\(\int e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx\) [319]

Optimal result

Integrand size = 13, antiderivative size = 144 \[ \int e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx=\frac {46 \sqrt {-\frac {1-x}{x}} (1+x)^{3/2}}{21 \left (1+\frac {1}{x}\right )^{3/2}}+\frac {92 \sqrt {-\frac {1-x}{x}} (1+x)^{3/2}}{21 \left (1+\frac {1}{x}\right )^{3/2} x}+\frac {8 \sqrt {-\frac {1-x}{x}} x (1+x)^{3/2}}{7 \left (1+\frac {1}{x}\right )^{3/2}}+\frac {2 \sqrt {-\frac {1-x}{x}} x^2 (1+x)^{3/2}}{7 \left (1+\frac {1}{x}\right )^{3/2}} \]

[Out]

46/21*(1+x)^(3/2)*((-1+x)/x)^(1/2)/(1+1/x)^(3/2)+92/21*(1+x)^(3/2)*((-1+x)/x)^(1/2)/(1+1/x)^(3/2)/x+8/7*x*(1+x
)^(3/2)*((-1+x)/x)^(1/2)/(1+1/x)^(3/2)+2/7*x^2*(1+x)^(3/2)*((-1+x)/x)^(1/2)/(1+1/x)^(3/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {6311, 6316, 91, 79, 47, 37} \[ \int e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx=\frac {2 \sqrt {-\frac {1-x}{x}} (x+1)^{3/2} x^2}{7 \left (\frac {1}{x}+1\right )^{3/2}}+\frac {8 \sqrt {-\frac {1-x}{x}} (x+1)^{3/2} x}{7 \left (\frac {1}{x}+1\right )^{3/2}}+\frac {46 \sqrt {-\frac {1-x}{x}} (x+1)^{3/2}}{21 \left (\frac {1}{x}+1\right )^{3/2}}+\frac {92 \sqrt {-\frac {1-x}{x}} (x+1)^{3/2}}{21 \left (\frac {1}{x}+1\right )^{3/2} x} \]

[In]

Int[E^ArcCoth[x]*x*(1 + x)^(3/2),x]

[Out]

(46*Sqrt[-((1 - x)/x)]*(1 + x)^(3/2))/(21*(1 + x^(-1))^(3/2)) + (92*Sqrt[-((1 - x)/x)]*(1 + x)^(3/2))/(21*(1 +
x^(-1))^(3/2)*x) + (8*Sqrt[-((1 - x)/x)]*x*(1 + x)^(3/2))/(7*(1 + x^(-1))^(3/2)) + (2*Sqrt[-((1 - x)/x)]*x^2*
(1 + x)^(3/2))/(7*(1 + x^(-1))^(3/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 79

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

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 6311

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

Rule 6316

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> Dist[(-c^p)*x^m*(1/x)^m, S
ubst[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, m,
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*} \text {integral}& = \frac {(1+x)^{3/2} \int e^{\coth ^{-1}(x)} \left (1+\frac {1}{x}\right )^{3/2} x^{5/2} \, dx}{\left (1+\frac {1}{x}\right )^{3/2} x^{3/2}} \\ & = -\frac {\left (\left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}\right ) \text {Subst}\left (\int \frac {(1+x)^2}{\sqrt {1-x} x^{9/2}} \, dx,x,\frac {1}{x}\right )}{\left (1+\frac {1}{x}\right )^{3/2}} \\ & = \frac {2 \sqrt {-\frac {1-x}{x}} x^2 (1+x)^{3/2}}{7 \left (1+\frac {1}{x}\right )^{3/2}}-\frac {\left (2 \left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}\right ) \text {Subst}\left (\int \frac {10+\frac {7 x}{2}}{\sqrt {1-x} x^{7/2}} \, dx,x,\frac {1}{x}\right )}{7 \left (1+\frac {1}{x}\right )^{3/2}} \\ & = \frac {8 \sqrt {-\frac {1-x}{x}} x (1+x)^{3/2}}{7 \left (1+\frac {1}{x}\right )^{3/2}}+\frac {2 \sqrt {-\frac {1-x}{x}} x^2 (1+x)^{3/2}}{7 \left (1+\frac {1}{x}\right )^{3/2}}-\frac {\left (23 \left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x^{5/2}} \, dx,x,\frac {1}{x}\right )}{7 \left (1+\frac {1}{x}\right )^{3/2}} \\ & = \frac {46 \sqrt {-\frac {1-x}{x}} (1+x)^{3/2}}{21 \left (1+\frac {1}{x}\right )^{3/2}}+\frac {8 \sqrt {-\frac {1-x}{x}} x (1+x)^{3/2}}{7 \left (1+\frac {1}{x}\right )^{3/2}}+\frac {2 \sqrt {-\frac {1-x}{x}} x^2 (1+x)^{3/2}}{7 \left (1+\frac {1}{x}\right )^{3/2}}-\frac {\left (46 \left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x^{3/2}} \, dx,x,\frac {1}{x}\right )}{21 \left (1+\frac {1}{x}\right )^{3/2}} \\ & = \frac {46 \sqrt {-\frac {1-x}{x}} (1+x)^{3/2}}{21 \left (1+\frac {1}{x}\right )^{3/2}}+\frac {92 \sqrt {-\frac {1-x}{x}} (1+x)^{3/2}}{21 \left (1+\frac {1}{x}\right )^{3/2} x}+\frac {8 \sqrt {-\frac {1-x}{x}} x (1+x)^{3/2}}{7 \left (1+\frac {1}{x}\right )^{3/2}}+\frac {2 \sqrt {-\frac {1-x}{x}} x^2 (1+x)^{3/2}}{7 \left (1+\frac {1}{x}\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.32 \[ \int e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx=\frac {2 \sqrt {\frac {-1+x}{x}} \sqrt {1+x} \left (46+23 x+12 x^2+3 x^3\right )}{21 \sqrt {1+\frac {1}{x}}} \]

[In]

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

[Out]

(2*Sqrt[(-1 + x)/x]*Sqrt[1 + x]*(46 + 23*x + 12*x^2 + 3*x^3))/(21*Sqrt[1 + x^(-1)])

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.26

method	result	size
gosper	\(\frac {2 \left (x -1\right ) \left (3 x^{3}+12 x^{2}+23 x +46\right )}{21 \sqrt {1+x}\, \sqrt {\frac {x -1}{1+x}}}\)	\(37\)
default	\(\frac {2 \left (x -1\right ) \left (3 x^{3}+12 x^{2}+23 x +46\right )}{21 \sqrt {1+x}\, \sqrt {\frac {x -1}{1+x}}}\)	\(37\)
risch	\(\frac {2 \left (x -1\right ) \left (3 x^{3}+12 x^{2}+23 x +46\right )}{21 \sqrt {1+x}\, \sqrt {\frac {x -1}{1+x}}}\)	\(37\)

[In]

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

[Out]

2/21*(x-1)*(3*x^3+12*x^2+23*x+46)/(1+x)^(1/2)/((x-1)/(1+x))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.23 \[ \int e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx=\frac {2}{21} \, {\left (3 \, x^{3} + 12 \, x^{2} + 23 \, x + 46\right )} \sqrt {x + 1} \sqrt {\frac {x - 1}{x + 1}} \]

[In]

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

[Out]

2/21*(3*x^3 + 12*x^2 + 23*x + 46)*sqrt(x + 1)*sqrt((x - 1)/(x + 1))

Sympy [F]

\[ \int e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx=\int \frac {x \left (x + 1\right )^{\frac {3}{2}}}{\sqrt {\frac {x - 1}{x + 1}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.19 \[ \int e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx=\frac {2 \, {\left (3 \, x^{4} + 9 \, x^{3} + 11 \, x^{2} + 23 \, x - 46\right )}}{21 \, \sqrt {x - 1}} \]

[In]

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

[Out]

2/21*(3*x^4 + 9*x^3 + 11*x^2 + 23*x - 46)/sqrt(x - 1)

Giac [F(-2)]

Exception generated. \[ \int e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 4.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.33 \[ \int e^{\coth ^{-1}(x)} x (1+x)^{3/2} \, dx=\sqrt {\frac {x-1}{x+1}}\,\left (\frac {46\,x\,\sqrt {x+1}}{21}+\frac {92\,\sqrt {x+1}}{21}+\frac {8\,x^2\,\sqrt {x+1}}{7}+\frac {2\,x^3\,\sqrt {x+1}}{7}\right ) \]

[In]

int((x*(x + 1)^(3/2))/((x - 1)/(x + 1))^(1/2),x)

[Out]

((x - 1)/(x + 1))^(1/2)*((46*x*(x + 1)^(1/2))/21 + (92*(x + 1)^(1/2))/21 + (8*x^2*(x + 1)^(1/2))/7 + (2*x^3*(x
+ 1)^(1/2))/7)