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

Optimal. Leaf size=105 \[ -\frac {4 a^2 (4 a x+3)}{3 c \sqrt {1-a^2 x^2}}-\frac {8 a^2 (a x+1)}{3 c \left (1-a^2 x^2\right )^{3/2}}+\frac {a \sqrt {1-a^2 x^2}}{c x}+\frac {4 a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c} \]

[Out]

-8/3*a^2*(a*x+1)/c/(-a^2*x^2+1)^(3/2)+4*a^2*arctanh((-a^2*x^2+1)^(1/2))/c-4/3*a^2*(4*a*x+3)/c/(-a^2*x^2+1)^(1/
2)+a*(-a^2*x^2+1)^(1/2)/c/x

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Rubi [A]  time = 0.32, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {6131, 6128, 852, 1805, 807, 266, 63, 208} \[ -\frac {4 a^2 (4 a x+3)}{3 c \sqrt {1-a^2 x^2}}-\frac {8 a^2 (a x+1)}{3 c \left (1-a^2 x^2\right )^{3/2}}+\frac {a \sqrt {1-a^2 x^2}}{c x}+\frac {4 a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 852

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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 \frac {e^{3 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right ) x^3} \, dx &=-\frac {a \int \frac {e^{3 \tanh ^{-1}(a x)}}{x^2 (1-a x)} \, dx}{c}\\ &=-\frac {a \int \frac {\left (1-a^2 x^2\right )^{3/2}}{x^2 (1-a x)^4} \, dx}{c}\\ &=-\frac {a \int \frac {(1+a x)^4}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx}{c}\\ &=-\frac {8 a^2 (1+a x)}{3 c \left (1-a^2 x^2\right )^{3/2}}+\frac {a \int \frac {-3-12 a x-13 a^2 x^2}{x^2 \left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c}\\ &=-\frac {8 a^2 (1+a x)}{3 c \left (1-a^2 x^2\right )^{3/2}}-\frac {4 a^2 (3+4 a x)}{3 c \sqrt {1-a^2 x^2}}-\frac {a \int \frac {3+12 a x}{x^2 \sqrt {1-a^2 x^2}} \, dx}{3 c}\\ &=-\frac {8 a^2 (1+a x)}{3 c \left (1-a^2 x^2\right )^{3/2}}-\frac {4 a^2 (3+4 a x)}{3 c \sqrt {1-a^2 x^2}}+\frac {a \sqrt {1-a^2 x^2}}{c x}-\frac {\left (4 a^2\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx}{c}\\ &=-\frac {8 a^2 (1+a x)}{3 c \left (1-a^2 x^2\right )^{3/2}}-\frac {4 a^2 (3+4 a x)}{3 c \sqrt {1-a^2 x^2}}+\frac {a \sqrt {1-a^2 x^2}}{c x}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )}{c}\\ &=-\frac {8 a^2 (1+a x)}{3 c \left (1-a^2 x^2\right )^{3/2}}-\frac {4 a^2 (3+4 a x)}{3 c \sqrt {1-a^2 x^2}}+\frac {a \sqrt {1-a^2 x^2}}{c x}+\frac {4 \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )}{c}\\ &=-\frac {8 a^2 (1+a x)}{3 c \left (1-a^2 x^2\right )^{3/2}}-\frac {4 a^2 (3+4 a x)}{3 c \sqrt {1-a^2 x^2}}+\frac {a \sqrt {1-a^2 x^2}}{c x}+\frac {4 a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 92, normalized size = 0.88 \[ \frac {a \left (-19 a^3 x^3+7 a^2 x^2+12 a x (a x-1) \sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+23 a x-3\right )}{3 c x (a x-1) \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-3 + 23*a*x + 7*a^2*x^2 - 19*a^3*x^3 + 12*a*x*(-1 + a*x)*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]]))/(3
*c*x*(-1 + a*x)*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 0.43, size = 120, normalized size = 1.14 \[ -\frac {20 \, a^{4} x^{3} - 40 \, a^{3} x^{2} + 20 \, a^{2} x + 12 \, {\left (a^{4} x^{3} - 2 \, a^{3} x^{2} + a^{2} x\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (19 \, a^{3} x^{2} - 26 \, a^{2} x + 3 \, a\right )} \sqrt {-a^{2} x^{2} + 1}}{3 \, {\left (a^{2} c x^{3} - 2 \, a c x^{2} + c x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(20*a^4*x^3 - 40*a^3*x^2 + 20*a^2*x + 12*(a^4*x^3 - 2*a^3*x^2 + a^2*x)*log((sqrt(-a^2*x^2 + 1) - 1)/x) -
(19*a^3*x^2 - 26*a^2*x + 3*a)*sqrt(-a^2*x^2 + 1))/(a^2*c*x^3 - 2*a*c*x^2 + c*x)

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giac [B]  time = 0.22, size = 217, normalized size = 2.07 \[ \frac {4 \, a^{3} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c {\left | a \right |}} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a}{2 \, c x {\left | a \right |}} + \frac {{\left (3 \, a^{3} - \frac {89 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a}{x} + \frac {153 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a x^{2}} - \frac {99 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{3} x^{3}}\right )} a^{2} x}{6 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{3} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a^3*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c*abs(a)) + 1/2*(sqrt(-a^2*x^2 + 1)*abs(a
) + a)*a/(c*x*abs(a)) + 1/6*(3*a^3 - 89*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a/x + 153*(sqrt(-a^2*x^2 + 1)*abs(a) +
a)^2/(a*x^2) - 99*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^3*x^3))*a^2*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*c*((sqr
t(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^3*abs(a))

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maple [A]  time = 0.04, size = 165, normalized size = 1.57 \[ \frac {a \left (-\frac {a^{2} x}{\sqrt {-a^{2} x^{2}+1}}-4 a \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )+\frac {1}{x \sqrt {-a^{2} x^{2}+1}}+8 a \left (\frac {1}{3 a \left (x -\frac {1}{a}\right ) \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}+\frac {-2 a^{2} \left (x -\frac {1}{a}\right )-2 a}{3 a \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/c*(-a^2*x/(-a^2*x^2+1)^(1/2)-4*a*(1/(-a^2*x^2+1)^(1/2)-arctanh(1/(-a^2*x^2+1)^(1/2)))+1/x/(-a^2*x^2+1)^(1/2)
+8*a*(1/3/a/(x-1/a)/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+1/3/a*(-2*a^2*(x-1/a)-2*a)/(-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 \[ \int \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (c - \frac {c}{a x}\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.87, size = 138, normalized size = 1.31 \[ \frac {a\,\sqrt {1-a^2\,x^2}}{c\,x}-\frac {4\,a^4\,\sqrt {1-a^2\,x^2}}{3\,\left (c\,a^4\,x^2-2\,c\,a^3\,x+c\,a^2\right )}+\frac {16\,a^3\,\sqrt {1-a^2\,x^2}}{3\,\left (\frac {c\,\sqrt {-a^2}}{a}-c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {a^2\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*(1 - a^2*x^2)^(1/2))/(c*x) - (4*a^4*(1 - a^2*x^2)^(1/2))/(3*(a^2*c + a^4*c*x^2 - 2*a^3*c*x)) - (a^2*atan((1
- a^2*x^2)^(1/2)*1i)*4i)/c + (16*a^3*(1 - a^2*x^2)^(1/2))/(3*((c*(-a^2)^(1/2))/a - c*x*(-a^2)^(1/2))*(-a^2)^(
1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)/(c-c/a/x)/x**3,x)

[Out]

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

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