3.459 \(\int \frac{\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=266 \[ \frac{\left (-7 a^2 b^2+12 a^4-2 b^4\right ) \sin (c+d x)}{3 b^4 d \left (a^2-b^2\right )}+\frac{2 a^4 \left (4 a^2-5 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d \left (a^2-b^2\right )}-\frac{a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b^3 d \left (a^2-b^2\right )}-\frac{a x \left (4 a^2+b^2\right )}{b^5} \]

[Out]

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

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Rubi [A]  time = 0.722972, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2792, 3049, 3023, 2735, 2659, 205} \[ \frac{\left (-7 a^2 b^2+12 a^4-2 b^4\right ) \sin (c+d x)}{3 b^4 d \left (a^2-b^2\right )}+\frac{2 a^4 \left (4 a^2-5 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d \left (a^2-b^2\right )}-\frac{a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b^3 d \left (a^2-b^2\right )}-\frac{a x \left (4 a^2+b^2\right )}{b^5} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + b*Cos[c + d*x])^2,x]

[Out]

-((a*(4*a^2 + b^2)*x)/b^5) + (2*a^4*(4*a^2 - 5*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a -
b)^(3/2)*b^5*(a + b)^(3/2)*d) + ((12*a^4 - 7*a^2*b^2 - 2*b^4)*Sin[c + d*x])/(3*b^4*(a^2 - b^2)*d) - (a*(2*a^2
- b^2)*Cos[c + d*x]*Sin[c + d*x])/(b^3*(a^2 - b^2)*d) + ((4*a^2 - b^2)*Cos[c + d*x]^2*Sin[c + d*x])/(3*b^2*(a^
2 - b^2)*d) - (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(b*(a^2 - b^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 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))^2} \, dx &=-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\cos ^2(c+d x) \left (3 a^2-a b \cos (c+d x)-\left (4 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{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (-2 a \left (4 a^2-b^2\right )+b \left (a^2+2 b^2\right ) \cos (c+d x)+6 a \left (2 a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac{a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{6 a^2 \left (2 a^2-b^2\right )-2 a b \left (2 a^2+b^2\right ) \cos (c+d x)-2 \left (12 a^4-7 a^2 b^2-2 b^4\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )}\\ &=\frac{\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{6 a^2 b \left (2 a^2-b^2\right )+6 a \left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^4 \left (a^2-b^2\right )}\\ &=-\frac{a \left (4 a^2+b^2\right ) x}{b^5}+\frac{\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a^4 \left (4 a^2-5 b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^5 \left (a^2-b^2\right )}\\ &=-\frac{a \left (4 a^2+b^2\right ) x}{b^5}+\frac{\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (2 a^4 \left (4 a^2-5 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^5 \left (a^2-b^2\right ) d}\\ &=-\frac{a \left (4 a^2+b^2\right ) x}{b^5}+\frac{2 a^4 \left (4 a^2-5 b^2\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^5 (a+b)^{3/2} d}+\frac{\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.879634, size = 176, normalized size = 0.66 \[ \frac{9 b \left (4 a^2+b^2\right ) \sin (c+d x)+\frac{24 a^4 \left (4 a^2-5 b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{12 a^5 b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-6 a b^2 \sin (2 (c+d x))-12 a (2 a-i b) (2 a+i b) (c+d x)+b^3 \sin (3 (c+d x))}{12 b^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + b*Cos[c + d*x])^2,x]

[Out]

(-12*a*(2*a - I*b)*(2*a + I*b)*(c + d*x) + (24*a^4*(4*a^2 - 5*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^
2 + b^2]])/(-a^2 + b^2)^(3/2) + 9*b*(4*a^2 + b^2)*Sin[c + d*x] + (12*a^5*b*Sin[c + d*x])/((a - b)*(a + b)*(a +
b*Cos[c + d*x])) - 6*a*b^2*Sin[2*(c + d*x)] + b^3*Sin[3*(c + d*x)])/(12*b^5*d)

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Maple [A]  time = 0.098, size = 504, normalized size = 1.9 \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))^2,x)

[Out]

6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*a^2+2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2
*c)^5*a+2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5+12/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*
x+1/2*c)^3*a^2+4/3/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*ta
n(1/2*d*x+1/2*c)*a^2-2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*a+2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^
3*tan(1/2*d*x+1/2*c)-8/d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^3-2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a+2/d*a^5/b^4/(
a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)+8/d*a^6/b^5/(a-b)/(a+b)/((a-b)
*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-10/d*a^4/b^3/(a-b)/(a+b)/((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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.6388, size = 1644, normalized size = 6.18 \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))^2,x, algorithm="fricas")

[Out]

[-1/6*(6*(4*a^7*b - 7*a^5*b^3 + 2*a^3*b^5 + a*b^7)*d*x*cos(d*x + c) + 6*(4*a^8 - 7*a^6*b^2 + 2*a^4*b^4 + a^2*b
^6)*d*x + 3*(4*a^7 - 5*a^5*b^2 + (4*a^6*b - 5*a^4*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)) - 2*(12*a^7*b - 19*a^5*b^3 + 5*a^3*b^5 + 2*a*b^7 + (a^4*b^4 - 2*a^2*b^
6 + b^8)*cos(d*x + c)^3 - 2*(a^5*b^3 - 2*a^3*b^5 + a*b^7)*cos(d*x + c)^2 + 2*(3*a^6*b^2 - 5*a^4*b^4 + a^2*b^6
+ b^8)*cos(d*x + c))*sin(d*x + c))/((a^4*b^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c) + (a^5*b^5 - 2*a^3*b^7 + a*b^9
)*d), -1/3*(3*(4*a^7*b - 7*a^5*b^3 + 2*a^3*b^5 + a*b^7)*d*x*cos(d*x + c) + 3*(4*a^8 - 7*a^6*b^2 + 2*a^4*b^4 +
a^2*b^6)*d*x - 3*(4*a^7 - 5*a^5*b^2 + (4*a^6*b - 5*a^4*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))) - (12*a^7*b - 19*a^5*b^3 + 5*a^3*b^5 + 2*a*b^7 + (a^4*b^4 - 2*a^2*b^6
+ b^8)*cos(d*x + c)^3 - 2*(a^5*b^3 - 2*a^3*b^5 + a*b^7)*cos(d*x + c)^2 + 2*(3*a^6*b^2 - 5*a^4*b^4 + a^2*b^6 +
b^8)*cos(d*x + c))*sin(d*x + c))/((a^4*b^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c) + (a^5*b^5 - 2*a^3*b^7 + a*b^9)
*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.31688, size = 450, normalized size = 1.69 \begin{align*} \frac{\frac{6 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (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 )}} - \frac{6 \,{\left (4 \, a^{6} - 5 \, a^{4} 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 )}}{{\left (a^{2} b^{5} - b^{7}\right )} \sqrt{a^{2} - b^{2}}} - \frac{3 \,{\left (4 \, a^{3} + a b^{2}\right )}{\left (d x + c\right )}}{b^{5}} + \frac{2 \,{\left (9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, b^{2} \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^{4}}}{3 \, d} \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))^2,x, algorithm="giac")

[Out]

1/3*(6*a^5*tan(1/2*d*x + 1/2*c)/((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)
) - 6*(4*a^6 - 5*a^4*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)))/((a^2*b^5 - b^7)*sqrt(a^2 - b^2)) - 3*(4*a^3 + a*b^2)*(d*x + c)/b^
5 + 2*(9*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*b^2*tan(1/2*d*x + 1/2*c)^5 + 18*a^2*tan
(1/2*d*x + 1/2*c)^3 + 2*b^2*tan(1/2*d*x + 1/2*c)^3 + 9*a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) +
3*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^4))/d