3.12.12 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx\) [1112]

Optimal. Leaf size=52 \[ \frac {2 (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6286, 1819, 272, 65, 214} \begin {gather*} \frac {2 (a x+1)}{\sqrt {c-a^2 c x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 214

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

Rule 272

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 1819

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 6286

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 \sqrt {c-a^2 c x^2}} \, dx &=c \int \frac {(1+a x)^2}{x \left (c-a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac {2 (1+a x)}{\sqrt {c-a^2 c x^2}}+\int \frac {1}{x \sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {2 (1+a x)}{\sqrt {c-a^2 c x^2}}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \sqrt {c-a^2 c x}} \, dx,x,x^2\right )\\ &=\frac {2 (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\text {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^2 c}\\ &=\frac {2 (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\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 = 0.18, size = 66, normalized size = 1.27 \begin {gather*} \frac {2 \sqrt {c-a^2 c x^2}}{c-a c x}+\frac {\log (x)}{\sqrt {c}}-\frac {\log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right )}{\sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 80, normalized size = 1.54

method	result	size
default	\(-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{\sqrt {c}}-\frac {2 \sqrt {-c \,a^{2} \left (x -\frac {1}{a}\right )^{2}-2 c a \left (x -\frac {1}{a}\right )}}{a c \left (x -\frac {1}{a}\right )}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 0.38, size = 147, normalized size = 2.83 \begin {gather*} \left [\frac {{\left (a x - 1\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 ) - 4 \, \sqrt {-a^{2} c x^{2} + c}}{2 \, {\left (a c x - c\right )}}, -\frac {{\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + 2 \, \sqrt {-a^{2} c x^{2} + c}}{a c x - c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {a x}{a x^{2} \sqrt {- a^{2} c x^{2} + c} - x \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {1}{a x^{2} \sqrt {- a^{2} c x^{2} + c} - x \sqrt {- a^{2} c x^{2} + c}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} -\int \frac {{\left (a\,x+1\right )}^2}{x\,\sqrt {c-a^2\,c\,x^2}\,\left (a^2\,x^2-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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