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

Optimal. Leaf size=74 \[ \frac {4 (a x+1)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {5 a x+3}{3 c^2 \sqrt {1-a^2 x^2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {6128, 852, 1805, 823, 12, 266, 63, 208} \[ \frac {4 (a x+1)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {5 a x+3}{3 c^2 \sqrt {1-a^2 x^2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 78, normalized size = 1.05 \[ \frac {5 a^2 x^2-3 (a x-1) \sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-2 a x-7}{3 c^2 (a x-1) \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7 - 2*a*x + 5*a^2*x^2 - 3*(-1 + a*x)*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])/(3*c^2*(-1 + a*x)*Sqrt[1
- a^2*x^2])

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fricas [A]  time = 0.50, size = 93, normalized size = 1.26 \[ \frac {7 \, a^{2} x^{2} - 14 \, a x + 3 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt {-a^{2} x^{2} + 1} {\left (5 \, a x - 7\right )} + 7}{3 \, {\left (a^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(7*a^2*x^2 - 14*a*x + 3*(a^2*x^2 - 2*a*x + 1)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - sqrt(-a^2*x^2 + 1)*(5*a*x
- 7) + 7)/(a^2*c^2*x^2 - 2*a*c^2*x + c^2)

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giac [C]  time = 0.28, size = 245, normalized size = 3.31 \[ \frac {{\left (\frac {{\left (3 \, \log \relax (2) - 6 \, \log \left (i + 1\right ) + 10 i\right )} \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (c)}{c} + \frac {6 \, \log \left (\sqrt {-\frac {2 \, c}{a c x - c} - 1} + 1\right )}{c \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (c)} - \frac {6 \, \log \left ({\left | \sqrt {-\frac {2 \, c}{a c x - c} - 1} - 1 \right |}\right )}{c \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (c)} - \frac {2 \, {\left (c^{2} {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{a c x - c}\right )^{2} \mathrm {sgn}\relax (a)^{2} \mathrm {sgn}\relax (c)^{2} + 6 \, c^{2} \sqrt {-\frac {2 \, c}{a c x - c} - 1} \mathrm {sgn}\left (\frac {1}{a c x - c}\right )^{2} \mathrm {sgn}\relax (a)^{2} \mathrm {sgn}\relax (c)^{2}\right )}}{c^{3} \mathrm {sgn}\left (\frac {1}{a c x - c}\right )^{3} \mathrm {sgn}\relax (a)^{3} \mathrm {sgn}\relax (c)^{3}}\right )} a}{6 \, c {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((3*log(2) - 6*log(I + 1) + 10*I)*sgn(1/(a*c*x - c))*sgn(a)*sgn(c)/c + 6*log(sqrt(-2*c/(a*c*x - c) - 1) +
1)/(c*sgn(1/(a*c*x - c))*sgn(a)*sgn(c)) - 6*log(abs(sqrt(-2*c/(a*c*x - c) - 1) - 1))/(c*sgn(1/(a*c*x - c))*sgn
(a)*sgn(c)) - 2*(c^2*(-2*c/(a*c*x - c) - 1)^(3/2)*sgn(1/(a*c*x - c))^2*sgn(a)^2*sgn(c)^2 + 6*c^2*sqrt(-2*c/(a*
c*x - c) - 1)*sgn(1/(a*c*x - c))^2*sgn(a)^2*sgn(c)^2)/(c^3*sgn(1/(a*c*x - c))^3*sgn(a)^3*sgn(c)^3))*a/(c*abs(a
))

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maple [B]  time = 0.04, size = 147, normalized size = 1.99 \[ \frac {-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{a}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.81, size = 119, normalized size = 1.61 \[ \frac {2\,a^2\,\sqrt {1-a^2\,x^2}}{3\,\left (a^4\,c^2\,x^2-2\,a^3\,c^2\,x+a^2\,c^2\right )}-\frac {\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{c^2}+\frac {5\,a\,\sqrt {1-a^2\,x^2}}{3\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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