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

Optimal. Leaf size=166 \[ -\frac{a^4 \sin (c+d x)}{b^3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{2 a^3 \left (3 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \sin (c+d x)}{\sqrt{b^2-a^2} (\cos (c+d x)+1)}\right )}{b^4 d \left (b^2-a^2\right )^{3/2}}+\frac{x \left (6 a^2+b^2\right )}{2 b^4}-\frac{2 a \sin (c+d x)}{b^3 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 b^2 d} \]

[Out]

((6*a^2 + b^2)*x)/(2*b^4) - (2*a^3*(3*a^2 - 4*b^2)*ArcTanh[((a - b)*Sin[c + d*x])/(Sqrt[-a^2 + b^2]*(1 + Cos[c
+ d*x]))])/(b^4*(-a^2 + b^2)^(3/2)*d) - (2*a*Sin[c + d*x])/(b^3*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*b^2*d) -
(a^4*Sin[c + d*x])/(b^3*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.428566, antiderivative size = 213, normalized size of antiderivative = 1.28, 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, 3049, 3023, 2735, 2659, 205} \[ -\frac{a \left (3 a^2-2 b^2\right ) \sin (c+d x)}{b^3 d \left (a^2-b^2\right )}-\frac{2 a^3 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\left (3 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b^2 d \left (a^2-b^2\right )}+\frac{x \left (6 a^2+b^2\right )}{2 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.732477, size = 144, normalized size = 0.87 \[ \frac{2 \left (6 a^2+b^2\right ) (c+d x)-\frac{8 a^3 \left (3 a^2-4 b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-\frac{4 a^4 b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-8 a b \sin (c+d x)+b^2 \sin (2 (c+d x))}{4 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.091, size = 358, normalized size = 2.2 \begin{align*} -4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}a}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{1}{{b}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-4\,{\frac{a\tan \left ( 1/2\,dx+c/2 \right ) }{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{{b}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{2}}{d{b}^{4}}}+{\frac{1}{{b}^{2}d}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{{a}^{4}\tan \left ( 1/2\,dx+c/2 \right ) }{d{b}^{3} \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}-6\,{\frac{{a}^{5}}{d{b}^{4} \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }+8\,{\frac{{a}^{3}}{{b}^{2}d \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \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))^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.39211, size = 1428, normalized size = 8.6 \begin{align*} \left [\frac{{\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}\right )} d x \cos \left (d x + c\right ) +{\left (6 \, a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} d x -{\left (3 \, a^{6} - 4 \, a^{4} b^{2} +{\left (3 \, a^{5} b - 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) -{\left (6 \, a^{6} b - 10 \, a^{4} b^{3} + 4 \, a^{2} b^{5} -{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right ) +{\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8}\right )} d\right )}}, \frac{{\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}\right )} d x \cos \left (d x + c\right ) +{\left (6 \, a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} d x - 2 \,{\left (3 \, a^{6} - 4 \, a^{4} b^{2} +{\left (3 \, a^{5} b - 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) -{\left (6 \, a^{6} b - 10 \, a^{4} b^{3} + 4 \, a^{2} b^{5} -{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right ) +{\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8}\right )} d\right )}}\right ] \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))^2,x, algorithm="fricas")

[Out]

[1/2*((6*a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*d*x*cos(d*x + c) + (6*a^7 - 11*a^5*b^2 + 4*a^3*b^4 + a*b^6)*d*x
- (3*a^6 - 4*a^4*b^2 + (3*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)) - (6*a^6*b - 10*a^4*b^3 + 4*a^2*b^5 - (a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^
2 + 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c) +
(a^5*b^4 - 2*a^3*b^6 + a*b^8)*d), 1/2*((6*a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*d*x*cos(d*x + c) + (6*a^7 - 11
*a^5*b^2 + 4*a^3*b^4 + a*b^6)*d*x - 2*(3*a^6 - 4*a^4*b^2 + (3*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))) - (6*a^6*b - 10*a^4*b^3 + 4*a^2*b^5 - (a^4*b^3 -
2*a^2*b^5 + b^7)*cos(d*x + c)^2 + 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5 - 2*a
^2*b^7 + b^9)*d*cos(d*x + c) + (a^5*b^4 - 2*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/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.48344, size = 354, normalized size = 2.13 \begin{align*} -\frac{\frac{4 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{2} b^{3} - b^{5}\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{4 \,{\left (3 \, 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 )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} - \frac{{\left (6 \, a^{2} + b^{2}\right )}{\left (d x + c\right )}}{b^{4}} + \frac{2 \,{\left (4 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, 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 )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, 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))^2,x, algorithm="giac")

[Out]

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