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

Optimal. Leaf size=171 \[ -\frac{3 a^2 x^3 \sqrt{a x+1} \left (c-\frac{c}{a x}\right )^{5/2}}{(1-a x)^{5/2}}-\frac{3 a^{3/2} x^{5/2} \left (c-\frac{c}{a x}\right )^{5/2} \sinh ^{-1}\left (\sqrt{a} \sqrt{x}\right )}{(1-a x)^{5/2}}+\frac{4 a x^2 (a x+1)^{3/2} \left (c-\frac{c}{a x}\right )^{5/2}}{(1-a x)^{5/2}}-\frac{2 x (a x+1)^{3/2} \left (c-\frac{c}{a x}\right )^{5/2}}{3 (1-a x)^{5/2}} \]

[Out]

(-3*a^2*(c - c/(a*x))^(5/2)*x^3*Sqrt[1 + a*x])/(1 - a*x)^(5/2) - (2*(c - c/(a*x))^(5/2)*x*(1 + a*x)^(3/2))/(3*
(1 - a*x)^(5/2)) + (4*a*(c - c/(a*x))^(5/2)*x^2*(1 + a*x)^(3/2))/(1 - a*x)^(5/2) - (3*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 = 0.157882, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {6134, 6129, 89, 78, 50, 54, 215} \[ -\frac{3 a^2 x^3 \sqrt{a x+1} \left (c-\frac{c}{a x}\right )^{5/2}}{(1-a x)^{5/2}}-\frac{3 a^{3/2} x^{5/2} \left (c-\frac{c}{a x}\right )^{5/2} \sinh ^{-1}\left (\sqrt{a} \sqrt{x}\right )}{(1-a x)^{5/2}}+\frac{4 a x^2 (a x+1)^{3/2} \left (c-\frac{c}{a x}\right )^{5/2}}{(1-a x)^{5/2}}-\frac{2 x (a x+1)^{3/2} \left (c-\frac{c}{a x}\right )^{5/2}}{3 (1-a x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2*(c - c/(a*x))^(5/2)*x^3*Sqrt[1 + a*x])/(1 - a*x)^(5/2) - (2*(c - c/(a*x))^(5/2)*x*(1 + a*x)^(3/2))/(3*
(1 - a*x)^(5/2)) + (4*a*(c - c/(a*x))^(5/2)*x^2*(1 + a*x)^(3/2))/(1 - a*x)^(5/2) - (3*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 6129

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

Rule 89

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

Rule 78

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

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^{\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^{\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{(1-a x)^2 \sqrt{1+a x}}{x^{5/2}} \, dx}{(1-a x)^{5/2}}\\ &=-\frac{2 \left (c-\frac{c}{a x}\right )^{5/2} x (1+a x)^{3/2}}{3 (1-a x)^{5/2}}+\frac{\left (2 \left (c-\frac{c}{a x}\right )^{5/2} x^{5/2}\right ) \int \frac{\sqrt{1+a x} \left (-3 a+\frac{3 a^2 x}{2}\right )}{x^{3/2}} \, dx}{3 (1-a x)^{5/2}}\\ &=-\frac{2 \left (c-\frac{c}{a x}\right )^{5/2} x (1+a x)^{3/2}}{3 (1-a x)^{5/2}}+\frac{4 a \left (c-\frac{c}{a x}\right )^{5/2} x^2 (1+a x)^{3/2}}{(1-a x)^{5/2}}-\frac{\left (3 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{3 a^2 \left (c-\frac{c}{a x}\right )^{5/2} x^3 \sqrt{1+a x}}{(1-a x)^{5/2}}-\frac{2 \left (c-\frac{c}{a x}\right )^{5/2} x (1+a x)^{3/2}}{3 (1-a x)^{5/2}}+\frac{4 a \left (c-\frac{c}{a x}\right )^{5/2} x^2 (1+a x)^{3/2}}{(1-a x)^{5/2}}-\frac{\left (3 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{3 a^2 \left (c-\frac{c}{a x}\right )^{5/2} x^3 \sqrt{1+a x}}{(1-a x)^{5/2}}-\frac{2 \left (c-\frac{c}{a x}\right )^{5/2} x (1+a x)^{3/2}}{3 (1-a x)^{5/2}}+\frac{4 a \left (c-\frac{c}{a x}\right )^{5/2} x^2 (1+a x)^{3/2}}{(1-a x)^{5/2}}-\frac{\left (3 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{3 a^2 \left (c-\frac{c}{a x}\right )^{5/2} x^3 \sqrt{1+a x}}{(1-a x)^{5/2}}-\frac{2 \left (c-\frac{c}{a x}\right )^{5/2} x (1+a x)^{3/2}}{3 (1-a x)^{5/2}}+\frac{4 a \left (c-\frac{c}{a x}\right )^{5/2} x^2 (1+a x)^{3/2}}{(1-a x)^{5/2}}-\frac{3 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 [A]  time = 0.0639127, size = 87, normalized size = 0.51 \[ \frac{c^2 \sqrt{c-\frac{c}{a x}} \left (\sqrt{a x+1} \left (3 a^2 x^2+10 a x-2\right )-9 a^{3/2} x^{3/2} \sinh ^{-1}\left (\sqrt{a} \sqrt{x}\right )\right )}{3 a^2 x \sqrt{1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Sqrt[c - c/(a*x)]*(Sqrt[1 + a*x]*(-2 + 10*a*x + 3*a^2*x^2) - 9*a^(3/2)*x^(3/2)*ArcSinh[Sqrt[a]*Sqrt[x]]))
/(3*a^2*x*Sqrt[1 - a*x])

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Maple [A]  time = 0.144, 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}+9\,\arctan \left ( 1/2\,{\frac{2\,ax+1}{\sqrt{a}\sqrt{- \left ( ax+1 \right ) x}}} \right ){x}^{2}{a}^{2}+20\,{a}^{3/2}x\sqrt{- \left ( ax+1 \right ) x}-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)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(5/2),x)

[Out]

-1/6*(c*(a*x-1)/a/x)^(1/2)/x*c^2/a^(5/2)*(-a^2*x^2+1)^(1/2)*(6*a^(5/2)*x^2*(-(a*x+1)*x)^(1/2)+9*arctan(1/2/a^(
1/2)*(2*a*x+1)/(-(a*x+1)*x)^(1/2))*x^2*a^2+20*a^(3/2)*x*(-(a*x+1)*x)^(1/2)-4*a^(1/2)*(-(a*x+1)*x)^(1/2))/(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 )}{\left (c - \frac{c}{a x}\right )}^{\frac{5}{2}}}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.77078, size = 676, normalized size = 3.95 \begin{align*} \left [\frac{9 \,{\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} + 10 \, 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{9 \,{\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} + 10 \, 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)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(9*(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 + 10*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), 1/6*(9*(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 + 10*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- c \left (-1 + \frac{1}{a x}\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)*(c-c/a/x)**(5/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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