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

Optimal. Leaf size=187 \[ \frac{\sqrt{-a-b x+1} (a+b x+1)^{3/2} \left (22 a^2+2 (11-10 a) b x-54 a+29\right )}{8 b^4}+\frac{3 \left (-8 a^3+36 a^2-44 a+17\right ) \sqrt{-a-b x+1} \sqrt{a+b x+1}}{8 b^4}-\frac{3 \left (-8 a^3+36 a^2-44 a+17\right ) \sin ^{-1}(a+b x)}{8 b^4}+\frac{9 x^2 \sqrt{-a-b x+1} (a+b x+1)^{3/2}}{4 b^2}+\frac{2 x^3 (a+b x+1)^{3/2}}{b \sqrt{-a-b x+1}} \]

[Out]

(3*(17 - 44*a + 36*a^2 - 8*a^3)*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(8*b^4) + (2*x^3*(1 + a + b*x)^(3/2))/(b*
Sqrt[1 - a - b*x]) + (9*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)*(29 - 54*a + 22*a^2 + 2*(11 - 10*a)*b*x))/(8*b^4) - (3*(17 - 44*a + 36*a^2 - 8*a^3)*ArcSin[a + b*x])/(8*
b^4)

________________________________________________________________________________________

Rubi [A]  time = 0.182205, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {6163, 97, 153, 147, 50, 53, 619, 216} \[ \frac{\sqrt{-a-b x+1} (a+b x+1)^{3/2} \left (22 a^2+2 (11-10 a) b x-54 a+29\right )}{8 b^4}+\frac{3 \left (-8 a^3+36 a^2-44 a+17\right ) \sqrt{-a-b x+1} \sqrt{a+b x+1}}{8 b^4}-\frac{3 \left (-8 a^3+36 a^2-44 a+17\right ) \sin ^{-1}(a+b x)}{8 b^4}+\frac{9 x^2 \sqrt{-a-b x+1} (a+b x+1)^{3/2}}{4 b^2}+\frac{2 x^3 (a+b x+1)^{3/2}}{b \sqrt{-a-b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a + b*x])*x^3,x]

[Out]

(3*(17 - 44*a + 36*a^2 - 8*a^3)*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(8*b^4) + (2*x^3*(1 + a + b*x)^(3/2))/(b*
Sqrt[1 - a - b*x]) + (9*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)*(29 - 54*a + 22*a^2 + 2*(11 - 10*a)*b*x))/(8*b^4) - (3*(17 - 44*a + 36*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 97

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

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 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^{3 \tanh ^{-1}(a+b x)} x^3 \, dx &=\int \frac{x^3 (1+a+b x)^{3/2}}{(1-a-b x)^{3/2}} \, dx\\ &=\frac{2 x^3 (1+a+b x)^{3/2}}{b \sqrt{1-a-b x}}-\frac{2 \int \frac{x^2 \sqrt{1+a+b x} \left (3 (1+a)+\frac{9 b x}{2}\right )}{\sqrt{1-a-b x}} \, dx}{b}\\ &=\frac{2 x^3 (1+a+b x)^{3/2}}{b \sqrt{1-a-b x}}+\frac{9 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 (-9 (1-a) (1+a) b-\frac{3}{2} (11-10 a) b^2 x\right )}{\sqrt{1-a-b x}} \, dx}{2 b^3}\\ &=\frac{2 x^3 (1+a+b x)^{3/2}}{b \sqrt{1-a-b x}}+\frac{9 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 (29-54 a+22 a^2+2 (11-10 a) b x\right )}{8 b^4}-\frac{\left (3 \left (17-44 a+36 a^2-8 a^3\right )\right ) \int \frac{\sqrt{1+a+b x}}{\sqrt{1-a-b x}} \, dx}{8 b^3}\\ &=\frac{3 \left (17-44 a+36 a^2-8 a^3\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{8 b^4}+\frac{2 x^3 (1+a+b x)^{3/2}}{b \sqrt{1-a-b x}}+\frac{9 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 (29-54 a+22 a^2+2 (11-10 a) b x\right )}{8 b^4}-\frac{\left (3 \left (17-44 a+36 a^2-8 a^3\right )\right ) \int \frac{1}{\sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx}{8 b^3}\\ &=\frac{3 \left (17-44 a+36 a^2-8 a^3\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{8 b^4}+\frac{2 x^3 (1+a+b x)^{3/2}}{b \sqrt{1-a-b x}}+\frac{9 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 (29-54 a+22 a^2+2 (11-10 a) b x\right )}{8 b^4}-\frac{\left (3 \left (17-44 a+36 a^2-8 a^3\right )\right ) \int \frac{1}{\sqrt{(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx}{8 b^3}\\ &=\frac{3 \left (17-44 a+36 a^2-8 a^3\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{8 b^4}+\frac{2 x^3 (1+a+b x)^{3/2}}{b \sqrt{1-a-b x}}+\frac{9 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 (29-54 a+22 a^2+2 (11-10 a) b x\right )}{8 b^4}+\frac{\left (3 \left (17-44 a+36 a^2-8 a^3\right )\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{3 \left (17-44 a+36 a^2-8 a^3\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{8 b^4}+\frac{2 x^3 (1+a+b x)^{3/2}}{b \sqrt{1-a-b x}}+\frac{9 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 (29-54 a+22 a^2+2 (11-10 a) b x\right )}{8 b^4}-\frac{3 \left (17-44 a+36 a^2-8 a^3\right ) \sin ^{-1}(a+b x)}{8 b^4}\\ \end{align*}

Mathematica [A]  time = 0.284744, size = 203, normalized size = 1.09 \[ \frac{-\frac{\sqrt{b} \sqrt{a+b x+1} \left (a^2 (22 b x-233)-2 a^4+78 a^3+a \left (-10 b^2 x^2-54 b x+237\right )+2 b^4 x^4+6 b^3 x^3+11 b^2 x^2+29 b x-80\right )}{\sqrt{-a-b x+1}}+24 a \left (2 a^2+11\right ) \sqrt{-b} \sinh ^{-1}\left (\frac{\sqrt{-b} \sqrt{-a-b x+1}}{\sqrt{2} \sqrt{b}}\right )+6 \left (36 a^2+17\right ) \sqrt{-b} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{-a-b x+1}}{\sqrt{2} \sqrt{-b}}\right )}{8 b^{9/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-((Sqrt[b]*Sqrt[1 + a + b*x]*(-80 + 78*a^3 - 2*a^4 + 29*b*x + 11*b^2*x^2 + 6*b^3*x^3 + 2*b^4*x^4 + a^2*(-233
+ 22*b*x) + a*(237 - 54*b*x - 10*b^2*x^2)))/Sqrt[1 - a - b*x]) + 24*a*(11 + 2*a^2)*Sqrt[-b]*ArcSinh[(Sqrt[-b]*
Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])] + 6*(17 + 36*a^2)*Sqrt[-b]*ArcSinh[(Sqrt[b]*Sqrt[1 - a - b*x])/(Sqrt[2]*
Sqrt[-b])])/(8*b^(9/2))

________________________________________________________________________________________

Maple [B]  time = 0.088, size = 756, normalized size = 4. \begin{align*} \text{result too large to display} \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^3,x)

[Out]

-157/8/b^4*a/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/4*b*x^5/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/4*a*x^4/(-b^2*x^2-2*a*b
*x-a^2+1)^(1/2)-17/8/b*x^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+51/8/b^3*x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/4/b^4*a^
5/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+155/8/b^4*a^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+3*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))-1/2*a^2/b^4/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-19/2*a^4/b^4/(-b^2
*x^2-2*a*b*x-a^2+1)^(1/2)+33/2*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))+1/
2/b*a*x^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-3/2*x^2/b^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a^2-x^4/(-b^2*x^2-2*a*b*x-
a^2+1)^(1/2)-53/2*a/b^3*x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-25/2*a^3/b^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x-27/2*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))-51/8/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))-5*x^2/b^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+265/8/b^3*a^2*x/(-b^2*
x^2-2*a*b*x-a^2+1)^(1/2)+53/8/b^2*a*x^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/4/b^3*a^4*x/(-b^2*x^2-2*a*b*x-a^2+1)^
(1/2)+10/b^4/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)

________________________________________________________________________________________

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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.59418, size = 456, normalized size = 2.44 \begin{align*} -\frac{3 \,{\left (8 \, a^{4} - 44 \, a^{3} +{\left (8 \, a^{3} - 36 \, a^{2} + 44 \, a - 17\right )} b x + 80 \, a^{2} - 61 \, a + 17\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^{4} x^{4} + 6 \, b^{3} x^{3} -{\left (10 \, a - 11\right )} b^{2} x^{2} - 2 \, a^{4} + 78 \, a^{3} +{\left (22 \, a^{2} - 54 \, a + 29\right )} b x - 233 \, a^{2} + 237 \, a - 80\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{8 \,{\left (b^{5} x +{\left (a - 1\right )} b^{4}\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^3,x, algorithm="fricas")

[Out]

-1/8*(3*(8*a^4 - 44*a^3 + (8*a^3 - 36*a^2 + 44*a - 17)*b*x + 80*a^2 - 61*a + 17)*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^4*x^4 + 6*b^3*x^3 - (10*a - 11)*b^2*x^2 - 2*a^4 +
78*a^3 + (22*a^2 - 54*a + 29)*b*x - 233*a^2 + 237*a - 80)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/(b^5*x + (a - 1
)*b^4)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \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**3,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.26128, size = 261, normalized size = 1.4 \begin{align*} \frac{1}{8} \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left ({\left (2 \, x{\left (\frac{x}{b} - \frac{a b^{11} - 4 \, b^{11}}{b^{13}}\right )} + \frac{2 \, a^{2} b^{10} - 20 \, a b^{10} + 19 \, b^{10}}{b^{13}}\right )} x - \frac{2 \, a^{3} b^{9} - 44 \, a^{2} b^{9} + 93 \, a b^{9} - 48 \, b^{9}}{b^{13}}\right )} - \frac{3 \,{\left (8 \, a^{3} - 36 \, a^{2} + 44 \, a - 17\right )} \arcsin \left (-b x - a\right ) \mathrm{sgn}\left (b\right )}{8 \, b^{3}{\left | b \right |}} - \frac{8 \,{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )}}{b^{3}{\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^3,x, algorithm="giac")

[Out]

1/8*sqrt(-(b*x + a)^2 + 1)*((2*x*(x/b - (a*b^11 - 4*b^11)/b^13) + (2*a^2*b^10 - 20*a*b^10 + 19*b^10)/b^13)*x -
(2*a^3*b^9 - 44*a^2*b^9 + 93*a*b^9 - 48*b^9)/b^13) - 3/8*(8*a^3 - 36*a^2 + 44*a - 17)*arcsin(-b*x - a)*sgn(b)
/(b^3*abs(b)) - 8*(a^3 - 3*a^2 + 3*a - 1)/(b^3*((sqrt(-(b*x + a)^2 + 1)*abs(b) + b)/(b^2*x + a*b) - 1)*abs(b))