3.40 \(\int \frac {\sec ^4(x)}{a+b \cos ^2(x)} \, dx\)

Optimal. Leaf size=56 \[ -\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a+b}}+\frac {(a-b) \tan (x)}{a^2}+\frac {\tan ^3(x)}{3 a} \]

[Out]

-b^2*arctan(cot(x)*(a+b)^(1/2)/a^(1/2))/a^(5/2)/(a+b)^(1/2)+(a-b)*tan(x)/a^2+1/3*tan(x)^3/a

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Rubi [A]  time = 0.09, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3187, 461, 205} \[ -\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a+b}}+\frac {(a-b) \tan (x)}{a^2}+\frac {\tan ^3(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^4/(a + b*Cos[x]^2),x]

[Out]

-((b^2*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(a^(5/2)*Sqrt[a + b])) + ((a - b)*Tan[x])/a^2 + Tan[x]^3/(3*a)

Rule 205

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

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3187

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

Rubi steps

\begin {align*} \int \frac {\sec ^4(x)}{a+b \cos ^2(x)} \, dx &=-\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^4 \left (a+(a+b) x^2\right )} \, dx,x,\cot (x)\right )\\ &=-\operatorname {Subst}\left (\int \left (\frac {1}{a x^4}+\frac {a-b}{a^2 x^2}+\frac {b^2}{a^2 \left (a+(a+b) x^2\right )}\right ) \, dx,x,\cot (x)\right )\\ &=\frac {(a-b) \tan (x)}{a^2}+\frac {\tan ^3(x)}{3 a}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\cot (x)\right )}{a^2}\\ &=-\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a+b}}+\frac {(a-b) \tan (x)}{a^2}+\frac {\tan ^3(x)}{3 a}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 55, normalized size = 0.98 \[ \frac {b^2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a+b}}\right )}{a^{5/2} \sqrt {a+b}}+\frac {\tan (x) \left (a \sec ^2(x)+2 a-3 b\right )}{3 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^4/(a + b*Cos[x]^2),x]

[Out]

(b^2*ArcTan[(Sqrt[a]*Tan[x])/Sqrt[a + b]])/(a^(5/2)*Sqrt[a + b]) + ((2*a - 3*b + a*Sec[x]^2)*Tan[x])/(3*a^2)

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fricas [B]  time = 0.95, size = 276, normalized size = 4.93 \[ \left [-\frac {3 \, \sqrt {-a^{2} - a b} b^{2} \cos \relax (x)^{3} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \relax (x)^{4} - 2 \, {\left (4 \, a^{2} + 3 \, a b\right )} \cos \relax (x)^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \relax (x)^{3} - a \cos \relax (x)\right )} \sqrt {-a^{2} - a b} \sin \relax (x) + a^{2}}{b^{2} \cos \relax (x)^{4} + 2 \, a b \cos \relax (x)^{2} + a^{2}}\right ) - 4 \, {\left (a^{3} + a^{2} b + {\left (2 \, a^{3} - a^{2} b - 3 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{12 \, {\left (a^{4} + a^{3} b\right )} \cos \relax (x)^{3}}, -\frac {3 \, \sqrt {a^{2} + a b} b^{2} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \relax (x)^{2} - a}{2 \, \sqrt {a^{2} + a b} \cos \relax (x) \sin \relax (x)}\right ) \cos \relax (x)^{3} - 2 \, {\left (a^{3} + a^{2} b + {\left (2 \, a^{3} - a^{2} b - 3 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{6 \, {\left (a^{4} + a^{3} b\right )} \cos \relax (x)^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(a+b*cos(x)^2),x, algorithm="fricas")

[Out]

[-1/12*(3*sqrt(-a^2 - a*b)*b^2*cos(x)^3*log(((8*a^2 + 8*a*b + b^2)*cos(x)^4 - 2*(4*a^2 + 3*a*b)*cos(x)^2 + 4*(
(2*a + b)*cos(x)^3 - a*cos(x))*sqrt(-a^2 - a*b)*sin(x) + a^2)/(b^2*cos(x)^4 + 2*a*b*cos(x)^2 + a^2)) - 4*(a^3
+ a^2*b + (2*a^3 - a^2*b - 3*a*b^2)*cos(x)^2)*sin(x))/((a^4 + a^3*b)*cos(x)^3), -1/6*(3*sqrt(a^2 + a*b)*b^2*ar
ctan(1/2*((2*a + b)*cos(x)^2 - a)/(sqrt(a^2 + a*b)*cos(x)*sin(x)))*cos(x)^3 - 2*(a^3 + a^2*b + (2*a^3 - a^2*b
- 3*a*b^2)*cos(x)^2)*sin(x))/((a^4 + a^3*b)*cos(x)^3)]

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giac [A]  time = 0.17, size = 71, normalized size = 1.27 \[ \frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \relax (x)}{\sqrt {a^{2} + a b}}\right )\right )} b^{2}}{\sqrt {a^{2} + a b} a^{2}} + \frac {a^{2} \tan \relax (x)^{3} + 3 \, a^{2} \tan \relax (x) - 3 \, a b \tan \relax (x)}{3 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(a+b*cos(x)^2),x, algorithm="giac")

[Out]

(pi*floor(x/pi + 1/2)*sgn(a) + arctan(a*tan(x)/sqrt(a^2 + a*b)))*b^2/(sqrt(a^2 + a*b)*a^2) + 1/3*(a^2*tan(x)^3
+ 3*a^2*tan(x) - 3*a*b*tan(x))/a^3

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maple [A]  time = 0.11, size = 51, normalized size = 0.91 \[ \frac {\tan ^{3}\relax (x )}{3 a}+\frac {\tan \relax (x )}{a}-\frac {\tan \relax (x ) b}{a^{2}}+\frac {b^{2} \arctan \left (\frac {a \tan \relax (x )}{\sqrt {\left (a +b \right ) a}}\right )}{a^{2} \sqrt {\left (a +b \right ) a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^4/(a+b*cos(x)^2),x)

[Out]

1/3*tan(x)^3/a+tan(x)/a-1/a^2*tan(x)*b+b^2/a^2/((a+b)*a)^(1/2)*arctan(a*tan(x)/((a+b)*a)^(1/2))

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maxima [A]  time = 0.87, size = 48, normalized size = 0.86 \[ \frac {b^{2} \arctan \left (\frac {a \tan \relax (x)}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{2}} + \frac {a \tan \relax (x)^{3} + 3 \, {\left (a - b\right )} \tan \relax (x)}{3 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(a+b*cos(x)^2),x, algorithm="maxima")

[Out]

b^2*arctan(a*tan(x)/sqrt((a + b)*a))/(sqrt((a + b)*a)*a^2) + 1/3*(a*tan(x)^3 + 3*(a - b)*tan(x))/a^2

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mupad [B]  time = 2.31, size = 51, normalized size = 0.91 \[ \frac {{\mathrm {tan}\relax (x)}^3}{3\,a}-\mathrm {tan}\relax (x)\,\left (\frac {a+b}{a^2}-\frac {2}{a}\right )+\frac {b^2\,\mathrm {atan}\left (\frac {\sqrt {a}\,\mathrm {tan}\relax (x)}{\sqrt {a+b}}\right )}{a^{5/2}\,\sqrt {a+b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^4*(a + b*cos(x)^2)),x)

[Out]

tan(x)^3/(3*a) - tan(x)*((a + b)/a^2 - 2/a) + (b^2*atan((a^(1/2)*tan(x))/(a + b)^(1/2)))/(a^(5/2)*(a + b)^(1/2
))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\relax (x )}}{a + b \cos ^{2}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**4/(a+b*cos(x)**2),x)

[Out]

Integral(sec(x)**4/(a + b*cos(x)**2), x)

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