3.88 \(\int \frac{e^{\frac{5}{2} \tanh ^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=165 \[ \frac{287 a^3 \sqrt [4]{a x+1}}{24 \sqrt [4]{1-a x}}-\frac{61 a^2 \sqrt [4]{a x+1}}{24 x \sqrt [4]{1-a x}}-\frac{55}{8} a^3 \tan ^{-1}\left (\frac{\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac{55}{8} a^3 \tanh ^{-1}\left (\frac{\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac{13 a \sqrt [4]{a x+1}}{12 x^2 \sqrt [4]{1-a x}}-\frac{\sqrt [4]{a x+1}}{3 x^3 \sqrt [4]{1-a x}} \]

[Out]

(287*a^3*(1 + a*x)^(1/4))/(24*(1 - a*x)^(1/4)) - (1 + a*x)^(1/4)/(3*x^3*(1 - a*x)^(1/4)) - (13*a*(1 + a*x)^(1/
4))/(12*x^2*(1 - a*x)^(1/4)) - (61*a^2*(1 + a*x)^(1/4))/(24*x*(1 - a*x)^(1/4)) - (55*a^3*ArcTan[(1 + a*x)^(1/4
)/(1 - a*x)^(1/4)])/8 - (55*a^3*ArcTanh[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)])/8

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Rubi [A]  time = 0.0787501, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {6126, 98, 151, 155, 12, 93, 212, 206, 203} \[ \frac{287 a^3 \sqrt [4]{a x+1}}{24 \sqrt [4]{1-a x}}-\frac{61 a^2 \sqrt [4]{a x+1}}{24 x \sqrt [4]{1-a x}}-\frac{55}{8} a^3 \tan ^{-1}\left (\frac{\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac{55}{8} a^3 \tanh ^{-1}\left (\frac{\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac{13 a \sqrt [4]{a x+1}}{12 x^2 \sqrt [4]{1-a x}}-\frac{\sqrt [4]{a x+1}}{3 x^3 \sqrt [4]{1-a x}} \]

Antiderivative was successfully verified.

[In]

Int[E^((5*ArcTanh[a*x])/2)/x^4,x]

[Out]

(287*a^3*(1 + a*x)^(1/4))/(24*(1 - a*x)^(1/4)) - (1 + a*x)^(1/4)/(3*x^3*(1 - a*x)^(1/4)) - (13*a*(1 + a*x)^(1/
4))/(12*x^2*(1 - a*x)^(1/4)) - (61*a^2*(1 + a*x)^(1/4))/(24*x*(1 - a*x)^(1/4)) - (55*a^3*ArcTan[(1 + a*x)^(1/4
)/(1 - a*x)^(1/4)])/8 - (55*a^3*ArcTanh[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)])/8

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 98

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

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 155

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1])))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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^4} \, dx &=\int \frac{(1+a x)^{5/4}}{x^4 (1-a x)^{5/4}} \, dx\\ &=-\frac{\sqrt [4]{1+a x}}{3 x^3 \sqrt [4]{1-a x}}-\frac{1}{3} \int \frac{-\frac{13 a}{2}-6 a^2 x}{x^3 (1-a x)^{5/4} (1+a x)^{3/4}} \, dx\\ &=-\frac{\sqrt [4]{1+a x}}{3 x^3 \sqrt [4]{1-a x}}-\frac{13 a \sqrt [4]{1+a x}}{12 x^2 \sqrt [4]{1-a x}}+\frac{1}{6} \int \frac{\frac{61 a^2}{4}+13 a^3 x}{x^2 (1-a x)^{5/4} (1+a x)^{3/4}} \, dx\\ &=-\frac{\sqrt [4]{1+a x}}{3 x^3 \sqrt [4]{1-a x}}-\frac{13 a \sqrt [4]{1+a x}}{12 x^2 \sqrt [4]{1-a x}}-\frac{61 a^2 \sqrt [4]{1+a x}}{24 x \sqrt [4]{1-a x}}-\frac{1}{6} \int \frac{-\frac{165 a^3}{8}-\frac{61 a^4 x}{4}}{x (1-a x)^{5/4} (1+a x)^{3/4}} \, dx\\ &=\frac{287 a^3 \sqrt [4]{1+a x}}{24 \sqrt [4]{1-a x}}-\frac{\sqrt [4]{1+a x}}{3 x^3 \sqrt [4]{1-a x}}-\frac{13 a \sqrt [4]{1+a x}}{12 x^2 \sqrt [4]{1-a x}}-\frac{61 a^2 \sqrt [4]{1+a x}}{24 x \sqrt [4]{1-a x}}+\frac{\int \frac{165 a^4}{16 x \sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx}{3 a}\\ &=\frac{287 a^3 \sqrt [4]{1+a x}}{24 \sqrt [4]{1-a x}}-\frac{\sqrt [4]{1+a x}}{3 x^3 \sqrt [4]{1-a x}}-\frac{13 a \sqrt [4]{1+a x}}{12 x^2 \sqrt [4]{1-a x}}-\frac{61 a^2 \sqrt [4]{1+a x}}{24 x \sqrt [4]{1-a x}}+\frac{1}{16} \left (55 a^3\right ) \int \frac{1}{x \sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx\\ &=\frac{287 a^3 \sqrt [4]{1+a x}}{24 \sqrt [4]{1-a x}}-\frac{\sqrt [4]{1+a x}}{3 x^3 \sqrt [4]{1-a x}}-\frac{13 a \sqrt [4]{1+a x}}{12 x^2 \sqrt [4]{1-a x}}-\frac{61 a^2 \sqrt [4]{1+a x}}{24 x \sqrt [4]{1-a x}}+\frac{1}{4} \left (55 a^3\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{287 a^3 \sqrt [4]{1+a x}}{24 \sqrt [4]{1-a x}}-\frac{\sqrt [4]{1+a x}}{3 x^3 \sqrt [4]{1-a x}}-\frac{13 a \sqrt [4]{1+a x}}{12 x^2 \sqrt [4]{1-a x}}-\frac{61 a^2 \sqrt [4]{1+a x}}{24 x \sqrt [4]{1-a x}}-\frac{1}{8} \left (55 a^3\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}{8} \left (55 a^3\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{287 a^3 \sqrt [4]{1+a x}}{24 \sqrt [4]{1-a x}}-\frac{\sqrt [4]{1+a x}}{3 x^3 \sqrt [4]{1-a x}}-\frac{13 a \sqrt [4]{1+a x}}{12 x^2 \sqrt [4]{1-a x}}-\frac{61 a^2 \sqrt [4]{1+a x}}{24 x \sqrt [4]{1-a x}}-\frac{55}{8} a^3 \tan ^{-1}\left (\frac{\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )-\frac{55}{8} a^3 \tanh ^{-1}\left (\frac{\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0282137, size = 91, normalized size = 0.55 \[ \frac{110 a^3 x^3 (a x-1) \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},\frac{1-a x}{a x+1}\right )+287 a^4 x^4+226 a^3 x^3-87 a^2 x^2-34 a x-8}{24 x^3 \sqrt [4]{1-a x} (a x+1)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((5*ArcTanh[a*x])/2)/x^4,x]

[Out]

(-8 - 34*a*x - 87*a^2*x^2 + 226*a^3*x^3 + 287*a^4*x^4 + 110*a^3*x^3*(-1 + a*x)*Hypergeometric2F1[3/4, 1, 7/4,
(1 - a*x)/(1 + a*x)])/(24*x^3*(1 - a*x)^(1/4)*(1 + a*x)^(3/4))

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Maple [F]  time = 0.117, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \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^4,x)

[Out]

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^4,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^{4}}\,{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^4,x, algorithm="maxima")

[Out]

integrate(((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)/x^4, x)

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Fricas [A]  time = 2.0049, size = 360, normalized size = 2.18 \begin{align*} -\frac{330 \, a^{3} x^{3} \arctan \left (\sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}}\right ) + 165 \, a^{3} x^{3} \log \left (\sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) - 165 \, a^{3} x^{3} \log \left (\sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) - 2 \,{\left (287 \, a^{3} x^{3} - 61 \, a^{2} x^{2} - 26 \, a x - 8\right )} \sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}}}{48 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

-1/48*(330*a^3*x^3*arctan(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) + 165*a^3*x^3*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x
- 1)) + 1) - 165*a^3*x^3*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - 1) - 2*(287*a^3*x^3 - 61*a^2*x^2 - 26*a*x
- 8)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)))/x^3

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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**4,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^{4}}\,{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^4,x, algorithm="giac")

[Out]

integrate(((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)/x^4, x)