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

Optimal. Leaf size=44 \[ \frac {(1-a) \sqrt {a+b x+1}}{b^2 \sqrt {-a-b x+1}}-\frac {\sin ^{-1}(a+b x)}{b^2} \]

[Out]

-arcsin(b*x+a)/b^2+(1-a)*(b*x+a+1)^(1/2)/b^2/(-b*x-a+1)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {6164, 78, 53, 619, 216} \[ \frac {(1-a) \sqrt {a+b x+1}}{b^2 \sqrt {-a-b x+1}}-\frac {\sin ^{-1}(a+b x)}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a)*Sqrt[1 + a + b*x])/(b^2*Sqrt[1 - a - b*x]) - ArcSin[a + b*x]/b^2

Rule 53

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 6164

Int[E^(ArcTanh[(a_) + (b_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(c/
(1 - a^2))^p, Int[u*(1 - a - b*x)^(p - n/2)*(1 + a + b*x)^(p + n/2), x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]
&& EqQ[b*d - 2*a*e, 0] && EqQ[b^2*c + e*(1 - a^2), 0] && (IntegerQ[p] || GtQ[c/(1 - a^2), 0])

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 49, normalized size = 1.11 \[ -\frac {\sin ^{-1}(a+b x)-\frac {(a-1) \sqrt {-a^2-2 a b x-b^2 x^2+1}}{a+b x-1}}{b^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((-(((-1 + a)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2])/(-1 + a + b*x)) + ArcSin[a + b*x])/b^2)

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fricas [B]  time = 0.57, size = 98, normalized size = 2.23 \[ \frac {{\left (b x + a - 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a - 1\right )}}{b^{3} x + {\left (a - 1\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x + a - 1)*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 - 1)) + sqrt(-b^2*
x^2 - 2*a*b*x - a^2 + 1)*(a - 1))/(b^3*x + (a - 1)*b^2)

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giac [A]  time = 0.28, size = 66, normalized size = 1.50 \[ \frac {\arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{b {\left | b \right |}} - \frac {2 \, {\left (a - 1\right )}}{b {\left (\frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} - 1\right )} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(-b*x - a)*sgn(b)/(b*abs(b)) - 2*(a - 1)/(b*((sqrt(-(b*x + a)^2 + 1)*abs(b) + b)/(b^2*x + a*b) - 1)*abs(
b))

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maple [B]  time = 0.04, size = 160, normalized size = 3.64 \[ \frac {x}{b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b \sqrt {b^{2}}}+\frac {1}{b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {a x}{b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {a^{2}}{b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/b/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)
)+1/b^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-a/b/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x-a^2/b^2/(-b^2*x^2-2*a*b*x-a^2+1)^(
1/2)

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maxima [B]  time = 0.46, size = 120, normalized size = 2.73 \[ \frac {b^{2} {\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b^{4} x + a b^{3} - b^{3}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{4} x + a b^{3} - b^{3}} - \frac {\arcsin \left (b x + a\right )}{b^{3}}\right )}}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.65, size = 229, normalized size = 5.20 \[ \frac {\left (\frac {a^2\,x}{b}+\frac {a\,\left (a^2-1\right )}{b^2}\right )\,\sqrt {1-{\left (a+b\,x\right )}^2}}{a^2+2\,a\,b\,x+b^2\,x^2-1}+\frac {b\,\ln \left (\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}-\frac {x\,b^2+a\,b}{\sqrt {-b^2}}\right )}{{\left (-b^2\right )}^{3/2}}+\frac {\left (\frac {a^2-1}{b^2}+\frac {a\,x}{b}\right )\,\sqrt {1-{\left (a+b\,x\right )}^2}}{a^2+2\,a\,b\,x+b^2\,x^2-1}+\frac {x\,\left (b^2\,\left (a^2-1\right )-2\,a^2\,b^2\right )-a\,b\,\left (a^2-1\right )}{b\,\left (b^2\,\left (a^2-1\right )-a^2\,b^2\right )\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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