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

Optimal. Leaf size=105 \[ -\frac{11 x \sqrt{c-\frac{c}{a x}}}{8 a^2}-\frac{11 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{8 a^3}-\frac{1}{3} x^3 \sqrt{c-\frac{c}{a x}}-\frac{11 x^2 \sqrt{c-\frac{c}{a x}}}{12 a} \]

[Out]

(-11*Sqrt[c - c/(a*x)]*x)/(8*a^2) - (11*Sqrt[c - c/(a*x)]*x^2)/(12*a) - (Sqrt[c - c/(a*x)]*x^3)/3 - (11*Sqrt[c
]*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(8*a^3)

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Rubi [A]  time = 0.217048, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {6133, 25, 514, 446, 78, 51, 63, 208} \[ -\frac{11 x \sqrt{c-\frac{c}{a x}}}{8 a^2}-\frac{11 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{8 a^3}-\frac{1}{3} x^3 \sqrt{c-\frac{c}{a x}}-\frac{11 x^2 \sqrt{c-\frac{c}{a x}}}{12 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*Sqrt[c - c/(a*x)]*x)/(8*a^2) - (11*Sqrt[c - c/(a*x)]*x^2)/(12*a) - (Sqrt[c - c/(a*x)]*x^3)/3 - (11*Sqrt[c
]*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(8*a^3)

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 446

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

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 e^{2 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a x}} x^2 \, dx &=\int \frac{\sqrt{c-\frac{c}{a x}} x^2 (1+a x)}{1-a x} \, dx\\ &=-\frac{c \int \frac{x (1+a x)}{\sqrt{c-\frac{c}{a x}}} \, dx}{a}\\ &=-\frac{c \int \frac{\left (a+\frac{1}{x}\right ) x^2}{\sqrt{c-\frac{c}{a x}}} \, dx}{a}\\ &=\frac{c \operatorname{Subst}\left (\int \frac{a+x}{x^4 \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=-\frac{1}{3} \sqrt{c-\frac{c}{a x}} x^3+\frac{(11 c) \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{6 a}\\ &=-\frac{11 \sqrt{c-\frac{c}{a x}} x^2}{12 a}-\frac{1}{3} \sqrt{c-\frac{c}{a x}} x^3+\frac{(11 c) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{8 a^2}\\ &=-\frac{11 \sqrt{c-\frac{c}{a x}} x}{8 a^2}-\frac{11 \sqrt{c-\frac{c}{a x}} x^2}{12 a}-\frac{1}{3} \sqrt{c-\frac{c}{a x}} x^3+\frac{(11 c) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{16 a^3}\\ &=-\frac{11 \sqrt{c-\frac{c}{a x}} x}{8 a^2}-\frac{11 \sqrt{c-\frac{c}{a x}} x^2}{12 a}-\frac{1}{3} \sqrt{c-\frac{c}{a x}} x^3-\frac{11 \operatorname{Subst}\left (\int \frac{1}{a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-\frac{c}{a x}}\right )}{8 a^2}\\ &=-\frac{11 \sqrt{c-\frac{c}{a x}} x}{8 a^2}-\frac{11 \sqrt{c-\frac{c}{a x}} x^2}{12 a}-\frac{1}{3} \sqrt{c-\frac{c}{a x}} x^3-\frac{11 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{8 a^3}\\ \end{align*}

Mathematica [C]  time = 0.0348243, size = 50, normalized size = 0.48 \[ -\frac{\sqrt{c-\frac{c}{a x}} \left (11 \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},1-\frac{1}{a x}\right )+a^3 x^3\right )}{3 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Sqrt[c - c/(a*x)]*(a^3*x^3 + 11*Hypergeometric2F1[1/2, 3, 3/2, 1 - 1/(a*x)]))/(3*a^3)

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Maple [A]  time = 0.128, size = 155, normalized size = 1.5 \begin{align*} -{\frac{x}{48}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( 16\, \left ( a{x}^{2}-x \right ) ^{3/2}{a}^{5/2}+60\,\sqrt{a{x}^{2}-x}{a}^{5/2}x-30\,\sqrt{a{x}^{2}-x}{a}^{3/2}+96\,{a}^{3/2}\sqrt{ \left ( ax-1 \right ) x}+48\,a\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) -15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}-x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) a \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(c*(a*x-1)/a/x)^(1/2)*x*(16*(a*x^2-x)^(3/2)*a^(5/2)+60*(a*x^2-x)^(1/2)*a^(5/2)*x-30*(a*x^2-x)^(1/2)*a^(3
/2)+96*a^(3/2)*((a*x-1)*x)^(1/2)+48*a*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))-15*ln(1/2*(2*(a*x^
2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*a)/((a*x-1)*x)^(1/2)/a^(7/2)

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

[Out]

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

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Fricas [A]  time = 1.883, size = 373, normalized size = 3.55 \begin{align*} \left [-\frac{2 \,{\left (8 \, a^{3} x^{3} + 22 \, a^{2} x^{2} + 33 \, a x\right )} \sqrt{\frac{a c x - c}{a x}} - 33 \, \sqrt{c} \log \left (-2 \, a c x + 2 \, a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} + c\right )}{48 \, a^{3}}, -\frac{{\left (8 \, a^{3} x^{3} + 22 \, a^{2} x^{2} + 33 \, a x\right )} \sqrt{\frac{a c x - c}{a x}} - 33 \, \sqrt{-c} \arctan \left (\frac{\sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{c}\right )}{24 \, a^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(2*(8*a^3*x^3 + 22*a^2*x^2 + 33*a*x)*sqrt((a*c*x - c)/(a*x)) - 33*sqrt(c)*log(-2*a*c*x + 2*a*sqrt(c)*x*
sqrt((a*c*x - c)/(a*x)) + c))/a^3, -1/24*((8*a^3*x^3 + 22*a^2*x^2 + 33*a*x)*sqrt((a*c*x - c)/(a*x)) - 33*sqrt(
-c)*arctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c))/a^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{2} \sqrt{c - \frac{c}{a x}}}{a x - 1}\, dx - \int \frac{a x^{3} \sqrt{c - \frac{c}{a x}}}{a x - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28809, size = 173, normalized size = 1.65 \begin{align*} -\frac{1}{24} \, \sqrt{a^{2} c x^{2} - a c x}{\left (2 \, x{\left (\frac{4 \, x{\left | a \right |}}{a^{2} \mathrm{sgn}\left (x\right )} + \frac{11 \,{\left | a \right |}}{a^{3} \mathrm{sgn}\left (x\right )}\right )} + \frac{33 \,{\left | a \right |}}{a^{4} \mathrm{sgn}\left (x\right )}\right )} - \frac{11 \, \sqrt{c} \log \left ({\left | a \right |} \sqrt{{\left | c \right |}}\right ) \mathrm{sgn}\left (x\right )}{16 \, a^{3}} + \frac{11 \, \sqrt{c} \log \left ({\left | -2 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}{\left | a \right |} + a \sqrt{c} \right |}\right )}{16 \, a^{3} \mathrm{sgn}\left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*sqrt(a^2*c*x^2 - a*c*x)*(2*x*(4*x*abs(a)/(a^2*sgn(x)) + 11*abs(a)/(a^3*sgn(x))) + 33*abs(a)/(a^4*sgn(x))
) - 11/16*sqrt(c)*log(abs(a)*sqrt(abs(c)))*sgn(x)/a^3 + 11/16*sqrt(c)*log(abs(-2*(sqrt(a^2*c)*x - sqrt(a^2*c*x
^2 - a*c*x))*abs(a) + a*sqrt(c)))/(a^3*sgn(x))