3.413 \(\int e^{-\tanh ^{-1}(a x)} x \sqrt {c-a c x} \, dx\)

Optimal. Leaf size=101 \[ -\frac {2 \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}{5 a^2 c}-\frac {2 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{5 a^2}-\frac {8 c \sqrt {1-a^2 x^2}}{5 a^2 \sqrt {c-a c x}} \]

[Out]

-2/5*(-a*c*x+c)^(3/2)*(-a^2*x^2+1)^(1/2)/a^2/c-8/5*c*(-a^2*x^2+1)^(1/2)/a^2/(-a*c*x+c)^(1/2)-2/5*(-a*c*x+c)^(1
/2)*(-a^2*x^2+1)^(1/2)/a^2

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Rubi [A]  time = 0.12, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6128, 795, 657, 649} \[ -\frac {2 \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}{5 a^2 c}-\frac {2 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{5 a^2}-\frac {8 c \sqrt {1-a^2 x^2}}{5 a^2 \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*c*Sqrt[1 - a^2*x^2])/(5*a^2*Sqrt[c - a*c*x]) - (2*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(5*a^2) - (2*(c - a*c
*x)^(3/2)*Sqrt[1 - a^2*x^2])/(5*a^2*c)

Rule 649

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rule 657

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*Simplify[m + p])/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^
2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

Rule 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 6128

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[c^n,
Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c +
d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1, 0]) && IntegerQ[2*p]

Rubi steps

\begin {align*} \int e^{-\tanh ^{-1}(a x)} x \sqrt {c-a c x} \, dx &=\frac {\int \frac {x (c-a c x)^{3/2}}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=-\frac {2 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}{5 a^2 c}-\frac {3 \int \frac {(c-a c x)^{3/2}}{\sqrt {1-a^2 x^2}} \, dx}{5 a c}\\ &=-\frac {2 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}{5 a^2}-\frac {2 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}{5 a^2 c}-\frac {4 \int \frac {\sqrt {c-a c x}}{\sqrt {1-a^2 x^2}} \, dx}{5 a}\\ &=-\frac {8 c \sqrt {1-a^2 x^2}}{5 a^2 \sqrt {c-a c x}}-\frac {2 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}{5 a^2}-\frac {2 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}{5 a^2 c}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 46, normalized size = 0.46 \[ -\frac {2 c \sqrt {1-a^2 x^2} \left (a^2 x^2-3 a x+6\right )}{5 a^2 \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*Sqrt[1 - a^2*x^2]*(6 - 3*a*x + a^2*x^2))/(5*a^2*Sqrt[c - a*c*x])

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fricas [A]  time = 0.41, size = 49, normalized size = 0.49 \[ \frac {2 \, {\left (a^{2} x^{2} - 3 \, a x + 6\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{5 \, {\left (a^{3} x - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(a^2*x^2 - 3*a*x + 6)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^3*x - a^2)

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giac [A]  time = 0.95, size = 65, normalized size = 0.64 \[ \frac {8 \, \sqrt {2} {\left | c \right |}}{5 \, a^{2} \sqrt {c}} - \frac {4 \, \sqrt {a c x + c} {\left | c \right |}}{a^{2} c} - \frac {2 \, {\left ({\left (a c x + c\right )}^{\frac {5}{2}} {\left | c \right |} - 5 \, {\left (a c x + c\right )}^{\frac {3}{2}} c {\left | c \right |}\right )}}{5 \, a^{2} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/5*sqrt(2)*abs(c)/(a^2*sqrt(c)) - 4*sqrt(a*c*x + c)*abs(c)/(a^2*c) - 2/5*((a*c*x + c)^(5/2)*abs(c) - 5*(a*c*x
+ c)^(3/2)*c*abs(c))/(a^2*c^3)

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maple [A]  time = 0.03, size = 47, normalized size = 0.47 \[ \frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-a c x +c}\, \left (a^{2} x^{2}-3 a x +6\right )}{5 \left (a x -1\right ) a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(1/2)*(a^2*x^2-3*a*x+6)/(a*x-1)/a^2

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maxima [A]  time = 0.37, size = 50, normalized size = 0.50 \[ -\frac {2 \, {\left (a^{2} \sqrt {c} x^{2} - 3 \, a \sqrt {c} x + 6 \, \sqrt {c}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{5 \, {\left (a^{3} x - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(a^2*sqrt(c)*x^2 - 3*a*sqrt(c)*x + 6*sqrt(c))*sqrt(a*x + 1)*(a*x - 1)/(a^3*x - a^2)

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mupad [B]  time = 0.90, size = 75, normalized size = 0.74 \[ \frac {\sqrt {c-a\,c\,x}\,\left (\frac {12\,\sqrt {1-a^2\,x^2}}{5\,a^3}-\frac {6\,x\,\sqrt {1-a^2\,x^2}}{5\,a^2}+\frac {2\,x^2\,\sqrt {1-a^2\,x^2}}{5\,a}\right )}{x-\frac {1}{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c - a*c*x)^(1/2)*((12*(1 - a^2*x^2)^(1/2))/(5*a^3) - (6*x*(1 - a^2*x^2)^(1/2))/(5*a^2) + (2*x^2*(1 - a^2*x^2
)^(1/2))/(5*a)))/(x - 1/a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(-c*(a*x - 1))*sqrt(-(a*x - 1)*(a*x + 1))/(a*x + 1), x)

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