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

Optimal. Leaf size=102 \[ -\frac{5 a^2 c^3 \sqrt{1-a^2 x^2}}{8 x^2}+\frac{2 a c^3 \left (1-a^2 x^2\right )^{3/2}}{3 x^3}-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}+\frac{5}{8} a^4 c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

[Out]

(-5*a^2*c^3*Sqrt[1 - a^2*x^2])/(8*x^2) - (c^3*(1 - a^2*x^2)^(3/2))/(4*x^4) + (2*a*c^3*(1 - a^2*x^2)^(3/2))/(3*
x^3) + (5*a^4*c^3*ArcTanh[Sqrt[1 - a^2*x^2]])/8

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Rubi [A]  time = 0.163448, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {6128, 1807, 807, 266, 47, 63, 208} \[ -\frac{5 a^2 c^3 \sqrt{1-a^2 x^2}}{8 x^2}+\frac{2 a c^3 \left (1-a^2 x^2\right )^{3/2}}{3 x^3}-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}+\frac{5}{8} a^4 c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a^2*c^3*Sqrt[1 - a^2*x^2])/(8*x^2) - (c^3*(1 - a^2*x^2)^(3/2))/(4*x^4) + (2*a*c^3*(1 - a^2*x^2)^(3/2))/(3*
x^3) + (5*a^4*c^3*ArcTanh[Sqrt[1 - a^2*x^2]])/8

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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 \frac{e^{\tanh ^{-1}(a x)} (c-a c x)^3}{x^5} \, dx &=c \int \frac{(c-a c x)^2 \sqrt{1-a^2 x^2}}{x^5} \, dx\\ &=-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}-\frac{1}{4} c \int \frac{\left (8 a c^2-5 a^2 c^2 x\right ) \sqrt{1-a^2 x^2}}{x^4} \, dx\\ &=-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}+\frac{2 a c^3 \left (1-a^2 x^2\right )^{3/2}}{3 x^3}+\frac{1}{4} \left (5 a^2 c^3\right ) \int \frac{\sqrt{1-a^2 x^2}}{x^3} \, dx\\ &=-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}+\frac{2 a c^3 \left (1-a^2 x^2\right )^{3/2}}{3 x^3}+\frac{1}{8} \left (5 a^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-a^2 x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{5 a^2 c^3 \sqrt{1-a^2 x^2}}{8 x^2}-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}+\frac{2 a c^3 \left (1-a^2 x^2\right )^{3/2}}{3 x^3}-\frac{1}{16} \left (5 a^4 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{5 a^2 c^3 \sqrt{1-a^2 x^2}}{8 x^2}-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}+\frac{2 a c^3 \left (1-a^2 x^2\right )^{3/2}}{3 x^3}+\frac{1}{8} \left (5 a^2 c^3\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 )\\ &=-\frac{5 a^2 c^3 \sqrt{1-a^2 x^2}}{8 x^2}-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}+\frac{2 a c^3 \left (1-a^2 x^2\right )^{3/2}}{3 x^3}+\frac{5}{8} a^4 c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0322196, size = 99, normalized size = 0.97 \[ \frac{c^3 \left (16 a^5 x^5+9 a^4 x^4-32 a^3 x^3-3 a^2 x^2+15 a^4 x^4 \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+16 a x-6\right )}{24 x^4 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*(-6 + 16*a*x - 3*a^2*x^2 - 32*a^3*x^3 + 9*a^4*x^4 + 16*a^5*x^5 + 15*a^4*x^4*Sqrt[1 - a^2*x^2]*ArcTanh[Sqr
t[1 - a^2*x^2]]))/(24*x^4*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.046, size = 144, normalized size = 1.4 \begin{align*} -{c}^{3} \left ({\frac{1}{4\,{x}^{4}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{3\,{a}^{2}}{4} \left ( -{\frac{1}{2\,{x}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{a}^{2}}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) } \right ) }+2\,{\frac{{a}^{3}\sqrt{-{a}^{2}{x}^{2}+1}}{x}}-{a}^{4}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) +2\,a \left ( -1/3\,{\frac{\sqrt{-{a}^{2}{x}^{2}+1}}{{x}^{3}}}-2/3\,{\frac{{a}^{2}\sqrt{-{a}^{2}{x}^{2}+1}}{x}} \right ) \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)*(-a*c*x+c)^3/x^5,x)

[Out]

-c^3*(1/4*(-a^2*x^2+1)^(1/2)/x^4-3/4*a^2*(-1/2*(-a^2*x^2+1)^(1/2)/x^2-1/2*a^2*arctanh(1/(-a^2*x^2+1)^(1/2)))+2
*a^3*(-a^2*x^2+1)^(1/2)/x-a^4*arctanh(1/(-a^2*x^2+1)^(1/2))+2*a*(-1/3*(-a^2*x^2+1)^(1/2)/x^3-2/3*a^2*(-a^2*x^2
+1)^(1/2)/x))

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Maxima [A]  time = 1.43689, size = 165, normalized size = 1.62 \begin{align*} \frac{5}{8} \, a^{4} c^{3} \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) - \frac{2 \, \sqrt{-a^{2} x^{2} + 1} a^{3} c^{3}}{3 \, x} - \frac{3 \, \sqrt{-a^{2} x^{2} + 1} a^{2} c^{3}}{8 \, x^{2}} + \frac{2 \, \sqrt{-a^{2} x^{2} + 1} a c^{3}}{3 \, x^{3}} - \frac{\sqrt{-a^{2} x^{2} + 1} c^{3}}{4 \, x^{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)*(-a*c*x+c)^3/x^5,x, algorithm="maxima")

[Out]

5/8*a^4*c^3*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) - 2/3*sqrt(-a^2*x^2 + 1)*a^3*c^3/x - 3/8*sqrt(-a^2*x^2
+ 1)*a^2*c^3/x^2 + 2/3*sqrt(-a^2*x^2 + 1)*a*c^3/x^3 - 1/4*sqrt(-a^2*x^2 + 1)*c^3/x^4

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Fricas [A]  time = 1.56625, size = 182, normalized size = 1.78 \begin{align*} -\frac{15 \, a^{4} c^{3} x^{4} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) +{\left (16 \, a^{3} c^{3} x^{3} + 9 \, a^{2} c^{3} x^{2} - 16 \, a c^{3} x + 6 \, c^{3}\right )} \sqrt{-a^{2} x^{2} + 1}}{24 \, x^{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)*(-a*c*x+c)^3/x^5,x, algorithm="fricas")

[Out]

-1/24*(15*a^4*c^3*x^4*log((sqrt(-a^2*x^2 + 1) - 1)/x) + (16*a^3*c^3*x^3 + 9*a^2*c^3*x^2 - 16*a*c^3*x + 6*c^3)*
sqrt(-a^2*x^2 + 1))/x^4

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Sympy [C]  time = 8.71955, size = 347, normalized size = 3.4 \begin{align*} - a^{4} c^{3} \left (\begin{cases} - \operatorname{acosh}{\left (\frac{1}{a x} \right )} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname{asin}{\left (\frac{1}{a x} \right )} & \text{otherwise} \end{cases}\right ) + 2 a^{3} c^{3} \left (\begin{cases} - \frac{i \sqrt{a^{2} x^{2} - 1}}{x} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac{\sqrt{- a^{2} x^{2} + 1}}{x} & \text{otherwise} \end{cases}\right ) - 2 a c^{3} \left (\begin{cases} - \frac{2 i a^{2} \sqrt{a^{2} x^{2} - 1}}{3 x} - \frac{i \sqrt{a^{2} x^{2} - 1}}{3 x^{3}} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac{2 a^{2} \sqrt{- a^{2} x^{2} + 1}}{3 x} - \frac{\sqrt{- a^{2} x^{2} + 1}}{3 x^{3}} & \text{otherwise} \end{cases}\right ) + c^{3} \left (\begin{cases} - \frac{3 a^{4} \operatorname{acosh}{\left (\frac{1}{a x} \right )}}{8} + \frac{3 a^{3}}{8 x \sqrt{-1 + \frac{1}{a^{2} x^{2}}}} - \frac{a}{8 x^{3} \sqrt{-1 + \frac{1}{a^{2} x^{2}}}} - \frac{1}{4 a x^{5} \sqrt{-1 + \frac{1}{a^{2} x^{2}}}} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac{3 i a^{4} \operatorname{asin}{\left (\frac{1}{a x} \right )}}{8} - \frac{3 i a^{3}}{8 x \sqrt{1 - \frac{1}{a^{2} x^{2}}}} + \frac{i a}{8 x^{3} \sqrt{1 - \frac{1}{a^{2} x^{2}}}} + \frac{i}{4 a x^{5} \sqrt{1 - \frac{1}{a^{2} x^{2}}}} & \text{otherwise} \end{cases}\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)*(-a*c*x+c)**3/x**5,x)

[Out]

-a**4*c**3*Piecewise((-acosh(1/(a*x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True)) + 2*a**3*c**3*Piecewise
((-I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True)) - 2*a*c**3*Piecewise((-2*I*a
**2*sqrt(a**2*x**2 - 1)/(3*x) - I*sqrt(a**2*x**2 - 1)/(3*x**3), Abs(a**2*x**2) > 1), (-2*a**2*sqrt(-a**2*x**2
+ 1)/(3*x) - sqrt(-a**2*x**2 + 1)/(3*x**3), True)) + c**3*Piecewise((-3*a**4*acosh(1/(a*x))/8 + 3*a**3/(8*x*sq
rt(-1 + 1/(a**2*x**2))) - a/(8*x**3*sqrt(-1 + 1/(a**2*x**2))) - 1/(4*a*x**5*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(a
**2*x**2) > 1), (3*I*a**4*asin(1/(a*x))/8 - 3*I*a**3/(8*x*sqrt(1 - 1/(a**2*x**2))) + I*a/(8*x**3*sqrt(1 - 1/(a
**2*x**2))) + I/(4*a*x**5*sqrt(1 - 1/(a**2*x**2))), True))

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Giac [B]  time = 1.16439, size = 405, normalized size = 3.97 \begin{align*} \frac{{\left (3 \, a^{5} c^{3} - \frac{16 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{3} c^{3}}{x} + \frac{24 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} a c^{3}}{x^{2}} + \frac{48 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} c^{3}}{a x^{3}}\right )} a^{8} x^{4}}{192 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}{\left | a \right |}} + \frac{5 \, a^{5} c^{3} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{8 \,{\left | a \right |}} - \frac{\frac{48 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{5} c^{3}{\left | a \right |}}{x} + \frac{24 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} a^{3} c^{3}{\left | a \right |}}{x^{2}} - \frac{16 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} a c^{3}{\left | a \right |}}{x^{3}} + \frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4} c^{3}{\left | a \right |}}{a x^{4}}}{192 \, a^{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)*(-a*c*x+c)^3/x^5,x, algorithm="giac")

[Out]

1/192*(3*a^5*c^3 - 16*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^3*c^3/x + 24*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a*c^3/x
^2 + 48*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c^3/(a*x^3))*a^8*x^4/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*abs(a)) + 5/
8*a^5*c^3*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - 1/192*(48*(sqrt(-a^2*x^2 + 1)
*abs(a) + a)*a^5*c^3*abs(a)/x + 24*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^3*c^3*abs(a)/x^2 - 16*(sqrt(-a^2*x^2 +
1)*abs(a) + a)^3*a*c^3*abs(a)/x^3 + 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*c^3*abs(a)/(a*x^4))/a^4