3.837 \(\int e^{3 \tanh ^{-1}(a+b x)} x \, dx\)

Optimal. Leaf size=121 \[ \frac {(1-a) (a+b x+1)^{5/2}}{b^2 \sqrt {-a-b x+1}}+\frac {(3-2 a) \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b^2}+\frac {3 (3-2 a) \sqrt {-a-b x+1} \sqrt {a+b x+1}}{2 b^2}-\frac {3 (3-2 a) \sin ^{-1}(a+b x)}{2 b^2} \]

[Out]

-3/2*(3-2*a)*arcsin(b*x+a)/b^2+(1-a)*(b*x+a+1)^(5/2)/b^2/(-b*x-a+1)^(1/2)+1/2*(3-2*a)*(b*x+a+1)^(3/2)*(-b*x-a+
1)^(1/2)/b^2+3/2*(3-2*a)*(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/b^2

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Rubi [A]  time = 0.10, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6163, 78, 50, 53, 619, 216} \[ \frac {(1-a) (a+b x+1)^{5/2}}{b^2 \sqrt {-a-b x+1}}+\frac {(3-2 a) \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b^2}+\frac {3 (3-2 a) \sqrt {-a-b x+1} \sqrt {a+b x+1}}{2 b^2}-\frac {3 (3-2 a) \sin ^{-1}(a+b x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a + b*x])*x,x]

[Out]

(3*(3 - 2*a)*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(2*b^2) + ((3 - 2*a)*Sqrt[1 - a - b*x]*(1 + a + b*x)^(3/2))/
(2*b^2) + ((1 - a)*(1 + a + b*x)^(5/2))/(b^2*Sqrt[1 - a - b*x]) - (3*(3 - 2*a)*ArcSin[a + b*x])/(2*b^2)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

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Mathematica [A]  time = 0.16, size = 142, normalized size = 1.17 \[ \frac {\frac {\sqrt {b} \sqrt {a+b x+1} \left (a^2-15 a-b^2 x^2-5 b x+14\right )}{\sqrt {-a-b x+1}}+12 a \sqrt {-b} \sinh ^{-1}\left (\frac {\sqrt {-b} \sqrt {-a-b x+1}}{\sqrt {2} \sqrt {b}}\right )+18 \sqrt {-b} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {-a-b x+1}}{\sqrt {2} \sqrt {-b}}\right )}{2 b^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcTanh[a + b*x])*x,x]

[Out]

((Sqrt[b]*Sqrt[1 + a + b*x]*(14 - 15*a + a^2 - 5*b*x - b^2*x^2))/Sqrt[1 - a - b*x] + 12*a*Sqrt[-b]*ArcSinh[(Sq
rt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])] + 18*Sqrt[-b]*ArcSinh[(Sqrt[b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[-
b])])/(2*b^(5/2))

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fricas [A]  time = 0.73, size = 131, normalized size = 1.08 \[ -\frac {3 \, {\left ({\left (2 \, a - 3\right )} b x + 2 \, a^{2} - 5 \, a + 3\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - {\left (b^{2} x^{2} - a^{2} + 5 \, b x + 15 \, a - 14\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \, {\left (b^{3} x + {\left (a - 1\right )} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*((2*a - 3)*b*x + 2*a^2 - 5*a + 3)*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)) - (b^2*x^2 - a^2 + 5*b*x + 15*a - 14)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/(b^3*x + (a - 1)*b^2)

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giac [A]  time = 0.25, size = 109, normalized size = 0.90 \[ \frac {1}{2} \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (\frac {x}{b} - \frac {a b^{2} - 6 \, b^{2}}{b^{4}}\right )} - \frac {3 \, {\left (2 \, a - 3\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{2 \, b {\left | b \right |}} - \frac {8 \, {\left (a - 1\right )}}{b {\left (\frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} - 1\right )} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(-(b*x + a)^2 + 1)*(x/b - (a*b^2 - 6*b^2)/b^4) - 3/2*(2*a - 3)*arcsin(-b*x - a)*sgn(b)/(b*abs(b)) - 8*
(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 = 0.05, size = 381, normalized size = 3.15 \[ -\frac {10 a x}{b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {3 a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b \sqrt {b^{2}}}+\frac {a^{2} x}{2 b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {3 x^{2}}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {7}{b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {9 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b \sqrt {b^{2}}}-\frac {7 a^{2}}{b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {a}{2 b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {9 x}{2 b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {b \,x^{3}}{2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {a \,x^{2}}{2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {a^{3}}{2 b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*a/b/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+3*a/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))+1/2/b*a^2*x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-3*x^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+7/b^2/(-b^2*x^2-2*a*b*
x-a^2+1)^(1/2)-9/2/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))-7*a^2/b^2/(-b^2*x^
2-2*a*b*x-a^2+1)^(1/2)-1/2/b^2*a/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+9/2/b*x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/2*b*x
^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/2*a*x^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/2/b^2*a^3/(-b^2*x^2-2*a*b*x-a^2+1
)^(1/2)

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maxima [B]  time = 0.43, size = 1137, normalized size = 9.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*a^4*b*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - 31/2*(a^2 - 1)*a^2*b*x/((a^2*b^2 -
(a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - 1/2*b*x^3/sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1) + 5/2*(a^2
- 1)*a^3/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + 6*(a^2*b + 2*a*b + b)*a^2*x/((a^2*b^
2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - 18*(a*b^2 + b^2)*a^3*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqr
t(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) + 3/2*(a^2 - 1)^2*b*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x -
a^2 + 1)) - (a^3 + 3*a^2 + 3*a + 1)*a*b*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + 5/2
*a*x^2/sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1) - 3/2*(a^2 - 1)^2*a/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b
*x - a^2 + 1)) - (a^3 + 3*a^2 + 3*a + 1)*a^2/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) -
3*(a^2*b + 2*a*b + b)*(a^2 - 1)*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + 15*(a*b^2 +
b^2)*(a^2 - 1)*a*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) + 15/2*a^2*arcsin(-(b^2*x
+ a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^2 - 3*(a*b^2 + b^2)*(a^2 - 1)*a^2/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^
2*x^2 - 2*a*b*x - a^2 + 1)*b^2) + 3*(a^2*b + 2*a*b + b)*(a^2 - 1)*a/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 -
2*a*b*x - a^2 + 1)*b) - 3*(a*b^2 + b^2)*x^2/(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^2) - 3/2*(a^2 - 1)*arcsin(-
(b^2*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^2 + 5*(a^2 - 1)*a/(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^2) - 9*
(a*b^2 + b^2)*a*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^4 + 3*(a^2*b + 2*a*b + b)*arcsin(-(b^2*
x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^3 + (a^3 + 3*a^2 + 3*a + 1)/(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^2)
- 6*(a*b^2 + b^2)*(a^2 - 1)/(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\left (a+b\,x+1\right )}^3}{{\left (1-{\left (a+b\,x\right )}^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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