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

Optimal. Leaf size=140 \[ \frac{c^4 \sqrt{1-a^2 x^2}}{a}-\frac{32 c^4 \sqrt{1-a^2 x^2}}{3 a^2 x}+\frac{5 c^4 \sqrt{1-a^2 x^2}}{2 a^3 x^2}-\frac{c^4 \sqrt{1-a^2 x^2}}{3 a^4 x^3}+\frac{25 c^4 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 a}+\frac{5 c^4 \sin ^{-1}(a x)}{a} \]

[Out]

(c^4*Sqrt[1 - a^2*x^2])/a - (c^4*Sqrt[1 - a^2*x^2])/(3*a^4*x^3) + (5*c^4*Sqrt[1 - a^2*x^2])/(2*a^3*x^2) - (32*
c^4*Sqrt[1 - a^2*x^2])/(3*a^2*x) + (5*c^4*ArcSin[a*x])/a + (25*c^4*ArcTanh[Sqrt[1 - a^2*x^2]])/(2*a)

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Rubi [A]  time = 0.419824, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {6131, 6128, 1807, 1809, 844, 216, 266, 63, 208} \[ \frac{c^4 \sqrt{1-a^2 x^2}}{a}-\frac{32 c^4 \sqrt{1-a^2 x^2}}{3 a^2 x}+\frac{5 c^4 \sqrt{1-a^2 x^2}}{2 a^3 x^2}-\frac{c^4 \sqrt{1-a^2 x^2}}{3 a^4 x^3}+\frac{25 c^4 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 a}+\frac{5 c^4 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))^4/E^ArcTanh[a*x],x]

[Out]

(c^4*Sqrt[1 - a^2*x^2])/a - (c^4*Sqrt[1 - a^2*x^2])/(3*a^4*x^3) + (5*c^4*Sqrt[1 - a^2*x^2])/(2*a^3*x^2) - (32*
c^4*Sqrt[1 - a^2*x^2])/(3*a^2*x) + (5*c^4*ArcSin[a*x])/a + (25*c^4*ArcTanh[Sqrt[1 - a^2*x^2]])/(2*a)

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 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 844

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

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]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int e^{-\tanh ^{-1}(a x)} \left (c-\frac{c}{a x}\right )^4 \, dx &=\frac{c^4 \int \frac{e^{-\tanh ^{-1}(a x)} (1-a x)^4}{x^4} \, dx}{a^4}\\ &=\frac{c^4 \int \frac{(1-a x)^5}{x^4 \sqrt{1-a^2 x^2}} \, dx}{a^4}\\ &=-\frac{c^4 \sqrt{1-a^2 x^2}}{3 a^4 x^3}-\frac{c^4 \int \frac{15 a-32 a^2 x+30 a^3 x^2-15 a^4 x^3+3 a^5 x^4}{x^3 \sqrt{1-a^2 x^2}} \, dx}{3 a^4}\\ &=-\frac{c^4 \sqrt{1-a^2 x^2}}{3 a^4 x^3}+\frac{5 c^4 \sqrt{1-a^2 x^2}}{2 a^3 x^2}+\frac{c^4 \int \frac{64 a^2-75 a^3 x+30 a^4 x^2-6 a^5 x^3}{x^2 \sqrt{1-a^2 x^2}} \, dx}{6 a^4}\\ &=-\frac{c^4 \sqrt{1-a^2 x^2}}{3 a^4 x^3}+\frac{5 c^4 \sqrt{1-a^2 x^2}}{2 a^3 x^2}-\frac{32 c^4 \sqrt{1-a^2 x^2}}{3 a^2 x}-\frac{c^4 \int \frac{75 a^3-30 a^4 x+6 a^5 x^2}{x \sqrt{1-a^2 x^2}} \, dx}{6 a^4}\\ &=\frac{c^4 \sqrt{1-a^2 x^2}}{a}-\frac{c^4 \sqrt{1-a^2 x^2}}{3 a^4 x^3}+\frac{5 c^4 \sqrt{1-a^2 x^2}}{2 a^3 x^2}-\frac{32 c^4 \sqrt{1-a^2 x^2}}{3 a^2 x}+\frac{c^4 \int \frac{-75 a^5+30 a^6 x}{x \sqrt{1-a^2 x^2}} \, dx}{6 a^6}\\ &=\frac{c^4 \sqrt{1-a^2 x^2}}{a}-\frac{c^4 \sqrt{1-a^2 x^2}}{3 a^4 x^3}+\frac{5 c^4 \sqrt{1-a^2 x^2}}{2 a^3 x^2}-\frac{32 c^4 \sqrt{1-a^2 x^2}}{3 a^2 x}+\left (5 c^4\right ) \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx-\frac{\left (25 c^4\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{2 a}\\ &=\frac{c^4 \sqrt{1-a^2 x^2}}{a}-\frac{c^4 \sqrt{1-a^2 x^2}}{3 a^4 x^3}+\frac{5 c^4 \sqrt{1-a^2 x^2}}{2 a^3 x^2}-\frac{32 c^4 \sqrt{1-a^2 x^2}}{3 a^2 x}+\frac{5 c^4 \sin ^{-1}(a x)}{a}-\frac{\left (25 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{4 a}\\ &=\frac{c^4 \sqrt{1-a^2 x^2}}{a}-\frac{c^4 \sqrt{1-a^2 x^2}}{3 a^4 x^3}+\frac{5 c^4 \sqrt{1-a^2 x^2}}{2 a^3 x^2}-\frac{32 c^4 \sqrt{1-a^2 x^2}}{3 a^2 x}+\frac{5 c^4 \sin ^{-1}(a x)}{a}+\frac{\left (25 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )}{2 a^3}\\ &=\frac{c^4 \sqrt{1-a^2 x^2}}{a}-\frac{c^4 \sqrt{1-a^2 x^2}}{3 a^4 x^3}+\frac{5 c^4 \sqrt{1-a^2 x^2}}{2 a^3 x^2}-\frac{32 c^4 \sqrt{1-a^2 x^2}}{3 a^2 x}+\frac{5 c^4 \sin ^{-1}(a x)}{a}+\frac{25 c^4 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.239494, size = 85, normalized size = 0.61 \[ \frac{c^4 \left (\frac{\sqrt{1-a^2 x^2} \left (6 a^3 x^3-64 a^2 x^2+15 a x-2\right )}{a^3 x^3}+75 \log \left (\sqrt{1-a^2 x^2}+1\right )-75 \log (a x)+30 \sin ^{-1}(a x)\right )}{6 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^4*((Sqrt[1 - a^2*x^2]*(-2 + 15*a*x - 64*a^2*x^2 + 6*a^3*x^3))/(a^3*x^3) + 30*ArcSin[a*x] - 75*Log[a*x] + 75
*Log[1 + Sqrt[1 - a^2*x^2]]))/(6*a)

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Maple [A]  time = 0.052, size = 232, normalized size = 1.7 \begin{align*} -11\,{\frac{{c}^{4} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{3/2}}{{a}^{2}x}}-11\,{c}^{4}x\sqrt{-{a}^{2}{x}^{2}+1}-11\,{\frac{{c}^{4}}{\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+{\frac{25\,{c}^{4}}{2\,a}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }-{\frac{25\,{c}^{4}}{2\,a}\sqrt{-{a}^{2}{x}^{2}+1}}+16\,{\frac{{c}^{4}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}{a}}+16\,{\frac{{c}^{4}}{\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}} \right ) }+{\frac{5\,{c}^{4}}{2\,{x}^{2}{a}^{3}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{{c}^{4}}{3\,{a}^{4}{x}^{3}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11*c^4/a^2/x*(-a^2*x^2+1)^(3/2)-11*c^4*x*(-a^2*x^2+1)^(1/2)-11*c^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2
+1)^(1/2))+25/2*c^4/a*arctanh(1/(-a^2*x^2+1)^(1/2))-25/2*c^4*(-a^2*x^2+1)^(1/2)/a+16*c^4/a*(-a^2*(x+1/a)^2+2*a
*(x+1/a))^(1/2)+16*c^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))+5/2*c^4*(-a^2*x^2+
1)^(3/2)/x^2/a^3-1/3*c^4*(-a^2*x^2+1)^(3/2)/a^4/x^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^4/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.13532, size = 286, normalized size = 2.04 \begin{align*} -\frac{60 \, a^{3} c^{4} x^{3} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) + 75 \, a^{3} c^{4} x^{3} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) - 6 \, a^{3} c^{4} x^{3} -{\left (6 \, a^{3} c^{4} x^{3} - 64 \, a^{2} c^{4} x^{2} + 15 \, a c^{4} x - 2 \, c^{4}\right )} \sqrt{-a^{2} x^{2} + 1}}{6 \, a^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^4/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-1/6*(60*a^3*c^4*x^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + 75*a^3*c^4*x^3*log((sqrt(-a^2*x^2 + 1) - 1)/x) -
6*a^3*c^4*x^3 - (6*a^3*c^4*x^3 - 64*a^2*c^4*x^2 + 15*a*c^4*x - 2*c^4)*sqrt(-a^2*x^2 + 1))/(a^4*x^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.18686, size = 354, normalized size = 2.53 \begin{align*} \frac{{\left (c^{4} - \frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c^{4}}{a^{2} x} + \frac{129 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c^{4}}{a^{4} x^{2}}\right )} a^{6} x^{3}}{24 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}{\left | a \right |}} + \frac{5 \, c^{4} \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} + \frac{25 \, c^{4} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{2 \,{\left | a \right |}} + \frac{\sqrt{-a^{2} x^{2} + 1} c^{4}}{a} - \frac{\frac{129 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c^{4}}{x} - \frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c^{4}}{a^{2} x^{2}} + \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} c^{4}}{a^{4} x^{3}}}{24 \, a^{2}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^4/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

1/24*(c^4 - 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^4/(a^2*x) + 129*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^4/(a^4*x^
2))*a^6*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*abs(a)) + 5*c^4*arcsin(a*x)*sgn(a)/abs(a) + 25/2*c^4*log(1/2*ab
s(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + sqrt(-a^2*x^2 + 1)*c^4/a - 1/24*(129*(sqrt(-a^2*x
^2 + 1)*abs(a) + a)*c^4/x - 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^4/(a^2*x^2) + (sqrt(-a^2*x^2 + 1)*abs(a) +
a)^3*c^4/(a^4*x^3))/(a^2*abs(a))