3.7.91 \(\int \frac {\csc ^{-1}(x)}{x^2 (-1+x^2)^{5/2}} \, dx\) [691]

Optimal. Leaf size=70 \[ -\frac {1}{\sqrt {x^2}}+\frac {\sqrt {x^2}}{6 \left (-1+x^2\right )}-\frac {11}{6} \coth ^{-1}\left (\sqrt {x^2}\right )+\frac {\left (3-12 x^2+8 x^4\right ) \csc ^{-1}(x)}{3 x \left (-1+x^2\right )^{3/2}} \]

[Out]

-11/6*arccoth((x^2)^(1/2))+1/3*(8*x^4-12*x^2+3)*arccsc(x)/x/(x^2-1)^(3/2)-1/(x^2)^(1/2)+1/6*(x^2)^(1/2)/(x^2-1
)

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Rubi [A]
time = 0.06, antiderivative size = 91, normalized size of antiderivative = 1.30, number of steps used = 5, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {277, 198, 197, 5347, 12, 1273, 464, 212} \begin {gather*} -\frac {1}{\sqrt {x^2}}-\frac {\sqrt {x^2}}{6 \left (1-x^2\right )}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {4 x \csc ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}+\frac {\csc ^{-1}(x)}{x \left (x^2-1\right )^{3/2}}-\frac {11 x \tanh ^{-1}(x)}{6 \sqrt {x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCsc[x]/(x^2*(-1 + x^2)^(5/2)),x]

[Out]

-(1/Sqrt[x^2]) - Sqrt[x^2]/(6*(1 - x^2)) + ArcCsc[x]/(x*(-1 + x^2)^(3/2)) - (4*x*ArcCsc[x])/(3*(-1 + x^2)^(3/2
)) + (8*x*ArcCsc[x])/(3*Sqrt[-1 + x^2]) - (11*x*ArcTanh[x])/(6*Sqrt[x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 1273

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(-d)^(m
/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*
p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x],
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 5347

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCsc[c*x], u, x] + Dist[b*c*(x/Sqrt[c^2*x^2]), Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

\begin {align*} \int \frac {\csc ^{-1}(x)}{x^2 \left (-1+x^2\right )^{5/2}} \, dx &=\frac {\csc ^{-1}(x)}{x \left (-1+x^2\right )^{3/2}}-\frac {4 x \csc ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {-1+x^2}}+\frac {x \int \frac {3-12 x^2+8 x^4}{3 x^2 \left (1-x^2\right )^2} \, dx}{\sqrt {x^2}}\\ &=\frac {\csc ^{-1}(x)}{x \left (-1+x^2\right )^{3/2}}-\frac {4 x \csc ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {-1+x^2}}+\frac {x \int \frac {3-12 x^2+8 x^4}{x^2 \left (1-x^2\right )^2} \, dx}{3 \sqrt {x^2}}\\ &=-\frac {\sqrt {x^2}}{6 \left (1-x^2\right )}+\frac {\csc ^{-1}(x)}{x \left (-1+x^2\right )^{3/2}}-\frac {4 x \csc ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {-1+x^2}}-\frac {x \int \frac {-6+17 x^2}{x^2 \left (1-x^2\right )} \, dx}{6 \sqrt {x^2}}\\ &=-\frac {1}{\sqrt {x^2}}-\frac {\sqrt {x^2}}{6 \left (1-x^2\right )}+\frac {\csc ^{-1}(x)}{x \left (-1+x^2\right )^{3/2}}-\frac {4 x \csc ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {-1+x^2}}-\frac {(11 x) \int \frac {1}{1-x^2} \, dx}{6 \sqrt {x^2}}\\ &=-\frac {1}{\sqrt {x^2}}-\frac {\sqrt {x^2}}{6 \left (1-x^2\right )}+\frac {\csc ^{-1}(x)}{x \left (-1+x^2\right )^{3/2}}-\frac {4 x \csc ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {-1+x^2}}-\frac {11 x \tanh ^{-1}(x)}{6 \sqrt {x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 79, normalized size = 1.13 \begin {gather*} \frac {4 \left (3-12 x^2+8 x^4\right ) \csc ^{-1}(x)+\sqrt {1-\frac {1}{x^2}} x \left (12-10 x^2+11 x \left (-1+x^2\right ) \log (1-x)-11 x \left (-1+x^2\right ) \log (1+x)\right )}{12 x \left (-1+x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCsc[x]/(x^2*(-1 + x^2)^(5/2)),x]

[Out]

(4*(3 - 12*x^2 + 8*x^4)*ArcCsc[x] + Sqrt[1 - x^(-2)]*x*(12 - 10*x^2 + 11*x*(-1 + x^2)*Log[1 - x] - 11*x*(-1 +
x^2)*Log[1 + x]))/(12*x*(-1 + x^2)^(3/2))

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Maple [C] Result contains complex when optimal does not.
time = 0.63, size = 702, normalized size = 10.03

method	result	size
default	\(-\frac {3 i x^{2}-4 i-4 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +\sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}}{4 \left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -1\right ) x \sqrt {x^{2}-1}}+\frac {\left (x^{2}-2+2 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \right ) \mathrm {arccsc}\left (x \right )}{4 x \sqrt {x^{2}-1}}+\frac {x \,\mathrm {arccsc}\left (x \right )}{2 \sqrt {x^{2}-1}}+\frac {\left (x^{2}-2-2 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \right ) \mathrm {arccsc}\left (x \right )}{4 \sqrt {x^{2}-1}\, x}+\frac {x^{3}}{4 \sqrt {x^{2}-1}\, \left (i x^{2}-2 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -2 i\right )}+\frac {2 \sqrt {x^{2}-1}\, x^{3} \mathrm {arccsc}\left (x \right )}{3 \left (x^{4}-2 x^{2}+1\right )}-\frac {x^{5} \left (\sqrt {\frac {x^{2}-1}{x^{2}}}\, x +i\right )}{24 \sqrt {x^{2}-1}\, \left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{5}-5 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}-3 x^{4}+4 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +7 x^{2}-4\right )}+\frac {x \sqrt {x^{2}-1}\, \left (x^{2}-2-2 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \right ) \mathrm {arccsc}\left (x \right )}{2 x^{4}-4 x^{2}+2}+\frac {x \left (5 i x^{4}-20 i x^{2}-12 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}+\sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{5}+16 i+16 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \right )}{24 \sqrt {x^{2}-1}\, \left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{5}-5 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}-3 x^{4}+4 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +7 x^{2}-4\right )}+\frac {\sqrt {x^{2}-1}\, x \left (x^{2}-2+2 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \right ) \mathrm {arccsc}\left (x \right )}{2 x^{4}-4 x^{2}+2}+\frac {11 \left (\sqrt {\frac {x^{2}-1}{x^{2}}}\, x +i\right ) \ln \left (\frac {i}{x}+\sqrt {1-\frac {1}{x^{2}}}-i\right )}{12 \sqrt {x^{2}-1}}+\frac {11 \left (\sqrt {\frac {x^{2}-1}{x^{2}}}\, x -i\right ) \ln \left (\frac {i}{x}+\sqrt {1-\frac {1}{x^{2}}}-i\right )}{12 \sqrt {x^{2}-1}}-\frac {11 \left (\sqrt {\frac {x^{2}-1}{x^{2}}}\, x +i\right ) \ln \left (\frac {i}{x}+\sqrt {1-\frac {1}{x^{2}}}+i\right )}{12 \sqrt {x^{2}-1}}-\frac {11 \left (\sqrt {\frac {x^{2}-1}{x^{2}}}\, x -i\right ) \ln \left (\frac {i}{x}+\sqrt {1-\frac {1}{x^{2}}}+i\right )}{12 \sqrt {x^{2}-1}}\)	\(702\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccsc(x)/x^2/(x^2-1)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/(I*((x^2-1)/x^2)^(1/2)*x-1)/x/(x^2-1)^(1/2)*(3*I*x^2-4*I-4*((x^2-1)/x^2)^(1/2)*x+((x^2-1)/x^2)^(1/2)*x^3)
+1/4/x/(x^2-1)^(1/2)*(x^2-2+2*I*((x^2-1)/x^2)^(1/2)*x)*arccsc(x)+1/2*x*arccsc(x)/(x^2-1)^(1/2)+1/4/(x^2-1)^(1/
2)*(x^2-2-2*I*((x^2-1)/x^2)^(1/2)*x)*arccsc(x)/x+1/4*x^3/(x^2-1)^(1/2)/(I*x^2-2*((x^2-1)/x^2)^(1/2)*x-2*I)+2/3
*(x^2-1)^(1/2)*x^3/(x^4-2*x^2+1)*arccsc(x)-1/24*x^5*(((x^2-1)/x^2)^(1/2)*x+I)/(x^2-1)^(1/2)/(I*((x^2-1)/x^2)^(
1/2)*x^5-5*I*((x^2-1)/x^2)^(1/2)*x^3-3*x^4+4*I*((x^2-1)/x^2)^(1/2)*x+7*x^2-4)+1/2*x*(x^2-1)^(1/2)*(x^2-2-2*I*(
(x^2-1)/x^2)^(1/2)*x)*arccsc(x)/(x^4-2*x^2+1)+1/24*x*(5*I*x^4-20*I*x^2-12*((x^2-1)/x^2)^(1/2)*x^3+((x^2-1)/x^2
)^(1/2)*x^5+16*I+16*((x^2-1)/x^2)^(1/2)*x)/(x^2-1)^(1/2)/(I*((x^2-1)/x^2)^(1/2)*x^5-5*I*((x^2-1)/x^2)^(1/2)*x^
3-3*x^4+4*I*((x^2-1)/x^2)^(1/2)*x+7*x^2-4)+1/2*(x^2-1)^(1/2)*x*(x^2-2+2*I*((x^2-1)/x^2)^(1/2)*x)*arccsc(x)/(x^
4-2*x^2+1)+11/12/(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*x+I)*ln(I/x+(1-1/x^2)^(1/2)-I)+11/12/(x^2-1)^(1/2)*(((x^2-
1)/x^2)^(1/2)*x-I)*ln(I/x+(1-1/x^2)^(1/2)-I)-11/12/(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*x+I)*ln(I/x+(1-1/x^2)^(1
/2)+I)-11/12/(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*x-I)*ln(I/x+(1-1/x^2)^(1/2)+I)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (56) = 112\).
time = 4.48, size = 123, normalized size = 1.76 \begin {gather*} \frac {32 \, x^{4} \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right ) - {\left (x^{3} - x\right )} \sqrt {x + 1} \sqrt {x - 1} {\left (\frac {2 \, {\left (5 \, x^{2} - 6\right )}}{x^{3} - x} + 11 \, \log \left (x + 1\right ) - 11 \, \log \left (x - 1\right )\right )} - 48 \, x^{2} \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right ) + 12 \, \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right )}{12 \, {\left (x^{3} - x\right )} \sqrt {x + 1} \sqrt {x - 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(x)/x^2/(x^2-1)^(5/2),x, algorithm="maxima")

[Out]

1/12*(32*x^4*arctan2(1, sqrt(x + 1)*sqrt(x - 1)) - (x^3 - x)*sqrt(x + 1)*sqrt(x - 1)*(2*(5*x^2 - 6)/(x^3 - x)
+ 11*log(x + 1) - 11*log(x - 1)) - 48*x^2*arctan2(1, sqrt(x + 1)*sqrt(x - 1)) + 12*arctan2(1, sqrt(x + 1)*sqrt
(x - 1)))/((x^3 - x)*sqrt(x + 1)*sqrt(x - 1))

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Fricas [A]
time = 0.66, size = 81, normalized size = 1.16 \begin {gather*} -\frac {10 \, x^{4} - 4 \, {\left (8 \, x^{4} - 12 \, x^{2} + 3\right )} \sqrt {x^{2} - 1} \operatorname {arccsc}\left (x\right ) - 22 \, x^{2} + 11 \, {\left (x^{5} - 2 \, x^{3} + x\right )} \log \left (x + 1\right ) - 11 \, {\left (x^{5} - 2 \, x^{3} + x\right )} \log \left (x - 1\right ) + 12}{12 \, {\left (x^{5} - 2 \, x^{3} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(x)/x^2/(x^2-1)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(10*x^4 - 4*(8*x^4 - 12*x^2 + 3)*sqrt(x^2 - 1)*arccsc(x) - 22*x^2 + 11*(x^5 - 2*x^3 + x)*log(x + 1) - 11
*(x^5 - 2*x^3 + x)*log(x - 1) + 12)/(x^5 - 2*x^3 + x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsc(x)/x**2/(x**2-1)**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6190 deep

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Giac [A]
time = 0.80, size = 105, normalized size = 1.50 \begin {gather*} \frac {1}{3} \, {\left (\frac {{\left (5 \, x^{2} - 6\right )} x}{{\left (x^{2} - 1\right )}^{\frac {3}{2}}} + \frac {6}{{\left (x - \sqrt {x^{2} - 1}\right )}^{2} + 1}\right )} \arcsin \left (\frac {1}{x}\right ) + \frac {2 \, \arctan \left (-x + \sqrt {x^{2} - 1}\right )}{\mathrm {sgn}\left (x\right )} - \frac {11 \, \log \left ({\left | x + 1 \right |}\right )}{12 \, \mathrm {sgn}\left (x\right )} + \frac {11 \, \log \left ({\left | x - 1 \right |}\right )}{12 \, \mathrm {sgn}\left (x\right )} - \frac {5 \, x^{2} - 6}{6 \, {\left (x^{3} - x\right )} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(x)/x^2/(x^2-1)^(5/2),x, algorithm="giac")

[Out]

1/3*((5*x^2 - 6)*x/(x^2 - 1)^(3/2) + 6/((x - sqrt(x^2 - 1))^2 + 1))*arcsin(1/x) + 2*arctan(-x + sqrt(x^2 - 1))
/sgn(x) - 11/12*log(abs(x + 1))/sgn(x) + 11/12*log(abs(x - 1))/sgn(x) - 1/6*(5*x^2 - 6)/((x^3 - x)*sgn(x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {asin}\left (\frac {1}{x}\right )}{x^2\,{\left (x^2-1\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(1/x)/(x^2*(x^2 - 1)^(5/2)),x)

[Out]

int(asin(1/x)/(x^2*(x^2 - 1)^(5/2)), x)

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