3.859 \(\int \frac {e^{-2 \tanh ^{-1}(a+b x)}}{x^4} \, dx\)

Optimal. Leaf size=70 \[ -\frac {2 b^3 \log (x)}{(a+1)^4}+\frac {2 b^3 \log (a+b x+1)}{(a+1)^4}-\frac {2 b^2}{(a+1)^3 x}+\frac {b}{(a+1)^2 x^2}-\frac {1-a}{3 (a+1) x^3} \]

[Out]

1/3*(-1+a)/(1+a)/x^3+b/(1+a)^2/x^2-2*b^2/(1+a)^3/x-2*b^3*ln(x)/(1+a)^4+2*b^3*ln(b*x+a+1)/(1+a)^4

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Rubi [A]  time = 0.06, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6163, 77} \[ -\frac {2 b^2}{(a+1)^3 x}-\frac {2 b^3 \log (x)}{(a+1)^4}+\frac {2 b^3 \log (a+b x+1)}{(a+1)^4}+\frac {b}{(a+1)^2 x^2}-\frac {1-a}{3 (a+1) x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a + b*x])*x^4),x]

[Out]

-(1 - a)/(3*(1 + a)*x^3) + b/((1 + a)^2*x^2) - (2*b^2)/((1 + a)^3*x) - (2*b^3*Log[x])/(1 + a)^4 + (2*b^3*Log[1
+ a + b*x])/(1 + a)^4

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 6163

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1
+ a*c + b*c*x)^(n/2))/(1 - a*c - b*c*x)^(n/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rubi steps

\begin {align*} \int \frac {e^{-2 \tanh ^{-1}(a+b x)}}{x^4} \, dx &=\int \frac {1-a-b x}{x^4 (1+a+b x)} \, dx\\ &=\int \left (\frac {1-a}{(1+a) x^4}-\frac {2 b}{(1+a)^2 x^3}+\frac {2 b^2}{(1+a)^3 x^2}-\frac {2 b^3}{(1+a)^4 x}+\frac {2 b^4}{(1+a)^4 (1+a+b x)}\right ) \, dx\\ &=-\frac {1-a}{3 (1+a) x^3}+\frac {b}{(1+a)^2 x^2}-\frac {2 b^2}{(1+a)^3 x}-\frac {2 b^3 \log (x)}{(1+a)^4}+\frac {2 b^3 \log (1+a+b x)}{(1+a)^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 70, normalized size = 1.00 \[ -\frac {2 b^3 \log (x)}{(a+1)^4}+\frac {2 b^3 \log (a+b x+1)}{(a+1)^4}-\frac {2 b^2}{(a+1)^3 x}+\frac {b}{(a+1)^2 x^2}-\frac {1-a}{3 (a+1) x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a + b*x])*x^4),x]

[Out]

-1/3*(1 - a)/((1 + a)*x^3) + b/((1 + a)^2*x^2) - (2*b^2)/((1 + a)^3*x) - (2*b^3*Log[x])/(1 + a)^4 + (2*b^3*Log
[1 + a + b*x])/(1 + a)^4

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fricas [A]  time = 0.58, size = 86, normalized size = 1.23 \[ \frac {6 \, b^{3} x^{3} \log \left (b x + a + 1\right ) - 6 \, b^{3} x^{3} \log \relax (x) - 6 \, {\left (a + 1\right )} b^{2} x^{2} + a^{4} + 2 \, a^{3} + 3 \, {\left (a^{2} + 2 \, a + 1\right )} b x - 2 \, a - 1}{3 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a+1)^2*(1-(b*x+a)^2)/x^4,x, algorithm="fricas")

[Out]

1/3*(6*b^3*x^3*log(b*x + a + 1) - 6*b^3*x^3*log(x) - 6*(a + 1)*b^2*x^2 + a^4 + 2*a^3 + 3*(a^2 + 2*a + 1)*b*x -
2*a - 1)/((a^4 + 4*a^3 + 6*a^2 + 4*a + 1)*x^3)

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giac [B]  time = 0.44, size = 159, normalized size = 2.27 \[ -\frac {2 \, b^{4} \log \left ({\left | -\frac {a}{b x + a + 1} - \frac {1}{b x + a + 1} + 1 \right |}\right )}{a^{4} b + 4 \, a^{3} b + 6 \, a^{2} b + 4 \, a b + b} - \frac {\frac {a b^{3} - 10 \, b^{3}}{a + 1} - \frac {3 \, {\left (a b^{4} - 8 \, b^{4}\right )}}{{\left (b x + a + 1\right )} b} + \frac {3 \, {\left (a^{2} b^{5} - 4 \, a b^{5} - 5 \, b^{5}\right )}}{{\left (b x + a + 1\right )}^{2} b^{2}}}{3 \, {\left (a + 1\right )}^{3} {\left (\frac {a}{b x + a + 1} + \frac {1}{b x + a + 1} - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a+1)^2*(1-(b*x+a)^2)/x^4,x, algorithm="giac")

[Out]

-2*b^4*log(abs(-a/(b*x + a + 1) - 1/(b*x + a + 1) + 1))/(a^4*b + 4*a^3*b + 6*a^2*b + 4*a*b + b) - 1/3*((a*b^3
- 10*b^3)/(a + 1) - 3*(a*b^4 - 8*b^4)/((b*x + a + 1)*b) + 3*(a^2*b^5 - 4*a*b^5 - 5*b^5)/((b*x + a + 1)^2*b^2))
/((a + 1)^3*(a/(b*x + a + 1) + 1/(b*x + a + 1) - 1)^3)

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maple [A]  time = 0.03, size = 75, normalized size = 1.07 \[ -\frac {1}{3 \left (1+a \right ) x^{3}}+\frac {a}{3 \left (1+a \right ) x^{3}}+\frac {b}{\left (1+a \right )^{2} x^{2}}-\frac {2 b^{3} \ln \relax (x )}{\left (1+a \right )^{4}}-\frac {2 b^{2}}{\left (1+a \right )^{3} x}+\frac {2 b^{3} \ln \left (b x +a +1\right )}{\left (1+a \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/(1+a)/x^3+1/3/(1+a)/x^3*a+b/(1+a)^2/x^2-2*b^3*ln(x)/(1+a)^4-2*b^2/(1+a)^3/x+2*b^3*ln(b*x+a+1)/(1+a)^4

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maxima [A]  time = 0.31, size = 108, normalized size = 1.54 \[ \frac {2 \, b^{3} \log \left (b x + a + 1\right )}{a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1} - \frac {2 \, b^{3} \log \relax (x)}{a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1} - \frac {6 \, b^{2} x^{2} - a^{3} - 3 \, {\left (a + 1\right )} b x - a^{2} + a + 1}{3 \, {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a+1)^2*(1-(b*x+a)^2)/x^4,x, algorithm="maxima")

[Out]

2*b^3*log(b*x + a + 1)/(a^4 + 4*a^3 + 6*a^2 + 4*a + 1) - 2*b^3*log(x)/(a^4 + 4*a^3 + 6*a^2 + 4*a + 1) - 1/3*(6
*b^2*x^2 - a^3 - 3*(a + 1)*b*x - a^2 + a + 1)/((a^3 + 3*a^2 + 3*a + 1)*x^3)

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mupad [B]  time = 0.10, size = 83, normalized size = 1.19 \[ \frac {\frac {a-1}{3\,\left (a+1\right )}-\frac {2\,b^2\,x^2}{{\left (a+1\right )}^3}+\frac {b\,x}{{\left (a+1\right )}^2}}{x^3}+\frac {4\,b^3\,\mathrm {atanh}\left (\frac {a^4+4\,a^3+6\,a^2+4\,a+1}{{\left (a+1\right )}^4}+\frac {2\,b\,x}{a+1}\right )}{{\left (a+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a - 1)/(3*(a + 1)) - (2*b^2*x^2)/(a + 1)^3 + (b*x)/(a + 1)^2)/x^3 + (4*b^3*atanh((4*a + 6*a^2 + 4*a^3 + a^4
+ 1)/(a + 1)^4 + (2*b*x)/(a + 1)))/(a + 1)^4

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sympy [B]  time = 0.53, size = 262, normalized size = 3.74 \[ - \frac {2 b^{3} \log {\left (x + \frac {- \frac {2 a^{5} b^{3}}{\left (a + 1\right )^{4}} - \frac {10 a^{4} b^{3}}{\left (a + 1\right )^{4}} - \frac {20 a^{3} b^{3}}{\left (a + 1\right )^{4}} - \frac {20 a^{2} b^{3}}{\left (a + 1\right )^{4}} + 2 a b^{3} - \frac {10 a b^{3}}{\left (a + 1\right )^{4}} + 2 b^{3} - \frac {2 b^{3}}{\left (a + 1\right )^{4}}}{4 b^{4}} \right )}}{\left (a + 1\right )^{4}} + \frac {2 b^{3} \log {\left (x + \frac {\frac {2 a^{5} b^{3}}{\left (a + 1\right )^{4}} + \frac {10 a^{4} b^{3}}{\left (a + 1\right )^{4}} + \frac {20 a^{3} b^{3}}{\left (a + 1\right )^{4}} + \frac {20 a^{2} b^{3}}{\left (a + 1\right )^{4}} + 2 a b^{3} + \frac {10 a b^{3}}{\left (a + 1\right )^{4}} + 2 b^{3} + \frac {2 b^{3}}{\left (a + 1\right )^{4}}}{4 b^{4}} \right )}}{\left (a + 1\right )^{4}} - \frac {- a^{3} - a^{2} + a + 6 b^{2} x^{2} + x \left (- 3 a b - 3 b\right ) + 1}{x^{3} \left (3 a^{3} + 9 a^{2} + 9 a + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a+1)**2*(1-(b*x+a)**2)/x**4,x)

[Out]

-2*b**3*log(x + (-2*a**5*b**3/(a + 1)**4 - 10*a**4*b**3/(a + 1)**4 - 20*a**3*b**3/(a + 1)**4 - 20*a**2*b**3/(a
+ 1)**4 + 2*a*b**3 - 10*a*b**3/(a + 1)**4 + 2*b**3 - 2*b**3/(a + 1)**4)/(4*b**4))/(a + 1)**4 + 2*b**3*log(x +
(2*a**5*b**3/(a + 1)**4 + 10*a**4*b**3/(a + 1)**4 + 20*a**3*b**3/(a + 1)**4 + 20*a**2*b**3/(a + 1)**4 + 2*a*b
**3 + 10*a*b**3/(a + 1)**4 + 2*b**3 + 2*b**3/(a + 1)**4)/(4*b**4))/(a + 1)**4 - (-a**3 - a**2 + a + 6*b**2*x**
2 + x*(-3*a*b - 3*b) + 1)/(x**3*(3*a**3 + 9*a**2 + 9*a + 3))

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