3.564 \(\int \frac{\cos (c+d x) (A+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=128 \[ -\frac{2 a \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (2 a^2 C+b^2 (2 A+C)\right )}{2 b^3}-\frac{a C \sin (c+d x)}{b^2 d}+\frac{C \sin (c+d x) \cos (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.258365, antiderivative size = 126, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {3050, 3023, 2735, 2659, 205} \[ -\frac{2 a \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (\frac{2 a^2 C}{b^2}+2 A+C\right )}{2 b}-\frac{a C \sin (c+d x)}{b^2 d}+\frac{C \sin (c+d x) \cos (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 (c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=\frac{C \cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\int \frac{a C+b (2 A+C) \cos (c+d x)-2 a C \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b}\\ &=-\frac{a C \sin (c+d x)}{b^2 d}+\frac{C \cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\int \frac{a b C+\left (2 a^2 C+b^2 (2 A+C)\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^2}\\ &=\frac{\left (2 a^2 C+b^2 (2 A+C)\right ) x}{2 b^3}-\frac{a C \sin (c+d x)}{b^2 d}+\frac{C \cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (a \left (A b^2+a^2 C\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^3}\\ &=\frac{\left (2 a^2 C+b^2 (2 A+C)\right ) x}{2 b^3}-\frac{a C \sin (c+d x)}{b^2 d}+\frac{C \cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (2 a \left (A b^2+a^2 C\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^3 d}\\ &=\frac{\left (2 a^2 C+b^2 (2 A+C)\right ) x}{2 b^3}-\frac{2 a \left (A b^2+a^2 C\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^3 \sqrt{a+b} d}-\frac{a C \sin (c+d x)}{b^2 d}+\frac{C \cos (c+d x) \sin (c+d x)}{2 b d}\\ \end{align*}

Mathematica [A]  time = 0.403958, size = 117, normalized size = 0.91 \[ \frac{2 (c+d x) \left (C \left (2 a^2+b^2\right )+2 A b^2\right )+\frac{8 a \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-4 a b C \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.033, size = 296, normalized size = 2.3 \begin{align*} -2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}aC}{d{b}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{C}{db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) aC}{d{b}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{C}{db}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{db}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{2}C}{d{b}^{3}}}+{\frac{C}{db}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{aA}{db\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{{a}^{3}C}{d{b}^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \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)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)

[Out]

-2/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)^3*a*C-1/d/b/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*
c)^3*C-2/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)*a*C+1/d/b/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+
1/2*c)*C+2/d/b*arctan(tan(1/2*d*x+1/2*c))*A+2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a^2*C+1/d/b*arctan(tan(1/2*d*x+
1/2*c))*C-2/d*a/b/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A-2/d*a^3/b^3/((a+b
)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C

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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)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.67143, size = 832, normalized size = 6.5 \begin{align*} \left [\frac{{\left (2 \, C a^{4} +{\left (2 \, A - C\right )} a^{2} b^{2} -{\left (2 \, A + C\right )} b^{4}\right )} d x -{\left (C a^{3} + A a b^{2}\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 (2 \, C a^{3} b - 2 \, C a b^{3} -{\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{3} - b^{5}\right )} d}, \frac{{\left (2 \, C a^{4} +{\left (2 \, A - C\right )} a^{2} b^{2} -{\left (2 \, A + C\right )} b^{4}\right )} d x - 2 \,{\left (C a^{3} + A a b^{2}\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 (2 \, C a^{3} b - 2 \, C a b^{3} -{\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

[1/2*((2*C*a^4 + (2*A - C)*a^2*b^2 - (2*A + C)*b^4)*d*x - (C*a^3 + A*a*b^2)*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*C*a^3*b - 2*C*a*b^3 - (C*a^2*b^2 - C*b^4)*cos(d*x + c))*si
n(d*x + c))/((a^2*b^3 - b^5)*d), 1/2*((2*C*a^4 + (2*A - C)*a^2*b^2 - (2*A + C)*b^4)*d*x - 2*(C*a^3 + A*a*b^2)*
sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (2*C*a^3*b - 2*C*a*b^3 - (C*a^2
*b^2 - C*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^3 - b^5)*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)*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.22571, size = 269, normalized size = 2.1 \begin{align*} \frac{\frac{{\left (2 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )}{\left (d x + c\right )}}{b^{3}} + \frac{4 \,{\left (C a^{3} + A a 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 )}}{\sqrt{a^{2} - b^{2}} b^{3}} - \frac{2 \,{\left (2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C 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^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

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