3.195 \(\int \sqrt{a+a \cos (x)} (A+B \sec (x)) \, dx\)

Optimal. Leaf size=44 \[ \frac{2 a A \sin (x)}{\sqrt{a \cos (x)+a}}+2 \sqrt{a} B \tanh ^{-1}\left (\frac{\sqrt{a} \sin (x)}{\sqrt{a \cos (x)+a}}\right ) \]

[Out]

2*Sqrt[a]*B*ArcTanh[(Sqrt[a]*Sin[x])/Sqrt[a + a*Cos[x]]] + (2*a*A*Sin[x])/Sqrt[a + a*Cos[x]]

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Rubi [A]  time = 0.160331, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2828, 2981, 2773, 206} \[ \frac{2 a A \sin (x)}{\sqrt{a \cos (x)+a}}+2 \sqrt{a} B \tanh ^{-1}\left (\frac{\sqrt{a} \sin (x)}{\sqrt{a \cos (x)+a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Cos[x]]*(A + B*Sec[x]),x]

[Out]

2*Sqrt[a]*B*ArcTanh[(Sqrt[a]*Sin[x])/Sqrt[a + a*Cos[x]]] + (2*a*A*Sin[x])/Sqrt[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 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]

Rule 2773

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0370911, size = 47, normalized size = 1.07 \[ \sec \left (\frac{x}{2}\right ) \sqrt{a (\cos (x)+1)} \left (2 A \sin \left (\frac{x}{2}\right )+\sqrt{2} B \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a*(1 + Cos[x])]*Sec[x/2]*(Sqrt[2]*B*ArcTanh[Sqrt[2]*Sin[x/2]] + 2*A*Sin[x/2])

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Maple [B]  time = 2.702, size = 152, normalized size = 3.5 \begin{align*}{\cos \left ({\frac{x}{2}} \right ) \sqrt{a \left ( \sin \left ({\frac{x}{2}} \right ) \right ) ^{2}} \left ( 2\,A\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}\sqrt{a}+B\ln \left ( -4\,{\frac{\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}-a\sqrt{2}\cos \left ( x/2 \right ) +2\,a}{-2\,\cos \left ( x/2 \right ) +\sqrt{2}}} \right ) a+B\ln \left ( 4\,{\frac{a\sqrt{2}\cos \left ( x/2 \right ) +\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}+2\,a}{2\,\cos \left ( x/2 \right ) +\sqrt{2}}} \right ) a \right ){\frac{1}{\sqrt{a}}} \left ( \sin \left ({\frac{x}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{x}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^(1/2)*cos(1/2*x)*(a*sin(1/2*x)^2)^(1/2)*(2*A*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)*a^(1/2)+B*ln(-4/(-2*cos(1/2*x)
+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)-a*2^(1/2)*cos(1/2*x)+2*a))*a+B*ln(4/(2*cos(1/2*x)+2^(1/2))*(
a*2^(1/2)*cos(1/2*x)+a^(1/2)*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)+2*a))*a)/sin(1/2*x)/(cos(1/2*x)^2*a)^(1/2)

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Maxima [A]  time = 1.66974, size = 18, normalized size = 0.41 \begin{align*} 2 \, \sqrt{2} A \sqrt{a} \sin \left (\frac{1}{2} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^(1/2)*(A+B*sec(x)),x, algorithm="maxima")

[Out]

2*sqrt(2)*A*sqrt(a)*sin(1/2*x)

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Fricas [B]  time = 2.43054, size = 252, normalized size = 5.73 \begin{align*} \frac{{\left (B \cos \left (x\right ) + B\right )} \sqrt{a} \log \left (\frac{a \cos \left (x\right )^{3} - 7 \, a \cos \left (x\right )^{2} - 4 \, \sqrt{a \cos \left (x\right ) + a} \sqrt{a}{\left (\cos \left (x\right ) - 2\right )} \sin \left (x\right ) + 8 \, a}{\cos \left (x\right )^{3} + \cos \left (x\right )^{2}}\right ) + 4 \, \sqrt{a \cos \left (x\right ) + a} A \sin \left (x\right )}{2 \,{\left (\cos \left (x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^(1/2)*(A+B*sec(x)),x, algorithm="fricas")

[Out]

1/2*((B*cos(x) + B)*sqrt(a)*log((a*cos(x)^3 - 7*a*cos(x)^2 - 4*sqrt(a*cos(x) + a)*sqrt(a)*(cos(x) - 2)*sin(x)
+ 8*a)/(cos(x)^3 + cos(x)^2)) + 4*sqrt(a*cos(x) + a)*A*sin(x))/(cos(x) + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 3.22241, size = 155, normalized size = 3.52 \begin{align*} \frac{2 \, \sqrt{2} A a \tan \left (\frac{1}{2} \, x\right )}{\sqrt{a \tan \left (\frac{1}{2} \, x\right )^{2} + a}} + \frac{B a^{\frac{3}{2}} \log \left (\frac{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, x\right ) - \sqrt{a \tan \left (\frac{1}{2} \, x\right )^{2} + a}\right )}^{2} - 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, x\right ) - \sqrt{a \tan \left (\frac{1}{2} \, x\right )^{2} + a}\right )}^{2} + 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^(1/2)*(A+B*sec(x)),x, algorithm="giac")

[Out]

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