3.352 \(\int e^{-3 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx\)

Optimal. Leaf size=182 \[ \frac {316 \sqrt {\frac {1}{a x}+1} \sqrt {c-a c x}}{15 a^2 \sqrt {1-\frac {1}{a x}}}-\frac {158 \sqrt {c-a c x}}{15 a^2 \sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}}+\frac {2 x^2 \sqrt {c-a c x}}{5 \sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}}-\frac {32 x \sqrt {c-a c x}}{15 a \sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}} \]

[Out]

-158/15*(-a*c*x+c)^(1/2)/a^2/(1-1/a/x)^(1/2)/(1+1/a/x)^(1/2)-32/15*x*(-a*c*x+c)^(1/2)/a/(1-1/a/x)^(1/2)/(1+1/a
/x)^(1/2)+2/5*x^2*(-a*c*x+c)^(1/2)/(1-1/a/x)^(1/2)/(1+1/a/x)^(1/2)+316/15*(1+1/a/x)^(1/2)*(-a*c*x+c)^(1/2)/a^2
/(1-1/a/x)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6176, 6181, 89, 78, 45, 37} \[ \frac {316 \sqrt {\frac {1}{a x}+1} \sqrt {c-a c x}}{15 a^2 \sqrt {1-\frac {1}{a x}}}-\frac {158 \sqrt {c-a c x}}{15 a^2 \sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}}+\frac {2 x^2 \sqrt {c-a c x}}{5 \sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}}-\frac {32 x \sqrt {c-a c x}}{15 a \sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sqrt[c - a*c*x])/E^(3*ArcCoth[a*x]),x]

[Out]

(-158*Sqrt[c - a*c*x])/(15*a^2*Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)]) + (316*Sqrt[1 + 1/(a*x)]*Sqrt[c - a*c*x])/
(15*a^2*Sqrt[1 - 1/(a*x)]) - (32*x*Sqrt[c - a*c*x])/(15*a*Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)]) + (2*x^2*Sqrt[c
- a*c*x])/(5*Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)])

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 45

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 78

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] || LtQ
[p, n]))))

Rule 89

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 6176

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 6181

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> -Dist[c^p*x^m*(1/x)^m, Sub
st[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*} \int e^{-3 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx &=\frac {\sqrt {c-a c x} \int e^{-3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a x}} x^{3/2} \, dx}{\sqrt {1-\frac {1}{a x}} \sqrt {x}}\\ &=-\frac {\left (\sqrt {\frac {1}{x}} \sqrt {c-a c x}\right ) \operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^2}{x^{7/2} \left (1+\frac {x}{a}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a x}}}\\ &=\frac {2 x^2 \sqrt {c-a c x}}{5 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}-\frac {\left (2 \sqrt {\frac {1}{x}} \sqrt {c-a c x}\right ) \operatorname {Subst}\left (\int \frac {-\frac {8}{a}+\frac {5 x}{2 a^2}}{x^{5/2} \left (1+\frac {x}{a}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{5 \sqrt {1-\frac {1}{a x}}}\\ &=-\frac {32 x \sqrt {c-a c x}}{15 a \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}+\frac {2 x^2 \sqrt {c-a c x}}{5 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}-\frac {\left (79 \sqrt {\frac {1}{x}} \sqrt {c-a c x}\right ) \operatorname {Subst}\left (\int \frac {1}{x^{3/2} \left (1+\frac {x}{a}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{15 a^2 \sqrt {1-\frac {1}{a x}}}\\ &=-\frac {158 \sqrt {c-a c x}}{15 a^2 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}-\frac {32 x \sqrt {c-a c x}}{15 a \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}+\frac {2 x^2 \sqrt {c-a c x}}{5 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}-\frac {\left (158 \sqrt {\frac {1}{x}} \sqrt {c-a c x}\right ) \operatorname {Subst}\left (\int \frac {1}{x^{3/2} \sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{15 a^2 \sqrt {1-\frac {1}{a x}}}\\ &=-\frac {158 \sqrt {c-a c x}}{15 a^2 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}+\frac {316 \sqrt {1+\frac {1}{a x}} \sqrt {c-a c x}}{15 a^2 \sqrt {1-\frac {1}{a x}}}-\frac {32 x \sqrt {c-a c x}}{15 a \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}+\frac {2 x^2 \sqrt {c-a c x}}{5 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 57, normalized size = 0.31 \[ \frac {2 \left (3 a^3 x^3-16 a^2 x^2+79 a x+158\right ) \sqrt {c-a c x}}{15 a^3 x \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Sqrt[c - a*c*x])/E^(3*ArcCoth[a*x]),x]

[Out]

(2*Sqrt[c - a*c*x]*(158 + 79*a*x - 16*a^2*x^2 + 3*a^3*x^3))/(15*a^3*Sqrt[1 - 1/(a^2*x^2)]*x)

________________________________________________________________________________________

fricas [A]  time = 0.72, size = 61, normalized size = 0.34 \[ \frac {2 \, {\left (3 \, a^{3} x^{3} - 16 \, a^{2} x^{2} + 79 \, a x + 158\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{15 \, {\left (a^{3} x - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^3*x^3 - 16*a^2*x^2 + 79*a*x + 158)*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1))/(a^3*x - a^2)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(a*x+1)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [A]  time = 0.04, size = 64, normalized size = 0.35 \[ \frac {2 \left (a x +1\right ) \left (3 x^{3} a^{3}-16 a^{2} x^{2}+79 a x +158\right ) \sqrt {-a c x +c}\, \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{15 a^{2} \left (a x -1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(a*x+1)*(3*a^3*x^3-16*a^2*x^2+79*a*x+158)*(-a*c*x+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/a^2/(a*x-1)^2

________________________________________________________________________________________

maxima [A]  time = 0.34, size = 91, normalized size = 0.50 \[ \frac {2 \, {\left (3 \, a^{4} \sqrt {-c} x^{4} - 13 \, a^{3} \sqrt {-c} x^{3} + 63 \, a^{2} \sqrt {-c} x^{2} + 237 \, a \sqrt {-c} x + 158 \, \sqrt {-c}\right )} {\left (a x - 1\right )}^{2}}{15 \, {\left (a^{4} x^{2} - 2 \, a^{3} x + a^{2}\right )} {\left (a x + 1\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^4*sqrt(-c)*x^4 - 13*a^3*sqrt(-c)*x^3 + 63*a^2*sqrt(-c)*x^2 + 237*a*sqrt(-c)*x + 158*sqrt(-c))*(a*x -
1)^2/((a^4*x^2 - 2*a^3*x + a^2)*(a*x + 1)^(3/2))

________________________________________________________________________________________

mupad [B]  time = 1.33, size = 58, normalized size = 0.32 \[ \frac {2\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (3\,a^3\,x^3-16\,a^2\,x^2+79\,a\,x+158\right )}{15\,a^2\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(c - a*c*x)^(1/2)*((a*x - 1)/(a*x + 1))^(1/2)*(79*a*x - 16*a^2*x^2 + 3*a^3*x^3 + 158))/(15*a^2*(a*x - 1))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________