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

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

[Out]

5*c^4*arcsin(a*x)/a+25/2*c^4*arctanh((-a^2*x^2+1)^(1/2))/a+c^4*(-a^2*x^2+1)^(1/2)/a-1/3*c^4*(-a^2*x^2+1)^(1/2)
/a^4/x^3+5/2*c^4*(-a^2*x^2+1)^(1/2)/a^3/x^2-32/3*c^4*(-a^2*x^2+1)^(1/2)/a^2/x

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Rubi [A]  time = 0.42, antiderivative size = 140, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.25, size = 85, normalized size = 0.61 \[ \frac {c^4 \left (75 \log \left (\sqrt {1-a^2 x^2}+1\right )+\frac {\sqrt {1-a^2 x^2} \left (6 a^3 x^3-64 a^2 x^2+15 a x-2\right )}{a^3 x^3}-75 \log (a x)+30 \sin ^{-1}(a x)\right )}{6 a} \]

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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fricas [A]  time = 0.43, size = 132, normalized size = 0.94 \[ -\frac {60 \, a^{3} c^{4} x^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + 75 \, a^{3} c^{4} x^{3} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - 6 \, a^{3} c^{4} x^{3} - {\left (6 \, a^{3} c^{4} x^{3} - 64 \, a^{2} c^{4} x^{2} + 15 \, a c^{4} x - 2 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, a^{4} x^{3}} \]

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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giac [B]  time = 0.20, size = 262, normalized size = 1.87 \[ \frac {{\left (c^{4} - \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{4}}{a^{2} x} + \frac {129 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c^{4}}{a^{4} x^{2}}\right )} a^{6} x^{3}}{24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} {\left | a \right |}} + \frac {5 \, c^{4} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} + \frac {25 \, c^{4} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, {\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{4}}{a} - \frac {\frac {129 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{4}}{x} - \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c^{4}}{a^{2} x^{2}} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c^{4}}{a^{4} x^{3}}}{24 \, a^{2} {\left | a \right |}} \]

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))

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maple [A]  time = 0.05, size = 232, normalized size = 1.66 \[ -\frac {25 c^{4} \sqrt {-a^{2} x^{2}+1}}{2 a}+\frac {25 c^{4} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a}-\frac {11 c^{4} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{a^{2} x}-11 c^{4} x \sqrt {-a^{2} x^{2}+1}-\frac {11 c^{4} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}-\frac {c^{4} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 a^{4} x^{3}}+\frac {5 c^{4} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{2 x^{2} a^{3}}+\frac {16 c^{4} \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{a}+\frac {16 c^{4} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}} \]

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]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 4 \, a c^{4} {\left (\frac {\arcsin \left (a x\right )}{a^{2}} + \frac {\log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )}{a^{2}}\right )} + c^{4} {\left (\frac {\arcsin \left (a x\right )}{a} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a}\right )} + \int \frac {{\left (6 \, a^{2} c^{4} x^{2} - 4 \, a c^{4} x + c^{4}\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{a^{5} x^{5} + a^{4} x^{4}}\,{d x} \]

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]

4*a*c^4*(arcsin(a*x)/a^2 + log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x))/a^2) + c^4*(arcsin(a*x)/a + sqrt(-a^2*x
^2 + 1)/a) + integrate((6*a^2*c^4*x^2 - 4*a*c^4*x + c^4)*sqrt(a*x + 1)*sqrt(-a*x + 1)/(a^5*x^5 + a^4*x^4), x)

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mupad [B]  time = 0.85, size = 136, normalized size = 0.97 \[ \frac {5\,c^4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}+\frac {c^4\,\sqrt {1-a^2\,x^2}}{a}-\frac {32\,c^4\,\sqrt {1-a^2\,x^2}}{3\,a^2\,x}+\frac {5\,c^4\,\sqrt {1-a^2\,x^2}}{2\,a^3\,x^2}-\frac {c^4\,\sqrt {1-a^2\,x^2}}{3\,a^4\,x^3}-\frac {c^4\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,25{}\mathrm {i}}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*c^4*asinh(x*(-a^2)^(1/2)))/(-a^2)^(1/2) - (c^4*atan((1 - a^2*x^2)^(1/2)*1i)*25i)/(2*a) + (c^4*(1 - a^2*x^2)
^(1/2))/a - (32*c^4*(1 - a^2*x^2)^(1/2))/(3*a^2*x) + (5*c^4*(1 - a^2*x^2)^(1/2))/(2*a^3*x^2) - (c^4*(1 - a^2*x
^2)^(1/2))/(3*a^4*x^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {c^{4} \left (\int \frac {\sqrt {- a^{2} x^{2} + 1}}{a x^{5} + x^{4}}\, dx + \int \left (- \frac {4 a x \sqrt {- a^{2} x^{2} + 1}}{a x^{5} + x^{4}}\right )\, dx + \int \frac {6 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1}}{a x^{5} + x^{4}}\, dx + \int \left (- \frac {4 a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1}}{a x^{5} + x^{4}}\right )\, dx + \int \frac {a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1}}{a x^{5} + x^{4}}\, dx\right )}{a^{4}} \]

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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