3.254 \(\int e^{-\tanh ^{-1}(a x)} (c-a c x)^{5/2} \, dx\)

Optimal. Leaf size=136 \[ \frac{256 c^3 \sqrt{1-a^2 x^2}}{35 a \sqrt{c-a c x}}+\frac{64 c^2 \sqrt{1-a^2 x^2} \sqrt{c-a c x}}{35 a}+\frac{24 c \sqrt{1-a^2 x^2} (c-a c x)^{3/2}}{35 a}+\frac{2 \sqrt{1-a^2 x^2} (c-a c x)^{5/2}}{7 a} \]

[Out]

(256*c^3*Sqrt[1 - a^2*x^2])/(35*a*Sqrt[c - a*c*x]) + (64*c^2*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(35*a) + (24*c
*(c - a*c*x)^(3/2)*Sqrt[1 - a^2*x^2])/(35*a) + (2*(c - a*c*x)^(5/2)*Sqrt[1 - a^2*x^2])/(7*a)

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Rubi [A]  time = 0.106038, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6127, 657, 649} \[ \frac{256 c^3 \sqrt{1-a^2 x^2}}{35 a \sqrt{c-a c x}}+\frac{64 c^2 \sqrt{1-a^2 x^2} \sqrt{c-a c x}}{35 a}+\frac{24 c \sqrt{1-a^2 x^2} (c-a c x)^{3/2}}{35 a}+\frac{2 \sqrt{1-a^2 x^2} (c-a c x)^{5/2}}{7 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^(5/2)/E^ArcTanh[a*x],x]

[Out]

(256*c^3*Sqrt[1 - a^2*x^2])/(35*a*Sqrt[c - a*c*x]) + (64*c^2*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(35*a) + (24*c
*(c - a*c*x)^(3/2)*Sqrt[1 - a^2*x^2])/(35*a) + (2*(c - a*c*x)^(5/2)*Sqrt[1 - a^2*x^2])/(7*a)

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]

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 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]

Rubi steps

\begin{align*} \int e^{-\tanh ^{-1}(a x)} (c-a c x)^{5/2} \, dx &=\frac{\int \frac{(c-a c x)^{7/2}}{\sqrt{1-a^2 x^2}} \, dx}{c}\\ &=\frac{2 (c-a c x)^{5/2} \sqrt{1-a^2 x^2}}{7 a}+\frac{12}{7} \int \frac{(c-a c x)^{5/2}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{24 c (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}{35 a}+\frac{2 (c-a c x)^{5/2} \sqrt{1-a^2 x^2}}{7 a}+\frac{1}{35} (96 c) \int \frac{(c-a c x)^{3/2}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{64 c^2 \sqrt{c-a c x} \sqrt{1-a^2 x^2}}{35 a}+\frac{24 c (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}{35 a}+\frac{2 (c-a c x)^{5/2} \sqrt{1-a^2 x^2}}{7 a}+\frac{1}{35} \left (128 c^2\right ) \int \frac{\sqrt{c-a c x}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{256 c^3 \sqrt{1-a^2 x^2}}{35 a \sqrt{c-a c x}}+\frac{64 c^2 \sqrt{c-a c x} \sqrt{1-a^2 x^2}}{35 a}+\frac{24 c (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}{35 a}+\frac{2 (c-a c x)^{5/2} \sqrt{1-a^2 x^2}}{7 a}\\ \end{align*}

Mathematica [A]  time = 0.034936, size = 57, normalized size = 0.42 \[ -\frac{2 c^3 \sqrt{1-a^2 x^2} \left (5 a^3 x^3-27 a^2 x^2+71 a x-177\right )}{35 a \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - a*c*x)^(5/2)/E^ArcTanh[a*x],x]

[Out]

(-2*c^3*Sqrt[1 - a^2*x^2]*(-177 + 71*a*x - 27*a^2*x^2 + 5*a^3*x^3))/(35*a*Sqrt[c - a*c*x])

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Maple [A]  time = 0.03, size = 56, normalized size = 0.4 \begin{align*}{\frac{10\,{x}^{3}{a}^{3}-54\,{a}^{2}{x}^{2}+142\,ax-354}{35\, \left ( ax-1 \right ) ^{3}a}\sqrt{-{a}^{2}{x}^{2}+1} \left ( -acx+c \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(5/2)*(5*a^3*x^3-27*a^2*x^2+71*a*x-177)/(a*x-1)^3/a

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Maxima [A]  time = 1.01309, size = 81, normalized size = 0.6 \begin{align*} -\frac{2 \,{\left (5 \, a^{3} c^{\frac{5}{2}} x^{3} - 27 \, a^{2} c^{\frac{5}{2}} x^{2} + 71 \, a c^{\frac{5}{2}} x - 177 \, c^{\frac{5}{2}}\right )} \sqrt{a x + 1}{\left (a x - 1\right )}}{35 \,{\left (a^{2} x - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(5/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-2/35*(5*a^3*c^(5/2)*x^3 - 27*a^2*c^(5/2)*x^2 + 71*a*c^(5/2)*x - 177*c^(5/2))*sqrt(a*x + 1)*(a*x - 1)/(a^2*x -
a)

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Fricas [A]  time = 2.11936, size = 149, normalized size = 1.1 \begin{align*} \frac{2 \,{\left (5 \, a^{3} c^{2} x^{3} - 27 \, a^{2} c^{2} x^{2} + 71 \, a c^{2} x - 177 \, c^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{35 \,{\left (a^{2} x - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(5/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

2/35*(5*a^3*c^2*x^3 - 27*a^2*c^2*x^2 + 71*a*c^2*x - 177*c^2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^2*x - a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- c \left (a x - 1\right )\right )^{\frac{5}{2}} \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}{a x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)**(5/2)/(a*x+1)*(-a**2*x**2+1)**(1/2),x)

[Out]

Integral((-c*(a*x - 1))**(5/2)*sqrt(-(a*x - 1)*(a*x + 1))/(a*x + 1), x)

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Giac [A]  time = 1.26309, size = 97, normalized size = 0.71 \begin{align*} -\frac{256 \, \sqrt{2} c^{\frac{3}{2}}{\left | c \right |}}{35 \, a} - \frac{2 \,{\left (5 \,{\left (a c x + c\right )}^{\frac{7}{2}} - 42 \,{\left (a c x + c\right )}^{\frac{5}{2}} c + 140 \,{\left (a c x + c\right )}^{\frac{3}{2}} c^{2} - 280 \, \sqrt{a c x + c} c^{3}\right )}{\left | c \right |}}{35 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(5/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

-256/35*sqrt(2)*c^(3/2)*abs(c)/a - 2/35*(5*(a*c*x + c)^(7/2) - 42*(a*c*x + c)^(5/2)*c + 140*(a*c*x + c)^(3/2)*
c^2 - 280*sqrt(a*c*x + c)*c^3)*abs(c)/(a*c^2)