\(\int (-1+\csc ^2(x))^{3/2} \, dx\) [22]

Optimal result

Integrand size = 10, antiderivative size = 30 \[ \int \left (-1+\csc ^2(x)\right )^{3/2} \, dx=-\frac {1}{2} \cot ^2(x)^{3/2} \tan (x)-\sqrt {\cot ^2(x)} \log (\sin (x)) \tan (x) \]

[Out]

-1/2*(cot(x)^2)^(3/2)*tan(x)-ln(sin(x))*(cot(x)^2)^(1/2)*tan(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4206, 3739, 3554, 3556} \[ \int \left (-1+\csc ^2(x)\right )^{3/2} \, dx=-\frac {1}{2} \tan (x) \cot ^2(x)^{3/2}-\tan (x) \sqrt {\cot ^2(x)} \log (\sin (x)) \]

[In]

Int[(-1 + Csc[x]^2)^(3/2),x]

[Out]

-1/2*((Cot[x]^2)^(3/2)*Tan[x]) - Sqrt[Cot[x]^2]*Log[Sin[x]]*Tan[x]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3556

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

Rule 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[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 4206

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

Rubi steps \begin{align*} \text {integral}& = \int \cot ^2(x)^{3/2} \, dx \\ & = \left (\sqrt {\cot ^2(x)} \tan (x)\right ) \int \cot ^3(x) \, dx \\ & = -\frac {1}{2} \cot ^2(x)^{3/2} \tan (x)-\left (\sqrt {\cot ^2(x)} \tan (x)\right ) \int \cot (x) \, dx \\ & = -\frac {1}{2} \cot ^2(x)^{3/2} \tan (x)-\sqrt {\cot ^2(x)} \log (\sin (x)) \tan (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \left (-1+\csc ^2(x)\right )^{3/2} \, dx=-\frac {1}{2} \sqrt {\cot ^2(x)} \left (\cot ^2(x)+2 (\log (\cos (x))+\log (\tan (x)))\right ) \tan (x) \]

[In]

Integrate[(-1 + Csc[x]^2)^(3/2),x]

[Out]

-1/2*(Sqrt[Cot[x]^2]*(Cot[x]^2 + 2*(Log[Cos[x]] + Log[Tan[x]]))*Tan[x])

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.55 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30

method	result	size
default	\(-\frac {\operatorname {csgn}\left (\cot \left (x \right )\right ) \left (4 \ln \left (\csc \left (x \right )-\cot \left (x \right )\right )-4 \ln \left (\frac {2}{\cos \left (x \right )+1}\right )+\cot \left (x \right )^{2}+\csc \left (x \right )^{2}\right ) \sqrt {4}}{8}\)	\(39\)
risch	\(\frac {\sqrt {-\frac {\left ({\mathrm e}^{2 i x}+1\right )^{2}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}\, \left (i {\mathrm e}^{4 i x} \ln \left ({\mathrm e}^{2 i x}-1\right )-2 i {\mathrm e}^{2 i x} \ln \left ({\mathrm e}^{2 i x}-1\right )+{\mathrm e}^{4 i x} x -2 i {\mathrm e}^{2 i x}+i \ln \left ({\mathrm e}^{2 i x}-1\right )-2 \,{\mathrm e}^{2 i x} x +x \right )}{\left ({\mathrm e}^{2 i x}+1\right ) \left ({\mathrm e}^{2 i x}-1\right )}\)	\(110\)

[In]

int((csc(x)^2-1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/8*csgn(cot(x))*(4*ln(csc(x)-cot(x))-4*ln(2/(cos(x)+1))+cot(x)^2+csc(x)^2)*4^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \left (-1+\csc ^2(x)\right )^{3/2} \, dx=\frac {2 \, {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (x\right )\right ) - 1}{2 \, {\left (\cos \left (x\right )^{2} - 1\right )}} \]

[In]

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

[Out]

1/2*(2*(cos(x)^2 - 1)*log(1/2*sin(x)) - 1)/(cos(x)^2 - 1)

Sympy [F]

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

[In]

integrate((-1+csc(x)**2)**(3/2),x)

[Out]

Integral((csc(x)**2 - 1)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \left (-1+\csc ^2(x)\right )^{3/2} \, dx=-\frac {1}{2 \, \tan \left (x\right )^{2}} + \frac {1}{2} \, \log \left (\tan \left (x\right )^{2} + 1\right ) - \log \left (\tan \left (x\right )\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \left (-1+\csc ^2(x)\right )^{3/2} \, dx=\frac {1}{8} \, \tan \left (\frac {1}{2} \, x\right )^{2} - \frac {4 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 1}{8 \, \tan \left (\frac {1}{2} \, x\right )^{2}} - \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right ) + \frac {1}{2} \, \log \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right ) \]

[In]

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

[Out]

1/8*tan(1/2*x)^2 - 1/8*(4*tan(1/2*x)^2 - 1)/tan(1/2*x)^2 - log(tan(1/2*x)^2 + 1) + 1/2*log(tan(1/2*x)^2)

Mupad [F(-1)]

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

[In]

int((1/sin(x)^2 - 1)^(3/2),x)

[Out]

int((1/sin(x)^2 - 1)^(3/2), x)