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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 187, normalized size of antiderivative = 1.00, 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} \[ \frac {(-a-b x+1)^{3/2} \sqrt {a+b x+1} \left (22 a^2-2 (10 a+11) b x+54 a+29\right )}{8 b^4}+\frac {3 \left (8 a^3+36 a^2+44 a+17\right ) \sqrt {-a-b x+1} \sqrt {a+b x+1}}{8 b^4}+\frac {3 \left (8 a^3+36 a^2+44 a+17\right ) \sin ^{-1}(a+b x)}{8 b^4}+\frac {9 x^2 (-a-b x+1)^{3/2} \sqrt {a+b x+1}}{4 b^2}-\frac {2 x^3 (-a-b x+1)^{3/2}}{b \sqrt {a+b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/E^(3*ArcTanh[a + b*x]),x]

[Out]

(-2*x^3*(1 - a - b*x)^(3/2))/(b*Sqrt[1 + a + b*x]) + (3*(17 + 44*a + 36*a^2 + 8*a^3)*Sqrt[1 - a - b*x]*Sqrt[1
+ a + b*x])/(8*b^4) + (9*x^2*(1 - a - b*x)^(3/2)*Sqrt[1 + a + b*x])/(4*b^2) + ((1 - a - b*x)^(3/2)*Sqrt[1 + a
+ b*x]*(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 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 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 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 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 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^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 \left (3 (1-a)-\frac {9 b x}{2}\right ) \sqrt {1-a-b x}}{\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 (1-a-b x)^{3/2} \sqrt {1+a+b x}}{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 (1-a-b x)^{3/2} \sqrt {1+a+b x}}{4 b^2}+\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x} \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 {2 x^3 (1-a-b x)^{3/2}}{b \sqrt {1+a+b x}}+\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 {9 x^2 (1-a-b x)^{3/2} \sqrt {1+a+b x}}{4 b^2}+\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x} \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 {2 x^3 (1-a-b x)^{3/2}}{b \sqrt {1+a+b x}}+\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 {9 x^2 (1-a-b x)^{3/2} \sqrt {1+a+b x}}{4 b^2}+\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x} \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 {2 x^3 (1-a-b x)^{3/2}}{b \sqrt {1+a+b x}}+\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 {9 x^2 (1-a-b x)^{3/2} \sqrt {1+a+b x}}{4 b^2}+\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x} \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 {2 x^3 (1-a-b x)^{3/2}}{b \sqrt {1+a+b x}}+\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 {9 x^2 (1-a-b x)^{3/2} \sqrt {1+a+b x}}{4 b^2}+\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x} \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 = 0.21, size = 231, normalized size = 1.24 \[ \frac {6 \left (8 a^3+36 a^2+44 a+17\right ) \sqrt {b} \sqrt {-a^2-2 a b x-b^2 x^2+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {-a-b x+1}}{\sqrt {2} \sqrt {-b}}\right )+\sqrt {-b} \left (-2 a^5-2 a^4 (b x+38)-5 a^3 (20 b x+31)-a^2 \left (12 b^2 x^2+265 b x+4\right )+a \left (2 b^4 x^4+4 b^3 x^3-53 b^2 x^2-212 b x+157\right )+2 b^5 x^5-8 b^4 x^4+17 b^3 x^3-40 b^2 x^2-51 b x+80\right )}{8 (-b)^{9/2} \sqrt {-((a+b x-1) (a+b x+1))}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[-b]*(80 - 2*a^5 - 51*b*x - 40*b^2*x^2 + 17*b^3*x^3 - 8*b^4*x^4 + 2*b^5*x^5 - 2*a^4*(38 + b*x) - 5*a^3*(3
1 + 20*b*x) - a^2*(4 + 265*b*x + 12*b^2*x^2) + a*(157 - 212*b*x - 53*b^2*x^2 + 4*b^3*x^3 + 2*b^4*x^4)) + 6*(17
 + 44*a + 36*a^2 + 8*a^3)*Sqrt[b]*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSinh[(Sqrt[b]*Sqrt[1 - a - b*x])/(Sqrt[
2]*Sqrt[-b])])/(8*(-b)^(9/2)*Sqrt[-((-1 + a + b*x)*(1 + a + b*x))])

________________________________________________________________________________________

fricas [A]  time = 0.82, size = 191, normalized size = 1.02 \[ -\frac {3 \, {\left (8 \, a^{4} + 44 \, a^{3} + {\left (8 \, a^{3} + 36 \, a^{2} + 44 \, a + 17\right )} b x + 80 \, a^{2} + 61 \, a + 17\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + {\left (2 \, b^{4} x^{4} - 6 \, b^{3} x^{3} + {\left (10 \, a + 11\right )} b^{2} x^{2} - 2 \, a^{4} - 78 \, a^{3} - {\left (22 \, a^{2} + 54 \, a + 29\right )} b x - 233 \, a^{2} - 237 \, a - 80\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{8 \, {\left (b^{5} x + {\left (a + 1\right )} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

giac [A]  time = 0.23, size = 211, normalized size = 1.13 \[ -\frac {1}{8} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left ({\left (2 \, x {\left (\frac {x}{b} - \frac {a b^{11} + 4 \, b^{11}}{b^{13}}\right )} + \frac {2 \, a^{2} b^{10} + 20 \, a b^{10} + 19 \, b^{10}}{b^{13}}\right )} x - \frac {2 \, a^{3} b^{9} + 44 \, a^{2} b^{9} + 93 \, a b^{9} + 48 \, b^{9}}{b^{13}}\right )} - \frac {3 \, {\left (8 \, a^{3} + 36 \, a^{2} + 44 \, a + 17\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{8 \, b^{3} {\left | b \right |}} - \frac {8 \, {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )}}{b^{3} {\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} + 1\right )} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(-b^2*x^2 - 2*a*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^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(
b^2*x + a*b) + 1)*abs(b))

________________________________________________________________________________________

maple [B]  time = 0.05, size = 1271, normalized size = 6.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/2/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x*a^2+33/2/b^3*a*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2
)*x+3/b^7/(x+1/b+a/b)^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)*a^2+3/b^7/(x+1/b+a/b)^3*(-(x+(1+a)/b)^2*b^2
+2*b*(x+(1+a)/b))^(5/2)*a+2/b^6/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)*a^3+9/b^6/(x+1/b+a/b)
^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)*a^2+12/b^6/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5
/2)*a+3/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x*a^3+1/b^7/(x+1/b+a/b)^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1
+a)/b))^(5/2)*a^3+33/2/b^3/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))
^(1/2))*a+27/2/b^3/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))*
a^2+3/b^3/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))*a^3+1/b^7
/(x+1/b+a/b)^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)+5/b^6/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)
/b))^(5/2)+2/b^4*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)*a^3+9/b^4*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/
2)*a^2+3/b^4*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a^4+6/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x
+11/b^4*a*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)+27/2/b^4*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a^3+3
3/2/b^4*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a^2+6/b^4*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a+6/b^
3/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))+1/4/b^3*x*(-b^2*x
^2-2*a*b*x-a^2+1)^(3/2)+1/4/b^4*a*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)+3/8*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))+4/b^4*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)

________________________________________________________________________________________

maxima [C]  time = 0.43, size = 985, normalized size = 5.27 \[ -\frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} a^{3}}{b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4} + 2 \, b^{5} x + 2 \, a b^{4} + b^{4}} - \frac {3 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} a^{2}}{b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4} + 2 \, b^{5} x + 2 \, a b^{4} + b^{4}} + \frac {3 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} a^{2}}{2 \, {\left (b^{5} x + a b^{4} + b^{4}\right )}} + \frac {6 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{3}}{b^{5} x + a b^{4} + b^{4}} - \frac {3 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} a}{b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4} + 2 \, b^{5} x + 2 \, a b^{4} + b^{4}} + \frac {3 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} a}{b^{5} x + a b^{4} + b^{4}} + \frac {18 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2}}{b^{5} x + a b^{4} + b^{4}} - \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4} + 2 \, b^{5} x + 2 \, a b^{4} + b^{4}} + \frac {3 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{2 \, {\left (b^{5} x + a b^{4} + b^{4}\right )}} + \frac {18 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b^{5} x + a b^{4} + b^{4}} + \frac {3 \, a^{3} \arcsin \left (b x + a\right )}{b^{4}} + \frac {6 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{5} x + a b^{4} + b^{4}} + \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, b^{3}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 4 \, b x + 4 \, a + 3} a x}{2 \, b^{3}} + \frac {27 \, a^{2} \arcsin \left (b x + a\right )}{2 \, b^{4}} - \frac {3 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} a}{4 \, b^{4}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 4 \, b x + 4 \, a + 3} a^{2}}{2 \, b^{4}} + \frac {9 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2}}{2 \, b^{4}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 4 \, b x + 4 \, a + 3} x}{2 \, b^{3}} + \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x}{8 \, b^{3}} + \frac {3 i \, a \arcsin \left (b x + a + 2\right )}{2 \, b^{4}} + \frac {18 \, a \arcsin \left (b x + a\right )}{b^{4}} - \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{b^{4}} - \frac {9 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 4 \, b x + 4 \, a + 3} a}{2 \, b^{4}} + \frac {75 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{8 \, b^{4}} + \frac {3 i \, \arcsin \left (b x + a + 2\right )}{2 \, b^{4}} + \frac {63 \, \arcsin \left (b x + a\right )}{8 \, b^{4}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 4 \, b x + 4 \, a + 3}}{b^{4}} + \frac {9 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*a^3/(b^6*x^2 + 2*a*b^5*x + a^2*b^4 + 2*b^5*x + 2*a*b^4 + b^4) - 3*(-b^2*
x^2 - 2*a*b*x - a^2 + 1)^(3/2)*a^2/(b^6*x^2 + 2*a*b^5*x + a^2*b^4 + 2*b^5*x + 2*a*b^4 + b^4) + 3/2*(-b^2*x^2 -
 2*a*b*x - a^2 + 1)^(3/2)*a^2/(b^5*x + a*b^4 + b^4) + 6*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a^3/(b^5*x + a*b^4
+ b^4) - 3*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*a/(b^6*x^2 + 2*a*b^5*x + a^2*b^4 + 2*b^5*x + 2*a*b^4 + b^4) +
3*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*a/(b^5*x + a*b^4 + b^4) + 18*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a^2/(b^
5*x + a*b^4 + b^4) - (-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/(b^6*x^2 + 2*a*b^5*x + a^2*b^4 + 2*b^5*x + 2*a*b^4 +
 b^4) + 3/2*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/(b^5*x + a*b^4 + b^4) + 18*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)
*a/(b^5*x + a*b^4 + b^4) + 3*a^3*arcsin(b*x + a)/b^4 + 6*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)/(b^5*x + a*b^4 + b
^4) + 1/4*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*x/b^3 - 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 4*b*x + 4*a + 3)*a*x
/b^3 + 27/2*a^2*arcsin(b*x + a)/b^4 - 3/4*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*a/b^4 - 3/2*sqrt(b^2*x^2 + 2*a*
b*x + a^2 + 4*b*x + 4*a + 3)*a^2/b^4 + 9/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a^2/b^4 - 3/2*sqrt(b^2*x^2 + 2*a
*b*x + a^2 + 4*b*x + 4*a + 3)*x/b^3 + 3/8*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*x/b^3 + 3/2*I*a*arcsin(b*x + a +
2)/b^4 + 18*a*arcsin(b*x + a)/b^4 - (-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/b^4 - 9/2*sqrt(b^2*x^2 + 2*a*b*x + a^
2 + 4*b*x + 4*a + 3)*a/b^4 + 75/8*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a/b^4 + 3/2*I*arcsin(b*x + a + 2)/b^4 + 6
3/8*arcsin(b*x + a)/b^4 - 3*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 4*b*x + 4*a + 3)/b^4 + 9/2*sqrt(-b^2*x^2 - 2*a*b*x
- a^2 + 1)/b^4

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________