3.983 \(\int \frac{e^{\tanh ^{-1}(a x)}}{x (c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=252 \[ \frac{\sqrt{1-a^2 x^2}}{2 c^2 (1-a x) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 c^2 (a x+1) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2} \log (x)}{c^2 \sqrt{c-a^2 c x^2}}-\frac{11 \sqrt{1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt{c-a^2 c x^2}}-\frac{5 \sqrt{1-a^2 x^2} \log (a x+1)}{16 c^2 \sqrt{c-a^2 c x^2}} \]

[Out]

Sqrt[1 - a^2*x^2]/(8*c^2*(1 - a*x)^2*Sqrt[c - a^2*c*x^2]) + Sqrt[1 - a^2*x^2]/(2*c^2*(1 - a*x)*Sqrt[c - a^2*c*
x^2]) + Sqrt[1 - a^2*x^2]/(8*c^2*(1 + a*x)*Sqrt[c - a^2*c*x^2]) + (Sqrt[1 - a^2*x^2]*Log[x])/(c^2*Sqrt[c - a^2
*c*x^2]) - (11*Sqrt[1 - a^2*x^2]*Log[1 - a*x])/(16*c^2*Sqrt[c - a^2*c*x^2]) - (5*Sqrt[1 - a^2*x^2]*Log[1 + a*x
])/(16*c^2*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.243718, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {6153, 6150, 88} \[ \frac{\sqrt{1-a^2 x^2}}{2 c^2 (1-a x) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 c^2 (a x+1) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2} \log (x)}{c^2 \sqrt{c-a^2 c x^2}}-\frac{11 \sqrt{1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt{c-a^2 c x^2}}-\frac{5 \sqrt{1-a^2 x^2} \log (a x+1)}{16 c^2 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(x*(c - a^2*c*x^2)^(5/2)),x]

[Out]

Sqrt[1 - a^2*x^2]/(8*c^2*(1 - a*x)^2*Sqrt[c - a^2*c*x^2]) + Sqrt[1 - a^2*x^2]/(2*c^2*(1 - a*x)*Sqrt[c - a^2*c*
x^2]) + Sqrt[1 - a^2*x^2]/(8*c^2*(1 + a*x)*Sqrt[c - a^2*c*x^2]) + (Sqrt[1 - a^2*x^2]*Log[x])/(c^2*Sqrt[c - a^2
*c*x^2]) - (11*Sqrt[1 - a^2*x^2]*Log[1 - a*x])/(16*c^2*Sqrt[c - a^2*c*x^2]) - (5*Sqrt[1 - a^2*x^2]*Log[1 + a*x
])/(16*c^2*Sqrt[c - a^2*c*x^2])

Rule 6153

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

Rule 6150

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx &=\frac{\sqrt{1-a^2 x^2} \int \frac{e^{\tanh ^{-1}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2 \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \frac{1}{x (1-a x)^3 (1+a x)^2} \, dx}{c^2 \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \left (\frac{1}{x}-\frac{a}{4 (-1+a x)^3}+\frac{a}{2 (-1+a x)^2}-\frac{11 a}{16 (-1+a x)}-\frac{a}{8 (1+a x)^2}-\frac{5 a}{16 (1+a x)}\right ) \, dx}{c^2 \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{2 c^2 (1-a x) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 c^2 (1+a x) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2} \log (x)}{c^2 \sqrt{c-a^2 c x^2}}-\frac{11 \sqrt{1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt{c-a^2 c x^2}}-\frac{5 \sqrt{1-a^2 x^2} \log (1+a x)}{16 c^2 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0725986, size = 86, normalized size = 0.34 \[ \frac{\sqrt{1-a^2 x^2} \left (\frac{8}{1-a x}+\frac{2}{a x+1}+\frac{2}{(a x-1)^2}-11 \log (1-a x)-5 \log (a x+1)+16 \log (x)\right )}{16 c^2 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(x*(c - a^2*c*x^2)^(5/2)),x]

[Out]

(Sqrt[1 - a^2*x^2]*(8/(1 - a*x) + 2/(-1 + a*x)^2 + 2/(1 + a*x) + 16*Log[x] - 11*Log[1 - a*x] - 5*Log[1 + a*x])
)/(16*c^2*Sqrt[c - a^2*c*x^2])

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Maple [A]  time = 0.093, size = 193, normalized size = 0.8 \begin{align*} -{\frac{16\,{a}^{3}\ln \left ( x \right ){x}^{3}-5\,{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) -11\,\ln \left ( ax-1 \right ){x}^{3}{a}^{3}-16\,{a}^{2}\ln \left ( x \right ){x}^{2}+5\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}+11\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}-6\,{a}^{2}{x}^{2}-16\,a\ln \left ( x \right ) x+5\,ax\ln \left ( ax+1 \right ) +11\,\ln \left ( ax-1 \right ) xa-2\,ax+16\,\ln \left ( x \right ) -5\,\ln \left ( ax+1 \right ) -11\,\ln \left ( ax-1 \right ) +12}{ \left ( 16\,{a}^{2}{x}^{2}-16 \right ){c}^{3} \left ( ax-1 \right ) ^{2} \left ( ax+1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x)

[Out]

-1/16*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(16*a^3*ln(x)*x^3-5*a^3*x^3*ln(a*x+1)-11*ln(a*x-1)*x^3*a^3-16*
a^2*ln(x)*x^2+5*ln(a*x+1)*a^2*x^2+11*ln(a*x-1)*a^2*x^2-6*a^2*x^2-16*a*ln(x)*x+5*a*x*ln(a*x+1)+11*ln(a*x-1)*x*a
-2*a*x+16*ln(x)-5*ln(a*x+1)-11*ln(a*x-1)+12)/(a^2*x^2-1)/c^3/(a*x-1)^2/(a*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}} \sqrt{-a^{2} x^{2} + 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{a^{7} c^{3} x^{8} - a^{6} c^{3} x^{7} - 3 \, a^{5} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{5} + 3 \, a^{3} c^{3} x^{4} - 3 \, a^{2} c^{3} x^{3} - a c^{3} x^{2} + c^{3} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^7*c^3*x^8 - a^6*c^3*x^7 - 3*a^5*c^3*x^6 + 3*a^4*c^3*x^5 +
3*a^3*c^3*x^4 - 3*a^2*c^3*x^3 - a*c^3*x^2 + c^3*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{x \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x/(-a**2*c*x**2+c)**(5/2),x)

[Out]

Integral((a*x + 1)/(x*sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(5/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}} \sqrt{-a^{2} x^{2} + 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)*x), x)