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

Optimal. Leaf size=94 \[ -\frac{2 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a b d \sqrt{a-b} \sqrt{a+b}}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C x}{b} \]

[Out]

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

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Rubi [A]  time = 0.137051, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3057, 2659, 205, 3770} \[ -\frac{2 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a b d \sqrt{a-b} \sqrt{a+b}}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3057

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(C*x)/(b*d), x] + (Dist[(A*b^2 - a*b*B + a
^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)), Int[
1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 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]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx &=\frac{C x}{b}+\frac{A \int \sec (c+d x) \, dx}{a}-\left (\frac{A b}{a}-B+\frac{a C}{b}\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx\\ &=\frac{C x}{b}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}-\frac{\left (2 \left (\frac{A b}{a}-B+\frac{a C}{b}\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 )}{d}\\ &=\frac{C x}{b}-\frac{2 \left (\frac{A b}{a}-B+\frac{a C}{b}\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b} d}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}\\ \end{align*}

Mathematica [C]  time = 0.608345, size = 256, normalized size = 2.72 \[ \frac{2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \left (2 (\sin (c)+i \cos (c)) \left (a (a C-b B)+A b^2\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (b \cos (c)-a)+b \sin (c)\right )}{\sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2}}\right )+\sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2} \left (a C d x-A b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+A b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{a b d \sqrt{\left (b^2-a^2\right ) (\cos (2 c)-i \sin (2 c))} (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.067, size = 202, normalized size = 2.2 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{db}}-{\frac{A}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-2\,{\frac{Ab}{ad\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{B}{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{aC}{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 ) }+{\frac{A}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c)),x)

[Out]

2/d/b*arctan(tan(1/2*d*x+1/2*c))*C-1/a/d*A*ln(tan(1/2*d*x+1/2*c)-1)-2/d/a*b/((a+b)*(a-b))^(1/2)*arctan((a-b)*t
an(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*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))*B-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))*C+1/a/d*A*ln(tan(1/
2*d*x+1/2*c)+1)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 19.9063, size = 807, normalized size = 8.59 \begin{align*} \left [\frac{2 \,{\left (C a^{3} - C a b^{2}\right )} d x -{\left (C a^{2} - B a b + 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 (A a^{2} b - A b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{2} b - A b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (a^{3} b - a b^{3}\right )} d}, \frac{2 \,{\left (C a^{3} - C a b^{2}\right )} d x - 2 \,{\left (C a^{2} - B a b + 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 (A a^{2} b - A b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{2} b - A b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (a^{3} b - a b^{3}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(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 - B*a*b + 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)) + (A*a^2*b - A*b^3)*log(sin(d*x + c) + 1) - (A*a^2*b - A*b^3)*log(-sin(d*x + c) + 1
))/((a^3*b - a*b^3)*d), 1/2*(2*(C*a^3 - C*a*b^2)*d*x - 2*(C*a^2 - B*a*b + A*b^2)*sqrt(a^2 - b^2)*arctan(-(a*co
s(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + (A*a^2*b - A*b^3)*log(sin(d*x + c) + 1) - (A*a^2*b - A*b^3)*
log(-sin(d*x + c) + 1))/((a^3*b - a*b^3)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \cos{\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}}{a + b \cos{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)/(a+b*cos(d*x+c)),x)

[Out]

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)/(a + b*cos(c + d*x)), x)

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Giac [A]  time = 1.23375, size = 200, normalized size = 2.13 \begin{align*} \frac{\frac{{\left (d x + c\right )} C}{b} + \frac{A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{2 \,{\left (C a^{2} - B a b + 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}} a b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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