3.1145 \(\int \frac{e^{3 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x^5} \, dx\)

Optimal. Leaf size=115 \[ -\frac{3 a^3 c \sqrt{1-a^2 x^2}}{x}-\frac{15 a^2 c \sqrt{1-a^2 x^2}}{8 x^2}-\frac{a c \sqrt{1-a^2 x^2}}{x^3}-\frac{c \sqrt{1-a^2 x^2}}{4 x^4}-\frac{15}{8} a^4 c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

[Out]

-(c*Sqrt[1 - a^2*x^2])/(4*x^4) - (a*c*Sqrt[1 - a^2*x^2])/x^3 - (15*a^2*c*Sqrt[1 - a^2*x^2])/(8*x^2) - (3*a^3*c
*Sqrt[1 - a^2*x^2])/x - (15*a^4*c*ArcTanh[Sqrt[1 - a^2*x^2]])/8

________________________________________________________________________________________

Rubi [A]  time = 0.223379, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {6148, 1807, 835, 807, 266, 63, 208} \[ -\frac{3 a^3 c \sqrt{1-a^2 x^2}}{x}-\frac{15 a^2 c \sqrt{1-a^2 x^2}}{8 x^2}-\frac{a c \sqrt{1-a^2 x^2}}{x^3}-\frac{c \sqrt{1-a^2 x^2}}{4 x^4}-\frac{15}{8} a^4 c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2))/x^5,x]

[Out]

-(c*Sqrt[1 - a^2*x^2])/(4*x^4) - (a*c*Sqrt[1 - a^2*x^2])/x^3 - (15*a^2*c*Sqrt[1 - a^2*x^2])/(8*x^2) - (3*a^3*c
*Sqrt[1 - a^2*x^2])/x - (15*a^4*c*ArcTanh[Sqrt[1 - a^2*x^2]])/8

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 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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

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 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^{3 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )}{x^5} \, dx &=c \int \frac{(1+a x)^3}{x^5 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{c \sqrt{1-a^2 x^2}}{4 x^4}-\frac{1}{4} c \int \frac{-12 a-15 a^2 x-4 a^3 x^2}{x^4 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{c \sqrt{1-a^2 x^2}}{4 x^4}-\frac{a c \sqrt{1-a^2 x^2}}{x^3}+\frac{1}{12} c \int \frac{45 a^2+36 a^3 x}{x^3 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{c \sqrt{1-a^2 x^2}}{4 x^4}-\frac{a c \sqrt{1-a^2 x^2}}{x^3}-\frac{15 a^2 c \sqrt{1-a^2 x^2}}{8 x^2}-\frac{1}{24} c \int \frac{-72 a^3-45 a^4 x}{x^2 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{c \sqrt{1-a^2 x^2}}{4 x^4}-\frac{a c \sqrt{1-a^2 x^2}}{x^3}-\frac{15 a^2 c \sqrt{1-a^2 x^2}}{8 x^2}-\frac{3 a^3 c \sqrt{1-a^2 x^2}}{x}+\frac{1}{8} \left (15 a^4 c\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{c \sqrt{1-a^2 x^2}}{4 x^4}-\frac{a c \sqrt{1-a^2 x^2}}{x^3}-\frac{15 a^2 c \sqrt{1-a^2 x^2}}{8 x^2}-\frac{3 a^3 c \sqrt{1-a^2 x^2}}{x}+\frac{1}{16} \left (15 a^4 c\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{c \sqrt{1-a^2 x^2}}{4 x^4}-\frac{a c \sqrt{1-a^2 x^2}}{x^3}-\frac{15 a^2 c \sqrt{1-a^2 x^2}}{8 x^2}-\frac{3 a^3 c \sqrt{1-a^2 x^2}}{x}-\frac{1}{8} \left (15 a^2 c\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{c \sqrt{1-a^2 x^2}}{4 x^4}-\frac{a c \sqrt{1-a^2 x^2}}{x^3}-\frac{15 a^2 c \sqrt{1-a^2 x^2}}{8 x^2}-\frac{3 a^3 c \sqrt{1-a^2 x^2}}{x}-\frac{15}{8} a^4 c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\\ \end{align*}

Mathematica [C]  time = 0.110184, size = 97, normalized size = 0.84 \[ \frac{1}{2} a c \left (-2 a^3 \sqrt{1-a^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},1-a^2 x^2\right )-\frac{\sqrt{1-a^2 x^2} \left (6 a^2 x^2+3 a x+2\right )}{x^3}-3 a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2))/x^5,x]

[Out]

(a*c*(-((Sqrt[1 - a^2*x^2]*(2 + 3*a*x + 6*a^2*x^2))/x^3) - 3*a^3*ArcTanh[Sqrt[1 - a^2*x^2]] - 2*a^3*Sqrt[1 - a
^2*x^2]*Hypergeometric2F1[1/2, 3, 3/2, 1 - a^2*x^2]))/2

________________________________________________________________________________________

Maple [B]  time = 0.044, size = 231, normalized size = 2. \begin{align*} -c \left ({{a}^{5}x{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+{\frac{1}{4\,{x}^{4}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-{\frac{13\,{a}^{2}}{4} \left ( -{\frac{1}{2\,{x}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+{\frac{3\,{a}^{2}}{2} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}-{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) \right ) } \right ) }+2\,{a}^{3} \left ( -{\frac{1}{x\sqrt{-{a}^{2}{x}^{2}+1}}}+2\,{\frac{{a}^{2}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) +3\,{a}^{4} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}-{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) \right ) -3\,a \left ( -1/3\,{\frac{1}{{x}^{3}\sqrt{-{a}^{2}{x}^{2}+1}}}+4/3\,{a}^{2} \left ( -{\frac{1}{x\sqrt{-{a}^{2}{x}^{2}+1}}}+2\,{\frac{{a}^{2}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*(a^5*x/(-a^2*x^2+1)^(1/2)+1/4/x^4/(-a^2*x^2+1)^(1/2)-13/4*a^2*(-1/2/x^2/(-a^2*x^2+1)^(1/2)+3/2*a^2*(1/(-a^2
*x^2+1)^(1/2)-arctanh(1/(-a^2*x^2+1)^(1/2))))+2*a^3*(-1/x/(-a^2*x^2+1)^(1/2)+2*a^2*x/(-a^2*x^2+1)^(1/2))+3*a^4
*(1/(-a^2*x^2+1)^(1/2)-arctanh(1/(-a^2*x^2+1)^(1/2)))-3*a*(-1/3/x^3/(-a^2*x^2+1)^(1/2)+4/3*a^2*(-1/x/(-a^2*x^2
+1)^(1/2)+2*a^2*x/(-a^2*x^2+1)^(1/2))))

________________________________________________________________________________________

Maxima [A]  time = 0.96264, size = 201, normalized size = 1.75 \begin{align*} \frac{3 \, a^{5} c x}{\sqrt{-a^{2} x^{2} + 1}} - \frac{15}{8} \, a^{4} c \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{15 \, a^{4} c}{8 \, \sqrt{-a^{2} x^{2} + 1}} - \frac{2 \, a^{3} c}{\sqrt{-a^{2} x^{2} + 1} x} - \frac{13 \, a^{2} c}{8 \, \sqrt{-a^{2} x^{2} + 1} x^{2}} - \frac{a c}{\sqrt{-a^{2} x^{2} + 1} x^{3}} - \frac{c}{4 \, \sqrt{-a^{2} x^{2} + 1} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a^5*c*x/sqrt(-a^2*x^2 + 1) - 15/8*a^4*c*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) + 15/8*a^4*c/sqrt(-a^2*x
^2 + 1) - 2*a^3*c/(sqrt(-a^2*x^2 + 1)*x) - 13/8*a^2*c/(sqrt(-a^2*x^2 + 1)*x^2) - a*c/(sqrt(-a^2*x^2 + 1)*x^3)
- 1/4*c/(sqrt(-a^2*x^2 + 1)*x^4)

________________________________________________________________________________________

Fricas [A]  time = 2.58346, size = 166, normalized size = 1.44 \begin{align*} \frac{15 \, a^{4} c x^{4} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) -{\left (24 \, a^{3} c x^{3} + 15 \, a^{2} c x^{2} + 8 \, a c x + 2 \, c\right )} \sqrt{-a^{2} x^{2} + 1}}{8 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(15*a^4*c*x^4*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (24*a^3*c*x^3 + 15*a^2*c*x^2 + 8*a*c*x + 2*c)*sqrt(-a^2*x^
2 + 1))/x^4

________________________________________________________________________________________

Sympy [C]  time = 10.2415, size = 411, normalized size = 3.57 \begin{align*} a^{3} c \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 ) + 3 a^{2} c \left (\begin{cases} - \frac{a^{2} \operatorname{acosh}{\left (\frac{1}{a x} \right )}}{2} - \frac{a \sqrt{-1 + \frac{1}{a^{2} x^{2}}}}{2 x} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac{i a^{2} \operatorname{asin}{\left (\frac{1}{a x} \right )}}{2} - \frac{i a}{2 x \sqrt{1 - \frac{1}{a^{2} x^{2}}}} + \frac{i}{2 a x^{3} \sqrt{1 - \frac{1}{a^{2} x^{2}}}} & \text{otherwise} \end{cases}\right ) + 3 a c \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 \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)**3/(-a**2*x**2+1)**(3/2)*(-a**2*c*x**2+c)/x**5,x)

[Out]

a**3*c*Piecewise((-I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True)) + 3*a**2*c*P
iecewise((-a**2*acosh(1/(a*x))/2 - a*sqrt(-1 + 1/(a**2*x**2))/(2*x), 1/Abs(a**2*x**2) > 1), (I*a**2*asin(1/(a*
x))/2 - I*a/(2*x*sqrt(1 - 1/(a**2*x**2))) + I/(2*a*x**3*sqrt(1 - 1/(a**2*x**2))), True)) + 3*a*c*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*Piecewise((-3*a**4*acosh(1/(a*x))/8 + 3*a**3/(8*x*s
qrt(-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))

________________________________________________________________________________________

Giac [B]  time = 1.16481, size = 378, normalized size = 3.29 \begin{align*} \frac{{\left (a^{5} c + \frac{8 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{3} c}{x} + \frac{32 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} a c}{x^{2}} + \frac{104 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} c}{a x^{3}}\right )} a^{8} x^{4}}{64 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}{\left | a \right |}} - \frac{15 \, a^{5} c \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{104 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{5} c{\left | a \right |}}{x} + \frac{32 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} a^{3} c{\left | a \right |}}{x^{2}} + \frac{8 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} a c{\left | a \right |}}{x^{3}} + \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4} c{\left | a \right |}}{a x^{4}}}{64 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(a^5*c + 8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^3*c/x + 32*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a*c/x^2 + 104*(
sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c/(a*x^3))*a^8*x^4/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*abs(a)) - 15/8*a^5*c*lo
g(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - 1/64*(104*(sqrt(-a^2*x^2 + 1)*abs(a) + a)
*a^5*c*abs(a)/x + 32*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^3*c*abs(a)/x^2 + 8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*
a*c*abs(a)/x^3 + (sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*c*abs(a)/(a*x^4))/a^4