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

Optimal. Leaf size=43 \[ \frac {4 \sqrt {x+1}}{\sqrt {1-x}}-\sin ^{-1}(x)-\tanh ^{-1}\left (\sqrt {1-x} \sqrt {x+1}\right ) \]

[Out]

-arcsin(x)-arctanh((1-x)^(1/2)*(1+x)^(1/2))+4*(1+x)^(1/2)/(1-x)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {98, 21, 105, 41, 216, 92, 206} \[ \frac {4 \sqrt {x+1}}{\sqrt {1-x}}-\sin ^{-1}(x)-\tanh ^{-1}\left (\sqrt {1-x} \sqrt {x+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^(3/2)/((1 - x)^(3/2)*x),x]

[Out]

(4*Sqrt[1 + x])/Sqrt[1 - x] - ArcSin[x] - ArcTanh[Sqrt[1 - x]*Sqrt[1 + x]]

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 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

Rule 206

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

Rule 216

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

Rubi steps

\begin {align*} \int \frac {(1+x)^{3/2}}{(1-x)^{3/2} x} \, dx &=\frac {4 \sqrt {1+x}}{\sqrt {1-x}}-2 \int \frac {-\frac {1}{2}+\frac {x}{2}}{\sqrt {1-x} x \sqrt {1+x}} \, dx\\ &=\frac {4 \sqrt {1+x}}{\sqrt {1-x}}+\int \frac {\sqrt {1-x}}{x \sqrt {1+x}} \, dx\\ &=\frac {4 \sqrt {1+x}}{\sqrt {1-x}}-\int \frac {1}{\sqrt {1-x} \sqrt {1+x}} \, dx+\int \frac {1}{\sqrt {1-x} x \sqrt {1+x}} \, dx\\ &=\frac {4 \sqrt {1+x}}{\sqrt {1-x}}-\int \frac {1}{\sqrt {1-x^2}} \, dx-\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-x} \sqrt {1+x}\right )\\ &=\frac {4 \sqrt {1+x}}{\sqrt {1-x}}-\sin ^{-1}(x)-\tanh ^{-1}\left (\sqrt {1-x} \sqrt {1+x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 61, normalized size = 1.42 \[ \frac {2 \left (\sqrt {1-x^2} \sin ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )+2 x+2\right )}{\sqrt {1-x^2}}-\tanh ^{-1}\left (\sqrt {1-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)^(3/2)/((1 - x)^(3/2)*x),x]

[Out]

(2*(2 + 2*x + Sqrt[1 - x^2]*ArcSin[Sqrt[1 - x]/Sqrt[2]]))/Sqrt[1 - x^2] - ArcTanh[Sqrt[1 - x^2]]

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fricas [B]  time = 0.41, size = 74, normalized size = 1.72 \[ \frac {2 \, {\left (x - 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) + {\left (x - 1\right )} \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) + 4 \, x - 4 \, \sqrt {x + 1} \sqrt {-x + 1} - 4}{x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(x - 1)*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) + (x - 1)*log((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) + 4*x - 4*
sqrt(x + 1)*sqrt(-x + 1) - 4)/(x - 1)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,-4,0,%%%{
4,[2]%%%}] at parameters values [-93.616423693]Warning, choosing root of [1,0,-4,0,%%%{4,[2]%%%}] at parameter
s values [-17.8804557086]2*(-1/2*pi-atan(sqrt(x+1)*((-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1))^2-1)/(-2*sqrt(-
x+1)+2*sqrt(2))))-ln(abs(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))+2-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1)))+ln(
abs(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-2-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1)))+4*sqrt(x+1)*sqrt(-x+1)/(
-x+1)

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maple [A]  time = 0.02, size = 70, normalized size = 1.63 \[ \frac {\left (-x \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )-x \arcsin \relax (x )+\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )+\arcsin \relax (x )-4 \sqrt {-x^{2}+1}\right ) \sqrt {-x +1}\, \sqrt {x +1}}{\left (x -1\right ) \sqrt {-x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-arcsin(x)*x-arctanh(1/(-x^2+1)^(1/2))*x+arcsin(x)+arctanh(1/(-x^2+1)^(1/2))-4*(-x^2+1)^(1/2))*(-x+1)^(1/2)*(
x+1)^(1/2)/(x-1)/(-x^2+1)^(1/2)

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maxima [A]  time = 1.99, size = 53, normalized size = 1.23 \[ \frac {4 \, x}{\sqrt {-x^{2} + 1}} + \frac {4}{\sqrt {-x^{2} + 1}} - \arcsin \relax (x) - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x/sqrt(-x^2 + 1) + 4/sqrt(-x^2 + 1) - arcsin(x) - log(2*sqrt(-x^2 + 1)/abs(x) + 2/abs(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (x+1\right )}^{3/2}}{x\,{\left (1-x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x + 1\right )^{\frac {3}{2}}}{x \left (1 - x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x + 1)**(3/2)/(x*(1 - x)**(3/2)), x)

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