3.529 \(\int e^{3 \tanh ^{-1}(a x)} (c-\frac{c}{a x})^{5/2} \, dx\)

Optimal. Leaf size=176 \[ -\frac{a^2 x^3 \sqrt{a x+1} \left (c-\frac{c}{a x}\right )^{5/2}}{(1-a x)^{5/2}}-\frac{2 x \left (1-a^2 x^2\right )^{5/2} \left (c-\frac{c}{a x}\right )^{5/2}}{3 (1-a x)^5}-\frac{a^{3/2} x^{5/2} \left (c-\frac{c}{a x}\right )^{5/2} \sinh ^{-1}\left (\sqrt{a} \sqrt{x}\right )}{(1-a x)^{5/2}}+\frac{2 a x^2 (a x+1)^{3/2} \left (c-\frac{c}{a x}\right )^{5/2}}{3 (1-a x)^{5/2}} \]

[Out]

-((a^2*(c - c/(a*x))^(5/2)*x^3*Sqrt[1 + a*x])/(1 - a*x)^(5/2)) + (2*a*(c - c/(a*x))^(5/2)*x^2*(1 + a*x)^(3/2))
/(3*(1 - a*x)^(5/2)) - (2*(c - c/(a*x))^(5/2)*x*(1 - a^2*x^2)^(5/2))/(3*(1 - a*x)^5) - (a^(3/2)*(c - c/(a*x))^
(5/2)*x^(5/2)*ArcSinh[Sqrt[a]*Sqrt[x]])/(1 - a*x)^(5/2)

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Rubi [A]  time = 0.205523, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6134, 6128, 879, 848, 47, 50, 54, 215} \[ -\frac{a^2 x^3 \sqrt{a x+1} \left (c-\frac{c}{a x}\right )^{5/2}}{(1-a x)^{5/2}}-\frac{2 x \left (1-a^2 x^2\right )^{5/2} \left (c-\frac{c}{a x}\right )^{5/2}}{3 (1-a x)^5}-\frac{a^{3/2} x^{5/2} \left (c-\frac{c}{a x}\right )^{5/2} \sinh ^{-1}\left (\sqrt{a} \sqrt{x}\right )}{(1-a x)^{5/2}}+\frac{2 a x^2 (a x+1)^{3/2} \left (c-\frac{c}{a x}\right )^{5/2}}{3 (1-a x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a*x])*(c - c/(a*x))^(5/2),x]

[Out]

-((a^2*(c - c/(a*x))^(5/2)*x^3*Sqrt[1 + a*x])/(1 - a*x)^(5/2)) + (2*a*(c - c/(a*x))^(5/2)*x^2*(1 + a*x)^(3/2))
/(3*(1 - a*x)^(5/2)) - (2*(c - c/(a*x))^(5/2)*x*(1 - a^2*x^2)^(5/2))/(3*(1 - a*x)^5) - (a^(3/2)*(c - c/(a*x))^
(5/2)*x^(5/2)*ArcSinh[Sqrt[a]*Sqrt[x]])/(1 - a*x)^(5/2)

Rule 6134

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Dist[(x^p*(c + d/x)^p)/(1 + (c*
x)/d)^p, Int[(u*(1 + (c*x)/d)^p*E^(n*ArcTanh[a*x]))/x^p, x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c^2 - a^2*
d^2, 0] &&  !IntegerQ[p]

Rule 6128

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[c^n,
Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c +
d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1, 0]) && IntegerQ[2*p]

Rule 879

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(e*f
- d*g)*(d + e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + 1)*(e*f + d*g)), x] - Dist[(e*(e*f*
(p + 1) - d*g*(2*n + p + 3)))/(g*(n + 1)*(e*f + d*g)), Int[(d + e*x)^(m - 1)*(f + g*x)^(n + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&
EqQ[m + p - 1, 0] && LtQ[n, -1] && IntegerQ[2*p]

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, 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 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [C]  time = 0.0397677, size = 66, normalized size = 0.38 \[ -\frac{2 c^2 \sqrt{c-\frac{c}{a x}} \left ((a x+1)^{5/2}-a x \text{Hypergeometric2F1}\left (-\frac{3}{2},-\frac{1}{2},\frac{1}{2},-a x\right )\right )}{3 a^2 x \sqrt{1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcTanh[a*x])*(c - c/(a*x))^(5/2),x]

[Out]

(-2*c^2*Sqrt[c - c/(a*x)]*((1 + a*x)^(5/2) - a*x*Hypergeometric2F1[-3/2, -1/2, 1/2, -(a*x)]))/(3*a^2*x*Sqrt[1
- a*x])

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Maple [A]  time = 0.142, size = 136, normalized size = 0.8 \begin{align*}{\frac{{c}^{2}}{6\, \left ( ax-1 \right ) x}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}}\sqrt{-{a}^{2}{x}^{2}+1} \left ( 6\,{a}^{5/2}{x}^{2}\sqrt{- \left ( ax+1 \right ) x}+4\,{a}^{3/2}x\sqrt{- \left ( ax+1 \right ) x}-3\,\arctan \left ( 1/2\,{\frac{2\,ax+1}{\sqrt{a}\sqrt{- \left ( ax+1 \right ) x}}} \right ){x}^{2}{a}^{2}+4\,\sqrt{a}\sqrt{- \left ( ax+1 \right ) x} \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{- \left ( ax+1 \right ) x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^(5/2),x)

[Out]

1/6*(c*(a*x-1)/a/x)^(1/2)*c^2*(-a^2*x^2+1)^(1/2)*(6*a^(5/2)*x^2*(-(a*x+1)*x)^(1/2)+4*a^(3/2)*x*(-(a*x+1)*x)^(1
/2)-3*arctan(1/2/a^(1/2)*(2*a*x+1)/(-(a*x+1)*x)^(1/2))*x^2*a^2+4*a^(1/2)*(-(a*x+1)*x)^(1/2))/x/a^(5/2)/(a*x-1)
/(-(a*x+1)*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )}^{3}{\left (c - \frac{c}{a x}\right )}^{\frac{5}{2}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)^3*(c - c/(a*x))^(5/2)/(-a^2*x^2 + 1)^(3/2), x)

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Fricas [A]  time = 2.15303, size = 674, normalized size = 3.83 \begin{align*} \left [\frac{3 \,{\left (a^{2} c^{2} x^{2} - a c^{2} x\right )} \sqrt{-c} \log \left (-\frac{8 \, a^{3} c x^{3} - 7 \, a c x - 4 \,{\left (2 \, a^{2} x^{2} + a x\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-c} \sqrt{\frac{a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \,{\left (3 \, a^{2} c^{2} x^{2} + 2 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a c x - c}{a x}}}{12 \,{\left (a^{3} x^{2} - a^{2} x\right )}}, \frac{3 \,{\left (a^{2} c^{2} x^{2} - a c^{2} x\right )} \sqrt{c} \arctan \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1} a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \,{\left (3 \, a^{2} c^{2} x^{2} + 2 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a c x - c}{a x}}}{6 \,{\left (a^{3} x^{2} - a^{2} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(a^2*c^2*x^2 - a*c^2*x)*sqrt(-c)*log(-(8*a^3*c*x^3 - 7*a*c*x - 4*(2*a^2*x^2 + a*x)*sqrt(-a^2*x^2 + 1)
*sqrt(-c)*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) + 4*(3*a^2*c^2*x^2 + 2*a*c^2*x + 2*c^2)*sqrt(-a^2*x^2 + 1)*s
qrt((a*c*x - c)/(a*x)))/(a^3*x^2 - a^2*x), 1/6*(3*(a^2*c^2*x^2 - a*c^2*x)*sqrt(c)*arctan(2*sqrt(-a^2*x^2 + 1)*
a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x - c)) + 2*(3*a^2*c^2*x^2 + 2*a*c^2*x + 2*c^2)*sqrt(-a
^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^3*x^2 - a^2*x)]

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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*x+1)**3/(-a**2*x**2+1)**(3/2)*(c-c/a/x)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )}^{3}{\left (c - \frac{c}{a x}\right )}^{\frac{5}{2}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^(5/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)^3*(c - c/(a*x))^(5/2)/(-a^2*x^2 + 1)^(3/2), x)