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]

15/8*arctanh((1/x+1)^(1/2)*((-1+x)/x)^(1/2))+5/8*(1/x+1)^(3/2)*x^2*((-1+x)/x)^(1/2)+1/4*(1/x+1)^(5/2)*x^3*((-1
+x)/x)^(1/2)+1/4*(1/x+1)^(7/2)*x^4*((-1+x)/x)^(1/2)+15/8*x*(1/x+1)^(1/2)*((-1+x)/x)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 133, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.05, size = 52, normalized size = 0.39 \[ \frac {15}{8} \log \left (\left (\sqrt {1-\frac {1}{x^2}}+1\right ) x\right )+\frac {1}{8} \sqrt {1-\frac {1}{x^2}} x \left (2 x^3+8 x^2+15 x+24\right ) \]

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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fricas [A]  time = 0.46, size = 66, normalized size = 0.50 \[ \frac {1}{8} \, {\left (2 \, x^{4} + 10 \, x^{3} + 23 \, x^{2} + 39 \, x + 24\right )} \sqrt {\frac {x - 1}{x + 1}} + \frac {15}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {15}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

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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giac [A]  time = 0.14, size = 130, normalized size = 0.98 \[ -\frac {\frac {73 \, {\left (x - 1\right )} \sqrt {\frac {x - 1}{x + 1}}}{x + 1} - \frac {55 \, {\left (x - 1\right )}^{2} \sqrt {\frac {x - 1}{x + 1}}}{{\left (x + 1\right )}^{2}} + \frac {15 \, {\left (x - 1\right )}^{3} \sqrt {\frac {x - 1}{x + 1}}}{{\left (x + 1\right )}^{3}} - 49 \, \sqrt {\frac {x - 1}{x + 1}}}{4 \, {\left (\frac {x - 1}{x + 1} - 1\right )}^{4}} + \frac {15}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {15}{8} \, \log \left ({\left | \sqrt {\frac {x - 1}{x + 1}} - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.04, size = 79, normalized size = 0.59 \[ \frac {\left (-1+x \right ) \left (2 x \left (x^{2}-1\right )^{\frac {3}{2}}+8 \left (\left (1+x \right ) \left (-1+x \right )\right )^{\frac {3}{2}}+17 x \sqrt {x^{2}-1}+32 \sqrt {x^{2}-1}+15 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{8 \sqrt {\frac {-1+x}{1+x}}\, \sqrt {\left (1+x \right ) \left (-1+x \right )}} \]

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 = 0.31, size = 138, normalized size = 1.04 \[ \frac {15 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {7}{2}} - 55 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{2}} + 73 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} - 49 \, \sqrt {\frac {x - 1}{x + 1}}}{4 \, {\left (\frac {4 \, {\left (x - 1\right )}}{x + 1} - \frac {6 \, {\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac {4 \, {\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - \frac {{\left (x - 1\right )}^{4}}{{\left (x + 1\right )}^{4}} - 1\right )}} + \frac {15}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {15}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

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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mupad [B]  time = 0.05, size = 118, normalized size = 0.89 \[ \frac {15\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )}{4}+\frac {\frac {49\,\sqrt {\frac {x-1}{x+1}}}{4}-\frac {73\,{\left (\frac {x-1}{x+1}\right )}^{3/2}}{4}+\frac {55\,{\left (\frac {x-1}{x+1}\right )}^{5/2}}{4}-\frac {15\,{\left (\frac {x-1}{x+1}\right )}^{7/2}}{4}}{\frac {6\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}-\frac {4\,\left (x-1\right )}{x+1}-\frac {4\,{\left (x-1\right )}^3}{{\left (x+1\right )}^3}+\frac {{\left (x-1\right )}^4}{{\left (x+1\right )}^4}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*atanh(((x - 1)/(x + 1))^(1/2)))/4 + ((49*((x - 1)/(x + 1))^(1/2))/4 - (73*((x - 1)/(x + 1))^(3/2))/4 + (55
*((x - 1)/(x + 1))^(5/2))/4 - (15*((x - 1)/(x + 1))^(7/2))/4)/((6*(x - 1)^2)/(x + 1)^2 - (4*(x - 1))/(x + 1) -
(4*(x - 1)^3)/(x + 1)^3 + (x - 1)^4/(x + 1)^4 + 1)

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

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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