3.1.87 \(\int \frac {e^{\frac {5}{2} \tanh ^{-1}(a x)}}{x^3} \, dx\) [87]

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

[Out]

25/2*a^2*(a*x+1)^(1/4)/(-a*x+1)^(1/4)-5/4*a*(a*x+1)^(5/4)/x/(-a*x+1)^(1/4)-1/2*(a*x+1)^(9/4)/x^2/(-a*x+1)^(1/4
)-25/4*a^2*arctan((a*x+1)^(1/4)/(-a*x+1)^(1/4))-25/4*a^2*arctanh((a*x+1)^(1/4)/(-a*x+1)^(1/4))

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Rubi [A]
time = 0.03, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6261, 98, 96, 95, 218, 212, 209} \begin {gather*} -\frac {25}{4} a^2 \text {ArcTan}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )+\frac {25 a^2 \sqrt [4]{a x+1}}{2 \sqrt [4]{1-a x}}-\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}} \end {gather*}

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 95

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 96

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[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 98

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 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 6261

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]

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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 86, normalized size = 0.63 \begin {gather*} \frac {3 \left (-2-11 a x+34 a^2 x^2+43 a^3 x^3\right )+50 a^2 x^2 (-1+a x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {1-a x}{1+a x}\right )}{12 x^2 \sqrt [4]{1-a x} (1+a x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )^{\frac {5}{2}}}{x^{3}}\, dx\]

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 = 0.00, 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)/(-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 = 0.35, size = 145, normalized size = 1.07 \begin {gather*} -\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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{5/2}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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