3.470 \(\int \frac{\cos ^4(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=221 \[ \frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )}+\frac{3 a^2 \left (-5 a^2 b^2+2 a^4+4 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{3 a^3 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{3 a x}{b^4} \]

[Out]

(-3*a*x)/b^4 + (3*a^2*(2*a^4 - 5*a^2*b^2 + 4*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)
^(5/2)*b^4*(a + b)^(5/2)*d) + ((3*a^2 - 2*b^2)*Sin[c + d*x])/(2*b^3*(a^2 - b^2)*d) - (a^2*Cos[c + d*x]^2*Sin[c
+ d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + (3*a^3*(a^2 - 2*b^2)*Sin[c + d*x])/(2*b^3*(a^2 - b^2)^2*
d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.487519, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2792, 3031, 3023, 2735, 2659, 205} \[ \frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )}+\frac{3 a^2 \left (-5 a^2 b^2+2 a^4+4 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{3 a^3 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{3 a x}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*x)/b^4 + (3*a^2*(2*a^4 - 5*a^2*b^2 + 4*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)
^(5/2)*b^4*(a + b)^(5/2)*d) + ((3*a^2 - 2*b^2)*Sin[c + d*x])/(2*b^3*(a^2 - b^2)*d) - (a^2*Cos[c + d*x]^2*Sin[c
+ d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + (3*a^3*(a^2 - 2*b^2)*Sin[c + d*x])/(2*b^3*(a^2 - b^2)^2*
d*(a + b*Cos[c + d*x]))

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(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 - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3031

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

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)}{(a+b \cos (c+d x))^3} \, dx &=-\frac{a^2 \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\int \frac{\cos (c+d x) \left (2 a^2-2 a b \cos (c+d x)-\left (3 a^2-2 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{a^2 \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{3 a^3 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\int \frac{3 a^2 b \left (a^2-2 b^2\right )+a \left (3 a^2-4 b^2\right ) \left (a^2-b^2\right ) \cos (c+d x)-b \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{3 a^3 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\int \frac{3 a^2 b^2 \left (a^2-2 b^2\right )+6 a b \left (a^2-b^2\right )^2 \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a x}{b^4}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{3 a^3 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (3 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a x}{b^4}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{3 a^3 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (3 a^2 \left (2 a^4-5 a^2 b^2+4 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^4 \left (a^2-b^2\right )^2 d}\\ &=-\frac{3 a x}{b^4}+\frac{3 a^2 \left (2 a^4-5 a^2 b^2+4 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)^{5/2} b^4 (a+b)^{5/2} d}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{3 a^3 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.52214, size = 177, normalized size = 0.8 \[ \frac{\frac{a^3 b \left (5 a^2-8 b^2\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}-\frac{6 a^2 \left (-5 a^2 b^2+2 a^4+4 b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}-\frac{a^4 b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}-6 a (c+d x)+2 b \sin (c+d x)}{2 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.094, size = 679, normalized size = 3.1 \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/(a+b*cos(d*x+c))^3,x)

[Out]

2/d/b^3*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-6/d/b^4*a*arctan(tan(1/2*d*x+1/2*c))+4/d*a^5/b^3/(tan(1/2*
d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/d*a^4/b^2/(tan(1/2*d
*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-8/d*a^3/b/(tan(1/2*d*x+
1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+4/d*a^5/b^3/(tan(1/2*d*x+1
/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)+1/d*a^4/b^2/(tan(1/2*d*x+1/2*
c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)-8/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*
a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)+6/d*a^6/b^4/(a^4-2*a^2*b^2+b^4)/((a-b
)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-15/d*a^4/b^2/(a^4-2*a^2*b^2+b^4)/((a-b)*(a
+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+12/d*a^2/(a^4-2*a^2*b^2+b^4)/((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., 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/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.42042, size = 2209, normalized size = 10. \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/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(12*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*d*x*cos(d*x + c)^2 + 24*(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a
^2*b^7)*d*x*cos(d*x + c) + 12*(a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*x + 3*(2*a^8 - 5*a^6*b^2 + 4*a^4*b^4 +
(2*a^6*b^2 - 5*a^4*b^4 + 4*a^2*b^6)*cos(d*x + c)^2 + 2*(2*a^7*b - 5*a^5*b^3 + 4*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)*si
n(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^8*b - 17*a^6*b^3 + 13*a^4*
b^5 - 2*a^2*b^7 + 2*(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cos(d*x + c)^2 + (9*a^7*b^2 - 25*a^5*b^4 + 20*a^3*
b^6 - 4*a*b^8)*cos(d*x + c))*sin(d*x + c))/((a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c)^2 + 2*(a^
7*b^5 - 3*a^5*b^7 + 3*a^3*b^9 - a*b^11)*d*cos(d*x + c) + (a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d), -1/2
*(6*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*d*x*cos(d*x + c)^2 + 12*(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7
)*d*x*cos(d*x + c) + 6*(a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*x - 3*(2*a^8 - 5*a^6*b^2 + 4*a^4*b^4 + (2*a^6
*b^2 - 5*a^4*b^4 + 4*a^2*b^6)*cos(d*x + c)^2 + 2*(2*a^7*b - 5*a^5*b^3 + 4*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))) - (6*a^8*b - 17*a^6*b^3 + 13*a^4*b^5 - 2*a^2*b
^7 + 2*(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cos(d*x + c)^2 + (9*a^7*b^2 - 25*a^5*b^4 + 20*a^3*b^6 - 4*a*b^8
)*cos(d*x + c))*sin(d*x + c))/((a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c)^2 + 2*(a^7*b^5 - 3*a^5
*b^7 + 3*a^3*b^9 - a*b^11)*d*cos(d*x + c) + (a^8*b^4 - 3*a^6*b^6 + 3*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/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.45234, size = 478, normalized size = 2.16 \begin{align*} -\frac{\frac{3 \,{\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 4 \, 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^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sqrt{a^{2} - b^{2}}} - \frac{4 \, a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, 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} + 4 \, a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, 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 )}{{\left (a^{4} b^{3} - 2 \, 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 (d x + c\right )} a}{b^{4}} - \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-(3*(2*a^6 - 5*a^4*b^2 + 4*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^4*b^4 - 2*a^2*b^6 + b^8)*sqrt(a^2 - b^2)) - (4*a^6*t
an(1/2*d*x + 1/2*c)^3 - 5*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 + 4*a^6*tan(1/2*d*x + 1/2*c) + 5*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^4*b^3 - 2*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*(d*x + c)*a/b^4 - 2*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*b^3))/d