3.680 \(\int \frac{e^{2 \coth ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x^4} \, dx\)

Optimal. Leaf size=99 \[ \frac{5 a^2 \sqrt{c-a^2 c x^2}}{3 x}+\frac{a \sqrt{c-a^2 c x^2}}{x^2}+\frac{\sqrt{c-a^2 c x^2}}{3 x^3}+a^3 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right ) \]

[Out]

Sqrt[c - a^2*c*x^2]/(3*x^3) + (a*Sqrt[c - a^2*c*x^2])/x^2 + (5*a^2*Sqrt[c - a^2*c*x^2])/(3*x) + a^3*Sqrt[c]*Ar
cTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]]

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Rubi [A]  time = 0.37803, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {6167, 6151, 1807, 835, 807, 266, 63, 208} \[ \frac{5 a^2 \sqrt{c-a^2 c x^2}}{3 x}+\frac{a \sqrt{c-a^2 c x^2}}{x^2}+\frac{\sqrt{c-a^2 c x^2}}{3 x^3}+a^3 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcCoth[a*x])*Sqrt[c - a^2*c*x^2])/x^4,x]

[Out]

Sqrt[c - a^2*c*x^2]/(3*x^3) + (a*Sqrt[c - a^2*c*x^2])/x^2 + (5*a^2*Sqrt[c - a^2*c*x^2])/(3*x) + a^3*Sqrt[c]*Ar
cTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]]

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6151

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

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

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

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 \coth ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x^4} \, dx &=-\int \frac{e^{2 \tanh ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x^4} \, dx\\ &=-\left (c \int \frac{(1+a x)^2}{x^4 \sqrt{c-a^2 c x^2}} \, dx\right )\\ &=\frac{\sqrt{c-a^2 c x^2}}{3 x^3}+\frac{1}{3} \int \frac{-6 a c-5 a^2 c x}{x^3 \sqrt{c-a^2 c x^2}} \, dx\\ &=\frac{\sqrt{c-a^2 c x^2}}{3 x^3}+\frac{a \sqrt{c-a^2 c x^2}}{x^2}-\frac{\int \frac{10 a^2 c^2+6 a^3 c^2 x}{x^2 \sqrt{c-a^2 c x^2}} \, dx}{6 c}\\ &=\frac{\sqrt{c-a^2 c x^2}}{3 x^3}+\frac{a \sqrt{c-a^2 c x^2}}{x^2}+\frac{5 a^2 \sqrt{c-a^2 c x^2}}{3 x}-\left (a^3 c\right ) \int \frac{1}{x \sqrt{c-a^2 c x^2}} \, dx\\ &=\frac{\sqrt{c-a^2 c x^2}}{3 x^3}+\frac{a \sqrt{c-a^2 c x^2}}{x^2}+\frac{5 a^2 \sqrt{c-a^2 c x^2}}{3 x}-\frac{1}{2} \left (a^3 c\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c-a^2 c x}} \, dx,x,x^2\right )\\ &=\frac{\sqrt{c-a^2 c x^2}}{3 x^3}+\frac{a \sqrt{c-a^2 c x^2}}{x^2}+\frac{5 a^2 \sqrt{c-a^2 c x^2}}{3 x}+a \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2 c}} \, dx,x,\sqrt{c-a^2 c x^2}\right )\\ &=\frac{\sqrt{c-a^2 c x^2}}{3 x^3}+\frac{a \sqrt{c-a^2 c x^2}}{x^2}+\frac{5 a^2 \sqrt{c-a^2 c x^2}}{3 x}+a^3 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.115516, size = 82, normalized size = 0.83 \[ \frac{\left (5 a^2 x^2+3 a x+1\right ) \sqrt{c-a^2 c x^2}}{3 x^3}+a^3 \sqrt{c} \log \left (\sqrt{c} \sqrt{c-a^2 c x^2}+c\right )+a^3 \left (-\sqrt{c}\right ) \log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(2*ArcCoth[a*x])*Sqrt[c - a^2*c*x^2])/x^4,x]

[Out]

((1 + 3*a*x + 5*a^2*x^2)*Sqrt[c - a^2*c*x^2])/(3*x^3) - a^3*Sqrt[c]*Log[x] + a^3*Sqrt[c]*Log[c + Sqrt[c]*Sqrt[
c - a^2*c*x^2]]

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Maple [B]  time = 0.065, size = 261, normalized size = 2.6 \begin{align*} 2\,{\frac{{a}^{2} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{3/2}}{cx}}+2\,{a}^{4}x\sqrt{-{a}^{2}c{x}^{2}+c}+2\,{\frac{{a}^{4}c}{\sqrt{{a}^{2}c}}\arctan \left ({\frac{\sqrt{{a}^{2}c}x}{\sqrt{-{a}^{2}c{x}^{2}+c}}} \right ) }+\sqrt{c}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{-{a}^{2}c{x}^{2}+c} \right ) } \right ){a}^{3}-\sqrt{-{a}^{2}c{x}^{2}+c}{a}^{3}+2\,{a}^{3}\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }-2\,{\frac{{a}^{4}c}{\sqrt{{a}^{2}c}}\arctan \left ({\sqrt{{a}^{2}c}x{\frac{1}{\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}}} \right ) }+{\frac{a}{c{x}^{2}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{1}{3\,c{x}^{3}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(a*x-1)*(-a^2*c*x^2+c)^(1/2)/x^4,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} c x^{2} + c}{\left (a x + 1\right )}}{{\left (a x - 1\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*(a*x + 1)/((a*x - 1)*x^4), x)

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Fricas [A]  time = 1.6548, size = 370, normalized size = 3.74 \begin{align*} \left [\frac{3 \, a^{3} \sqrt{c} x^{3} \log \left (-\frac{a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c x^{2} + c} \sqrt{c} - 2 \, c}{x^{2}}\right ) + 2 \, \sqrt{-a^{2} c x^{2} + c}{\left (5 \, a^{2} x^{2} + 3 \, a x + 1\right )}}{6 \, x^{3}}, \frac{3 \, a^{3} \sqrt{-c} x^{3} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} \sqrt{-c}}{a^{2} c x^{2} - c}\right ) + \sqrt{-a^{2} c x^{2} + c}{\left (5 \, a^{2} x^{2} + 3 \, a x + 1\right )}}{3 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[1/6*(3*a^3*sqrt(c)*x^3*log(-(a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) + 2*sqrt(-a^2*c*x^2 + c)*
(5*a^2*x^2 + 3*a*x + 1))/x^3, 1/3*(3*a^3*sqrt(-c)*x^3*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/(a^2*c*x^2 - c)) +
sqrt(-a^2*c*x^2 + c)*(5*a^2*x^2 + 3*a*x + 1))/x^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- c \left (a x - 1\right ) \left (a x + 1\right )} \left (a x + 1\right )}{x^{4} \left (a x - 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)*(-a**2*c*x**2+c)**(1/2)/x**4,x)

[Out]

Integral(sqrt(-c*(a*x - 1)*(a*x + 1))*(a*x + 1)/(x**4*(a*x - 1)), x)

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Giac [B]  time = 1.17871, size = 338, normalized size = 3.41 \begin{align*} -\frac{2 \, a^{3} c \arctan \left (-\frac{\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} + \frac{2 \,{\left (3 \,{\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{5} a^{3} c - 3 \,{\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{4} a^{2} \sqrt{-c} c{\left | a \right |} + 12 \,{\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{2} a^{2} \sqrt{-c} c^{2}{\left | a \right |} - 3 \,{\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )} a^{3} c^{3} - 5 \, a^{2} \sqrt{-c} c^{3}{\left | a \right |}\right )}}{3 \,{\left ({\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm="giac")

[Out]

-2*a^3*c*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) + 2/3*(3*(sqrt(-a^2*c)*x - sqrt(-a
^2*c*x^2 + c))^5*a^3*c - 3*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^4*a^2*sqrt(-c)*c*abs(a) + 12*(sqrt(-a^2*c)*
x - sqrt(-a^2*c*x^2 + c))^2*a^2*sqrt(-c)*c^2*abs(a) - 3*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))*a^3*c^3 - 5*a^
2*sqrt(-c)*c^3*abs(a))/((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2 - c)^3