3.189 \(\int \frac{A+B \sec (x)}{a+a \cos (x)} \, dx\)

Optimal. Leaf size=25 \[ \frac{(A-B) \sin (x)}{a \cos (x)+a}+\frac{B \tanh ^{-1}(\sin (x))}{a} \]

[Out]

(B*ArcTanh[Sin[x]])/a + ((A - B)*Sin[x])/(a + a*Cos[x])

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Rubi [A]  time = 0.0928542, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2828, 2978, 12, 3770} \[ \frac{(A-B) \sin (x)}{a \cos (x)+a}+\frac{B \tanh ^{-1}(\sin (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[x])/(a + a*Cos[x]),x]

[Out]

(B*ArcTanh[Sin[x]])/a + ((A - B)*Sin[x])/(a + a*Cos[x])

Rule 2828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[((a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n)/Sin[e + f*x]^n, x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[n]

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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{A+B \sec (x)}{a+a \cos (x)} \, dx &=\int \frac{(B+A \cos (x)) \sec (x)}{a+a \cos (x)} \, dx\\ &=\frac{(A-B) \sin (x)}{a+a \cos (x)}+\frac{\int a B \sec (x) \, dx}{a^2}\\ &=\frac{(A-B) \sin (x)}{a+a \cos (x)}+\frac{B \int \sec (x) \, dx}{a}\\ &=\frac{B \tanh ^{-1}(\sin (x))}{a}+\frac{(A-B) \sin (x)}{a+a \cos (x)}\\ \end{align*}

Mathematica [B]  time = 0.0846522, size = 71, normalized size = 2.84 \[ -\frac{2 \cos \left (\frac{x}{2}\right ) \left ((B-A) \sin \left (\frac{x}{2}\right )+B \cos \left (\frac{x}{2}\right ) \left (\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )\right )}{a (\cos (x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[x/2]*(B*Cos[x/2]*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]]) + (-A + B)*Sin[x/2]))/(a*(1 + C
os[x]))

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Maple [A]  time = 0.03, size = 46, normalized size = 1.8 \begin{align*}{\frac{A}{a}\tan \left ({\frac{x}{2}} \right ) }-{\frac{B}{a}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{B}{a}\ln \left ( 1+\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{B}{a}\tan \left ({\frac{x}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(x))/(a+a*cos(x)),x)

[Out]

1/a*A*tan(1/2*x)-1/a*B*ln(tan(1/2*x)-1)+1/a*B*ln(1+tan(1/2*x))-1/a*B*tan(1/2*x)

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Maxima [B]  time = 0.992415, size = 85, normalized size = 3.4 \begin{align*} B{\left (\frac{\log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{a} - \frac{\sin \left (x\right )}{a{\left (\cos \left (x\right ) + 1\right )}}\right )} + \frac{A \sin \left (x\right )}{a{\left (\cos \left (x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(x))/(a+a*cos(x)),x, algorithm="maxima")

[Out]

B*(log(sin(x)/(cos(x) + 1) + 1)/a - log(sin(x)/(cos(x) + 1) - 1)/a - sin(x)/(a*(cos(x) + 1))) + A*sin(x)/(a*(c
os(x) + 1))

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Fricas [A]  time = 2.46082, size = 143, normalized size = 5.72 \begin{align*} \frac{{\left (B \cos \left (x\right ) + B\right )} \log \left (\sin \left (x\right ) + 1\right ) -{\left (B \cos \left (x\right ) + B\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left (A - B\right )} \sin \left (x\right )}{2 \,{\left (a \cos \left (x\right ) + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((B*cos(x) + B)*log(sin(x) + 1) - (B*cos(x) + B)*log(-sin(x) + 1) + 2*(A - B)*sin(x))/(a*cos(x) + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A}{\cos{\left (x \right )} + 1}\, dx + \int \frac{B \sec{\left (x \right )}}{\cos{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A/(cos(x) + 1), x) + Integral(B*sec(x)/(cos(x) + 1), x))/a

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Giac [A]  time = 1.17919, size = 62, normalized size = 2.48 \begin{align*} \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} + \frac{A \tan \left (\frac{1}{2} \, x\right ) - B \tan \left (\frac{1}{2} \, x\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(abs(tan(1/2*x) + 1))/a - B*log(abs(tan(1/2*x) - 1))/a + (A*tan(1/2*x) - B*tan(1/2*x))/a