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

Optimal. Leaf size=125 \[ -\frac{(a x+1)^2}{5 a c^5 \left (1-a^2 x^2\right )^{5/2}}+\frac{22 (a x+1)}{15 a c^5 \left (1-a^2 x^2\right )^{3/2}}-\frac{\sqrt{1-a^2 x^2}}{a c^5}-\frac{2 (23 a x+30)}{15 a c^5 \sqrt{1-a^2 x^2}}+\frac{2 \sin ^{-1}(a x)}{a c^5} \]

[Out]

-(1 + a*x)^2/(5*a*c^5*(1 - a^2*x^2)^(5/2)) + (22*(1 + a*x))/(15*a*c^5*(1 - a^2*x^2)^(3/2)) - (2*(30 + 23*a*x))
/(15*a*c^5*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(a*c^5) + (2*ArcSin[a*x])/(a*c^5)

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Rubi [A]  time = 0.317702, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {6131, 6128, 852, 1635, 1814, 641, 216} \[ -\frac{(a x+1)^2}{5 a c^5 \left (1-a^2 x^2\right )^{5/2}}+\frac{22 (a x+1)}{15 a c^5 \left (1-a^2 x^2\right )^{3/2}}-\frac{\sqrt{1-a^2 x^2}}{a c^5}-\frac{2 (23 a x+30)}{15 a c^5 \sqrt{1-a^2 x^2}}+\frac{2 \sin ^{-1}(a x)}{a c^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - c/(a*x))^5),x]

[Out]

-(1 + a*x)^2/(5*a*c^5*(1 - a^2*x^2)^(5/2)) + (22*(1 + a*x))/(15*a*c^5*(1 - a^2*x^2)^(3/2)) - (2*(30 + 23*a*x))
/(15*a*c^5*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(a*c^5) + (2*ArcSin[a*x])/(a*c^5)

Rule 6131

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

Rule 6128

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

Rule 852

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

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[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[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 216

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

Rubi steps

\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a x}\right )^5} \, dx &=-\frac{a^5 \int \frac{e^{-3 \tanh ^{-1}(a x)} x^5}{(1-a x)^5} \, dx}{c^5}\\ &=-\frac{a^5 \int \frac{x^5}{(1-a x)^2 \left (1-a^2 x^2\right )^{3/2}} \, dx}{c^5}\\ &=-\frac{a^5 \int \frac{x^5 (1+a x)^2}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^5}\\ &=-\frac{(1+a x)^2}{5 a c^5 \left (1-a^2 x^2\right )^{5/2}}+\frac{a^5 \int \frac{(1+a x) \left (\frac{2}{a^5}+\frac{5 x}{a^4}+\frac{5 x^2}{a^3}+\frac{5 x^3}{a^2}+\frac{5 x^4}{a}\right )}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{5 c^5}\\ &=-\frac{(1+a x)^2}{5 a c^5 \left (1-a^2 x^2\right )^{5/2}}+\frac{22 (1+a x)}{15 a c^5 \left (1-a^2 x^2\right )^{3/2}}-\frac{a^5 \int \frac{\frac{16}{a^5}+\frac{45 x}{a^4}+\frac{30 x^2}{a^3}+\frac{15 x^3}{a^2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{15 c^5}\\ &=-\frac{(1+a x)^2}{5 a c^5 \left (1-a^2 x^2\right )^{5/2}}+\frac{22 (1+a x)}{15 a c^5 \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (30+23 a x)}{15 a c^5 \sqrt{1-a^2 x^2}}+\frac{a^5 \int \frac{\frac{30}{a^5}+\frac{15 x}{a^4}}{\sqrt{1-a^2 x^2}} \, dx}{15 c^5}\\ &=-\frac{(1+a x)^2}{5 a c^5 \left (1-a^2 x^2\right )^{5/2}}+\frac{22 (1+a x)}{15 a c^5 \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (30+23 a x)}{15 a c^5 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{a c^5}+\frac{2 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c^5}\\ &=-\frac{(1+a x)^2}{5 a c^5 \left (1-a^2 x^2\right )^{5/2}}+\frac{22 (1+a x)}{15 a c^5 \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (30+23 a x)}{15 a c^5 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{a c^5}+\frac{2 \sin ^{-1}(a x)}{a c^5}\\ \end{align*}

Mathematica [A]  time = 0.215928, size = 76, normalized size = 0.61 \[ \frac{\frac{\sqrt{1-a^2 x^2} \left (-15 a^4 x^4+76 a^3 x^3-32 a^2 x^2-82 a x+56\right )}{(a x-1)^3 (a x+1)}+30 \sin ^{-1}(a x)}{15 a c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a*x))^5),x]

[Out]

((Sqrt[1 - a^2*x^2]*(56 - 82*a*x - 32*a^2*x^2 + 76*a^3*x^3 - 15*a^4*x^4))/((-1 + a*x)^3*(1 + a*x)) + 30*ArcSin
[a*x])/(15*a*c^5)

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Maple [B]  time = 0.086, size = 468, normalized size = 3.7 \begin{align*} -{\frac{187}{128\,a{c}^{5}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{7}{48\,{a}^{5}{c}^{5}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{a}^{-1} \right ) ^{-4}}+{\frac{31}{48\,{a}^{4}{c}^{5}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{a}^{-1} \right ) ^{-3}}-{\frac{139}{96\,{a}^{3}{c}^{5}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{561\,x}{256\,{c}^{5}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}+{\frac{561}{256\,{c}^{5}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{1}{32\,{a}^{4}{c}^{5} \left ( x+{a}^{-1} \right ) ^{3}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{9}{64\,{a}^{3}{c}^{5} \left ( x+{a}^{-1} \right ) ^{2}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{49\,x}{256\,{c}^{5}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{49}{256\,{c}^{5}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{40\,{a}^{6}{c}^{5}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{a}^{-1} \right ) ^{-5}}-{\frac{49}{384\,a{c}^{5}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-187/128/a/c^5*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+7/48/a^5/c^5/(x-1/a)^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5/2)+31
/48/a^4/c^5/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5/2)-139/96/a^3/c^5/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))
^(5/2)+561/256/c^5*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x+561/256/c^5/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-
1/a)^2-2*a*(x-1/a))^(1/2))-1/32/a^4/c^5/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-9/64/a^3/c^5/(x+1/a)^2*(-
a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-49/256/c^5*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-49/256/c^5/(a^2)^(1/2)*arctan
((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))+1/40/a^6/c^5/(x-1/a)^5*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5/2)-4
9/384/a/c^5*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.44363, size = 342, normalized size = 2.74 \begin{align*} -\frac{56 \, a^{4} x^{4} - 112 \, a^{3} x^{3} + 112 \, a x + 60 \,{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (15 \, a^{4} x^{4} - 76 \, a^{3} x^{3} + 32 \, a^{2} x^{2} + 82 \, a x - 56\right )} \sqrt{-a^{2} x^{2} + 1} - 56}{15 \,{\left (a^{5} c^{5} x^{4} - 2 \, a^{4} c^{5} x^{3} + 2 \, a^{2} c^{5} x - a c^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(56*a^4*x^4 - 112*a^3*x^3 + 112*a*x + 60*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) -
1)/(a*x)) + (15*a^4*x^4 - 76*a^3*x^3 + 32*a^2*x^2 + 82*a*x - 56)*sqrt(-a^2*x^2 + 1) - 56)/(a^5*c^5*x^4 - 2*a^4
*c^5*x^3 + 2*a^2*c^5*x - a*c^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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