3.120 \(\int \frac{1}{(a-a \sin ^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=42 \[ \frac{\tan (x)}{2 a \sqrt{a \cos ^2(x)}}+\frac{\cos (x) \tanh ^{-1}(\sin (x))}{2 a \sqrt{a \cos ^2(x)}} \]

[Out]

(ArcTanh[Sin[x]]*Cos[x])/(2*a*Sqrt[a*Cos[x]^2]) + Tan[x]/(2*a*Sqrt[a*Cos[x]^2])

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Rubi [A]  time = 0.0374599, antiderivative size = 42, normalized size of antiderivative = 1., 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 = {3176, 3204, 3207, 3770} \[ \frac{\tan (x)}{2 a \sqrt{a \cos ^2(x)}}+\frac{\cos (x) \tanh ^{-1}(\sin (x))}{2 a \sqrt{a \cos ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sin[x]^2)^(-3/2),x]

[Out]

(ArcTanh[Sin[x]]*Cos[x])/(2*a*Sqrt[a*Cos[x]^2]) + Tan[x]/(2*a*Sqrt[a*Cos[x]^2])

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3204

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

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 0.0651455, size = 91, normalized size = 2.17 \[ -\frac{\cos (x) \left (-2 \sin (x)+\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+\cos (2 x) \left (\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )}{4 \left (a \cos ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sin[x]^2)^(-3/2),x]

[Out]

-(Cos[x]*(Log[Cos[x/2] - Sin[x/2]] + Cos[2*x]*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]]) - Log[Cos[
x/2] + Sin[x/2]] - 2*Sin[x]))/(4*(a*Cos[x]^2)^(3/2))

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Maple [B]  time = 0.994, size = 70, normalized size = 1.7 \begin{align*}{\frac{1}{2\,\sin \left ( x \right ) \cos \left ( x \right ) }\sqrt{a \left ( \sin \left ( x \right ) \right ) ^{2}} \left ( \ln \left ( 2\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( x \right ) \right ) ^{2}}+a}{\cos \left ( x \right ) }} \right ) \left ( \cos \left ( x \right ) \right ) ^{2}a+\sqrt{a}\sqrt{a \left ( \sin \left ( x \right ) \right ) ^{2}} \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a \left ( \cos \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-a*sin(x)^2)^(3/2),x)

[Out]

1/2/a^(5/2)/cos(x)*(a*sin(x)^2)^(1/2)*(ln(2/cos(x)*(a^(1/2)*(a*sin(x)^2)^(1/2)+a))*cos(x)^2*a+a^(1/2)*(a*sin(x
)^2)^(1/2))/sin(x)/(a*cos(x)^2)^(1/2)

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Maxima [B]  time = 1.6278, size = 410, normalized size = 9.76 \begin{align*} \frac{4 \,{\left (\sin \left (3 \, x\right ) - \sin \left (x\right )\right )} \cos \left (4 \, x\right ) +{\left (2 \,{\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) + 1\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \sin \left (x\right ) + 1\right ) -{\left (2 \,{\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) + 1\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \sin \left (x\right ) + 1\right ) - 4 \,{\left (\cos \left (3 \, x\right ) - \cos \left (x\right )\right )} \sin \left (4 \, x\right ) + 4 \,{\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \sin \left (3 \, x\right ) - 8 \, \cos \left (3 \, x\right ) \sin \left (2 \, x\right ) + 8 \, \cos \left (x\right ) \sin \left (2 \, x\right ) - 8 \, \cos \left (2 \, x\right ) \sin \left (x\right ) - 4 \, \sin \left (x\right )}{4 \,{\left (a \cos \left (4 \, x\right )^{2} + 4 \, a \cos \left (2 \, x\right )^{2} + a \sin \left (4 \, x\right )^{2} + 4 \, a \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, a \sin \left (2 \, x\right )^{2} + 2 \,{\left (2 \, a \cos \left (2 \, x\right ) + a\right )} \cos \left (4 \, x\right ) + 4 \, a \cos \left (2 \, x\right ) + a\right )} \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(x)^2)^(3/2),x, algorithm="maxima")

[Out]

1/4*(4*(sin(3*x) - sin(x))*cos(4*x) + (2*(2*cos(2*x) + 1)*cos(4*x) + cos(4*x)^2 + 4*cos(2*x)^2 + sin(4*x)^2 +
4*sin(4*x)*sin(2*x) + 4*sin(2*x)^2 + 4*cos(2*x) + 1)*log(cos(x)^2 + sin(x)^2 + 2*sin(x) + 1) - (2*(2*cos(2*x)
+ 1)*cos(4*x) + cos(4*x)^2 + 4*cos(2*x)^2 + sin(4*x)^2 + 4*sin(4*x)*sin(2*x) + 4*sin(2*x)^2 + 4*cos(2*x) + 1)*
log(cos(x)^2 + sin(x)^2 - 2*sin(x) + 1) - 4*(cos(3*x) - cos(x))*sin(4*x) + 4*(2*cos(2*x) + 1)*sin(3*x) - 8*cos
(3*x)*sin(2*x) + 8*cos(x)*sin(2*x) - 8*cos(2*x)*sin(x) - 4*sin(x))/((a*cos(4*x)^2 + 4*a*cos(2*x)^2 + a*sin(4*x
)^2 + 4*a*sin(4*x)*sin(2*x) + 4*a*sin(2*x)^2 + 2*(2*a*cos(2*x) + a)*cos(4*x) + 4*a*cos(2*x) + a)*sqrt(a))

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Fricas [A]  time = 1.75947, size = 124, normalized size = 2.95 \begin{align*} -\frac{\sqrt{a \cos \left (x\right )^{2}}{\left (\cos \left (x\right )^{2} \log \left (-\frac{\sin \left (x\right ) - 1}{\sin \left (x\right ) + 1}\right ) - 2 \, \sin \left (x\right )\right )}}{4 \, a^{2} \cos \left (x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(x)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(a*cos(x)^2)*(cos(x)^2*log(-(sin(x) - 1)/(sin(x) + 1)) - 2*sin(x))/(a^2*cos(x)^3)

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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-a*sin(x)**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.23198, size = 59, normalized size = 1.4 \begin{align*} -\frac{\sqrt{a} \log \left ({\left | -\sqrt{a} \tan \left (x\right ) + \sqrt{a \tan \left (x\right )^{2} + a} \right |}\right ) - \sqrt{a \tan \left (x\right )^{2} + a} \tan \left (x\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(x)^2)^(3/2),x, algorithm="giac")

[Out]

-1/2*(sqrt(a)*log(abs(-sqrt(a)*tan(x) + sqrt(a*tan(x)^2 + a))) - sqrt(a*tan(x)^2 + a)*tan(x))/a^2