3.422 \(\int e^{-2 \tanh ^{-1}(a x)} \sqrt {c-a c x} \, dx\)

Optimal. Leaf size=76 \[ \frac {2 (c-a c x)^{3/2}}{3 a c}+\frac {4 \sqrt {c-a c x}}{a}-\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6130, 21, 50, 63, 206} \[ \frac {2 (c-a c x)^{3/2}}{3 a c}+\frac {4 \sqrt {c-a c x}}{a}-\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - a*c*x]/E^(2*ArcTanh[a*x]),x]

[Out]

(4*Sqrt[c - a*c*x])/a + (2*(c - a*c*x)^(3/2))/(3*a*c) - (4*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sq
rt[c])])/a

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 50

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6130

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 61, normalized size = 0.80 \[ -\frac {2 (a x-7) \sqrt {c-a c x}+12 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c - a*c*x]/E^(2*ArcTanh[a*x]),x]

[Out]

-1/3*(2*(-7 + a*x)*Sqrt[c - a*c*x] + 12*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/a

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fricas [A]  time = 0.52, size = 120, normalized size = 1.58 \[ \left [\frac {2 \, {\left (3 \, \sqrt {2} \sqrt {c} \log \left (\frac {a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) - \sqrt {-a c x + c} {\left (a x - 7\right )}\right )}}{3 \, a}, \frac {2 \, {\left (6 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) - \sqrt {-a c x + c} {\left (a x - 7\right )}\right )}}{3 \, a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/3*(3*sqrt(2)*sqrt(c)*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) - sqrt(-a*c*x + c)*(
a*x - 7))/a, 2/3*(6*sqrt(2)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) - sqrt(-a*c*x + c)*(a*x -
7))/a]

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giac [A]  time = 0.17, size = 77, normalized size = 1.01 \[ \frac {4 \, \sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a \sqrt {-c}} + \frac {2 \, {\left ({\left (-a c x + c\right )}^{\frac {3}{2}} a^{2} c^{2} + 6 \, \sqrt {-a c x + c} a^{2} c^{3}\right )}}{3 \, a^{3} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*sqrt(2)*c*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)) + 2/3*((-a*c*x + c)^(3/2)*a^2*c^2 + 6*s
qrt(-a*c*x + c)*a^2*c^3)/(a^3*c^3)

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maple [A]  time = 0.03, size = 59, normalized size = 0.78 \[ \frac {\frac {2 \left (-a c x +c \right )^{\frac {3}{2}}}{3}+4 c \sqrt {-a c x +c}-4 c^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 79, normalized size = 1.04 \[ \frac {2 \, {\left (3 \, \sqrt {2} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + {\left (-a c x + c\right )}^{\frac {3}{2}} + 6 \, \sqrt {-a c x + c} c\right )}}{3 \, a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.00, size = 61, normalized size = 0.80 \[ \frac {4\,\sqrt {c-a\,c\,x}}{a}+\frac {2\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a\,c}-\frac {4\,\sqrt {2}\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}}{2\,\sqrt {c}}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 8.08, size = 75, normalized size = 0.99 \[ - \frac {2 \left (- \frac {2 \sqrt {2} c^{2} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{\sqrt {- c}} - 2 c \sqrt {- a c x + c} - \frac {\left (- a c x + c\right )^{\frac {3}{2}}}{3}\right )}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-2*sqrt(2)*c**2*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sqrt(-c)))/sqrt(-c) - 2*c*sqrt(-a*c*x + c) - (-a*c*x + c)
**(3/2)/3)/(a*c)

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