3.3.84 \(\int e^{\coth ^{-1}(x)} (1+x)^2 \, dx\) [284]

Optimal. Leaf size=106 \[ \frac {5}{2} \sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}} x+\frac {5}{6} \left (1+\frac {1}{x}\right )^{3/2} \sqrt {\frac {-1+x}{x}} x^2+\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3+\frac {5}{2} \tanh ^{-1}\left (\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}}\right ) \]

[Out]

5/2*arctanh((1+1/x)^(1/2)*((-1+x)/x)^(1/2))+5/6*(1+1/x)^(3/2)*x^2*((-1+x)/x)^(1/2)+1/3*(1+1/x)^(5/2)*x^3*((-1+
x)/x)^(1/2)+5/2*x*(1+1/x)^(1/2)*((-1+x)/x)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6310, 6315, 96, 94, 212} \begin {gather*} \frac {1}{3} \left (\frac {1}{x}+1\right )^{5/2} \sqrt {\frac {x-1}{x}} x^3+\frac {5}{6} \left (\frac {1}{x}+1\right )^{3/2} \sqrt {\frac {x-1}{x}} x^2+\frac {5}{2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {x-1}{x}} x+\frac {5}{2} \tanh ^{-1}\left (\sqrt {\frac {1}{x}+1} \sqrt {\frac {x-1}{x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*(1 + x)^2,x]

[Out]

(5*Sqrt[1 + x^(-1)]*Sqrt[(-1 + x)/x]*x)/2 + (5*(1 + x^(-1))^(3/2)*Sqrt[(-1 + x)/x]*x^2)/6 + ((1 + x^(-1))^(5/2
)*Sqrt[(-1 + x)/x]*x^3)/3 + (5*ArcTanh[Sqrt[1 + x^(-1)]*Sqrt[(-1 + x)/x]])/2

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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -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 6310

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*x^p*(1 + c/(d*
x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !IntegerQ[n/2] && Inte
gerQ[p]

Rule 6315

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[-c^p, Subst[Int[(1 +
d*(x/c))^p*((1 + x/a)^(n/2)/(x^(m + 2)*(1 - x/a)^(n/2))), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ
[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegerQ[m]

Rubi steps

\begin {align*} \int e^{\coth ^{-1}(x)} (1+x)^2 \, dx &=\int e^{\coth ^{-1}(x)} \left (1+\frac {1}{x}\right )^2 x^2 \, dx\\ &=-\text {Subst}\left (\int \frac {(1+x)^{5/2}}{\sqrt {1-x} x^4} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3-\frac {5}{3} \text {Subst}\left (\int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^3} \, dx,x,\frac {1}{x}\right )\\ &=\frac {5}{6} \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2+\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3-\frac {5}{2} \text {Subst}\left (\int \frac {\sqrt {1+x}}{\sqrt {1-x} x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {5}{2} \sqrt {1+\frac {1}{x}} \sqrt {-\frac {1-x}{x}} x+\frac {5}{6} \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2+\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3-\frac {5}{2} \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x \sqrt {1+x}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {5}{2} \sqrt {1+\frac {1}{x}} \sqrt {-\frac {1-x}{x}} x+\frac {5}{6} \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2+\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3+\frac {5}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}}\right )\\ &=\frac {5}{2} \sqrt {1+\frac {1}{x}} \sqrt {-\frac {1-x}{x}} x+\frac {5}{6} \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2+\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3+\frac {5}{2} \tanh ^{-1}\left (\sqrt {1+\frac {1}{x}} \sqrt {-\frac {1-x}{x}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 47, normalized size = 0.44 \begin {gather*} \frac {1}{6} \sqrt {1-\frac {1}{x^2}} x \left (22+9 x+2 x^2\right )+\frac {5}{2} \log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCoth[x]*(1 + x)^2,x]

[Out]

(Sqrt[1 - x^(-2)]*x*(22 + 9*x + 2*x^2))/6 + (5*Log[(1 + Sqrt[1 - x^(-2)])*x])/2

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Maple [A]
time = 0.11, size = 69, normalized size = 0.65

method	result	size
risch	\(\frac {\left (2 x^{2}+9 x +22\right ) \left (-1+x \right )}{6 \sqrt {\frac {-1+x}{1+x}}}+\frac {5 \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (1+x \right ) \left (-1+x \right )}}{2 \sqrt {\frac {-1+x}{1+x}}\, \left (1+x \right )}\)	\(65\)
trager	\(\frac {\left (1+x \right ) \left (2 x^{2}+9 x +22\right ) \sqrt {-\frac {1-x}{1+x}}}{6}+\frac {5 \ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )}{2}\)	\(66\)
default	\(\frac {\left (-1+x \right ) \left (2 \left (\left (1+x \right ) \left (-1+x \right )\right )^{\frac {3}{2}}+9 x \sqrt {x^{2}-1}+24 \sqrt {x^{2}-1}+15 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{6 \sqrt {\frac {-1+x}{1+x}}\, \sqrt {\left (1+x \right ) \left (-1+x \right )}}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(-1+x)*(2*((1+x)*(-1+x))^(3/2)+9*x*(x^2-1)^(1/2)+24*(x^2-1)^(1/2)+15*ln(x+(x^2-1)^(1/2)))/((-1+x)/(1+x))^(
1/2)/((1+x)*(-1+x))^(1/2)

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Maxima [A]
time = 0.25, size = 112, normalized size = 1.06 \begin {gather*} -\frac {15 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{2}} - 40 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} + 33 \, \sqrt {\frac {x - 1}{x + 1}}}{3 \, {\left (\frac {3 \, {\left (x - 1\right )}}{x + 1} - \frac {3 \, {\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac {{\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - 1\right )}} + \frac {5}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {5}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(15*((x - 1)/(x + 1))^(5/2) - 40*((x - 1)/(x + 1))^(3/2) + 33*sqrt((x - 1)/(x + 1)))/(3*(x - 1)/(x + 1) -
3*(x - 1)^2/(x + 1)^2 + (x - 1)^3/(x + 1)^3 - 1) + 5/2*log(sqrt((x - 1)/(x + 1)) + 1) - 5/2*log(sqrt((x - 1)/
(x + 1)) - 1)

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Fricas [A]
time = 0.42, size = 61, normalized size = 0.58 \begin {gather*} \frac {1}{6} \, {\left (2 \, x^{3} + 11 \, x^{2} + 31 \, x + 22\right )} \sqrt {\frac {x - 1}{x + 1}} + \frac {5}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {5}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*x^3 + 11*x^2 + 31*x + 22)*sqrt((x - 1)/(x + 1)) + 5/2*log(sqrt((x - 1)/(x + 1)) + 1) - 5/2*log(sqrt((x
- 1)/(x + 1)) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.41, size = 60, normalized size = 0.57 \begin {gather*} \frac {1}{6} \, \sqrt {x^{2} - 1} {\left (x {\left (\frac {2 \, x}{\mathrm {sgn}\left (x + 1\right )} + \frac {9}{\mathrm {sgn}\left (x + 1\right )}\right )} + \frac {22}{\mathrm {sgn}\left (x + 1\right )}\right )} - \frac {5 \, \log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{2 \, \mathrm {sgn}\left (x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(x^2 - 1)*(x*(2*x/sgn(x + 1) + 9/sgn(x + 1)) + 22/sgn(x + 1)) - 5/2*log(abs(-x + sqrt(x^2 - 1)))/sgn(x
+ 1)

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Mupad [B]
time = 0.05, size = 94, normalized size = 0.89 \begin {gather*} 5\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )-\frac {11\,\sqrt {\frac {x-1}{x+1}}-\frac {40\,{\left (\frac {x-1}{x+1}\right )}^{3/2}}{3}+5\,{\left (\frac {x-1}{x+1}\right )}^{5/2}}{\frac {3\,\left (x-1\right )}{x+1}-\frac {3\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}+\frac {{\left (x-1\right )}^3}{{\left (x+1\right )}^3}-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*atanh(((x - 1)/(x + 1))^(1/2)) - (11*((x - 1)/(x + 1))^(1/2) - (40*((x - 1)/(x + 1))^(3/2))/3 + 5*((x - 1)/(
x + 1))^(5/2))/((3*(x - 1))/(x + 1) - (3*(x - 1)^2)/(x + 1)^2 + (x - 1)^3/(x + 1)^3 - 1)

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