3.87 \(\int \frac{e^{\frac{5}{2} \tanh ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=136 \[ \frac{25 a^2 \sqrt [4]{a x+1}}{2 \sqrt [4]{1-a x}}-\frac{25}{4} a^2 \tan ^{-1}\left (\frac{\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac{25}{4} a^2 \tanh ^{-1}\left (\frac{\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac{(a x+1)^{9/4}}{2 x^2 \sqrt [4]{1-a x}}-\frac{5 a (a x+1)^{5/4}}{4 x \sqrt [4]{1-a x}} \]

[Out]

(25*a^2*(1 + a*x)^(1/4))/(2*(1 - a*x)^(1/4)) - (5*a*(1 + a*x)^(5/4))/(4*x*(1 - a*x)^(1/4)) - (1 + a*x)^(9/4)/(
2*x^2*(1 - a*x)^(1/4)) - (25*a^2*ArcTan[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)])/4 - (25*a^2*ArcTanh[(1 + a*x)^(1/4)/
(1 - a*x)^(1/4)])/4

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Rubi [A]  time = 0.0485649, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6126, 96, 94, 93, 212, 206, 203} \[ \frac{25 a^2 \sqrt [4]{a x+1}}{2 \sqrt [4]{1-a x}}-\frac{25}{4} a^2 \tan ^{-1}\left (\frac{\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac{25}{4} a^2 \tanh ^{-1}\left (\frac{\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac{(a x+1)^{9/4}}{2 x^2 \sqrt [4]{1-a x}}-\frac{5 a (a x+1)^{5/4}}{4 x \sqrt [4]{1-a x}} \]

Antiderivative was successfully verified.

[In]

Int[E^((5*ArcTanh[a*x])/2)/x^3,x]

[Out]

(25*a^2*(1 + a*x)^(1/4))/(2*(1 - a*x)^(1/4)) - (5*a*(1 + a*x)^(5/4))/(4*x*(1 - a*x)^(1/4)) - (1 + a*x)^(9/4)/(
2*x^2*(1 - a*x)^(1/4)) - (25*a^2*ArcTan[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)])/4 - (25*a^2*ArcTanh[(1 + a*x)^(1/4)/
(1 - a*x)^(1/4)])/4

Rule 6126

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{e^{\frac{5}{2} \tanh ^{-1}(a x)}}{x^3} \, dx &=\int \frac{(1+a x)^{5/4}}{x^3 (1-a x)^{5/4}} \, dx\\ &=-\frac{(1+a x)^{9/4}}{2 x^2 \sqrt [4]{1-a x}}+\frac{1}{4} (5 a) \int \frac{(1+a x)^{5/4}}{x^2 (1-a x)^{5/4}} \, dx\\ &=-\frac{5 a (1+a x)^{5/4}}{4 x \sqrt [4]{1-a x}}-\frac{(1+a x)^{9/4}}{2 x^2 \sqrt [4]{1-a x}}+\frac{1}{8} \left (25 a^2\right ) \int \frac{\sqrt [4]{1+a x}}{x (1-a x)^{5/4}} \, dx\\ &=\frac{25 a^2 \sqrt [4]{1+a x}}{2 \sqrt [4]{1-a x}}-\frac{5 a (1+a x)^{5/4}}{4 x \sqrt [4]{1-a x}}-\frac{(1+a x)^{9/4}}{2 x^2 \sqrt [4]{1-a x}}+\frac{1}{8} \left (25 a^2\right ) \int \frac{1}{x \sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx\\ &=\frac{25 a^2 \sqrt [4]{1+a x}}{2 \sqrt [4]{1-a x}}-\frac{5 a (1+a x)^{5/4}}{4 x \sqrt [4]{1-a x}}-\frac{(1+a x)^{9/4}}{2 x^2 \sqrt [4]{1-a x}}+\frac{1}{2} \left (25 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^4} \, dx,x,\frac{\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )\\ &=\frac{25 a^2 \sqrt [4]{1+a x}}{2 \sqrt [4]{1-a x}}-\frac{5 a (1+a x)^{5/4}}{4 x \sqrt [4]{1-a x}}-\frac{(1+a x)^{9/4}}{2 x^2 \sqrt [4]{1-a x}}-\frac{1}{4} \left (25 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )-\frac{1}{4} \left (25 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )\\ &=\frac{25 a^2 \sqrt [4]{1+a x}}{2 \sqrt [4]{1-a x}}-\frac{5 a (1+a x)^{5/4}}{4 x \sqrt [4]{1-a x}}-\frac{(1+a x)^{9/4}}{2 x^2 \sqrt [4]{1-a x}}-\frac{25}{4} a^2 \tan ^{-1}\left (\frac{\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )-\frac{25}{4} a^2 \tanh ^{-1}\left (\frac{\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0235652, size = 86, normalized size = 0.63 \[ \frac{50 a^2 x^2 (a x-1) \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},\frac{1-a x}{a x+1}\right )+3 \left (43 a^3 x^3+34 a^2 x^2-11 a x-2\right )}{12 x^2 \sqrt [4]{1-a x} (a x+1)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((5*ArcTanh[a*x])/2)/x^3,x]

[Out]

(3*(-2 - 11*a*x + 34*a^2*x^2 + 43*a^3*x^3) + 50*a^2*x^2*(-1 + a*x)*Hypergeometric2F1[3/4, 1, 7/4, (1 - a*x)/(1
+ a*x)])/(12*x^2*(1 - a*x)^(1/4)*(1 + a*x)^(3/4))

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Maple [F]  time = 0.109, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) ^{{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x)

[Out]

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}\right )^{\frac{5}{2}}}{x^{3}}\,{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))^(5/2)/x^3,x, algorithm="maxima")

[Out]

integrate(((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)/x^3, x)

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Fricas [A]  time = 2.12449, size = 335, normalized size = 2.46 \begin{align*} -\frac{50 \, a^{2} x^{2} \arctan \left (\sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}}\right ) + 25 \, a^{2} x^{2} \log \left (\sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) - 25 \, a^{2} x^{2} \log \left (\sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) - 2 \,{\left (43 \, a^{2} x^{2} - 9 \, a x - 2\right )} \sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}}}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x, algorithm="fricas")

[Out]

-1/8*(50*a^2*x^2*arctan(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) + 25*a^2*x^2*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x -
1)) + 1) - 25*a^2*x^2*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - 1) - 2*(43*a^2*x^2 - 9*a*x - 2)*sqrt(-sqrt(-a^
2*x^2 + 1)/(a*x - 1)))/x^2

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x+1)/(-a**2*x**2+1)**(1/2))**(5/2)/x**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}\right )^{\frac{5}{2}}}{x^{3}}\,{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))^(5/2)/x^3,x, algorithm="giac")

[Out]

integrate(((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)/x^3, x)