3.254 \(\int \frac{(A+B \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=76 \[ \frac{A \tanh ^{-1}(\sin (c+d x))}{a d}-\frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d \sqrt{a-b} \sqrt{a+b}} \]

[Out]

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

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Rubi [A]  time = 0.115054, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {3001, 3770, 2659, 205} \[ \frac{A \tanh ^{-1}(\sin (c+d x))}{a d}-\frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d \sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3001

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

Rule 3770

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

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

Mathematica [A]  time = 0.154502, size = 112, normalized size = 1.47 \[ \frac{\frac{2 (A b-a B) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+A \left (\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.151, size = 135, normalized size = 1.8 \begin{align*} -{\frac{A}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{A}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{Ab}{da\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 ) }+2\,{\frac{B}{d\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((A+B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c)),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.65766, size = 689, normalized size = 9.07 \begin{align*} \left [\frac{{\left (B a - A b\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} - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{2} - A b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (a^{3} - a b^{2}\right )} d}, \frac{2 \,{\left (B a - A b\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} - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{2} - A b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (a^{3} - a b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

[1/2*((B*a - A*b)*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 -
A*b^2)*log(sin(d*x + c) + 1) - (A*a^2 - A*b^2)*log(-sin(d*x + c) + 1))/((a^3 - a*b^2)*d), 1/2*(2*(B*a - A*b)*s
qrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + (A*a^2 - A*b^2)*log(sin(d*x + c)
+ 1) - (A*a^2 - A*b^2)*log(-sin(d*x + c) + 1))/((a^3 - a*b^2)*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 )}\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))*sec(d*x+c)/(a+b*cos(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.48512, size = 171, normalized size = 2.25 \begin{align*} \frac{\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 (\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 (B a - A b\right )}}{\sqrt{a^{2} - b^{2}} a}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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*(pi*floor(1/2*(d*x + c)/p
i + 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)))*(B*a - A*
b)/(sqrt(a^2 - b^2)*a))/d