3.261 \(\int e^{-2 \tanh ^{-1}(a x)} (c-a c x)^{7/2} \, dx\)

Optimal. Leaf size=137 \[ \frac{16 c^2 (c-a c x)^{3/2}}{3 a}+\frac{32 c^3 \sqrt{c-a c x}}{a}-\frac{32 \sqrt{2} c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{a}+\frac{2 (c-a c x)^{9/2}}{9 a c}+\frac{4 (c-a c x)^{7/2}}{7 a}+\frac{8 c (c-a c x)^{5/2}}{5 a} \]

[Out]

(32*c^3*Sqrt[c - a*c*x])/a + (16*c^2*(c - a*c*x)^(3/2))/(3*a) + (8*c*(c - a*c*x)^(5/2))/(5*a) + (4*(c - a*c*x)
^(7/2))/(7*a) + (2*(c - a*c*x)^(9/2))/(9*a*c) - (32*Sqrt[2]*c^(7/2)*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])]
)/a

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Rubi [A]  time = 0.110809, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{16 c^2 (c-a c x)^{3/2}}{3 a}+\frac{32 c^3 \sqrt{c-a c x}}{a}-\frac{32 \sqrt{2} c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{a}+\frac{2 (c-a c x)^{9/2}}{9 a c}+\frac{4 (c-a c x)^{7/2}}{7 a}+\frac{8 c (c-a c x)^{5/2}}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^(7/2)/E^(2*ArcTanh[a*x]),x]

[Out]

(32*c^3*Sqrt[c - a*c*x])/a + (16*c^2*(c - a*c*x)^(3/2))/(3*a) + (8*c*(c - a*c*x)^(5/2))/(5*a) + (4*(c - a*c*x)
^(7/2))/(7*a) + (2*(c - a*c*x)^(9/2))/(9*a*c) - (32*Sqrt[2]*c^(7/2)*ArcTanh[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)^{7/2} \, dx &=\int \frac{(1-a x) (c-a c x)^{7/2}}{1+a x} \, dx\\ &=\frac{\int \frac{(c-a c x)^{9/2}}{1+a x} \, dx}{c}\\ &=\frac{2 (c-a c x)^{9/2}}{9 a c}+2 \int \frac{(c-a c x)^{7/2}}{1+a x} \, dx\\ &=\frac{4 (c-a c x)^{7/2}}{7 a}+\frac{2 (c-a c x)^{9/2}}{9 a c}+(4 c) \int \frac{(c-a c x)^{5/2}}{1+a x} \, dx\\ &=\frac{8 c (c-a c x)^{5/2}}{5 a}+\frac{4 (c-a c x)^{7/2}}{7 a}+\frac{2 (c-a c x)^{9/2}}{9 a c}+\left (8 c^2\right ) \int \frac{(c-a c x)^{3/2}}{1+a x} \, dx\\ &=\frac{16 c^2 (c-a c x)^{3/2}}{3 a}+\frac{8 c (c-a c x)^{5/2}}{5 a}+\frac{4 (c-a c x)^{7/2}}{7 a}+\frac{2 (c-a c x)^{9/2}}{9 a c}+\left (16 c^3\right ) \int \frac{\sqrt{c-a c x}}{1+a x} \, dx\\ &=\frac{32 c^3 \sqrt{c-a c x}}{a}+\frac{16 c^2 (c-a c x)^{3/2}}{3 a}+\frac{8 c (c-a c x)^{5/2}}{5 a}+\frac{4 (c-a c x)^{7/2}}{7 a}+\frac{2 (c-a c x)^{9/2}}{9 a c}+\left (32 c^4\right ) \int \frac{1}{(1+a x) \sqrt{c-a c x}} \, dx\\ &=\frac{32 c^3 \sqrt{c-a c x}}{a}+\frac{16 c^2 (c-a c x)^{3/2}}{3 a}+\frac{8 c (c-a c x)^{5/2}}{5 a}+\frac{4 (c-a c x)^{7/2}}{7 a}+\frac{2 (c-a c x)^{9/2}}{9 a c}-\frac{\left (64 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{2-\frac{x^2}{c}} \, dx,x,\sqrt{c-a c x}\right )}{a}\\ &=\frac{32 c^3 \sqrt{c-a c x}}{a}+\frac{16 c^2 (c-a c x)^{3/2}}{3 a}+\frac{8 c (c-a c x)^{5/2}}{5 a}+\frac{4 (c-a c x)^{7/2}}{7 a}+\frac{2 (c-a c x)^{9/2}}{9 a c}-\frac{32 \sqrt{2} c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.0736412, size = 88, normalized size = 0.64 \[ \frac{2 c^3 \left (\left (35 a^4 x^4-230 a^3 x^3+732 a^2 x^2-1754 a x+6257\right ) \sqrt{c-a c x}-5040 \sqrt{2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )\right )}{315 a} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - a*c*x)^(7/2)/E^(2*ArcTanh[a*x]),x]

[Out]

(2*c^3*(Sqrt[c - a*c*x]*(6257 - 1754*a*x + 732*a^2*x^2 - 230*a^3*x^3 + 35*a^4*x^4) - 5040*Sqrt[2]*Sqrt[c]*ArcT
anh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])]))/(315*a)

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Maple [A]  time = 0.035, size = 101, normalized size = 0.7 \begin{align*} 2\,{\frac{1}{ac} \left ( 1/9\, \left ( -acx+c \right ) ^{9/2}+2/7\, \left ( -acx+c \right ) ^{7/2}c+4/5\, \left ( -acx+c \right ) ^{5/2}{c}^{2}+8/3\,{c}^{3} \left ( -acx+c \right ) ^{3/2}+16\,\sqrt{-acx+c}{c}^{4}-16\,{c}^{9/2}\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)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x)

[Out]

2/c/a*(1/9*(-a*c*x+c)^(9/2)+2/7*(-a*c*x+c)^(7/2)*c+4/5*(-a*c*x+c)^(5/2)*c^2+8/3*c^3*(-a*c*x+c)^(3/2)+16*(-a*c*
x+c)^(1/2)*c^4-16*c^(9/2)*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)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.2586, size = 517, normalized size = 3.77 \begin{align*} \left [\frac{2 \,{\left (2520 \, \sqrt{2} c^{\frac{7}{2}} \log \left (\frac{a c x + 2 \, \sqrt{2} \sqrt{-a c x + c} \sqrt{c} - 3 \, c}{a x + 1}\right ) +{\left (35 \, a^{4} c^{3} x^{4} - 230 \, a^{3} c^{3} x^{3} + 732 \, a^{2} c^{3} x^{2} - 1754 \, a c^{3} x + 6257 \, c^{3}\right )} \sqrt{-a c x + c}\right )}}{315 \, a}, \frac{2 \,{\left (5040 \, \sqrt{2} \sqrt{-c} c^{3} \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c} \sqrt{-c}}{2 \, c}\right ) +{\left (35 \, a^{4} c^{3} x^{4} - 230 \, a^{3} c^{3} x^{3} + 732 \, a^{2} c^{3} x^{2} - 1754 \, a c^{3} x + 6257 \, c^{3}\right )} \sqrt{-a c x + c}\right )}}{315 \, a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="fricas")

[Out]

[2/315*(2520*sqrt(2)*c^(7/2)*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + (35*a^4*c^3*x
^4 - 230*a^3*c^3*x^3 + 732*a^2*c^3*x^2 - 1754*a*c^3*x + 6257*c^3)*sqrt(-a*c*x + c))/a, 2/315*(5040*sqrt(2)*sqr
t(-c)*c^3*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) + (35*a^4*c^3*x^4 - 230*a^3*c^3*x^3 + 732*a^2*c^3*x^
2 - 1754*a*c^3*x + 6257*c^3)*sqrt(-a*c*x + c))/a]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)**(7/2)/(a*x+1)**2*(-a**2*x**2+1),x)

[Out]

Timed out

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Giac [A]  time = 1.25569, size = 217, normalized size = 1.58 \begin{align*} \frac{32 \, \sqrt{2} c^{4} \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c}}{2 \, \sqrt{-c}}\right )}{a \sqrt{-c}} + \frac{2 \,{\left (35 \,{\left (a c x - c\right )}^{4} \sqrt{-a c x + c} a^{8} c^{8} - 90 \,{\left (a c x - c\right )}^{3} \sqrt{-a c x + c} a^{8} c^{9} + 252 \,{\left (a c x - c\right )}^{2} \sqrt{-a c x + c} a^{8} c^{10} + 840 \,{\left (-a c x + c\right )}^{\frac{3}{2}} a^{8} c^{11} + 5040 \, \sqrt{-a c x + c} a^{8} c^{12}\right )}}{315 \, a^{9} c^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(7/2)/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="giac")

[Out]

32*sqrt(2)*c^4*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)) + 2/315*(35*(a*c*x - c)^4*sqrt(-a*c*
x + c)*a^8*c^8 - 90*(a*c*x - c)^3*sqrt(-a*c*x + c)*a^8*c^9 + 252*(a*c*x - c)^2*sqrt(-a*c*x + c)*a^8*c^10 + 840
*(-a*c*x + c)^(3/2)*a^8*c^11 + 5040*sqrt(-a*c*x + c)*a^8*c^12)/(a^9*c^9)