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

Optimal. Leaf size=148 \[ \frac{\left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac{2 a^4 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{a x \left (2 a^2+b^2\right )}{2 b^4}-\frac{a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{\sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.326544, antiderivative size = 148, 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 = {2793, 3049, 3023, 2735, 2659, 205} \[ \frac{\left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac{2 a^4 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{a x \left (2 a^2+b^2\right )}{2 b^4}-\frac{a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{\sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2793

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

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

Mathematica [A]  time = 0.330407, size = 122, normalized size = 0.82 \[ \frac{-6 a \left (2 a^2+b^2\right ) (c+d x)+3 b \left (4 a^2+3 b^2\right ) \sin (c+d x)-\frac{24 a^4 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-3 a b^2 \sin (2 (c+d x))+b^3 \sin (3 (c+d x))}{12 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.086, size = 367, normalized size = 2.5 \begin{align*} 2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}{a}^{2}}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{a}{{b}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{bd \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{a}^{2}}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{4}{3\,bd} \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 ) ^{-3}}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ){a}^{2}}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{bd \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{a}{{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 ) ^{-3}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{3}}{d{b}^{4}}}-{\frac{a}{{b}^{2}d}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+2\,{\frac{{a}^{4}}{d{b}^{4}\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)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.23132, size = 860, normalized size = 5.81 \begin{align*} \left [-\frac{3 \, \sqrt{-a^{2} + b^{2}} a^{4} \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 ) + 3 \,{\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} d x -{\left (6 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 2 \,{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} b^{4} - b^{6}\right )} d}, \frac{6 \, \sqrt{a^{2} - b^{2}} a^{4} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 3 \,{\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} d x +{\left (6 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 2 \,{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} b^{4} - b^{6}\right )} d}\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)),x, algorithm="fricas")

[Out]

[-1/6*(3*sqrt(-a^2 + b^2)*a^4*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*c
os(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)) + 3*(2*a^5 - a^3
*b^2 - a*b^4)*d*x - (6*a^4*b - 2*a^2*b^3 - 4*b^5 + 2*(a^2*b^3 - b^5)*cos(d*x + c)^2 - 3*(a^3*b^2 - a*b^4)*cos(
d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d), 1/6*(6*sqrt(a^2 - b^2)*a^4*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2
- b^2)*sin(d*x + c))) - 3*(2*a^5 - a^3*b^2 - a*b^4)*d*x + (6*a^4*b - 2*a^2*b^3 - 4*b^5 + 2*(a^2*b^3 - b^5)*co
s(d*x + c)^2 - 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*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)),x)

[Out]

Timed out

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Giac [A]  time = 1.39326, size = 336, normalized size = 2.27 \begin{align*} -\frac{\frac{12 \,{\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 )} a^{4}}{\sqrt{a^{2} - b^{2}} b^{4}} + \frac{3 \,{\left (2 \, a^{3} + a b^{2}\right )}{\left (d x + c\right )}}{b^{4}} - \frac{2 \,{\left (6 \, 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} + 6 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, 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 ) + 6 \, 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^{3}}}{6 \, 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)),x, algorithm="giac")

[Out]

-1/6*(12*(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/(sqrt(a^2 - b^2)*b^4) + 3*(2*a^3 + a*b^2)*(d*x + c)/b^4 - 2*(6*a^2*tan(1/2*d*x
+ 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*a^2*tan(1/2*d*x + 1/2*c)^3 + 4*b
^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*b^2*tan(1/2*d*x + 1/2*
c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d