3.6 \(\int \frac{A+B \tan (x)}{a+b \cos (x)} \, dx\)

Optimal. Leaf size=65 \[ \frac{2 A \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{B \log (a+b \cos (x))}{a}-\frac{B \log (\cos (x))}{a} \]

[Out]

(2*A*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]) - (B*Log[Cos[x]])/a + (B*Log[a + b*
Cos[x]])/a

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Rubi [A]  time = 0.136945, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {4401, 2659, 205, 2721, 36, 29, 31} \[ \frac{2 A \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{B \log (a+b \cos (x))}{a}-\frac{B \log (\cos (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[x])/(a + b*Cos[x]),x]

[Out]

(2*A*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]) - (B*Log[Cos[x]])/a + (B*Log[a + b*
Cos[x]])/a

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

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 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.145063, size = 61, normalized size = 0.94 \[ \frac{B (\log (a+b \cos (x))-\log (\cos (x)))}{a}-\frac{2 A \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[x])/(a + b*Cos[x]),x]

[Out]

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

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Maple [B]  time = 0.033, size = 129, normalized size = 2. \begin{align*} -{\frac{B}{a}\ln \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{B}{a}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{B}{a-b}\ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}a- \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}b+a+b \right ) }-{\frac{Bb}{a \left ( a-b \right ) }\ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}a- \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}b+a+b \right ) }+2\,{\frac{A}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( x/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+B*tan(x))/(a+b*cos(x)),x)

[Out]

-B/a*ln(tan(1/2*x)+1)-B/a*ln(tan(1/2*x)-1)+1/(a-b)*ln(tan(1/2*x)^2*a-tan(1/2*x)^2*b+a+b)*B-1/a/(a-b)*ln(tan(1/
2*x)^2*a-tan(1/2*x)^2*b+a+b)*B*b+2*A/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*x)/((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((A+B*tan(x))/(a+b*cos(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.39537, size = 617, normalized size = 9.49 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} A a \log \left (\frac{2 \, a b \cos \left (x\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) -{\left (B a^{2} - B b^{2}\right )} \log \left (b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}\right ) + 2 \,{\left (B a^{2} - B b^{2}\right )} \log \left (-\cos \left (x\right )\right )}{2 \,{\left (a^{3} - a b^{2}\right )}}, \frac{2 \, \sqrt{a^{2} - b^{2}} A a \arctan \left (-\frac{a \cos \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right ) +{\left (B a^{2} - B b^{2}\right )} \log \left (b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}\right ) - 2 \,{\left (B a^{2} - B b^{2}\right )} \log \left (-\cos \left (x\right )\right )}{2 \,{\left (a^{3} - a b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(-a^2 + b^2)*A*a*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin
(x) - a^2 + 2*b^2)/(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2)) - (B*a^2 - B*b^2)*log(b^2*cos(x)^2 + 2*a*b*cos(x) + a^
2) + 2*(B*a^2 - B*b^2)*log(-cos(x)))/(a^3 - a*b^2), 1/2*(2*sqrt(a^2 - b^2)*A*a*arctan(-(a*cos(x) + b)/(sqrt(a^
2 - b^2)*sin(x))) + (B*a^2 - B*b^2)*log(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2) - 2*(B*a^2 - B*b^2)*log(-cos(x)))/(
a^3 - a*b^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(x))/(a+b*cos(x)),x)

[Out]

Integral((A + B*tan(x))/(a + b*cos(x)), x)

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Giac [B]  time = 1.17813, size = 163, normalized size = 2.51 \begin{align*} -\frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )} A}{\sqrt{a^{2} - b^{2}}} + \frac{B \log \left (-a \tan \left (\frac{1}{2} \, x\right )^{2} + b \tan \left (\frac{1}{2} \, x\right )^{2} - a - b\right )}{a} - \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right )}{a} - \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))*A/sqrt(
a^2 - b^2) + B*log(-a*tan(1/2*x)^2 + b*tan(1/2*x)^2 - a - b)/a - B*log(abs(tan(1/2*x) + 1))/a - B*log(abs(tan(
1/2*x) - 1))/a