3.514 \(\int e^{\tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} \, dx\)

Optimal. Leaf size=84 \[ \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{\sqrt {a} \sqrt {1-a x}}-\frac {c \sqrt {1-a^2 x^2}}{a \sqrt {c-\frac {c}{a x}}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 85, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6134, 6128, 848, 50, 54, 215} \[ \frac {x \sqrt {a x+1} \sqrt {c-\frac {c}{a x}}}{\sqrt {1-a x}}+\frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{\sqrt {a} \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.04, size = 69, normalized size = 0.82 \[ \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (\sqrt {a} \sqrt {x} \sqrt {a x+1}+\sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )\right )}{\sqrt {a} \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c - c/(a*x)]*Sqrt[x]*(Sqrt[a]*Sqrt[x]*Sqrt[1 + a*x] + ArcSinh[Sqrt[a]*Sqrt[x]]))/(Sqrt[a]*Sqrt[1 - a*x])

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fricas [A]  time = 0.59, size = 249, normalized size = 2.96 \[ \left [-\frac {4 \, \sqrt {-a^{2} x^{2} + 1} a x \sqrt {\frac {a c x - c}{a x}} - {\left (a x - 1\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x + 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right )}{4 \, {\left (a^{2} x - a\right )}}, -\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a x \sqrt {\frac {a c x - c}{a x}} + {\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right )}{2 \, {\left (a^{2} x - a\right )}}\right ] \]

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)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*(4*sqrt(-a^2*x^2 + 1)*a*x*sqrt((a*c*x - c)/(a*x)) - (a*x - 1)*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)))/(a^2*x - a), -1/2*(2*sqr
t(-a^2*x^2 + 1)*a*x*sqrt((a*c*x - c)/(a*x)) + (a*x - 1)*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)))/(a^2*x - a)]

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

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 89, normalized size = 1.06 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \sqrt {-a^{2} x^{2}+1}\, \left (-2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}+\arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right )\right )}{2 \left (a x -1\right ) \sqrt {-\left (a x +1\right ) x}\, \sqrt {a}} \]

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)^(1/2),x)

[Out]

1/2*(c*(a*x-1)/a/x)^(1/2)*x*(-a^2*x^2+1)^(1/2)*(-2*a^(1/2)*(-(a*x+1)*x)^(1/2)+arctan(1/2/a^(1/2)*(2*a*x+1)/(-(
a*x+1)*x)^(1/2)))/(a*x-1)/(-(a*x+1)*x)^(1/2)/a^(1/2)

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

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)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)**(1/2),x)

[Out]

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

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