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

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

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 176, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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*}

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Mathematica [C]  time = 0.04, size = 66, normalized size = 0.38 \[ -\frac {2 c^2 \sqrt {c-\frac {c}{a x}} \left ((a x+1)^{5/2}-a x \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-a x\right )\right )}{3 a^2 x \sqrt {1-a x}} \]

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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fricas [A]  time = 0.53, size = 330, normalized size = 1.88 \[ \left [\frac {3 \, {\left (a^{2} c^{2} x^{2} - a c^{2} x\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (3 \, a^{2} c^{2} x^{2} + 2 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{12 \, {\left (a^{3} x^{2} - a^{2} x\right )}}, \frac {3 \, {\left (a^{2} c^{2} x^{2} - a c^{2} x\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \, {\left (3 \, a^{2} c^{2} x^{2} + 2 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{6 \, {\left (a^{3} x^{2} - a^{2} x\right )}}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a x}\right )}^{\frac {5}{2}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.05, size = 136, normalized size = 0.77 \[ -\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c^{2} \sqrt {-a^{2} x^{2}+1}\, \left (-6 a^{\frac {5}{2}} x^{2} \sqrt {-\left (a x +1\right ) x}+3 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right ) x^{2} a^{2}-4 a^{\frac {3}{2}} x \sqrt {-\left (a x +1\right ) x}-4 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}\right )}{6 x \,a^{\frac {5}{2}} \left (a x -1\right ) \sqrt {-\left (a x +1\right ) x}} \]

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)/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)+3*arctan(1/2/a^
(1/2)*(2*a*x+1)/(-(a*x+1)*x)^(1/2))*x^2*a^2-4*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a x}\right )}^{\frac {5}{2}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-\frac {c}{a\,x}\right )}^{5/2}\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (-1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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]

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

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