3.65 \(\int \frac{1}{(-1-\cos ^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=56 \[ \frac{\sin (x) \cos (x)}{2 \sqrt{-\cos ^2(x)-1}}+\frac{\sqrt{-\cos ^2(x)-1} E\left (\left .x+\frac{\pi }{2}\right |-1\right )}{2 \sqrt{\cos ^2(x)+1}} \]

[Out]

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

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Rubi [A]  time = 0.03296, antiderivative size = 56, 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} \[ \frac{\sin (x) \cos (x)}{2 \sqrt{-\cos ^2(x)-1}}+\frac{\sqrt{-\cos ^2(x)-1} E\left (\left .x+\frac{\pi }{2}\right |-1\right )}{2 \sqrt{\cos ^2(x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(-1 - Cos[x]^2)^(-3/2),x]

[Out]

(Sqrt[-1 - Cos[x]^2]*EllipticE[Pi/2 + x, -1])/(2*Sqrt[1 + Cos[x]^2]) + (Cos[x]*Sin[x])/(2*Sqrt[-1 - 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 (-1-\cos ^2(x)\right )^{3/2}} \, dx &=\frac{\cos (x) \sin (x)}{2 \sqrt{-1-\cos ^2(x)}}-\frac{1}{2} \int \frac{1+\cos ^2(x)}{\sqrt{-1-\cos ^2(x)}} \, dx\\ &=\frac{\cos (x) \sin (x)}{2 \sqrt{-1-\cos ^2(x)}}+\frac{1}{2} \int \sqrt{-1-\cos ^2(x)} \, dx\\ &=\frac{\cos (x) \sin (x)}{2 \sqrt{-1-\cos ^2(x)}}+\frac{\sqrt{-1-\cos ^2(x)} \int \sqrt{1+\cos ^2(x)} \, dx}{2 \sqrt{1+\cos ^2(x)}}\\ &=\frac{\sqrt{-1-\cos ^2(x)} E\left (\left .\frac{\pi }{2}+x\right |-1\right )}{2 \sqrt{1+\cos ^2(x)}}+\frac{\cos (x) \sin (x)}{2 \sqrt{-1-\cos ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0413073, size = 43, normalized size = 0.77 \[ \frac{\sin (2 x)-2 \sqrt{\cos (2 x)+3} E\left (x\left |\frac{1}{2}\right .\right )}{2 \sqrt{2} \sqrt{-\cos (2 x)-3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 - Cos[x]^2)^(-3/2),x]

[Out]

(-2*Sqrt[3 + Cos[2*x]]*EllipticE[x, 1/2] + Sin[2*x])/(2*Sqrt[2]*Sqrt[-3 - Cos[2*x]])

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Maple [A]  time = 1.184, size = 101, normalized size = 1.8 \begin{align*} -{\frac{1}{2\,\sin \left ( x \right ) }\sqrt{ \left ( \sin \left ( x \right ) \right ) ^{4}-2\, \left ( \sin \left ( x \right ) \right ) ^{2}} \left ( 2\,i\sqrt{ \left ( \sin \left ( x \right ) \right ) ^{2}}\sqrt{- \left ( \sin \left ( x \right ) \right ) ^{2}+2}{\it EllipticF} \left ( i\cos \left ( x \right ) ,i \right ) -i\sqrt{ \left ( \sin \left ( x \right ) \right ) ^{2}}\sqrt{- \left ( \sin \left ( x \right ) \right ) ^{2}+2}{\it EllipticE} \left ( i\cos \left ( x \right ) ,i \right ) -\cos \left ( x \right ) \left ( \sin \left ( x \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{ \left ( \cos \left ( x \right ) \right ) ^{4}-1}}}{\frac{1}{\sqrt{-1- \left ( \cos \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-1-cos(x)^2)^(3/2),x)

[Out]

-1/2*(sin(x)^4-2*sin(x)^2)^(1/2)*(2*I*(sin(x)^2)^(1/2)*(-sin(x)^2+2)^(1/2)*EllipticF(I*cos(x),I)-I*(sin(x)^2)^
(1/2)*(-sin(x)^2+2)^(1/2)*EllipticE(I*cos(x),I)-cos(x)*sin(x)^2)/(cos(x)^4-1)^(1/2)/sin(x)/(-1-cos(x)^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1-cos(x)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((-cos(x)^2 - 1)^(-3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left (e^{\left (4 i \, x\right )} + 6 \, e^{\left (2 i \, x\right )} + 1\right )}{\rm integral}\left (\frac{e^{\left (2 i \, x\right )} + 3}{2 \, \sqrt{e^{\left (4 i \, x\right )} + 6 \, e^{\left (2 i \, x\right )} + 1}}, x\right ) - \sqrt{e^{\left (4 i \, x\right )} + 6 \, e^{\left (2 i \, x\right )} + 1}{\left (e^{\left (3 i \, x\right )} + 3 \, e^{\left (i \, x\right )}\right )}}{2 \,{\left (e^{\left (4 i \, x\right )} + 6 \, e^{\left (2 i \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1-cos(x)^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(2*(e^(4*I*x) + 6*e^(2*I*x) + 1)*integral(1/2*(e^(2*I*x) + 3)/sqrt(e^(4*I*x) + 6*e^(2*I*x) + 1), x) - sqrt
(e^(4*I*x) + 6*e^(2*I*x) + 1)*(e^(3*I*x) + 3*e^(I*x)))/(e^(4*I*x) + 6*e^(2*I*x) + 1)

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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/(-1-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 (-\cos \left (x\right )^{2} - 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1-cos(x)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((-cos(x)^2 - 1)^(-3/2), x)