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

Optimal. Leaf size=162 \[ \frac{a^6 x^7 (a x+1)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{a^4 x^5 (8 a x+7)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{a^2 x^3 (48 a x+35)}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (64 a x+35)}{35 c^4 \sqrt{1-a^2 x^2}}-\frac{128 \sqrt{1-a^2 x^2}}{35 a c^4}+\frac{\sin ^{-1}(a x)}{a c^4} \]

[Out]

(a^6*x^7*(1 + a*x))/(7*c^4*(1 - a^2*x^2)^(7/2)) - (a^4*x^5*(7 + 8*a*x))/(35*c^4*(1 - a^2*x^2)^(5/2)) + (a^2*x^
3*(35 + 48*a*x))/(105*c^4*(1 - a^2*x^2)^(3/2)) - (x*(35 + 64*a*x))/(35*c^4*Sqrt[1 - a^2*x^2]) - (128*Sqrt[1 -
a^2*x^2])/(35*a*c^4) + ArcSin[a*x]/(a*c^4)

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Rubi [A]  time = 0.223133, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6157, 6148, 819, 641, 216} \[ \frac{a^6 x^7 (a x+1)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{a^4 x^5 (8 a x+7)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{a^2 x^3 (48 a x+35)}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (64 a x+35)}{35 c^4 \sqrt{1-a^2 x^2}}-\frac{128 \sqrt{1-a^2 x^2}}{35 a c^4}+\frac{\sin ^{-1}(a x)}{a c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^6*x^7*(1 + a*x))/(7*c^4*(1 - a^2*x^2)^(7/2)) - (a^4*x^5*(7 + 8*a*x))/(35*c^4*(1 - a^2*x^2)^(5/2)) + (a^2*x^
3*(35 + 48*a*x))/(105*c^4*(1 - a^2*x^2)^(3/2)) - (x*(35 + 64*a*x))/(35*c^4*Sqrt[1 - a^2*x^2]) - (128*Sqrt[1 -
a^2*x^2])/(35*a*c^4) + ArcSin[a*x]/(a*c^4)

Rule 6157

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

Rule 6148

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

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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^{\tanh ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^4} \, dx &=\frac{a^8 \int \frac{e^{\tanh ^{-1}(a x)} x^8}{\left (1-a^2 x^2\right )^4} \, dx}{c^4}\\ &=\frac{a^8 \int \frac{x^8 (1+a x)}{\left (1-a^2 x^2\right )^{9/2}} \, dx}{c^4}\\ &=\frac{a^6 x^7 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{a^6 \int \frac{x^6 (7+8 a x)}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{7 c^4}\\ &=\frac{a^6 x^7 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{a^4 x^5 (7+8 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{a^4 \int \frac{x^4 (35+48 a x)}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{35 c^4}\\ &=\frac{a^6 x^7 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{a^4 x^5 (7+8 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{a^2 x^3 (35+48 a x)}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{a^2 \int \frac{x^2 (105+192 a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{105 c^4}\\ &=\frac{a^6 x^7 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{a^4 x^5 (7+8 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{a^2 x^3 (35+48 a x)}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (35+64 a x)}{35 c^4 \sqrt{1-a^2 x^2}}+\frac{\int \frac{105+384 a x}{\sqrt{1-a^2 x^2}} \, dx}{105 c^4}\\ &=\frac{a^6 x^7 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{a^4 x^5 (7+8 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{a^2 x^3 (35+48 a x)}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (35+64 a x)}{35 c^4 \sqrt{1-a^2 x^2}}-\frac{128 \sqrt{1-a^2 x^2}}{35 a c^4}+\frac{\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c^4}\\ &=\frac{a^6 x^7 (1+a x)}{7 c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{a^4 x^5 (7+8 a x)}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{a^2 x^3 (35+48 a x)}{105 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (35+64 a x)}{35 c^4 \sqrt{1-a^2 x^2}}-\frac{128 \sqrt{1-a^2 x^2}}{35 a c^4}+\frac{\sin ^{-1}(a x)}{a c^4}\\ \end{align*}

Mathematica [A]  time = 0.0965116, size = 126, normalized size = 0.78 \[ \frac{105 a^7 x^7-281 a^6 x^6-559 a^5 x^5+965 a^4 x^4+715 a^3 x^3-1065 a^2 x^2+105 (a x-1)^3 (a x+1)^2 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)-279 a x+384}{105 a c^4 (a x-1)^3 (a x+1)^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(384 - 279*a*x - 1065*a^2*x^2 + 715*a^3*x^3 + 965*a^4*x^4 - 559*a^5*x^5 - 281*a^6*x^6 + 105*a^7*x^7 + 105*(-1
+ a*x)^3*(1 + a*x)^2*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(105*a*c^4*(-1 + a*x)^3*(1 + a*x)^2*Sqrt[1 - a^2*x^2])

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Maple [B]  time = 0.059, size = 341, normalized size = 2.1 \begin{align*} -{\frac{1}{a{c}^{4}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{1}{{c}^{4}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{17}{112\,{a}^{4}{c}^{4}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-3}}+{\frac{211}{336\,{a}^{3}{c}^{4}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{1657}{672\,{a}^{2}{c}^{4}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}}+{\frac{1}{56\,{a}^{5}{c}^{4}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-4}}+{\frac{7}{60\,{a}^{3}{c}^{4} \left ( x+{a}^{-1} \right ) ^{2}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{379}{480\,{a}^{2}{c}^{4} \left ( x+{a}^{-1} \right ) }\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{1}{80\,{a}^{4}{c}^{4} \left ( x+{a}^{-1} \right ) ^{3}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-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)/(c-c/a^2/x^2)^4,x)

[Out]

-(-a^2*x^2+1)^(1/2)/a/c^4+1/c^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+17/112/a^4/c^4/(x-1/a)^3*
(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+211/336/a^3/c^4/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+1657/672/a^2/c
^4/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+1/56/a^5/c^4/(x-1/a)^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+7/60/a
^3/c^4/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)-379/480/a^2/c^4/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)
-1/80/a^4/c^4/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{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)/(c-c/a^2/x^2)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.57646, size = 633, normalized size = 3.91 \begin{align*} -\frac{384 \, a^{7} x^{7} - 384 \, a^{6} x^{6} - 1152 \, a^{5} x^{5} + 1152 \, a^{4} x^{4} + 1152 \, a^{3} x^{3} - 1152 \, a^{2} x^{2} - 384 \, a x + 210 \,{\left (a^{7} x^{7} - a^{6} x^{6} - 3 \, a^{5} x^{5} + 3 \, a^{4} x^{4} + 3 \, a^{3} x^{3} - 3 \, a^{2} x^{2} - a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (105 \, a^{7} x^{7} - 281 \, a^{6} x^{6} - 559 \, a^{5} x^{5} + 965 \, a^{4} x^{4} + 715 \, a^{3} x^{3} - 1065 \, a^{2} x^{2} - 279 \, a x + 384\right )} \sqrt{-a^{2} x^{2} + 1} + 384}{105 \,{\left (a^{8} c^{4} x^{7} - a^{7} c^{4} x^{6} - 3 \, a^{6} c^{4} x^{5} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} - a^{2} c^{4} x + a c^{4}\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)/(c-c/a^2/x^2)^4,x, algorithm="fricas")

[Out]

-1/105*(384*a^7*x^7 - 384*a^6*x^6 - 1152*a^5*x^5 + 1152*a^4*x^4 + 1152*a^3*x^3 - 1152*a^2*x^2 - 384*a*x + 210*
(a^7*x^7 - a^6*x^6 - 3*a^5*x^5 + 3*a^4*x^4 + 3*a^3*x^3 - 3*a^2*x^2 - a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/
(a*x)) + (105*a^7*x^7 - 281*a^6*x^6 - 559*a^5*x^5 + 965*a^4*x^4 + 715*a^3*x^3 - 1065*a^2*x^2 - 279*a*x + 384)*
sqrt(-a^2*x^2 + 1) + 384)/(a^8*c^4*x^7 - a^7*c^4*x^6 - 3*a^6*c^4*x^5 + 3*a^5*c^4*x^4 + 3*a^4*c^4*x^3 - 3*a^3*c
^4*x^2 - a^2*c^4*x + a*c^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{8} \int \frac{x^{8}}{a^{7} x^{7} \sqrt{- a^{2} x^{2} + 1} - a^{6} x^{6} \sqrt{- a^{2} x^{2} + 1} - 3 a^{5} x^{5} \sqrt{- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} + 3 a^{3} x^{3} \sqrt{- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1} - a x \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx}{c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/(c-c/a**2/x**2)**4,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{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)/(c-c/a^2/x^2)^4,x, algorithm="giac")

[Out]

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