3.1113 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=77 \[ \frac {2 a (a x+1)}{\sqrt {c-a^2 c x^2}}-\frac {\sqrt {c-a^2 c x^2}}{c x}-\frac {2 a \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6151, 1805, 807, 266, 63, 208} \[ \frac {2 a (a x+1)}{\sqrt {c-a^2 c x^2}}-\frac {\sqrt {c-a^2 c x^2}}{c x}-\frac {2 a \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 78, normalized size = 1.01 \[ \frac {(1-3 a x) \sqrt {c-a^2 c x^2}}{c x (a x-1)}-\frac {2 a \log \left (\sqrt {c} \sqrt {c-a^2 c x^2}+c\right )}{\sqrt {c}}+\frac {2 a \log (x)}{\sqrt {c}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 178, normalized size = 2.31 \[ \left [\frac {{\left (a^{2} x^{2} - a x\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (3 \, a x - 1\right )}}{a c x^{2} - c x}, -\frac {2 \, {\left (a^{2} x^{2} - a x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + \sqrt {-a^{2} c x^{2} + c} {\left (3 \, a x - 1\right )}}{a c x^{2} - c x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [A]  time = 0.04, size = 99, normalized size = 1.29 \[ -\frac {2 a \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{\sqrt {c}}-\frac {\sqrt {-a^{2} c \,x^{2}+c}}{c x}-\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}{c \left (x -\frac {1}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {a x}{a x^{3} \sqrt {- a^{2} c x^{2} + c} - x^{2} \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {1}{a x^{3} \sqrt {- a^{2} c x^{2} + c} - x^{2} \sqrt {- a^{2} c x^{2} + c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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