3.523 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{\sqrt{c-\frac{c}{a x}}} \, dx\)

Optimal. Leaf size=71 \[ -\frac{x}{\sqrt{c-\frac{c}{a x}}}+\frac{5}{a \sqrt{c-\frac{c}{a x}}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a \sqrt{c}} \]

[Out]

5/(a*Sqrt[c - c/(a*x)]) - x/Sqrt[c - c/(a*x)] - (5*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(a*Sqrt[c])

________________________________________________________________________________________

Rubi [A]  time = 0.162234, antiderivative size = 71, 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 = {6133, 25, 514, 375, 78, 51, 63, 208} \[ -\frac{x}{\sqrt{c-\frac{c}{a x}}}+\frac{5}{a \sqrt{c-\frac{c}{a x}}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

5/(a*Sqrt[c - c/(a*x)]) - x/Sqrt[c - c/(a*x)] - (5*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(a*Sqrt[c])

Rule 6133

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

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
 d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

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 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [C]  time = 0.0263909, size = 44, normalized size = 0.62 \[ \frac{5 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},1-\frac{1}{a x}\right )-a x}{a \sqrt{c-\frac{c}{a x}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.144, size = 194, normalized size = 2.7 \begin{align*} -{\frac{x}{2\,c \left ( ax-1 \right ) ^{2}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( 10\,{a}^{5/2}\sqrt{ \left ( ax-1 \right ) x}{x}^{2}+5\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ){x}^{2}{a}^{2}-8\,{a}^{3/2} \left ( \left ( ax-1 \right ) x \right ) ^{3/2}-20\,{a}^{3/2}\sqrt{ \left ( ax-1 \right ) x}x-10\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) xa+10\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+5\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*(a*x-1)/a/x)^(1/2)*x*(10*a^(5/2)*((a*x-1)*x)^(1/2)*x^2+5*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/
a^(1/2))*x^2*a^2-8*a^(3/2)*((a*x-1)*x)^(3/2)-20*a^(3/2)*((a*x-1)*x)^(1/2)*x-10*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(
1/2)+2*a*x-1)/a^(1/2))*x*a+10*((a*x-1)*x)^(1/2)*a^(1/2)+5*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2)
))/((a*x-1)*x)^(1/2)/c/(a*x-1)^2/a^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.10614, size = 375, normalized size = 5.28 \begin{align*} \left [\frac{5 \,{\left (a x - 1\right )} \sqrt{c} \log \left (-2 \, a c x + 2 \, a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} + c\right ) - 2 \,{\left (a^{2} x^{2} - 5 \, a x\right )} \sqrt{\frac{a c x - c}{a x}}}{2 \,{\left (a^{2} c x - a c\right )}}, \frac{5 \,{\left (a x - 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{c}\right ) -{\left (a^{2} x^{2} - 5 \, a x\right )} \sqrt{\frac{a c x - c}{a x}}}{a^{2} c x - a c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(5*(a*x - 1)*sqrt(c)*log(-2*a*c*x + 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c) - 2*(a^2*x^2 - 5*a*x)*sqrt
((a*c*x - c)/(a*x)))/(a^2*c*x - a*c), (5*(a*x - 1)*sqrt(-c)*arctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) - (a^2*
x^2 - 5*a*x)*sqrt((a*c*x - c)/(a*x)))/(a^2*c*x - a*c)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{a x}{a x \sqrt{c - \frac{c}{a x}} - \sqrt{c - \frac{c}{a x}}}\, dx - \int \frac{1}{a x \sqrt{c - \frac{c}{a x}} - \sqrt{c - \frac{c}{a x}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.25152, size = 166, normalized size = 2.34 \begin{align*} a c{\left (\frac{5 \, \arctan \left (\frac{\sqrt{\frac{a c x - c}{a x}}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c} + \frac{4 \, c - \frac{5 \,{\left (a c x - c\right )}}{a x}}{{\left (c \sqrt{\frac{a c x - c}{a x}} - \frac{{\left (a c x - c\right )} \sqrt{\frac{a c x - c}{a x}}}{a x}\right )} a^{2} c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c*(5*arctan(sqrt((a*c*x - c)/(a*x))/sqrt(-c))/(a^2*sqrt(-c)*c) + (4*c - 5*(a*c*x - c)/(a*x))/((c*sqrt((a*c*x
 - c)/(a*x)) - (a*c*x - c)*sqrt((a*c*x - c)/(a*x))/(a*x))*a^2*c))