3.3.97 \(\int e^{\coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx\) [297]

Optimal. Leaf size=92 \[ -\frac {4 \left (1+\frac {1}{a x}\right )^{3/2} x \sqrt {c-a c x}}{15 a \sqrt {1-\frac {1}{a x}}}+\frac {2 \left (1+\frac {1}{a x}\right )^{3/2} x^2 \sqrt {c-a c x}}{5 \sqrt {1-\frac {1}{a x}}} \]

[Out]

-4/15*(1+1/a/x)^(3/2)*x*(-a*c*x+c)^(1/2)/a/(1-1/a/x)^(1/2)+2/5*(1+1/a/x)^(3/2)*x^2*(-a*c*x+c)^(1/2)/(1-1/a/x)^
(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {6311, 6316, 47, 37} \begin {gather*} \frac {2 x^2 \left (\frac {1}{a x}+1\right )^{3/2} \sqrt {c-a c x}}{5 \sqrt {1-\frac {1}{a x}}}-\frac {4 x \left (\frac {1}{a x}+1\right )^{3/2} \sqrt {c-a c x}}{15 a \sqrt {1-\frac {1}{a x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]*x*Sqrt[c - a*c*x],x]

[Out]

(-4*(1 + 1/(a*x))^(3/2)*x*Sqrt[c - a*c*x])/(15*a*Sqrt[1 - 1/(a*x)]) + (2*(1 + 1/(a*x))^(3/2)*x^2*Sqrt[c - a*c*
x])/(5*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 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 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*} \int e^{\coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx &=\frac {\sqrt {c-a c x} \int e^{\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 ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {x}{a}}}{x^{7/2}} \, dx,x,\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a x}}}\\ &=\frac {2 \left (1+\frac {1}{a x}\right )^{3/2} x^2 \sqrt {c-a c x}}{5 \sqrt {1-\frac {1}{a x}}}+\frac {\left (2 \sqrt {\frac {1}{x}} \sqrt {c-a c x}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {x}{a}}}{x^{5/2}} \, dx,x,\frac {1}{x}\right )}{5 a \sqrt {1-\frac {1}{a x}}}\\ &=-\frac {4 \left (1+\frac {1}{a x}\right )^{3/2} x \sqrt {c-a c x}}{15 a \sqrt {1-\frac {1}{a x}}}+\frac {2 \left (1+\frac {1}{a x}\right )^{3/2} x^2 \sqrt {c-a c x}}{5 \sqrt {1-\frac {1}{a x}}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 56, normalized size = 0.61 \begin {gather*} \frac {2 \sqrt {1+\frac {1}{a x}} (1+a x) (-2+3 a x) \sqrt {c-a c x}}{15 a^2 \sqrt {1-\frac {1}{a x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[a*x]*x*Sqrt[c - a*c*x],x]

[Out]

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

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Maple [A]
time = 0.12, size = 42, normalized size = 0.46

method	result	size
gosper	\(\frac {2 \left (a x +1\right ) \left (3 a x -2\right ) \sqrt {-a c x +c}}{15 a^{2} \sqrt {\frac {a x -1}{a x +1}}}\)	\(41\)
default	\(\frac {2 \sqrt {-c \left (a x -1\right )}\, \left (a x +1\right ) \left (3 a x -2\right )}{15 \sqrt {\frac {a x -1}{a x +1}}\, a^{2}}\)	\(42\)
risch	\(-\frac {2 c \left (a x -1\right ) \left (3 a^{2} x^{2}+a x -2\right )}{15 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a x -1\right )}\, a^{2}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 41, normalized size = 0.45 \begin {gather*} \frac {2 \, {\left (3 \, a^{2} \sqrt {-c} x^{2} + a \sqrt {-c} x - 2 \, \sqrt {-c}\right )} \sqrt {a x + 1}}{15 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 61, normalized size = 0.66 \begin {gather*} \frac {2 \, {\left (3 \, a^{3} x^{3} + 4 \, a^{2} x^{2} - a x - 2\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{15 \, {\left (a^{3} x - a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 20.07, size = 136, normalized size = 1.48 \begin {gather*} \frac {4 \sqrt {- a c x + c}}{15 a^{2} \sqrt {- \frac {a c x}{- a c x - c} + \frac {c}{- a c x - c}}} - \frac {14 \left (- a c x + c\right )^{\frac {3}{2}}}{15 a^{2} c \sqrt {- \frac {a c x}{- a c x - c} + \frac {c}{- a c x - c}}} + \frac {2 \left (- a c x + c\right )^{\frac {5}{2}}}{5 a^{2} c^{2} \sqrt {- \frac {a c x}{- a c x - c} + \frac {c}{- a c x - c}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*sqrt(-a*c*x + c)/(15*a**2*sqrt(-a*c*x/(-a*c*x - c) + c/(-a*c*x - c))) - 14*(-a*c*x + c)**(3/2)/(15*a**2*c*sq
rt(-a*c*x/(-a*c*x - c) + c/(-a*c*x - c))) + 2*(-a*c*x + c)**(5/2)/(5*a**2*c**2*sqrt(-a*c*x/(-a*c*x - c) + c/(-
a*c*x - c)))

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Giac [A]
time = 0.42, size = 81, normalized size = 0.88 \begin {gather*} \frac {2 \, c^{2} {\left (\frac {2 \, \sqrt {2} \sqrt {-c}}{a c} - \frac {3 \, {\left (a c x + c\right )}^{2} \sqrt {-a c x - c} + 5 \, {\left (-a c x - c\right )}^{\frac {3}{2}} c}{a c^{3}}\right )}}{15 \, a {\left | c \right |} \mathrm {sgn}\left (a x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*c^2*(2*sqrt(2)*sqrt(-c)/(a*c) - (3*(a*c*x + c)^2*sqrt(-a*c*x - c) + 5*(-a*c*x - c)^(3/2)*c)/(a*c^3))/(a*a
bs(c)*sgn(a*x + 1))

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Mupad [B]
time = 1.37, size = 49, normalized size = 0.53 \begin {gather*} \frac {2\,\sqrt {c-a\,c\,x}\,{\left (a\,x+1\right )}^2\,\left (3\,a\,x-2\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{15\,a^2\,\left (a\,x-1\right )} \end {gather*}

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))^(1/2),x)

[Out]

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

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