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

Optimal. Leaf size=97 \[ -\frac {2 (c-a c x)^{5/2}}{5 a^2 c^2}-\frac {2 (c-a c x)^{3/2}}{3 a^2 c}-\frac {4 \sqrt {c-a c x}}{a^2}+\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6130, 21, 80, 50, 63, 206} \[ -\frac {2 (c-a c x)^{5/2}}{5 a^2 c^2}-\frac {2 (c-a c x)^{3/2}}{3 a^2 c}-\frac {4 \sqrt {c-a c x}}{a^2}+\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[c - a*c*x])/a^2 - (2*(c - a*c*x)^(3/2))/(3*a^2*c) - (2*(c - a*c*x)^(5/2))/(5*a^2*c^2) + (4*Sqrt[2]*Sq
rt[c]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/a^2

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

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Mathematica [A]  time = 0.08, size = 70, normalized size = 0.72 \[ \frac {60 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )-2 \left (3 a^2 x^2-11 a x+38\right ) \sqrt {c-a c x}}{15 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - a*c*x]*(38 - 11*a*x + 3*a^2*x^2) + 60*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])
/(15*a^2)

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fricas [A]  time = 0.46, size = 137, normalized size = 1.41 \[ \left [\frac {2 \, {\left (15 \, \sqrt {2} \sqrt {c} \log \left (\frac {a c x - 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) - {\left (3 \, a^{2} x^{2} - 11 \, a x + 38\right )} \sqrt {-a c x + c}\right )}}{15 \, a^{2}}, -\frac {2 \, {\left (30 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) + {\left (3 \, a^{2} x^{2} - 11 \, a x + 38\right )} \sqrt {-a c x + c}\right )}}{15 \, a^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/15*(15*sqrt(2)*sqrt(c)*log((a*c*x - 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) - (3*a^2*x^2 - 11*
a*x + 38)*sqrt(-a*c*x + c))/a^2, -2/15*(30*sqrt(2)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) +
(3*a^2*x^2 - 11*a*x + 38)*sqrt(-a*c*x + c))/a^2]

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giac [A]  time = 0.27, size = 105, normalized size = 1.08 \[ -\frac {4 \, \sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a^{2} \sqrt {-c}} - \frac {2 \, {\left (3 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{8} c^{8} + 5 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{8} c^{9} + 30 \, \sqrt {-a c x + c} a^{8} c^{10}\right )}}{15 \, a^{10} c^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-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^2*sqrt(-c)) - 2/15*(3*(a*c*x - c)^2*sqrt(-a*c*x
+ c)*a^8*c^8 + 5*(-a*c*x + c)^(3/2)*a^8*c^9 + 30*sqrt(-a*c*x + c)*a^8*c^10)/(a^10*c^10)

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maple [A]  time = 0.04, size = 73, normalized size = 0.75 \[ -\frac {2 \left (\frac {\left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {c \left (-a c x +c \right )^{\frac {3}{2}}}{3}+2 \sqrt {-a c x +c}\, c^{2}-2 c^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{c^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/c^2/a^2*(1/5*(-a*c*x+c)^(5/2)+1/3*c*(-a*c*x+c)^(3/2)+2*(-a*c*x+c)^(1/2)*c^2-2*c^(5/2)*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 = 95, normalized size = 0.98 \[ -\frac {2 \, {\left (15 \, \sqrt {2} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + 3 \, {\left (-a c x + c\right )}^{\frac {5}{2}} + 5 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 30 \, \sqrt {-a c x + c} c^{2}\right )}}{15 \, a^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*sqrt(2)*c^(5/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2)*sqrt(c) + sqrt(-a*c*x + c))) + 3*
(-a*c*x + c)^(5/2) + 5*(-a*c*x + c)^(3/2)*c + 30*sqrt(-a*c*x + c)*c^2)/(a^2*c^2)

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mupad [B]  time = 0.86, size = 80, normalized size = 0.82 \[ -\frac {4\,\sqrt {c-a\,c\,x}}{a^2}-\frac {2\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a^2\,c}-\frac {2\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a^2\,c^2}-\frac {\sqrt {2}\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,4{}\mathrm {i}}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*(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 - (2*(c - a*c*x)^(3/2))/(3*a^2*c) - (2*(c - a*c*x)^(5/2))/(5*a^2*c^2) - (2^(1/2)*c
^(1/2)*atan((2^(1/2)*(c - a*c*x)^(1/2)*1i)/(2*c^(1/2)))*4i)/a^2

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sympy [A]  time = 11.87, size = 94, normalized size = 0.97 \[ \frac {2 \left (- \frac {2 \sqrt {2} c^{3} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{\sqrt {- c}} - 2 c^{2} \sqrt {- a c x + c} - \frac {c \left (- a c x + c\right )^{\frac {3}{2}}}{3} - \frac {\left (- a c x + c\right )^{\frac {5}{2}}}{5}\right )}{a^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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