3.263 \(\int e^{-2 \tanh ^{-1}(a x)} (c-a c x)^{3/2} \, dx\)

Optimal. Leaf size=95 \[ -\frac{8 \sqrt{2} c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{a}+\frac{2 (c-a c x)^{5/2}}{5 a c}+\frac{4 (c-a c x)^{3/2}}{3 a}+\frac{8 c \sqrt{c-a c x}}{a} \]

[Out]

(8*c*Sqrt[c - a*c*x])/a + (4*(c - a*c*x)^(3/2))/(3*a) + (2*(c - a*c*x)^(5/2))/(5*a*c) - (8*Sqrt[2]*c^(3/2)*Arc
Tanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/a

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Rubi [A]  time = 0.0792074, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6130, 21, 50, 63, 206} \[ -\frac{8 \sqrt{2} c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{a}+\frac{2 (c-a c x)^{5/2}}{5 a c}+\frac{4 (c-a c x)^{3/2}}{3 a}+\frac{8 c \sqrt{c-a c x}}{a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^(3/2)/E^(2*ArcTanh[a*x]),x]

[Out]

(8*c*Sqrt[c - a*c*x])/a + (4*(c - a*c*x)^(3/2))/(3*a) + (2*(c - a*c*x)^(5/2))/(5*a*c) - (8*Sqrt[2]*c^(3/2)*Arc
Tanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/a

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]
)

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])

Rubi steps

\begin{align*} \int e^{-2 \tanh ^{-1}(a x)} (c-a c x)^{3/2} \, dx &=\int \frac{(1-a x) (c-a c x)^{3/2}}{1+a x} \, dx\\ &=\frac{\int \frac{(c-a c x)^{5/2}}{1+a x} \, dx}{c}\\ &=\frac{2 (c-a c x)^{5/2}}{5 a c}+2 \int \frac{(c-a c x)^{3/2}}{1+a x} \, dx\\ &=\frac{4 (c-a c x)^{3/2}}{3 a}+\frac{2 (c-a c x)^{5/2}}{5 a c}+(4 c) \int \frac{\sqrt{c-a c x}}{1+a x} \, dx\\ &=\frac{8 c \sqrt{c-a c x}}{a}+\frac{4 (c-a c x)^{3/2}}{3 a}+\frac{2 (c-a c x)^{5/2}}{5 a c}+\left (8 c^2\right ) \int \frac{1}{(1+a x) \sqrt{c-a c x}} \, dx\\ &=\frac{8 c \sqrt{c-a c x}}{a}+\frac{4 (c-a c x)^{3/2}}{3 a}+\frac{2 (c-a c x)^{5/2}}{5 a c}-\frac{(16 c) \operatorname{Subst}\left (\int \frac{1}{2-\frac{x^2}{c}} \, dx,x,\sqrt{c-a c x}\right )}{a}\\ &=\frac{8 c \sqrt{c-a c x}}{a}+\frac{4 (c-a c x)^{3/2}}{3 a}+\frac{2 (c-a c x)^{5/2}}{5 a c}-\frac{8 \sqrt{2} c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.0417121, size = 71, normalized size = 0.75 \[ \frac{2 c \left (3 a^2 x^2-16 a x+73\right ) \sqrt{c-a c x}-120 \sqrt{2} c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{15 a} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - a*c*x)^(3/2)/E^(2*ArcTanh[a*x]),x]

[Out]

(2*c*Sqrt[c - a*c*x]*(73 - 16*a*x + 3*a^2*x^2) - 120*Sqrt[2]*c^(3/2)*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])
])/(15*a)

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Maple [A]  time = 0.033, size = 73, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{ac} \left ( 1/5\, \left ( -acx+c \right ) ^{5/2}+2/3\,c \left ( -acx+c \right ) ^{3/2}+4\,\sqrt{-acx+c}{c}^{2}-4\,{c}^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-acx+c}\sqrt{2}}{\sqrt{c}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/c/a*(1/5*(-a*c*x+c)^(5/2)+2/3*c*(-a*c*x+c)^(3/2)+4*(-a*c*x+c)^(1/2)*c^2-4*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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(3/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.37417, size = 379, normalized size = 3.99 \begin{align*} \left [\frac{2 \,{\left (30 \, \sqrt{2} c^{\frac{3}{2}} \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} c x^{2} - 16 \, a c x + 73 \, c\right )} \sqrt{-a c x + c}\right )}}{15 \, a}, \frac{2 \,{\left (60 \, \sqrt{2} \sqrt{-c} c \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c} \sqrt{-c}}{2 \, c}\right ) +{\left (3 \, a^{2} c x^{2} - 16 \, a c x + 73 \, c\right )} \sqrt{-a c x + c}\right )}}{15 \, a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(3/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="fricas")

[Out]

[2/15*(30*sqrt(2)*c^(3/2)*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + (3*a^2*c*x^2 - 1
6*a*c*x + 73*c)*sqrt(-a*c*x + c))/a, 2/15*(60*sqrt(2)*sqrt(-c)*c*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/
c) + (3*a^2*c*x^2 - 16*a*c*x + 73*c)*sqrt(-a*c*x + c))/a]

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Sympy [A]  time = 42.4353, size = 90, normalized size = 0.95 \begin{align*} \frac{8 \sqrt{2} c^{2} \operatorname{atan}{\left (\frac{\sqrt{2} \sqrt{- a c x + c}}{2 \sqrt{- c}} \right )}}{a \sqrt{- c}} + \frac{8 c \sqrt{- a c x + c}}{a} + \frac{4 \left (- a c x + c\right )^{\frac{3}{2}}}{3 a} + \frac{2 \left (- a c x + c\right )^{\frac{5}{2}}}{5 a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)**(3/2)/(a*x+1)**2*(-a**2*x**2+1),x)

[Out]

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

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Giac [A]  time = 1.25342, size = 144, normalized size = 1.52 \begin{align*} \frac{8 \, \sqrt{2} c^{2} \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c}}{2 \, \sqrt{-c}}\right )}{a \sqrt{-c}} + \frac{2 \,{\left (3 \,{\left (a c x - c\right )}^{2} \sqrt{-a c x + c} a^{4} c^{4} + 10 \,{\left (-a c x + c\right )}^{\frac{3}{2}} a^{4} c^{5} + 60 \, \sqrt{-a c x + c} a^{4} c^{6}\right )}}{15 \, a^{5} c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(3/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="giac")

[Out]

8*sqrt(2)*c^2*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)) + 2/15*(3*(a*c*x - c)^2*sqrt(-a*c*x +
c)*a^4*c^4 + 10*(-a*c*x + c)^(3/2)*a^4*c^5 + 60*sqrt(-a*c*x + c)*a^4*c^6)/(a^5*c^5)