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

Optimal. Leaf size=157 \[ -\frac {3 \sqrt {1-a x} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{a^{3/2} \sqrt {x} \sqrt {c-\frac {c}{a x}}}+\frac {2 \sqrt {2} \sqrt {1-a x} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {a x+1}}\right )}{a^{3/2} \sqrt {x} \sqrt {c-\frac {c}{a x}}}-\frac {\sqrt {1-a x} \sqrt {a x+1}}{a \sqrt {c-\frac {c}{a x}}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6134, 6129, 101, 157, 54, 215, 93, 206} \[ -\frac {3 \sqrt {1-a x} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{a^{3/2} \sqrt {x} \sqrt {c-\frac {c}{a x}}}+\frac {2 \sqrt {2} \sqrt {1-a x} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {a x+1}}\right )}{a^{3/2} \sqrt {x} \sqrt {c-\frac {c}{a x}}}-\frac {\sqrt {1-a x} \sqrt {a x+1}}{a \sqrt {c-\frac {c}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[1 - a*x]*Sqrt[1 + a*x])/(a*Sqrt[c - c/(a*x)])) - (3*Sqrt[1 - a*x]*ArcSinh[Sqrt[a]*Sqrt[x]])/(a^(3/2)*S
qrt[c - c/(a*x)]*Sqrt[x]) + (2*Sqrt[2]*Sqrt[1 - a*x]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[x])/Sqrt[1 + a*x]])/(a^(3/2
)*Sqrt[c - c/(a*x)]*Sqrt[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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 101

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

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 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 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 \frac {e^{\tanh ^{-1}(a x)}}{\sqrt {c-\frac {c}{a x}}} \, dx &=\frac {\sqrt {1-a x} \int \frac {e^{\tanh ^{-1}(a x)} \sqrt {x}}{\sqrt {1-a x}} \, dx}{\sqrt {c-\frac {c}{a x}} \sqrt {x}}\\ &=\frac {\sqrt {1-a x} \int \frac {\sqrt {x} \sqrt {1+a x}}{1-a x} \, dx}{\sqrt {c-\frac {c}{a x}} \sqrt {x}}\\ &=-\frac {\sqrt {1-a x} \sqrt {1+a x}}{a \sqrt {c-\frac {c}{a x}}}+\frac {\sqrt {1-a x} \int \frac {\frac {1}{2}+\frac {3 a x}{2}}{\sqrt {x} (1-a x) \sqrt {1+a x}} \, dx}{a \sqrt {c-\frac {c}{a x}} \sqrt {x}}\\ &=-\frac {\sqrt {1-a x} \sqrt {1+a x}}{a \sqrt {c-\frac {c}{a x}}}-\frac {\left (3 \sqrt {1-a x}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+a x}} \, dx}{2 a \sqrt {c-\frac {c}{a x}} \sqrt {x}}+\frac {\left (2 \sqrt {1-a x}\right ) \int \frac {1}{\sqrt {x} (1-a x) \sqrt {1+a x}} \, dx}{a \sqrt {c-\frac {c}{a x}} \sqrt {x}}\\ &=-\frac {\sqrt {1-a x} \sqrt {1+a x}}{a \sqrt {c-\frac {c}{a x}}}-\frac {\left (3 \sqrt {1-a x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+a x^2}} \, dx,x,\sqrt {x}\right )}{a \sqrt {c-\frac {c}{a x}} \sqrt {x}}+\frac {\left (4 \sqrt {1-a x}\right ) \operatorname {Subst}\left (\int \frac {1}{1-2 a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {1+a x}}\right )}{a \sqrt {c-\frac {c}{a x}} \sqrt {x}}\\ &=-\frac {\sqrt {1-a x} \sqrt {1+a x}}{a \sqrt {c-\frac {c}{a x}}}-\frac {3 \sqrt {1-a x} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{a^{3/2} \sqrt {c-\frac {c}{a x}} \sqrt {x}}+\frac {2 \sqrt {2} \sqrt {1-a x} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {1+a x}}\right )}{a^{3/2} \sqrt {c-\frac {c}{a x}} \sqrt {x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 105, normalized size = 0.67 \[ -\frac {\sqrt {1-a x} \left (\sqrt {a} \sqrt {x} \sqrt {a x+1}+3 \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )-2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {a x+1}}\right )\right )}{a^{3/2} \sqrt {x} \sqrt {c-\frac {c}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 447, normalized size = 2.85 \[ \left [-\frac {4 \, \sqrt {-a^{2} x^{2} + 1} a x \sqrt {\frac {a c x - c}{a x}} - 2 \, \sqrt {2} {\left (a c x - c\right )} \sqrt {-\frac {1}{c}} \log \left (-\frac {17 \, a^{3} x^{3} - 3 \, a^{2} x^{2} - 4 \, \sqrt {2} {\left (3 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {1}{c}} \sqrt {\frac {a c x - c}{a x}} - 13 \, a x - 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) + 3 \, {\left (a x - 1\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right )}{4 \, {\left (a^{2} c x - a c\right )}}, -\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a x \sqrt {\frac {a c x - c}{a x}} + 3 \, {\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - \frac {2 \, \sqrt {2} {\left (a c x - c\right )} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} a x \sqrt {\frac {a c x - c}{a x}}}{{\left (3 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt {c}}\right )}{\sqrt {c}}}{2 \, {\left (a^{2} c x - a c\right )}}\right ] \]

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)) - 2*sqrt(2)*(a*c*x - c)*sqrt(-1/c)*log(-(17*a^3*x^3 -
3*a^2*x^2 - 4*sqrt(2)*(3*a^2*x^2 + a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-1/c)*sqrt((a*c*x - c)/(a*x)) - 13*a*x - 1)/(a
^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)) + 3*(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*c*x - a*c), -1/2*(2*sqrt(-a^2*x^2 + 1)*a
*x*sqrt((a*c*x - c)/(a*x)) + 3*(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)) - 2*sqrt(2)*(a*c*x - c)*arctan(2*sqrt(2)*sqrt(-a^2*x^2 + 1)*a*x*sqrt((a*c*x - c)
/(a*x))/((3*a^2*x^2 - 2*a*x - 1)*sqrt(c)))/sqrt(c))/(a^2*c*x - a*c)]

________________________________________________________________________________________

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

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(-a^2*x^2 + 1)*sqrt(c - c/(a*x))), x)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 168, normalized size = 1.07 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \sqrt {-a^{2} x^{2}+1}\, \left (-2 \sqrt {-\left (a x +1\right ) x}\, a^{\frac {3}{2}} \sqrt {2}\, \sqrt {-\frac {1}{a}}+3 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right ) a \sqrt {2}\, \sqrt {-\frac {1}{a}}-4 \ln \left (\frac {2 \sqrt {2}\, \sqrt {-\frac {1}{a}}\, \sqrt {-\left (a x +1\right ) x}\, a -3 a x -1}{a x -1}\right ) \sqrt {a}\right ) \sqrt {2}}{4 a^{\frac {3}{2}} c \left (a x -1\right ) \sqrt {-\left (a x +1\right ) x}\, \sqrt {-\frac {1}{a}}} \]

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]

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

________________________________________________________________________________________

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

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(-a^2*x^2 + 1)*sqrt(c - c/(a*x))), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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

________________________________________________________________________________________