3.190 \(\int \frac{A+B \sec (x)}{(a+a \cos (x))^2} \, dx\)

Optimal. Leaf size=48 \[ \frac{(A-4 B) \sin (x)}{3 a^2 (\cos (x)+1)}+\frac{B \tanh ^{-1}(\sin (x))}{a^2}+\frac{(A-B) \sin (x)}{3 (a \cos (x)+a)^2} \]

[Out]

(B*ArcTanh[Sin[x]])/a^2 + ((A - 4*B)*Sin[x])/(3*a^2*(1 + Cos[x])) + ((A - B)*Sin[x])/(3*(a + a*Cos[x])^2)

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Rubi [A]  time = 0.184609, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2828, 2978, 12, 3770} \[ \frac{(A-4 B) \sin (x)}{3 a^2 (\cos (x)+1)}+\frac{B \tanh ^{-1}(\sin (x))}{a^2}+\frac{(A-B) \sin (x)}{3 (a \cos (x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*ArcTanh[Sin[x]])/a^2 + ((A - 4*B)*Sin[x])/(3*a^2*(1 + Cos[x])) + ((A - B)*Sin[x])/(3*(a + a*Cos[x])^2)

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))^2} \, dx &=\int \frac{(B+A \cos (x)) \sec (x)}{(a+a \cos (x))^2} \, dx\\ &=\frac{(A-B) \sin (x)}{3 (a+a \cos (x))^2}+\frac{\int \frac{(3 a B+a (A-B) \cos (x)) \sec (x)}{a+a \cos (x)} \, dx}{3 a^2}\\ &=\frac{(A-4 B) \sin (x)}{3 a^2 (1+\cos (x))}+\frac{(A-B) \sin (x)}{3 (a+a \cos (x))^2}+\frac{\int 3 a^2 B \sec (x) \, dx}{3 a^4}\\ &=\frac{(A-4 B) \sin (x)}{3 a^2 (1+\cos (x))}+\frac{(A-B) \sin (x)}{3 (a+a \cos (x))^2}+\frac{B \int \sec (x) \, dx}{a^2}\\ &=\frac{B \tanh ^{-1}(\sin (x))}{a^2}+\frac{(A-4 B) \sin (x)}{3 a^2 (1+\cos (x))}+\frac{(A-B) \sin (x)}{3 (a+a \cos (x))^2}\\ \end{align*}

Mathematica [A]  time = 0.203533, size = 76, normalized size = 1.58 \[ \frac{\sin (x) ((A-4 B) \cos (x)+2 A-5 B)-12 B \cos ^4\left (\frac{x}{2}\right ) \left (\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )}{3 a^2 (\cos (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*B*Cos[x/2]^4*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]]) + (2*A - 5*B + (A - 4*B)*Cos[x])*Sin[x
])/(3*a^2*(1 + Cos[x])^2)

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Maple [A]  time = 0.032, size = 71, normalized size = 1.5 \begin{align*}{\frac{A}{6\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{B}{6\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{A}{2\,{a}^{2}}\tan \left ({\frac{x}{2}} \right ) }-{\frac{3\,B}{2\,{a}^{2}}\tan \left ({\frac{x}{2}} \right ) }-{\frac{B}{{a}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{B}{{a}^{2}}\ln \left ( 1+\tan \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(x))/(a+a*cos(x))^2,x)

[Out]

1/6/a^2*tan(1/2*x)^3*A-1/6/a^2*tan(1/2*x)^3*B+1/2/a^2*A*tan(1/2*x)-3/2/a^2*B*tan(1/2*x)-1/a^2*B*ln(tan(1/2*x)-
1)+1/a^2*B*ln(1+tan(1/2*x))

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Maxima [B]  time = 1.02993, size = 126, normalized size = 2.62 \begin{align*} -\frac{1}{6} \, B{\left (\frac{\frac{9 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{\sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}}{a^{2}} - \frac{6 \, \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{a^{2}}\right )} + \frac{A{\left (\frac{3 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{\sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}\right )}}{6 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(x))/(a+a*cos(x))^2,x, algorithm="maxima")

[Out]

-1/6*B*((9*sin(x)/(cos(x) + 1) + sin(x)^3/(cos(x) + 1)^3)/a^2 - 6*log(sin(x)/(cos(x) + 1) + 1)/a^2 + 6*log(sin
(x)/(cos(x) + 1) - 1)/a^2) + 1/6*A*(3*sin(x)/(cos(x) + 1) + sin(x)^3/(cos(x) + 1)^3)/a^2

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Fricas [A]  time = 2.43617, size = 248, normalized size = 5.17 \begin{align*} \frac{3 \,{\left (B \cos \left (x\right )^{2} + 2 \, B \cos \left (x\right ) + B\right )} \log \left (\sin \left (x\right ) + 1\right ) - 3 \,{\left (B \cos \left (x\right )^{2} + 2 \, B \cos \left (x\right ) + B\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left ({\left (A - 4 \, B\right )} \cos \left (x\right ) + 2 \, A - 5 \, B\right )} \sin \left (x\right )}{6 \,{\left (a^{2} \cos \left (x\right )^{2} + 2 \, a^{2} \cos \left (x\right ) + a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(x))/(a+a*cos(x))^2,x, algorithm="fricas")

[Out]

1/6*(3*(B*cos(x)^2 + 2*B*cos(x) + B)*log(sin(x) + 1) - 3*(B*cos(x)^2 + 2*B*cos(x) + B)*log(-sin(x) + 1) + 2*((
A - 4*B)*cos(x) + 2*A - 5*B)*sin(x))/(a^2*cos(x)^2 + 2*a^2*cos(x) + a^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A}{\cos ^{2}{\left (x \right )} + 2 \cos{\left (x \right )} + 1}\, dx + \int \frac{B \sec{\left (x \right )}}{\cos ^{2}{\left (x \right )} + 2 \cos{\left (x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(x))/(a+a*cos(x))**2,x)

[Out]

(Integral(A/(cos(x)**2 + 2*cos(x) + 1), x) + Integral(B*sec(x)/(cos(x)**2 + 2*cos(x) + 1), x))/a**2

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Giac [A]  time = 1.19135, size = 104, normalized size = 2.17 \begin{align*} \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right )}{a^{2}} - \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right )}{a^{2}} + \frac{A a^{4} \tan \left (\frac{1}{2} \, x\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, x\right )^{3} + 3 \, A a^{4} \tan \left (\frac{1}{2} \, x\right ) - 9 \, B a^{4} \tan \left (\frac{1}{2} \, x\right )}{6 \, a^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(abs(tan(1/2*x) + 1))/a^2 - B*log(abs(tan(1/2*x) - 1))/a^2 + 1/6*(A*a^4*tan(1/2*x)^3 - B*a^4*tan(1/2*x)^3
+ 3*A*a^4*tan(1/2*x) - 9*B*a^4*tan(1/2*x))/a^6