3.32 \(\int \frac {\sec (x)}{a+b \cos ^2(x)} \, dx\)

Optimal. Leaf size=41 \[ \frac {\tanh ^{-1}(\sin (x))}{a}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}} \]

[Out]

arctanh(sin(x))/a-arctanh(sin(x)*b^(1/2)/(a+b)^(1/2))*b^(1/2)/a/(a+b)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3186, 391, 206, 208} \[ \frac {\tanh ^{-1}(\sin (x))}{a}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]/(a + b*Cos[x]^2),x]

[Out]

ArcTanh[Sin[x]]/a - (Sqrt[b]*ArcTanh[(Sqrt[b]*Sin[x])/Sqrt[a + b]])/(a*Sqrt[a + b])

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 208

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

Rule 391

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

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 38, normalized size = 0.93 \[ \frac {\tanh ^{-1}(\sin (x))-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]/(a + b*Cos[x]^2),x]

[Out]

(ArcTanh[Sin[x]] - (Sqrt[b]*ArcTanh[(Sqrt[b]*Sin[x])/Sqrt[a + b]])/Sqrt[a + b])/a

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fricas [A]  time = 0.60, size = 119, normalized size = 2.90 \[ \left [\frac {\sqrt {\frac {b}{a + b}} \log \left (-\frac {b \cos \relax (x)^{2} + 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \sin \relax (x) - a - 2 \, b}{b \cos \relax (x)^{2} + a}\right ) + \log \left (\sin \relax (x) + 1\right ) - \log \left (-\sin \relax (x) + 1\right )}{2 \, a}, \frac {2 \, \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \sin \relax (x)\right ) + \log \left (\sin \relax (x) + 1\right ) - \log \left (-\sin \relax (x) + 1\right )}{2 \, a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 57, normalized size = 1.39 \[ \frac {b \arctan \left (\frac {b \sin \relax (x)}{\sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a} + \frac {\log \left (\sin \relax (x) + 1\right )}{2 \, a} - \frac {\log \left (-\sin \relax (x) + 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*arctan(b*sin(x)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a) + 1/2*log(sin(x) + 1)/a - 1/2*log(-sin(x) + 1)/a

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maple [A]  time = 0.10, size = 47, normalized size = 1.15 \[ -\frac {b \arctanh \left (\frac {\sin \relax (x ) b}{\sqrt {\left (a +b \right ) b}}\right )}{a \sqrt {\left (a +b \right ) b}}-\frac {\ln \left (\sin \relax (x )-1\right )}{2 a}+\frac {\ln \left (\sin \relax (x )+1\right )}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a*b/((a+b)*b)^(1/2)*arctanh(sin(x)*b/((a+b)*b)^(1/2))-1/2/a*ln(sin(x)-1)+1/2/a*ln(sin(x)+1)

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maxima [A]  time = 0.92, size = 64, normalized size = 1.56 \[ \frac {b \log \left (\frac {b \sin \relax (x) - \sqrt {{\left (a + b\right )} b}}{b \sin \relax (x) + \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} a} + \frac {\log \left (\sin \relax (x) + 1\right )}{2 \, a} - \frac {\log \left (\sin \relax (x) - 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*log((b*sin(x) - sqrt((a + b)*b))/(b*sin(x) + sqrt((a + b)*b)))/(sqrt((a + b)*b)*a) + 1/2*log(sin(x) + 1)
/a - 1/2*log(sin(x) - 1)/a

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mupad [B]  time = 2.50, size = 414, normalized size = 10.10 \[ \frac {\mathrm {atanh}\left (\sin \relax (x)\right )}{a}+\frac {\mathrm {atan}\left (\frac {\frac {\left (2\,b^3\,\sin \relax (x)+\frac {\left (2\,a^2\,b^2-\frac {\sin \relax (x)\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^2+b\,a}+\frac {\left (2\,b^3\,\sin \relax (x)-\frac {\left (2\,a^2\,b^2+\frac {\sin \relax (x)\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^2+b\,a}}{\frac {\left (2\,b^3\,\sin \relax (x)+\frac {\left (2\,a^2\,b^2-\frac {\sin \relax (x)\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{a^2+b\,a}-\frac {\left (2\,b^3\,\sin \relax (x)-\frac {\left (2\,a^2\,b^2+\frac {\sin \relax (x)\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{a^2+b\,a}}\right )\,\sqrt {b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^2+b\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)*(a + b*cos(x)^2)),x)

[Out]

atanh(sin(x))/a + (atan((((2*b^3*sin(x) + ((2*a^2*b^2 - (sin(x)*(16*a^2*b^3 + 8*a^3*b^2)*(b*(a + b))^(1/2))/(4
*(a*b + a^2)))*(b*(a + b))^(1/2))/(2*(a*b + a^2)))*(b*(a + b))^(1/2)*1i)/(a*b + a^2) + ((2*b^3*sin(x) - ((2*a^
2*b^2 + (sin(x)*(16*a^2*b^3 + 8*a^3*b^2)*(b*(a + b))^(1/2))/(4*(a*b + a^2)))*(b*(a + b))^(1/2))/(2*(a*b + a^2)
))*(b*(a + b))^(1/2)*1i)/(a*b + a^2))/(((2*b^3*sin(x) + ((2*a^2*b^2 - (sin(x)*(16*a^2*b^3 + 8*a^3*b^2)*(b*(a +
b))^(1/2))/(4*(a*b + a^2)))*(b*(a + b))^(1/2))/(2*(a*b + a^2)))*(b*(a + b))^(1/2))/(a*b + a^2) - ((2*b^3*sin(
x) - ((2*a^2*b^2 + (sin(x)*(16*a^2*b^3 + 8*a^3*b^2)*(b*(a + b))^(1/2))/(4*(a*b + a^2)))*(b*(a + b))^(1/2))/(2*
(a*b + a^2)))*(b*(a + b))^(1/2))/(a*b + a^2)))*(b*(a + b))^(1/2)*1i)/(a*b + a^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec {\relax (x )}}{a + b \cos ^{2}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(x)/(a + b*cos(x)**2), x)

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