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

Optimal. Leaf size=106 \[ \frac{64 c^4 \left (1-a^2 x^2\right )^{3/2}}{105 a (c-a c x)^{3/2}}+\frac{16 c^3 \left (1-a^2 x^2\right )^{3/2}}{35 a \sqrt{c-a c x}}+\frac{2 c^2 \left (1-a^2 x^2\right )^{3/2} \sqrt{c-a c x}}{7 a} \]

[Out]

(64*c^4*(1 - a^2*x^2)^(3/2))/(105*a*(c - a*c*x)^(3/2)) + (16*c^3*(1 - a^2*x^2)^(3/2))/(35*a*Sqrt[c - a*c*x]) +
(2*c^2*Sqrt[c - a*c*x]*(1 - a^2*x^2)^(3/2))/(7*a)

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Rubi [A]  time = 0.0817861, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6127, 657, 649} \[ \frac{64 c^4 \left (1-a^2 x^2\right )^{3/2}}{105 a (c-a c x)^{3/2}}+\frac{16 c^3 \left (1-a^2 x^2\right )^{3/2}}{35 a \sqrt{c-a c x}}+\frac{2 c^2 \left (1-a^2 x^2\right )^{3/2} \sqrt{c-a c x}}{7 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*c^4*(1 - a^2*x^2)^(3/2))/(105*a*(c - a*c*x)^(3/2)) + (16*c^3*(1 - a^2*x^2)^(3/2))/(35*a*Sqrt[c - a*c*x]) +
(2*c^2*Sqrt[c - a*c*x]*(1 - a^2*x^2)^(3/2))/(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 &=c \int (c-a c x)^{3/2} \sqrt{1-a^2 x^2} \, dx\\ &=\frac{2 c^2 \sqrt{c-a c x} \left (1-a^2 x^2\right )^{3/2}}{7 a}+\frac{1}{7} \left (8 c^2\right ) \int \sqrt{c-a c x} \sqrt{1-a^2 x^2} \, dx\\ &=\frac{16 c^3 \left (1-a^2 x^2\right )^{3/2}}{35 a \sqrt{c-a c x}}+\frac{2 c^2 \sqrt{c-a c x} \left (1-a^2 x^2\right )^{3/2}}{7 a}+\frac{1}{35} \left (32 c^3\right ) \int \frac{\sqrt{1-a^2 x^2}}{\sqrt{c-a c x}} \, dx\\ &=\frac{64 c^4 \left (1-a^2 x^2\right )^{3/2}}{105 a (c-a c x)^{3/2}}+\frac{16 c^3 \left (1-a^2 x^2\right )^{3/2}}{35 a \sqrt{c-a c x}}+\frac{2 c^2 \sqrt{c-a c x} \left (1-a^2 x^2\right )^{3/2}}{7 a}\\ \end{align*}

Mathematica [A]  time = 0.0343594, size = 54, normalized size = 0.51 \[ \frac{2 c^2 (a x+1)^{3/2} \left (15 a^2 x^2-54 a x+71\right ) \sqrt{c-a c x}}{105 a \sqrt{1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*c^2*(1 + a*x)^(3/2)*Sqrt[c - a*c*x]*(71 - 54*a*x + 15*a^2*x^2))/(105*a*Sqrt[1 - a*x])

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Maple [A]  time = 0.031, size = 55, normalized size = 0.5 \begin{align*}{\frac{2\, \left ( 15\,{a}^{2}{x}^{2}-54\,ax+71 \right ) \left ( ax+1 \right ) ^{2}}{105\, \left ( ax-1 \right ) ^{2}a} \left ( -acx+c \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(a*x+1)^2*(15*a^2*x^2-54*a*x+71)*(-a*c*x+c)^(5/2)/a/(a*x-1)^2/(-a^2*x^2+1)^(1/2)

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Maxima [A]  time = 1.02198, size = 143, normalized size = 1.35 \begin{align*} \frac{2 \,{\left (3 \, a^{4} c^{\frac{5}{2}} x^{4} - 9 \, a^{3} c^{\frac{5}{2}} x^{3} + 11 \, a^{2} c^{\frac{5}{2}} x^{2} - 23 \, a c^{\frac{5}{2}} x - 46 \, c^{\frac{5}{2}}\right )}}{21 \, \sqrt{a x + 1} a} + \frac{2 \,{\left (3 \, a^{3} c^{\frac{5}{2}} x^{3} - 11 \, a^{2} c^{\frac{5}{2}} x^{2} + 29 \, a c^{\frac{5}{2}} x + 43 \, c^{\frac{5}{2}}\right )}}{15 \, \sqrt{a x + 1} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*(3*a^4*c^(5/2)*x^4 - 9*a^3*c^(5/2)*x^3 + 11*a^2*c^(5/2)*x^2 - 23*a*c^(5/2)*x - 46*c^(5/2))/(sqrt(a*x + 1)
*a) + 2/15*(3*a^3*c^(5/2)*x^3 - 11*a^2*c^(5/2)*x^2 + 29*a*c^(5/2)*x + 43*c^(5/2))/(sqrt(a*x + 1)*a)

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Fricas [A]  time = 1.68112, size = 151, normalized size = 1.42 \begin{align*} -\frac{2 \,{\left (15 \, a^{3} c^{2} x^{3} - 39 \, a^{2} c^{2} x^{2} + 17 \, a c^{2} x + 71 \, c^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{105 \,{\left (a^{2} x - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*a^3*c^2*x^3 - 39*a^2*c^2*x^2 + 17*a*c^2*x + 71*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}} \left (a x + 1\right )}{\sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.29471, size = 82, normalized size = 0.77 \begin{align*} -\frac{2 \,{\left (64 \, \sqrt{2} c^{\frac{3}{2}} - \frac{15 \,{\left (a c x + c\right )}^{\frac{7}{2}} - 84 \,{\left (a c x + c\right )}^{\frac{5}{2}} c + 140 \,{\left (a c x + c\right )}^{\frac{3}{2}} c^{2}}{c^{2}}\right )} c^{2}}{105 \, a{\left | c \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(64*sqrt(2)*c^(3/2) - (15*(a*c*x + c)^(7/2) - 84*(a*c*x + c)^(5/2)*c + 140*(a*c*x + c)^(3/2)*c^2)/c^2)*
c^2/(a*abs(c))