### 3.283 $$\int e^{\coth ^{-1}(x)} x (1+x)^2 \, dx$$

Optimal. Leaf size=133 $\frac{1}{4} \left (\frac{1}{x}+1\right )^{7/2} \sqrt{\frac{x-1}{x}} x^4+\frac{1}{4} \left (\frac{1}{x}+1\right )^{5/2} \sqrt{\frac{x-1}{x}} x^3+\frac{5}{8} \left (\frac{1}{x}+1\right )^{3/2} \sqrt{\frac{x-1}{x}} x^2+\frac{15}{8} \sqrt{\frac{1}{x}+1} \sqrt{\frac{x-1}{x}} x+\frac{15}{8} \tanh ^{-1}\left (\sqrt{\frac{1}{x}+1} \sqrt{\frac{x-1}{x}}\right )$

[Out]

(15*Sqrt[1 + x^(-1)]*Sqrt[(-1 + x)/x]*x)/8 + (5*(1 + x^(-1))^(3/2)*Sqrt[(-1 + x)/x]*x^2)/8 + ((1 + x^(-1))^(5/
2)*Sqrt[(-1 + x)/x]*x^3)/4 + ((1 + x^(-1))^(7/2)*Sqrt[(-1 + x)/x]*x^4)/4 + (15*ArcTanh[Sqrt[1 + x^(-1)]*Sqrt[(
-1 + x)/x]])/8

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Rubi [A]  time = 0.111144, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.546, Rules used = {6175, 6180, 96, 94, 92, 206} $\frac{1}{4} \left (\frac{1}{x}+1\right )^{7/2} \sqrt{\frac{x-1}{x}} x^4+\frac{1}{4} \left (\frac{1}{x}+1\right )^{5/2} \sqrt{\frac{x-1}{x}} x^3+\frac{5}{8} \left (\frac{1}{x}+1\right )^{3/2} \sqrt{\frac{x-1}{x}} x^2+\frac{15}{8} \sqrt{\frac{1}{x}+1} \sqrt{\frac{x-1}{x}} x+\frac{15}{8} \tanh ^{-1}\left (\sqrt{\frac{1}{x}+1} \sqrt{\frac{x-1}{x}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[E^ArcCoth[x]*x*(1 + x)^2,x]

[Out]

(15*Sqrt[1 + x^(-1)]*Sqrt[(-1 + x)/x]*x)/8 + (5*(1 + x^(-1))^(3/2)*Sqrt[(-1 + x)/x]*x^2)/8 + ((1 + x^(-1))^(5/
2)*Sqrt[(-1 + x)/x]*x^3)/4 + ((1 + x^(-1))^(7/2)*Sqrt[(-1 + x)/x]*x^4)/4 + (15*ArcTanh[Sqrt[1 + x^(-1)]*Sqrt[(
-1 + x)/x]])/8

Rule 6175

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 6180

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]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 94

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[p, 1] &&  !SumSimplerQ[m, 1])

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 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])

Rubi steps

\begin{align*} \int e^{\coth ^{-1}(x)} x (1+x)^2 \, dx &=\int e^{\coth ^{-1}(x)} \left (1+\frac{1}{x}\right )^2 x^3 \, dx\\ &=-\operatorname{Subst}\left (\int \frac{(1+x)^{5/2}}{\sqrt{1-x} x^5} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{4} \left (1+\frac{1}{x}\right )^{7/2} \sqrt{\frac{-1+x}{x}} x^4-\frac{3}{4} \operatorname{Subst}\left (\int \frac{(1+x)^{5/2}}{\sqrt{1-x} x^4} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{4} \left (1+\frac{1}{x}\right )^{5/2} \sqrt{-\frac{1-x}{x}} x^3+\frac{1}{4} \left (1+\frac{1}{x}\right )^{7/2} \sqrt{\frac{-1+x}{x}} x^4-\frac{5}{4} \operatorname{Subst}\left (\int \frac{(1+x)^{3/2}}{\sqrt{1-x} x^3} \, dx,x,\frac{1}{x}\right )\\ &=\frac{5}{8} \left (1+\frac{1}{x}\right )^{3/2} \sqrt{-\frac{1-x}{x}} x^2+\frac{1}{4} \left (1+\frac{1}{x}\right )^{5/2} \sqrt{-\frac{1-x}{x}} x^3+\frac{1}{4} \left (1+\frac{1}{x}\right )^{7/2} \sqrt{\frac{-1+x}{x}} x^4-\frac{15}{8} \operatorname{Subst}\left (\int \frac{\sqrt{1+x}}{\sqrt{1-x} x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{15}{8} \sqrt{1+\frac{1}{x}} \sqrt{-\frac{1-x}{x}} x+\frac{5}{8} \left (1+\frac{1}{x}\right )^{3/2} \sqrt{-\frac{1-x}{x}} x^2+\frac{1}{4} \left (1+\frac{1}{x}\right )^{5/2} \sqrt{-\frac{1-x}{x}} x^3+\frac{1}{4} \left (1+\frac{1}{x}\right )^{7/2} \sqrt{\frac{-1+x}{x}} x^4-\frac{15}{8} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} x \sqrt{1+x}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{15}{8} \sqrt{1+\frac{1}{x}} \sqrt{-\frac{1-x}{x}} x+\frac{5}{8} \left (1+\frac{1}{x}\right )^{3/2} \sqrt{-\frac{1-x}{x}} x^2+\frac{1}{4} \left (1+\frac{1}{x}\right )^{5/2} \sqrt{-\frac{1-x}{x}} x^3+\frac{1}{4} \left (1+\frac{1}{x}\right )^{7/2} \sqrt{\frac{-1+x}{x}} x^4+\frac{15}{8} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{1+\frac{1}{x}} \sqrt{\frac{-1+x}{x}}\right )\\ &=\frac{15}{8} \sqrt{1+\frac{1}{x}} \sqrt{-\frac{1-x}{x}} x+\frac{5}{8} \left (1+\frac{1}{x}\right )^{3/2} \sqrt{-\frac{1-x}{x}} x^2+\frac{1}{4} \left (1+\frac{1}{x}\right )^{5/2} \sqrt{-\frac{1-x}{x}} x^3+\frac{1}{4} \left (1+\frac{1}{x}\right )^{7/2} \sqrt{\frac{-1+x}{x}} x^4+\frac{15}{8} \tanh ^{-1}\left (\sqrt{1+\frac{1}{x}} \sqrt{-\frac{1-x}{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0441036, size = 52, normalized size = 0.39 $\frac{1}{8} \sqrt{1-\frac{1}{x^2}} x \left (2 x^3+8 x^2+15 x+24\right )+\frac{15}{8} \log \left (\left (\sqrt{1-\frac{1}{x^2}}+1\right ) x\right )$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCoth[x]*x*(1 + x)^2,x]

[Out]

(Sqrt[1 - x^(-2)]*x*(24 + 15*x + 8*x^2 + 2*x^3))/8 + (15*Log[(1 + Sqrt[1 - x^(-2)])*x])/8

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Maple [A]  time = 0.107, size = 79, normalized size = 0.6 \begin{align*}{\frac{-1+x}{8} \left ( 2\,x \left ({x}^{2}-1 \right ) ^{3/2}+8\, \left ( \left ( 1+x \right ) \left ( -1+x \right ) \right ) ^{3/2}+17\,x\sqrt{{x}^{2}-1}+32\,\sqrt{{x}^{2}-1}+15\,\ln \left ( x+\sqrt{{x}^{2}-1} \right ) \right ){\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}}{\frac{1}{\sqrt{ \left ( 1+x \right ) \left ( -1+x \right ) }}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(-1+x)*(2*x*(x^2-1)^(3/2)+8*((1+x)*(-1+x))^(3/2)+17*x*(x^2-1)^(1/2)+32*(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 = 1.03094, size = 186, normalized size = 1.4 \begin{align*} \frac{15 \, \left (\frac{x - 1}{x + 1}\right )^{\frac{7}{2}} - 55 \, \left (\frac{x - 1}{x + 1}\right )^{\frac{5}{2}} + 73 \, \left (\frac{x - 1}{x + 1}\right )^{\frac{3}{2}} - 49 \, \sqrt{\frac{x - 1}{x + 1}}}{4 \,{\left (\frac{4 \,{\left (x - 1\right )}}{x + 1} - \frac{6 \,{\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac{4 \,{\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - \frac{{\left (x - 1\right )}^{4}}{{\left (x + 1\right )}^{4}} - 1\right )}} + \frac{15}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) - \frac{15}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} - 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(15*((x - 1)/(x + 1))^(7/2) - 55*((x - 1)/(x + 1))^(5/2) + 73*((x - 1)/(x + 1))^(3/2) - 49*sqrt((x - 1)/(x
+ 1)))/(4*(x - 1)/(x + 1) - 6*(x - 1)^2/(x + 1)^2 + 4*(x - 1)^3/(x + 1)^3 - (x - 1)^4/(x + 1)^4 - 1) + 15/8*l
og(sqrt((x - 1)/(x + 1)) + 1) - 15/8*log(sqrt((x - 1)/(x + 1)) - 1)

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Fricas [A]  time = 1.88403, size = 190, normalized size = 1.43 \begin{align*} \frac{1}{8} \,{\left (2 \, x^{4} + 10 \, x^{3} + 23 \, x^{2} + 39 \, x + 24\right )} \sqrt{\frac{x - 1}{x + 1}} + \frac{15}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) - \frac{15}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} - 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*x^4 + 10*x^3 + 23*x^2 + 39*x + 24)*sqrt((x - 1)/(x + 1)) + 15/8*log(sqrt((x - 1)/(x + 1)) + 1) - 15/8*l
og(sqrt((x - 1)/(x + 1)) - 1)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14158, size = 176, normalized size = 1.32 \begin{align*} -\frac{\frac{73 \,{\left (x - 1\right )} \sqrt{\frac{x - 1}{x + 1}}}{x + 1} - \frac{55 \,{\left (x - 1\right )}^{2} \sqrt{\frac{x - 1}{x + 1}}}{{\left (x + 1\right )}^{2}} + \frac{15 \,{\left (x - 1\right )}^{3} \sqrt{\frac{x - 1}{x + 1}}}{{\left (x + 1\right )}^{3}} - 49 \, \sqrt{\frac{x - 1}{x + 1}}}{4 \,{\left (\frac{x - 1}{x + 1} - 1\right )}^{4}} + \frac{15}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) - \frac{15}{8} \, \log \left ({\left | \sqrt{\frac{x - 1}{x + 1}} - 1 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(73*(x - 1)*sqrt((x - 1)/(x + 1))/(x + 1) - 55*(x - 1)^2*sqrt((x - 1)/(x + 1))/(x + 1)^2 + 15*(x - 1)^3*s
qrt((x - 1)/(x + 1))/(x + 1)^3 - 49*sqrt((x - 1)/(x + 1)))/((x - 1)/(x + 1) - 1)^4 + 15/8*log(sqrt((x - 1)/(x
+ 1)) + 1) - 15/8*log(abs(sqrt((x - 1)/(x + 1)) - 1))