3.9.65 \(\int \frac {(1+x)^3}{x (1-x^2)^{3/2}} \, dx\) [865]

Optimal. Leaf size=35 \[ \frac {4 (1+x)}{\sqrt {1-x^2}}-\sin ^{-1}(x)-\tanh ^{-1}\left (\sqrt {1-x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1819, 858, 222, 272, 65, 212} \begin {gather*} -\text {ArcSin}(x)+\frac {4 (x+1)}{\sqrt {1-x^2}}-\tanh ^{-1}\left (\sqrt {1-x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x)^3/(x*(1 - x^2)^(3/2)),x]

[Out]

(4*(1 + x))/Sqrt[1 - x^2] - ArcSin[x] - ArcTanh[Sqrt[1 - x^2]]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]

Rule 272

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

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 59, normalized size = 1.69 \begin {gather*} -\frac {4 \sqrt {1-x^2}}{-1+x}+2 \tan ^{-1}\left (\frac {\sqrt {1-x^2}}{1+x}\right )-2 \tanh ^{-1}\left (\frac {\sqrt {1-x^2}}{1+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.62, size = 41, normalized size = 1.17

method	result	size
risch	\(\frac {4+4 x}{\sqrt {-x^{2}+1}}-\arcsin \left (x \right )-\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )\)	\(32\)
default	\(\frac {4 x}{\sqrt {-x^{2}+1}}-\arcsin \left (x \right )+\frac {4}{\sqrt {-x^{2}+1}}-\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )\)	\(41\)
trager	\(-\frac {4 \sqrt {-x^{2}+1}}{-1+x}+\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (x \RootOf \left (\textit {\_Z}^{2}+1\right )+\sqrt {-x^{2}+1}\right )+\ln \left (\frac {\sqrt {-x^{2}+1}-1}{x}\right )\)	\(60\)
meijerg	\(\frac {-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {-x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{2}+1}}{2}\right )+\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{2}}{\sqrt {\pi }}+\frac {3 x}{\sqrt {-x^{2}+1}}-\frac {3 \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {-x^{2}+1}}\right )}{\sqrt {\pi }}+\frac {i \left (-\frac {i \sqrt {\pi }\, x}{\sqrt {-x^{2}+1}}+i \sqrt {\pi }\, \arcsin \left (x \right )\right )}{\sqrt {\pi }}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 53, normalized size = 1.51 \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/x/(-x^2+1)^(3/2),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. 63 vs. \(2 (31) = 62\).
time = 0.35, size = 63, normalized size = 1.80 \begin {gather*} \frac {2 \, {\left (x - 1\right )} \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) + {\left (x - 1\right )} \log \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) + 4 \, x - 4 \, \sqrt {-x^{2} + 1} - 4}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(x - 1)*arctan((sqrt(-x^2 + 1) - 1)/x) + (x - 1)*log((sqrt(-x^2 + 1) - 1)/x) + 4*x - 4*sqrt(-x^2 + 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 )^{3}}{x \left (- \left (x - 1\right ) \left (x + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.41, size = 44, normalized size = 1.26 \begin {gather*} \frac {8}{\frac {\sqrt {-x^{2} + 1} - 1}{x} + 1} - \arcsin \left (x\right ) + \log \left (-\frac {\sqrt {-x^{2} + 1} - 1}{{\left | x \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.20, size = 37, normalized size = 1.06 \begin {gather*} \ln \left (\sqrt {\frac {1}{x^2}-1}-\sqrt {\frac {1}{x^2}}\right )-\mathrm {asin}\left (x\right )-\frac {4\,\sqrt {1-x^2}}{x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)^3/(x*(1 - x^2)^(3/2)),x)

[Out]

log((1/x^2 - 1)^(1/2) - (1/x^2)^(1/2)) - asin(x) - (4*(1 - x^2)^(1/2))/(x - 1)

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