\(\int \frac {1}{(a-a \sin ^2(x))^{3/2}} \, dx\) [120]

Optimal result

Integrand size = 13, antiderivative size = 42 \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}(\sin (x)) \cos (x)}{2 a \sqrt {a \cos ^2(x)}}+\frac {\tan (x)}{2 a \sqrt {a \cos ^2(x)}} \]

[Out]

1/2*arctanh(sin(x))*cos(x)/a/(a*cos(x)^2)^(1/2)+1/2*tan(x)/a/(a*cos(x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3255, 3283, 3286, 3855} \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{3/2}} \, dx=\frac {\cos (x) \text {arctanh}(\sin (x))}{2 a \sqrt {a \cos ^2(x)}}+\frac {\tan (x)}{2 a \sqrt {a \cos ^2(x)}} \]

[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 3255

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 3283

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 3286

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \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 {\text {arctanh}(\sin (x)) \cos (x)}{2 a \sqrt {a \cos ^2(x)}}+\frac {\tan (x)}{2 a \sqrt {a \cos ^2(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}(\sin (x)) \cos (x)+\tan (x)}{2 a \sqrt {a \cos ^2(x)}} \]

[In]

Integrate[(a - a*Sin[x]^2)^(-3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {\left (-\ln \left (1+\sin \left (x \right )\right )+\ln \left (\sin \left (x \right )-1\right )\right ) \left (\cos ^{2}\left (x \right )\right )-2 \sin \left (x \right )}{4 a \cos \left (x \right ) \sqrt {a \left (\cos ^{2}\left (x \right )\right )}}\)	\(41\)
risch	\(-\frac {i \left ({\mathrm e}^{2 i x}-1\right )}{a \left ({\mathrm e}^{2 i x}+1\right ) \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}-\frac {\ln \left ({\mathrm e}^{i x}-i\right ) \cos \left (x \right )}{a \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}+\frac {\ln \left ({\mathrm e}^{i x}+i\right ) \cos \left (x \right )}{a \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}\)	\(109\)

[In]

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

[Out]

-1/4/a*((-ln(1+sin(x))+ln(sin(x)-1))*cos(x)^2-2*sin(x))/cos(x)/(a*cos(x)^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{3/2}} \, dx=-\frac {\sqrt {a \cos \left (x\right )^{2}} {\left (\cos \left (x\right )^{2} \log \left (-\frac {\sin \left (x\right ) - 1}{\sin \left (x\right ) + 1}\right ) - 2 \, \sin \left (x\right )\right )}}{4 \, a^{2} \cos \left (x\right )^{3}} \]

[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)

Sympy [F]

\[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{3/2}} \, dx=\int \frac {1}{\left (- a \sin ^{2}{\left (x \right )} + a\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (34) = 68\).

Time = 0.42 (sec) , antiderivative size = 304, normalized size of antiderivative = 7.24 \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{3/2}} \, dx=\frac {4 \, {\left (\sin \left (3 \, x\right ) - \sin \left (x\right )\right )} \cos \left (4 \, x\right ) + {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) + 1\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \sin \left (x\right ) + 1\right ) - {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) + 1\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \sin \left (x\right ) + 1\right ) - 4 \, {\left (\cos \left (3 \, x\right ) - \cos \left (x\right )\right )} \sin \left (4 \, x\right ) + 4 \, {\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \sin \left (3 \, x\right ) - 8 \, \cos \left (3 \, x\right ) \sin \left (2 \, x\right ) + 8 \, \cos \left (x\right ) \sin \left (2 \, x\right ) - 8 \, \cos \left (2 \, x\right ) \sin \left (x\right ) - 4 \, \sin \left (x\right )}{4 \, {\left (a \cos \left (4 \, x\right )^{2} + 4 \, a \cos \left (2 \, x\right )^{2} + a \sin \left (4 \, x\right )^{2} + 4 \, a \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, a \sin \left (2 \, x\right )^{2} + 2 \, {\left (2 \, a \cos \left (2 \, x\right ) + a\right )} \cos \left (4 \, x\right ) + 4 \, a \cos \left (2 \, x\right ) + a\right )} \sqrt {a}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{3/2}} \, dx=-\frac {\frac {\log \left ({\left | -\sqrt {a} \tan \left (x\right ) + \sqrt {a \tan \left (x\right )^{2} + a} \right |}\right )}{\sqrt {a}} - \frac {\sqrt {a \tan \left (x\right )^{2} + a} \tan \left (x\right )}{a}}{2 \, a} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (a-a\,{\sin \left (x\right )}^2\right )}^{3/2}} \,d x \]

[In]

int(1/(a - a*sin(x)^2)^(3/2),x)

[Out]

int(1/(a - a*sin(x)^2)^(3/2), x)