3.66 \(\int \frac{1}{(a+b \cos ^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=78 \[ \frac{\sqrt{a+b \cos ^2(x)} E\left (x+\frac{\pi }{2}|-\frac{b}{a}\right )}{a (a+b) \sqrt{\frac{b \cos ^2(x)}{a}+1}}-\frac{b \sin (x) \cos (x)}{a (a+b) \sqrt{a+b \cos ^2(x)}} \]

[Out]

(Sqrt[a + b*Cos[x]^2]*EllipticE[Pi/2 + x, -(b/a)])/(a*(a + b)*Sqrt[1 + (b*Cos[x]^2)/a]) - (b*Cos[x]*Sin[x])/(a
*(a + b)*Sqrt[a + b*Cos[x]^2])

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Rubi [A]  time = 0.0474696, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3184, 21, 3178, 3177} \[ \frac{\sqrt{a+b \cos ^2(x)} E\left (x+\frac{\pi }{2}|-\frac{b}{a}\right )}{a (a+b) \sqrt{\frac{b \cos ^2(x)}{a}+1}}-\frac{b \sin (x) \cos (x)}{a (a+b) \sqrt{a+b \cos ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[x]^2)^(-3/2),x]

[Out]

(Sqrt[a + b*Cos[x]^2]*EllipticE[Pi/2 + x, -(b/a)])/(a*(a + b)*Sqrt[1 + (b*Cos[x]^2)/a]) - (b*Cos[x]*Sin[x])/(a
*(a + b)*Sqrt[a + b*Cos[x]^2])

Rule 3184

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
e + f*x]^2)^(p + 1))/(2*a*f*(p + 1)*(a + b)), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^(p
+ 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ
[a + b, 0] && LtQ[p, -1]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3178

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[1 + (b*Sin
[e + f*x]^2)/a], Int[Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rule 3177

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[e + f*x, -(b/a)])/f, x]
/; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \cos ^2(x)\right )^{3/2}} \, dx &=-\frac{b \cos (x) \sin (x)}{a (a+b) \sqrt{a+b \cos ^2(x)}}-\frac{\int \frac{-a-b \cos ^2(x)}{\sqrt{a+b \cos ^2(x)}} \, dx}{a (a+b)}\\ &=-\frac{b \cos (x) \sin (x)}{a (a+b) \sqrt{a+b \cos ^2(x)}}+\frac{\int \sqrt{a+b \cos ^2(x)} \, dx}{a (a+b)}\\ &=-\frac{b \cos (x) \sin (x)}{a (a+b) \sqrt{a+b \cos ^2(x)}}+\frac{\sqrt{a+b \cos ^2(x)} \int \sqrt{1+\frac{b \cos ^2(x)}{a}} \, dx}{a (a+b) \sqrt{1+\frac{b \cos ^2(x)}{a}}}\\ &=\frac{\sqrt{a+b \cos ^2(x)} E\left (\frac{\pi }{2}+x|-\frac{b}{a}\right )}{a (a+b) \sqrt{1+\frac{b \cos ^2(x)}{a}}}-\frac{b \cos (x) \sin (x)}{a (a+b) \sqrt{a+b \cos ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.175397, size = 75, normalized size = 0.96 \[ \frac{2 (a+b) \sqrt{\frac{2 a+b \cos (2 x)+b}{a+b}} E\left (x\left |\frac{b}{a+b}\right .\right )-\sqrt{2} b \sin (2 x)}{2 a (a+b) \sqrt{2 a+b \cos (2 x)+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[x]^2)^(-3/2),x]

[Out]

(2*(a + b)*Sqrt[(2*a + b + b*Cos[2*x])/(a + b)]*EllipticE[x, b/(a + b)] - Sqrt[2]*b*Sin[2*x])/(2*a*(a + b)*Sqr
t[2*a + b + b*Cos[2*x]])

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Maple [A]  time = 0.798, size = 73, normalized size = 0.9 \begin{align*} -{\frac{1}{ \left ( a+b \right ) a\sin \left ( x \right ) } \left ( \sqrt{ \left ( \sin \left ( x \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \sin \left ( x \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}a{\it EllipticE} \left ( \cos \left ( x \right ) ,\sqrt{-{\frac{b}{a}}} \right ) +b\cos \left ( x \right ) \left ( \sin \left ( x \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{a+b \left ( \cos \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(x)^2)^(3/2),x)

[Out]

-((sin(x)^2)^(1/2)*(-b/a*sin(x)^2+(a+b)/a)^(1/2)*a*EllipticE(cos(x),(-b/a)^(1/2))+b*cos(x)*sin(x)^2)/a/(a+b)/s
in(x)/(a+b*cos(x)^2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (x\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*cos(x)^2 + a)^(-3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \cos \left (x\right )^{2} + a}}{b^{2} \cos \left (x\right )^{4} + 2 \, a b \cos \left (x\right )^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*cos(x)^2 + a)/(b^2*cos(x)^4 + 2*a*b*cos(x)^2 + a^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (x\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*cos(x)^2 + a)^(-3/2), x)