3.9.64 \(\int \frac {(1+x)^{3/2}}{(1-x)^{3/2} x} \, dx\) [864]

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

[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 = {100, 21, 132, 41, 222, 94, 212} \begin {gather*} -\text {ArcSin}(x)+\frac {4 \sqrt {x+1}}{\sqrt {1-x}}-\tanh ^{-1}\left (\sqrt {1-x} \sqrt {x+1}\right ) \end {gather*}

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 94

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 100

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 132

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[b*d^(m
+ n)*f^p, Int[(a + b*x)^(m - 1)/(c + d*x)^m, x], x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandTo
Sum[(a + b*x)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n,
-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 222

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-\text {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 [C] Result contains complex when optimal does not.
time = 0.13, size = 61, normalized size = 1.42 \begin {gather*} -\frac {4 \sqrt {1-x^2}}{-1+x}-2 \tan ^{-1}\left (\frac {\sqrt {1+x}}{\sqrt {1-x}}\right )+2 i \tan ^{-1}\left (x+i \sqrt {1-x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[1 - x^2])/(-1 + x) - 2*ArcTan[Sqrt[1 + x]/Sqrt[1 - x]] + (2*I)*ArcTan[x + I*Sqrt[1 - x^2]]

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Maple [A]
time = 0.60, size = 70, normalized size = 1.63

method	result	size
default	\(\frac {\left (-\arcsin \left (x \right ) x -\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x +\arcsin \left (x \right )+\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )-4 \sqrt {-x^{2}+1}\right ) \sqrt {1-x}\, \sqrt {1+x}}{\left (-1+x \right ) \sqrt {-x^{2}+1}}\)	\(70\)
risch	\(\frac {4 \sqrt {1+x}\, \sqrt {\left (1-x \right ) \left (1+x \right )}}{\sqrt {-\left (1+x \right ) \left (-1+x \right )}\, \sqrt {1-x}}-\frac {\left (\arcsin \left (x \right )+\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )\right ) \sqrt {\left (1-x \right ) \left (1+x \right )}}{\sqrt {1-x}\, \sqrt {1+x}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))*(1-x)^(1/2)*(1
+x)^(1/2)/(-1+x)/(-x^2+1)^(1/2)

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Maxima [A]
time = 0.48, size = 53, normalized size = 1.23 \begin {gather*} \frac {4 \, x}{\sqrt {-x^{2} + 1}} + \frac {4}{\sqrt {-x^{2} + 1}} - \arcsin \left (x\right ) - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (35) = 70\).
time = 0.35, size = 74, normalized size = 1.72 \begin {gather*} \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} \end {gather*}

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

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

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 [5.38357630698]Warning, choosing root of [1,0,-4,0,%%%{4,[2]%%%}] at parameter
s values [81.11954429

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

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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