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

Optimal. Leaf size=66 \[ \frac {8 c \sqrt {1-a^2 x^2}}{3 a \sqrt {c-a c x}}+\frac {2 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6127, 657, 649} \[ \frac {8 c \sqrt {1-a^2 x^2}}{3 a \sqrt {c-a c x}}+\frac {2 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - a*c*x]/E^ArcTanh[a*x],x]

[Out]

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

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 6127

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 38, normalized size = 0.58 \[ -\frac {2 c (a x-5) \sqrt {1-a^2 x^2}}{3 a \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 39, normalized size = 0.59 \[ \frac {2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x - 5\right )}}{3 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 50, normalized size = 0.76 \[ -\frac {8 \, \sqrt {2} {\left | c \right |}}{3 \, a \sqrt {c}} - \frac {2 \, {\left (a c x + c\right )}^{\frac {3}{2}} {\left | c \right |}}{3 \, a c^{2}} + \frac {4 \, \sqrt {a c x + c} {\left | c \right |}}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 39, normalized size = 0.59 \[ \frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-a c x +c}\, \left (a x -5\right )}{3 \left (a x -1\right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 37, normalized size = 0.56 \[ -\frac {2 \, {\left (a \sqrt {c} x - 5 \, \sqrt {c}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{3 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.87, size = 56, normalized size = 0.85 \[ -\frac {\sqrt {c-a\,c\,x}\,\left (\frac {10\,\sqrt {1-a^2\,x^2}}{3\,a^2}-\frac {2\,x\,\sqrt {1-a^2\,x^2}}{3\,a}\right )}{x-\frac {1}{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c - a*c*x)^(1/2)*((10*(1 - a^2*x^2)^(1/2))/(3*a^2) - (2*x*(1 - a^2*x^2)^(1/2))/(3*a)))/(x - 1/a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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