3.836 \(\int e^{3 \tanh ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=168 \[ \frac{\left (6 a^2-18 a+11\right ) \sqrt{-a-b x+1} (a+b x+1)^{3/2}}{6 b^3}+\frac{\left (6 a^2-18 a+11\right ) \sqrt{-a-b x+1} \sqrt{a+b x+1}}{2 b^3}-\frac{\left (6 a^2-18 a+11\right ) \sin ^{-1}(a+b x)}{2 b^3}+\frac{\sqrt{-a-b x+1} (a+b x+1)^{5/2}}{3 b^3}+\frac{(1-a)^2 (a+b x+1)^{5/2}}{b^3 \sqrt{-a-b x+1}} \]

[Out]

((11 - 18*a + 6*a^2)*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(2*b^3) + ((11 - 18*a + 6*a^2)*Sqrt[1 - a - b*x]*(1
+ a + b*x)^(3/2))/(6*b^3) + ((1 - a)^2*(1 + a + b*x)^(5/2))/(b^3*Sqrt[1 - a - b*x]) + (Sqrt[1 - a - b*x]*(1 +
a + b*x)^(5/2))/(3*b^3) - ((11 - 18*a + 6*a^2)*ArcSin[a + b*x])/(2*b^3)

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Rubi [A]  time = 0.191308, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6163, 89, 80, 50, 53, 619, 216} \[ \frac{\left (6 a^2-18 a+11\right ) \sqrt{-a-b x+1} (a+b x+1)^{3/2}}{6 b^3}+\frac{\left (6 a^2-18 a+11\right ) \sqrt{-a-b x+1} \sqrt{a+b x+1}}{2 b^3}-\frac{\left (6 a^2-18 a+11\right ) \sin ^{-1}(a+b x)}{2 b^3}+\frac{\sqrt{-a-b x+1} (a+b x+1)^{5/2}}{3 b^3}+\frac{(1-a)^2 (a+b x+1)^{5/2}}{b^3 \sqrt{-a-b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a + b*x])*x^2,x]

[Out]

((11 - 18*a + 6*a^2)*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(2*b^3) + ((11 - 18*a + 6*a^2)*Sqrt[1 - a - b*x]*(1
+ a + b*x)^(3/2))/(6*b^3) + ((1 - a)^2*(1 + a + b*x)^(5/2))/(b^3*Sqrt[1 - a - b*x]) + (Sqrt[1 - a - b*x]*(1 +
a + b*x)^(5/2))/(3*b^3) - ((11 - 18*a + 6*a^2)*ArcSin[a + b*x])/(2*b^3)

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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^{3 \tanh ^{-1}(a+b x)} x^2 \, dx &=\int \frac{x^2 (1+a+b x)^{3/2}}{(1-a-b x)^{3/2}} \, dx\\ &=\frac{(1-a)^2 (1+a+b x)^{5/2}}{b^3 \sqrt{1-a-b x}}-\frac{\int \frac{(1+a+b x)^{3/2} \left ((3-2 a) (1-a) b+b^2 x\right )}{\sqrt{1-a-b x}} \, dx}{b^3}\\ &=\frac{(1-a)^2 (1+a+b x)^{5/2}}{b^3 \sqrt{1-a-b x}}+\frac{\sqrt{1-a-b x} (1+a+b x)^{5/2}}{3 b^3}-\frac{\left (11-18 a+6 a^2\right ) \int \frac{(1+a+b x)^{3/2}}{\sqrt{1-a-b x}} \, dx}{3 b^2}\\ &=\frac{\left (11-18 a+6 a^2\right ) \sqrt{1-a-b x} (1+a+b x)^{3/2}}{6 b^3}+\frac{(1-a)^2 (1+a+b x)^{5/2}}{b^3 \sqrt{1-a-b x}}+\frac{\sqrt{1-a-b x} (1+a+b x)^{5/2}}{3 b^3}-\frac{\left (11-18 a+6 a^2\right ) \int \frac{\sqrt{1+a+b x}}{\sqrt{1-a-b x}} \, dx}{2 b^2}\\ &=\frac{\left (11-18 a+6 a^2\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 b^3}+\frac{\left (11-18 a+6 a^2\right ) \sqrt{1-a-b x} (1+a+b x)^{3/2}}{6 b^3}+\frac{(1-a)^2 (1+a+b x)^{5/2}}{b^3 \sqrt{1-a-b x}}+\frac{\sqrt{1-a-b x} (1+a+b x)^{5/2}}{3 b^3}-\frac{\left (11-18 a+6 a^2\right ) \int \frac{1}{\sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx}{2 b^2}\\ &=\frac{\left (11-18 a+6 a^2\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 b^3}+\frac{\left (11-18 a+6 a^2\right ) \sqrt{1-a-b x} (1+a+b x)^{3/2}}{6 b^3}+\frac{(1-a)^2 (1+a+b x)^{5/2}}{b^3 \sqrt{1-a-b x}}+\frac{\sqrt{1-a-b x} (1+a+b x)^{5/2}}{3 b^3}-\frac{\left (11-18 a+6 a^2\right ) \int \frac{1}{\sqrt{(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx}{2 b^2}\\ &=\frac{\left (11-18 a+6 a^2\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 b^3}+\frac{\left (11-18 a+6 a^2\right ) \sqrt{1-a-b x} (1+a+b x)^{3/2}}{6 b^3}+\frac{(1-a)^2 (1+a+b x)^{5/2}}{b^3 \sqrt{1-a-b x}}+\frac{\sqrt{1-a-b x} (1+a+b x)^{5/2}}{3 b^3}+\frac{\left (11-18 a+6 a^2\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 )}{4 b^4}\\ &=\frac{\left (11-18 a+6 a^2\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 b^3}+\frac{\left (11-18 a+6 a^2\right ) \sqrt{1-a-b x} (1+a+b x)^{3/2}}{6 b^3}+\frac{(1-a)^2 (1+a+b x)^{5/2}}{b^3 \sqrt{1-a-b x}}+\frac{\sqrt{1-a-b x} (1+a+b x)^{5/2}}{3 b^3}-\frac{\left (11-18 a+6 a^2\right ) \sin ^{-1}(a+b x)}{2 b^3}\\ \end{align*}

Mathematica [A]  time = 0.217007, size = 170, normalized size = 1.01 \[ \frac{-\frac{\sqrt{b} \sqrt{a+b x+1} \left (2 a^3-53 a^2+a (103-16 b x)+2 b^3 x^3+7 b^2 x^2+19 b x-52\right )}{\sqrt{-a-b x+1}}+6 \left (6 a^2+11\right ) \sqrt{-b} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{-a-b x+1}}{\sqrt{2} \sqrt{-b}}\right )+108 a \sqrt{-b} \sinh ^{-1}\left (\frac{\sqrt{-b} \sqrt{-a-b x+1}}{\sqrt{2} \sqrt{b}}\right )}{6 b^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcTanh[a + b*x])*x^2,x]

[Out]

(-((Sqrt[b]*Sqrt[1 + a + b*x]*(-52 - 53*a^2 + 2*a^3 + 19*b*x + 7*b^2*x^2 + 2*b^3*x^3 + a*(103 - 16*b*x)))/Sqrt
[1 - a - b*x]) + 108*a*Sqrt[-b]*ArcSinh[(Sqrt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])] + 6*(11 + 6*a^2)*Sqrt[
-b]*ArcSinh[(Sqrt[b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[-b])])/(6*b^(7/2))

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Maple [B]  time = 0.051, size = 552, normalized size = 3.3 \begin{align*}{\frac{23\,{a}^{2}x}{2\,{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{3\,{x}^{3}}{2}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{b{x}^{4}}{3}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{{x}^{3}a}{3}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{{a}^{4}}{3\,{b}^{3}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{25\,{a}^{2}}{3\,{b}^{3}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{13\,{x}^{2}}{3\,b}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}+9\,{\frac{a}{{b}^{2}\sqrt{{b}^{2}}}\arctan \left ({\frac{\sqrt{{b}^{2}}}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}} \left ( x+{\frac{a}{b}} \right ) } \right ) }-{\frac{11}{2\,{b}^{2}}\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}}}}}-3\,{\frac{{a}^{2}}{{b}^{2}\sqrt{{b}^{2}}}\arctan \left ({\frac{\sqrt{{b}^{2}}}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}} \left ( x+{\frac{a}{b}} \right ) } \right ) }-{\frac{17\,a}{2\,{b}^{3}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}+{\frac{17\,{a}^{3}}{2\,{b}^{3}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{x{a}^{3}}{3\,{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{53\,ax}{3\,{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}+{\frac{26}{3\,{b}^{3}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}+{\frac{11\,x}{2\,{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}+{\frac{3\,a{x}^{2}}{2\,b}{\frac{1}{\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)^3/(1-(b*x+a)^2)^(3/2)*x^2,x)

[Out]

23/2*a^2/b^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x-3/2*x^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/3*b*x^4/(-b^2*x^2-2*a*b
*x-a^2+1)^(1/2)-1/3*a*x^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/3/b^3*a^4/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-25/3/b^3*a
^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-13/3/b*x^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+9*a/b^2/(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))-11/2/b^2/(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*a^2/b^2/(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))-17/2*a/b^3/(-b
^2*x^2-2*a*b*x-a^2+1)^(1/2)+17/2*a^3/b^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/3/b^2*a^3*x/(-b^2*x^2-2*a*b*x-a^2+1)
^(1/2)-53/3/b^2*a*x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+26/3/b^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+11/2*x/b^2/(-b^2*x^
2-2*a*b*x-a^2+1)^(1/2)+3/2/b*a*x^2/(-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)^3/(1-(b*x+a)^2)^(3/2)*x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.28028, size = 375, normalized size = 2.23 \begin{align*} \frac{3 \,{\left (6 \, a^{3} +{\left (6 \, a^{2} - 18 \, a + 11\right )} b x - 24 \, a^{2} + 29 \, a - 11\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 (2 \, b^{3} x^{3} + 7 \, b^{2} x^{2} + 2 \, a^{3} -{\left (16 \, a - 19\right )} b x - 53 \, a^{2} + 103 \, a - 52\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{6 \,{\left (b^{4} x +{\left (a - 1\right )} b^{3}\right )}} \end{align*}

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^2,x, algorithm="fricas")

[Out]

1/6*(3*(6*a^3 + (6*a^2 - 18*a + 11)*b*x - 24*a^2 + 29*a - 11)*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)) + (2*b^3*x^3 + 7*b^2*x^2 + 2*a^3 - (16*a - 19)*b*x - 53*a^2 + 103*a - 52)*s
qrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/(b^4*x + (a - 1)*b^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \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 \end{align*}

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**2,x)

[Out]

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

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Giac [A]  time = 1.18672, size = 200, normalized size = 1.19 \begin{align*} \frac{1}{6} \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (x{\left (\frac{2 \, x}{b} - \frac{2 \, a b^{6} - 9 \, b^{6}}{b^{8}}\right )} + \frac{2 \, a^{2} b^{5} - 27 \, a b^{5} + 28 \, b^{5}}{b^{8}}\right )} + \frac{{\left (6 \, a^{2} - 18 \, a + 11\right )} \arcsin \left (-b x - a\right ) \mathrm{sgn}\left (b\right )}{2 \, b^{2}{\left | b \right |}} + \frac{8 \,{\left (a^{2} - 2 \, a + 1\right )}}{b^{2}{\left (\frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left | b \right |} + b}{b^{2} x + a b} - 1\right )}{\left | b \right |}} \end{align*}

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^2,x, algorithm="giac")

[Out]

1/6*sqrt(-(b*x + a)^2 + 1)*(x*(2*x/b - (2*a*b^6 - 9*b^6)/b^8) + (2*a^2*b^5 - 27*a*b^5 + 28*b^5)/b^8) + 1/2*(6*
a^2 - 18*a + 11)*arcsin(-b*x - a)*sgn(b)/(b^2*abs(b)) + 8*(a^2 - 2*a + 1)/(b^2*((sqrt(-(b*x + a)^2 + 1)*abs(b)
+ b)/(b^2*x + a*b) - 1)*abs(b))