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

Optimal. Leaf size=86 \[ \frac{2 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{a C x}{b^2}+\frac{C \sin (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.125738, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3024, 2735, 2659, 205} \[ \frac{2 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{a C x}{b^2}+\frac{C \sin (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3024

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

Mathematica [A]  time = 0.210301, size = 82, normalized size = 0.95 \[ \frac{-\frac{2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-a C (c+d x)+b C \sin (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.028, size = 149, normalized size = 1.7 \begin{align*} 2\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{db \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{C\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) a}{d{b}^{2}}}+2\,{\frac{A}{d\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}^{2}C}{d{b}^{2}\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((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)

[Out]

2/d*C/b*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2+1)-2/d/b^2*C*arctan(tan(1/2*d*x+1/2*c))*a+2/d/((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/b^2/((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*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((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 = 1.56617, size = 640, normalized size = 7.44 \begin{align*} \left [-\frac{2 \,{\left (C a^{3} - C a b^{2}\right )} d x +{\left (C a^{2} + 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 ) - 2 \,{\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} d}, -\frac{{\left (C a^{3} - C a b^{2}\right )} d x -{\left (C a^{2} + 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 (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.23958, size = 184, normalized size = 2.14 \begin{align*} -\frac{\frac{{\left (d x + c\right )} C a}{b^{2}} - \frac{2 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} b} + \frac{2 \,{\left (C a^{2} + 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^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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