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

Optimal. Leaf size=68 \[ \frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {2} \sqrt {a \cos (x)+a}}\right )}{\sqrt {a}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {a \cos (x)+a}}\right )}{\sqrt {a}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2828, 2985, 2649, 206, 2773} \[ \frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {2} \sqrt {a \cos (x)+a}}\right )}{\sqrt {a}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {a \cos (x)+a}}\right )}{\sqrt {a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 2649

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

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 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 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(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/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 52, normalized size = 0.76 \[ \frac {2 \cos \left (\frac {x}{2}\right ) \left ((A-B) \tanh ^{-1}\left (\sin \left (\frac {x}{2}\right )\right )+\sqrt {2} B \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {x}{2}\right )\right )\right )}{\sqrt {a (\cos (x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.61, size = 116, normalized size = 1.71 \[ -\frac {\sqrt {2} {\left (A - B\right )} \sqrt {a} \log \left (-\frac {\cos \relax (x)^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \relax (x) + a} \sin \relax (x)}{\sqrt {a}} - 2 \, \cos \relax (x) - 3}{\cos \relax (x)^{2} + 2 \, \cos \relax (x) + 1}\right ) - B \sqrt {a} \log \left (\frac {a \cos \relax (x)^{3} - 7 \, a \cos \relax (x)^{2} - 4 \, \sqrt {a \cos \relax (x) + a} \sqrt {a} {\left (\cos \relax (x) - 2\right )} \sin \relax (x) + 8 \, a}{\cos \relax (x)^{3} + \cos \relax (x)^{2}}\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(2)*(A - B)*sqrt(a)*log(-(cos(x)^2 + 2*sqrt(2)*sqrt(a*cos(x) + a)*sin(x)/sqrt(a) - 2*cos(x) - 3)/(co
s(x)^2 + 2*cos(x) + 1)) - 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)))/a

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giac [B]  time = 0.70, size = 133, normalized size = 1.96 \[ -\frac {\sqrt {2} {\left (A \sqrt {a} - B \sqrt {a}\right )} \log \left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, x\right ) - \sqrt {a \tan \left (\frac {1}{2} \, x\right )^{2} + a}\right )}^{2}\right )}{2 \, a} + \frac {B \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, x\right ) - \sqrt {a \tan \left (\frac {1}{2} \, x\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{\sqrt {a}} - \frac {B \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, x\right ) - \sqrt {a \tan \left (\frac {1}{2} \, x\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.35, size = 192, normalized size = 2.82 \[ \frac {\cos \left (\frac {x}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}\, \left (\sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}+4 a}{\cos \left (\frac {x}{2}\right )}\right ) A -\sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}+4 a}{\cos \left (\frac {x}{2}\right )}\right ) B +B \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {x}{2}\right )+4 \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}+8 a}{2 \cos \left (\frac {x}{2}\right )+\sqrt {2}}\right )+B \ln \left (-\frac {4 \left (\sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}-a \sqrt {2}\, \cos \left (\frac {x}{2}\right )+2 a \right )}{-2 \cos \left (\frac {x}{2}\right )+\sqrt {2}}\right )\right )}{\sqrt {a}\, \sin \left (\frac {x}{2}\right ) \sqrt {\left (\cos ^{2}\left (\frac {x}{2}\right )\right ) a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.05, size = 58, normalized size = 0.85 \[ \frac {{\left (\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, x\right )^{2} + \sin \left (\frac {1}{2} \, x\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, x\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, x\right )^{2} + \sin \left (\frac {1}{2} \, x\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, x\right ) + 1\right )\right )} A}{2 \, \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {B}{\cos \relax (x)}}{\sqrt {a+a\,\cos \relax (x)}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sec {\relax (x )}}{\sqrt {a \left (\cos {\relax (x )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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