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

Optimal. Leaf size=156 \[ -\frac{\sqrt{-a-b x+1} (a+b x+1)^{3/2} \left (18 a^2+2 (1-6 a) b x-10 a+7\right )}{24 b^4}-\frac{\left (-8 a^3+12 a^2-12 a+3\right ) \sqrt{-a-b x+1} \sqrt{a+b x+1}}{8 b^4}+\frac{\left (-8 a^3+12 a^2-12 a+3\right ) \sin ^{-1}(a+b x)}{8 b^4}-\frac{x^2 \sqrt{-a-b x+1} (a+b x+1)^{3/2}}{4 b^2} \]

[Out]

-((3 - 12*a + 12*a^2 - 8*a^3)*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(8*b^4) - (x^2*Sqrt[1 - a - b*x]*(1 + a + b
*x)^(3/2))/(4*b^2) - (Sqrt[1 - a - b*x]*(1 + a + b*x)^(3/2)*(7 - 10*a + 18*a^2 + 2*(1 - 6*a)*b*x))/(24*b^4) +
((3 - 12*a + 12*a^2 - 8*a^3)*ArcSin[a + b*x])/(8*b^4)

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Rubi [A]  time = 0.159706, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {6163, 100, 147, 50, 53, 619, 216} \[ -\frac{\sqrt{-a-b x+1} (a+b x+1)^{3/2} \left (18 a^2+2 (1-6 a) b x-10 a+7\right )}{24 b^4}-\frac{\left (-8 a^3+12 a^2-12 a+3\right ) \sqrt{-a-b x+1} \sqrt{a+b x+1}}{8 b^4}+\frac{\left (-8 a^3+12 a^2-12 a+3\right ) \sin ^{-1}(a+b x)}{8 b^4}-\frac{x^2 \sqrt{-a-b x+1} (a+b x+1)^{3/2}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a + b*x]*x^3,x]

[Out]

-((3 - 12*a + 12*a^2 - 8*a^3)*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(8*b^4) - (x^2*Sqrt[1 - a - b*x]*(1 + a + b
*x)^(3/2))/(4*b^2) - (Sqrt[1 - a - b*x]*(1 + a + b*x)^(3/2)*(7 - 10*a + 18*a^2 + 2*(1 - 6*a)*b*x))/(24*b^4) +
((3 - 12*a + 12*a^2 - 8*a^3)*ArcSin[a + b*x])/(8*b^4)

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]

Rule 100

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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 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 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]

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(a+b x)} x^3 \, dx &=\int \frac{x^3 \sqrt{1+a+b x}}{\sqrt{1-a-b x}} \, dx\\ &=-\frac{x^2 \sqrt{1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac{\int \frac{x \sqrt{1+a+b x} \left (-2 \left (1-a^2\right )-(1-6 a) b x\right )}{\sqrt{1-a-b x}} \, dx}{4 b^2}\\ &=-\frac{x^2 \sqrt{1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac{\sqrt{1-a-b x} (1+a+b x)^{3/2} \left (7-10 a+18 a^2+2 (1-6 a) b x\right )}{24 b^4}+\frac{\left (3-12 a+12 a^2-8 a^3\right ) \int \frac{\sqrt{1+a+b x}}{\sqrt{1-a-b x}} \, dx}{8 b^3}\\ &=-\frac{\left (3-12 a+12 a^2-8 a^3\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{8 b^4}-\frac{x^2 \sqrt{1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac{\sqrt{1-a-b x} (1+a+b x)^{3/2} \left (7-10 a+18 a^2+2 (1-6 a) b x\right )}{24 b^4}+\frac{\left (3-12 a+12 a^2-8 a^3\right ) \int \frac{1}{\sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx}{8 b^3}\\ &=-\frac{\left (3-12 a+12 a^2-8 a^3\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{8 b^4}-\frac{x^2 \sqrt{1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac{\sqrt{1-a-b x} (1+a+b x)^{3/2} \left (7-10 a+18 a^2+2 (1-6 a) b x\right )}{24 b^4}+\frac{\left (3-12 a+12 a^2-8 a^3\right ) \int \frac{1}{\sqrt{(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx}{8 b^3}\\ &=-\frac{\left (3-12 a+12 a^2-8 a^3\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{8 b^4}-\frac{x^2 \sqrt{1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac{\sqrt{1-a-b x} (1+a+b x)^{3/2} \left (7-10 a+18 a^2+2 (1-6 a) b x\right )}{24 b^4}-\frac{\left (3-12 a+12 a^2-8 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4 b^2}}} \, dx,x,-2 a b-2 b^2 x\right )}{16 b^5}\\ &=-\frac{\left (3-12 a+12 a^2-8 a^3\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{8 b^4}-\frac{x^2 \sqrt{1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac{\sqrt{1-a-b x} (1+a+b x)^{3/2} \left (7-10 a+18 a^2+2 (1-6 a) b x\right )}{24 b^4}+\frac{\left (3-12 a+12 a^2-8 a^3\right ) \sin ^{-1}(a+b x)}{8 b^4}\\ \end{align*}

Mathematica [A]  time = 0.392637, size = 149, normalized size = 0.96 \[ -\frac{\sqrt{-a^2-2 a b x-b^2 x^2+1} \left (a^2 (6 b x+44)-6 a^3-a \left (6 b^2 x^2+20 b x+39\right )+6 b^3 x^3+8 b^2 x^2+9 b x+16\right )}{24 b^4}-\frac{\left (8 a^3-12 a^2+12 a-3\right ) \sqrt{-b} \sinh ^{-1}\left (\frac{\sqrt{-b} \sqrt{-a-b x+1}}{\sqrt{2} \sqrt{b}}\right )}{4 b^{9/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a + b*x]*x^3,x]

[Out]

-(Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*(16 - 6*a^3 + 9*b*x + 8*b^2*x^2 + 6*b^3*x^3 + a^2*(44 + 6*b*x) - a*(39 + 2
0*b*x + 6*b^2*x^2)))/(24*b^4) - ((-3 + 12*a - 12*a^2 + 8*a^3)*Sqrt[-b]*ArcSinh[(Sqrt[-b]*Sqrt[1 - a - b*x])/(S
qrt[2]*Sqrt[b])])/(4*b^(9/2))

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Maple [B]  time = 0.067, size = 487, normalized size = 3.1 \begin{align*} -{\frac{{x}^{3}}{4\,b}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}+{\frac{a{x}^{2}}{4\,{b}^{2}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}-{\frac{{a}^{2}x}{4\,{b}^{3}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}+{\frac{{a}^{3}}{4\,{b}^{4}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}+{\frac{3\,{a}^{2}}{2\,{b}^{3}}\arctan \left ({\sqrt{{b}^{2}} \left ( x+{\frac{a}{b}} \right ){\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{13\,a}{8\,{b}^{4}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}-{\frac{3\,x}{8\,{b}^{3}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}+{\frac{3}{8\,{b}^{3}}\arctan \left ({\sqrt{{b}^{2}} \left ( x+{\frac{a}{b}} \right ){\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{{x}^{2}}{3\,{b}^{2}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}+{\frac{5\,ax}{6\,{b}^{3}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}-{\frac{11\,{a}^{2}}{6\,{b}^{4}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}-{\frac{{a}^{3}}{{b}^{3}}\arctan \left ({\sqrt{{b}^{2}} \left ( x+{\frac{a}{b}} \right ){\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{3\,a}{2\,{b}^{3}}\arctan \left ({\sqrt{{b}^{2}} \left ( x+{\frac{a}{b}} \right ){\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{2}{3\,{b}^{4}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^3/b*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/4*a/b^2*x^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/4*a^2/b^3*x*(-b^2*x^2
-2*a*b*x-a^2+1)^(1/2)+1/4*a^3/b^4*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+3/2*a^2/b^3/(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))+13/8*a/b^4*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-3/8/b^3*x*(-b^2*x^2-2*a*b*x-a^
2+1)^(1/2)+3/8/b^3/(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))-1/3*x^2/b^2*(-b^2*x^
2-2*a*b*x-a^2+1)^(1/2)+5/6*a/b^3*x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-11/6*a^2/b^4*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-
a^3/b^3/(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))-3/2*a/b^3/(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))-2/3/b^4*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.14665, size = 338, normalized size = 2.17 \begin{align*} \frac{3 \,{\left (8 \, a^{3} - 12 \, a^{2} + 12 \, 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 (6 \, b^{3} x^{3} - 2 \,{\left (3 \, a - 4\right )} b^{2} x^{2} - 6 \, a^{3} +{\left (6 \, a^{2} - 20 \, a + 9\right )} b x + 44 \, a^{2} - 39 \, a + 16\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{24 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(8*a^3 - 12*a^2 + 12*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)) - (6*b^3*x^3 - 2*(3*a - 4)*b^2*x^2 - 6*a^3 + (6*a^2 - 20*a + 9)*b*x + 44*a^2 - 39*a + 16)*sqrt(-b^2*x
^2 - 2*a*b*x - a^2 + 1))/b^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (a + b x + 1\right )}{\sqrt{- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21645, size = 188, normalized size = 1.21 \begin{align*} -\frac{1}{24} \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left ({\left (2 \, x{\left (\frac{3 \, x}{b} - \frac{3 \, a b^{5} - 4 \, b^{5}}{b^{7}}\right )} + \frac{6 \, a^{2} b^{4} - 20 \, a b^{4} + 9 \, b^{4}}{b^{7}}\right )} x - \frac{6 \, a^{3} b^{3} - 44 \, a^{2} b^{3} + 39 \, a b^{3} - 16 \, b^{3}}{b^{7}}\right )} + \frac{{\left (8 \, a^{3} - 12 \, a^{2} + 12 \, a - 3\right )} \arcsin \left (-b x - a\right ) \mathrm{sgn}\left (b\right )}{8 \, b^{3}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*sqrt(-(b*x + a)^2 + 1)*((2*x*(3*x/b - (3*a*b^5 - 4*b^5)/b^7) + (6*a^2*b^4 - 20*a*b^4 + 9*b^4)/b^7)*x - (
6*a^3*b^3 - 44*a^2*b^3 + 39*a*b^3 - 16*b^3)/b^7) + 1/8*(8*a^3 - 12*a^2 + 12*a - 3)*arcsin(-b*x - a)*sgn(b)/(b^
3*abs(b))