(1 - Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^3,x]

[Out]

((-12*(-48*a*(c + d*x) - (6*(16*a^6 - 40*a^4*b^2 + 30*a^2*b^4 - 5*b^6)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt
[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + 16*b*Sin[c + d*x] - (b*(8*a^4 - 8*a^2*b^2 + b^4)*Sin[c + d*x])/((a - b)*(a
+ b)*(a + b*Cos[c + d*x])^2) + (a*b*(40*a^4 - 72*a^2*b^2 + 29*b^4)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*
Cos[c + d*x]))))/b^4 + 12*((-2*(2*a^2 + b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2
)^(5/2) + (b*(-4*a^2 + b^2 - 3*a*b*Cos[c + d*x])*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])^2)) +
(6*((-6*b^2*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + ((-(b*(2*a^2 + b^2)) + a
*(2*a^2 - 5*b^2)*Cos[c + d*x])*Sin[c + d*x])/(a + b*Cos[c + d*x])^2))/((a - b)^2*(a + b)^2) + ((12*(640*a^8 -
1792*a^6*b^2 + 1680*a^4*b^4 - 560*a^2*b^6 + 35*b^8)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^
2 + b^2)^(5/2) + (3840*a^9*c - 6912*a^7*b^2*c + 1728*a^5*b^4*c + 1920*a^3*b^6*c - 576*a*b^8*c + 3840*a^9*d*x -
6912*a^7*b^2*d*x + 1728*a^5*b^4*d*x + 1920*a^3*b^6*d*x - 576*a*b^8*d*x + 768*b*(10*a^2 - 3*b^2)*(a^3 - a*b^2)
^2*(c + d*x)*Cos[c + d*x] + 192*a*b^2*(10*a^2 - 3*b^2)*(a^2 - b^2)^2*(c + d*x)*Cos[2*(c + d*x)] - 3840*a^8*b*S
in[c + d*x] + 7872*a^6*b^3*Sin[c + d*x] - 4256*a^4*b^5*Sin[c + d*x] + 172*a^2*b^7*Sin[c + d*x] + 70*b^9*Sin[c
+ d*x] - 2880*a^7*b^2*Sin[2*(c + d*x)] + 6304*a^5*b^4*Sin[2*(c + d*x)] - 4022*a^3*b^6*Sin[2*(c + d*x)] + 607*a
*b^8*Sin[2*(c + d*x)] - 320*a^6*b^3*Sin[3*(c + d*x)] + 696*a^4*b^5*Sin[3*(c + d*x)] - 432*a^2*b^7*Sin[3*(c + d
*x)] + 56*b^9*Sin[3*(c + d*x)] + 40*a^5*b^4*Sin[4*(c + d*x)] - 80*a^3*b^6*Sin[4*(c + d*x)] + 40*a*b^8*Sin[4*(c
+ d*x)] - 8*a^4*b^5*Sin[5*(c + d*x)] + 16*a^2*b^7*Sin[5*(c + d*x)] - 8*b^9*Sin[5*(c + d*x)])/((a^2 - b^2)^2*(
a + b*Cos[c + d*x])^2))/b^6)/(384*d)

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Maple [B]  time = 0.039, size = 786, normalized size = 2.4 \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))^3,x)

[Out]

-12/d/b^5/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^5*a^2-3/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1
/2*c)^5*a-24/d/b^5/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^3*a^2+8/3/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^3*ta
n(1/2*d*x+1/2*c)^3-12/d/b^5/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)*a^2+3/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)
^3*tan(1/2*d*x+1/2*c)*a+20/d/b^6*arctan(tan(1/2*d*x+1/2*c))*a^3-3/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a-8/d*a^5/b
^5/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3+1/d*a^4/b^4/(a*tan(1/2*d*x
+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3+8/d*a^3/b^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2
*d*x+1/2*c)^2*b+a+b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3-8/d*a^5/b^5/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a
+b)^2/(a-b)*tan(1/2*d*x+1/2*c)-1/d*a^4/b^4/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)*tan(1/2
*d*x+1/2*c)+8/d*a^3/b^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)*tan(1/2*d*x+1/2*c)-20/d*a^
6/b^6/(a^2-b^2)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+33/d*a^4/b^4/(a^2-b^2
)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-12/d*a^2/b^2/(a^2-b^2)/((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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.16855, size = 2471, normalized size = 7.58 \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))^3,x, algorithm="fricas")

[Out]

[1/12*(6*(20*a^7*b^2 - 43*a^5*b^4 + 26*a^3*b^6 - 3*a*b^8)*d*x*cos(d*x + c)^2 + 12*(20*a^8*b - 43*a^6*b^3 + 26*
a^4*b^5 - 3*a^2*b^7)*d*x*cos(d*x + c) + 6*(20*a^9 - 43*a^7*b^2 + 26*a^5*b^4 - 3*a^3*b^6)*d*x + 3*(20*a^8 - 33*
a^6*b^2 + 12*a^4*b^4 + (20*a^6*b^2 - 33*a^4*b^4 + 12*a^2*b^6)*cos(d*x + c)^2 + 2*(20*a^7*b - 33*a^5*b^3 + 12*a
^3*b^5)*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*(60
*a^8*b - 119*a^6*b^3 + 61*a^4*b^5 - 2*a^2*b^7 + 2*(a^4*b^5 - 2*a^2*b^7 + b^9)*cos(d*x + c)^4 - 5*(a^5*b^4 - 2*
a^3*b^6 + a*b^8)*cos(d*x + c)^3 + 2*(10*a^6*b^3 - 21*a^4*b^5 + 12*a^2*b^7 - b^9)*cos(d*x + c)^2 + (90*a^7*b^2
- 181*a^5*b^4 + 95*a^3*b^6 - 4*a*b^8)*cos(d*x + c))*sin(d*x + c))/((a^4*b^8 - 2*a^2*b^10 + b^12)*d*cos(d*x + c
)^2 + 2*(a^5*b^7 - 2*a^3*b^9 + a*b^11)*d*cos(d*x + c) + (a^6*b^6 - 2*a^4*b^8 + a^2*b^10)*d), 1/6*(3*(20*a^7*b^
2 - 43*a^5*b^4 + 26*a^3*b^6 - 3*a*b^8)*d*x*cos(d*x + c)^2 + 6*(20*a^8*b - 43*a^6*b^3 + 26*a^4*b^5 - 3*a^2*b^7)
*d*x*cos(d*x + c) + 3*(20*a^9 - 43*a^7*b^2 + 26*a^5*b^4 - 3*a^3*b^6)*d*x - 3*(20*a^8 - 33*a^6*b^2 + 12*a^4*b^4
+ (20*a^6*b^2 - 33*a^4*b^4 + 12*a^2*b^6)*cos(d*x + c)^2 + 2*(20*a^7*b - 33*a^5*b^3 + 12*a^3*b^5)*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))) - (60*a^8*b - 119*a^6*b^3 + 61*
a^4*b^5 - 2*a^2*b^7 + 2*(a^4*b^5 - 2*a^2*b^7 + b^9)*cos(d*x + c)^4 - 5*(a^5*b^4 - 2*a^3*b^6 + a*b^8)*cos(d*x +
c)^3 + 2*(10*a^6*b^3 - 21*a^4*b^5 + 12*a^2*b^7 - b^9)*cos(d*x + c)^2 + (90*a^7*b^2 - 181*a^5*b^4 + 95*a^3*b^6
- 4*a*b^8)*cos(d*x + c))*sin(d*x + c))/((a^4*b^8 - 2*a^2*b^10 + b^12)*d*cos(d*x + c)^2 + 2*(a^5*b^7 - 2*a^3*b
^9 + a*b^11)*d*cos(d*x + c) + (a^6*b^6 - 2*a^4*b^8 + a^2*b^10)*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.7102, size = 587, normalized size = 1.8 \begin{align*} \frac{\frac{6 \,{\left (20 \, a^{6} - 33 \, a^{4} b^{2} + 12 \, a^{2} b^{4}\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^{2} b^{6} - b^{8}\right )} \sqrt{a^{2} - b^{2}}} - \frac{6 \,{\left (8 \, a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, a^{5} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a^{5} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 7 \, a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{2} b^{5} - b^{7}\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 )}^{2}} + \frac{3 \,{\left (20 \, a^{3} - 3 \, a b^{2}\right )}{\left (d x + c\right )}}{b^{6}} - \frac{2 \,{\left (36 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 72 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, a b \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 )}^{3} b^{5}}}{6 \, 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))^3,x, algorithm="giac")

[Out]

1/6*(6*(20*a^6 - 33*a^4*b^2 + 12*a^2*b^4)*(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^2*b^6 - b^8)*sqrt(a^2 - b^2)) - 6*(8*a^6*tan(1
/2*d*x + 1/2*c)^3 - 9*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 7*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 + 8*a^3*b^3*tan(1/2*d*x
+ 1/2*c)^3 + 8*a^6*tan(1/2*d*x + 1/2*c) + 9*a^5*b*tan(1/2*d*x + 1/2*c) - 7*a^4*b^2*tan(1/2*d*x + 1/2*c) - 8*a^
3*b^3*tan(1/2*d*x + 1/2*c))/((a^2*b^5 - b^7)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2)
+ 3*(20*a^3 - 3*a*b^2)*(d*x + c)/b^6 - 2*(36*a^2*tan(1/2*d*x + 1/2*c)^5 + 9*a*b*tan(1/2*d*x + 1/2*c)^5 + 72*a^
2*tan(1/2*d*x + 1/2*c)^3 - 8*b^2*tan(1/2*d*x + 1/2*c)^3 + 36*a^2*tan(1/2*d*x + 1/2*c) - 9*a*b*tan(1/2*d*x + 1/
2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^5))/d