3.469 \(\int \frac{\cos ^5(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=300 \[ -\frac{3 a \left (-7 a^2 b^2+4 a^4+2 b^4\right ) \sin (c+d x)}{2 b^4 d \left (a^2-b^2\right )^2}-\frac{a^3 \left (-29 a^2 b^2+12 a^4+20 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 \left (4 a^2-7 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{\left (-10 a^2 b^2+6 a^4+b^4\right ) \sin (c+d x) \cos (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2}+\frac{x \left (12 a^2+b^2\right )}{2 b^5} \]

[Out]

((12*a^2 + b^2)*x)/(2*b^5) - (a^3*(12*a^4 - 29*a^2*b^2 + 20*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a
+ b]])/((a - b)^(5/2)*b^5*(a + b)^(5/2)*d) - (3*a*(4*a^4 - 7*a^2*b^2 + 2*b^4)*Sin[c + d*x])/(2*b^4*(a^2 - b^2)
^2*d) + ((6*a^4 - 10*a^2*b^2 + b^4)*Cos[c + d*x]*Sin[c + d*x])/(2*b^3*(a^2 - b^2)^2*d) - (a^2*Cos[c + d*x]^3*S
in[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - (a^2*(4*a^2 - 7*b^2)*Cos[c + d*x]^2*Sin[c + d*x])/(2
*b^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.783223, antiderivative size = 300, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2792, 3047, 3049, 3023, 2735, 2659, 205} \[ -\frac{3 a \left (-7 a^2 b^2+4 a^4+2 b^4\right ) \sin (c+d x)}{2 b^4 d \left (a^2-b^2\right )^2}-\frac{a^3 \left (-29 a^2 b^2+12 a^4+20 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 \left (4 a^2-7 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{\left (-10 a^2 b^2+6 a^4+b^4\right ) \sin (c+d x) \cos (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2}+\frac{x \left (12 a^2+b^2\right )}{2 b^5} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + b*Cos[c + d*x])^3,x]

[Out]

((12*a^2 + b^2)*x)/(2*b^5) - (a^3*(12*a^4 - 29*a^2*b^2 + 20*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a
+ b]])/((a - b)^(5/2)*b^5*(a + b)^(5/2)*d) - (3*a*(4*a^4 - 7*a^2*b^2 + 2*b^4)*Sin[c + d*x])/(2*b^4*(a^2 - b^2)
^2*d) + ((6*a^4 - 10*a^2*b^2 + b^4)*Cos[c + d*x]*Sin[c + d*x])/(2*b^3*(a^2 - b^2)^2*d) - (a^2*Cos[c + d*x]^3*S
in[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - (a^2*(4*a^2 - 7*b^2)*Cos[c + d*x]^2*Sin[c + d*x])/(2
*b^2*(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 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + 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*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(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, B, 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 ^5(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=-\frac{a^2 \cos ^3(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 ^2(c+d x) \left (3 a^2-2 a b \cos (c+d x)-2 \left (2 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{a^2 \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (4 a^2-7 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{\cos (c+d x) \left (-2 a^2 \left (4 a^2-7 b^2\right )+a b \left (a^2-4 b^2\right ) \cos (c+d x)+2 \left (6 a^4-10 a^2 b^2+b^4\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 (6 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (4 a^2-7 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{2 a \left (6 a^4-10 a^2 b^2+b^4\right )-2 b \left (2 a^4-4 a^2 b^2-b^4\right ) \cos (c+d x)-6 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{4 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (6 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (4 a^2-7 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{2 a b \left (6 a^4-10 a^2 b^2+b^4\right )+2 \left (a^2-b^2\right )^2 \left (12 a^2+b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{4 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{\left (12 a^2+b^2\right ) x}{2 b^5}-\frac{3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (6 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (4 a^2-7 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (a^3 \left (12 a^4-29 a^2 b^2+20 b^4\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^5 \left (a^2-b^2\right )^2}\\ &=\frac{\left (12 a^2+b^2\right ) x}{2 b^5}-\frac{3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (6 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (4 a^2-7 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (a^3 \left (12 a^4-29 a^2 b^2+20 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^5 \left (a^2-b^2\right )^2 d}\\ &=\frac{\left (12 a^2+b^2\right ) x}{2 b^5}-\frac{a^3 \left (12 a^4-29 a^2 b^2+20 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^5 (a+b)^{5/2} d}-\frac{3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (6 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (4 a^2-7 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.99726, size = 199, normalized size = 0.66 \[ \frac{2 \left (12 a^2+b^2\right ) (c+d x)+\frac{2 a^4 b \left (10 b^2-7 a^2\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}+\frac{4 a^3 \left (-29 a^2 b^2+12 a^4+20 b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac{2 a^5 b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}-12 a b \sin (c+d x)+b^2 \sin (2 (c+d x))}{4 b^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + b*Cos[c + d*x])^3,x]

[Out]

(2*(12*a^2 + b^2)*(c + d*x) + (4*a^3*(12*a^4 - 29*a^2*b^2 + 20*b^4)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a
^2 + b^2]])/(-a^2 + b^2)^(5/2) - 12*a*b*Sin[c + d*x] + (2*a^5*b*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c +
d*x])^2) + (2*a^4*b*(-7*a^2 + 10*b^2)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])) + b^2*Sin[2*(c
+ d*x)])/(4*b^5*d)

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Maple [B]  time = 0.101, size = 802, normalized size = 2.7 \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)^5/(a+b*cos(d*x+c))^3,x)

[Out]

-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*a-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*
c)^3-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*a+1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/
2*c)+12/d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^2+1/d/b^3*arctan(tan(1/2*d*x+1/2*c))-6/d*a^6/b^4/(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^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+10/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)^3-6/d*a^6/b^4/(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^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)+10/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-t
an(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)-12/d*a^7/b^5/(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))+29/d*a^5/b^3/(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))-20/d*a^3/b/(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)^5/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.48709, size = 2525, normalized size = 8.42 \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)^5/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(2*(12*a^8*b^2 - 35*a^6*b^4 + 33*a^4*b^6 - 9*a^2*b^8 - b^10)*d*x*cos(d*x + c)^2 + 4*(12*a^9*b - 35*a^7*b^
3 + 33*a^5*b^5 - 9*a^3*b^7 - a*b^9)*d*x*cos(d*x + c) + 2*(12*a^10 - 35*a^8*b^2 + 33*a^6*b^4 - 9*a^4*b^6 - a^2*
b^8)*d*x - (12*a^9 - 29*a^7*b^2 + 20*a^5*b^4 + (12*a^7*b^2 - 29*a^5*b^4 + 20*a^3*b^6)*cos(d*x + c)^2 + 2*(12*a
^8*b - 29*a^6*b^3 + 20*a^4*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*(12*a^9*b - 33*a^7*b^3 + 27*a^5*b^5 - 6*a^3*b^7 - (a^6*b^4 - 3*a^4*b^6 + 3*a^2*b^8 - b^10
)*cos(d*x + c)^3 + 4*(a^7*b^3 - 3*a^5*b^5 + 3*a^3*b^7 - a*b^9)*cos(d*x + c)^2 + (18*a^8*b^2 - 50*a^6*b^4 + 43*
a^4*b^6 - 11*a^2*b^8)*cos(d*x + c))*sin(d*x + c))/((a^6*b^7 - 3*a^4*b^9 + 3*a^2*b^11 - b^13)*d*cos(d*x + c)^2
+ 2*(a^7*b^6 - 3*a^5*b^8 + 3*a^3*b^10 - a*b^12)*d*cos(d*x + c) + (a^8*b^5 - 3*a^6*b^7 + 3*a^4*b^9 - a^2*b^11)*
d), 1/2*((12*a^8*b^2 - 35*a^6*b^4 + 33*a^4*b^6 - 9*a^2*b^8 - b^10)*d*x*cos(d*x + c)^2 + 2*(12*a^9*b - 35*a^7*b
^3 + 33*a^5*b^5 - 9*a^3*b^7 - a*b^9)*d*x*cos(d*x + c) + (12*a^10 - 35*a^8*b^2 + 33*a^6*b^4 - 9*a^4*b^6 - a^2*b
^8)*d*x - (12*a^9 - 29*a^7*b^2 + 20*a^5*b^4 + (12*a^7*b^2 - 29*a^5*b^4 + 20*a^3*b^6)*cos(d*x + c)^2 + 2*(12*a^
8*b - 29*a^6*b^3 + 20*a^4*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))) - (12*a^9*b - 33*a^7*b^3 + 27*a^5*b^5 - 6*a^3*b^7 - (a^6*b^4 - 3*a^4*b^6 + 3*a^2*b^8 - b^10)*cos(d
*x + c)^3 + 4*(a^7*b^3 - 3*a^5*b^5 + 3*a^3*b^7 - a*b^9)*cos(d*x + c)^2 + (18*a^8*b^2 - 50*a^6*b^4 + 43*a^4*b^6
- 11*a^2*b^8)*cos(d*x + c))*sin(d*x + c))/((a^6*b^7 - 3*a^4*b^9 + 3*a^2*b^11 - b^13)*d*cos(d*x + c)^2 + 2*(a^
7*b^6 - 3*a^5*b^8 + 3*a^3*b^10 - a*b^12)*d*cos(d*x + c) + (a^8*b^5 - 3*a^6*b^7 + 3*a^4*b^9 - a^2*b^11)*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)**5/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.38984, size = 1030, normalized size = 3.43 \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)^5/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(2*(12*a^7 - 29*a^5*b^2 + 20*a^3*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^5 - 2*a^2*b^7 + b^9)*sqrt(a^2 - b^2)) - 2*
(12*a^7*tan(1/2*d*x + 1/2*c)^7 - 18*a^6*b*tan(1/2*d*x + 1/2*c)^7 - 17*a^5*b^2*tan(1/2*d*x + 1/2*c)^7 + 33*a^4*
b^3*tan(1/2*d*x + 1/2*c)^7 - 2*a^3*b^4*tan(1/2*d*x + 1/2*c)^7 - 13*a^2*b^5*tan(1/2*d*x + 1/2*c)^7 + 4*a*b^6*ta
n(1/2*d*x + 1/2*c)^7 + b^7*tan(1/2*d*x + 1/2*c)^7 + 36*a^7*tan(1/2*d*x + 1/2*c)^5 - 18*a^6*b*tan(1/2*d*x + 1/2
*c)^5 - 67*a^5*b^2*tan(1/2*d*x + 1/2*c)^5 + 29*a^4*b^3*tan(1/2*d*x + 1/2*c)^5 + 26*a^3*b^4*tan(1/2*d*x + 1/2*c
)^5 - 5*a^2*b^5*tan(1/2*d*x + 1/2*c)^5 - 4*a*b^6*tan(1/2*d*x + 1/2*c)^5 - 3*b^7*tan(1/2*d*x + 1/2*c)^5 + 36*a^
7*tan(1/2*d*x + 1/2*c)^3 + 18*a^6*b*tan(1/2*d*x + 1/2*c)^3 - 67*a^5*b^2*tan(1/2*d*x + 1/2*c)^3 - 29*a^4*b^3*ta
n(1/2*d*x + 1/2*c)^3 + 26*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 + 5*a^2*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*a*b^6*tan(1/2*
d*x + 1/2*c)^3 + 3*b^7*tan(1/2*d*x + 1/2*c)^3 + 12*a^7*tan(1/2*d*x + 1/2*c) + 18*a^6*b*tan(1/2*d*x + 1/2*c) -
17*a^5*b^2*tan(1/2*d*x + 1/2*c) - 33*a^4*b^3*tan(1/2*d*x + 1/2*c) - 2*a^3*b^4*tan(1/2*d*x + 1/2*c) + 13*a^2*b^
5*tan(1/2*d*x + 1/2*c) + 4*a*b^6*tan(1/2*d*x + 1/2*c) - b^7*tan(1/2*d*x + 1/2*c))/((a^4*b^4 - 2*a^2*b^6 + b^8)
*(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 + 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b)^2) + (12*a^2 + b^2
)*(d*x + c)/b^5)/d