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

Optimal. Leaf size=107 \[ -\frac {2 (a+1)^2 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(1-a) \sqrt {1-a^2}}+\frac {4 \sqrt {a+b x+1}}{(1-a) \sqrt {-a-b x+1}}-\sin ^{-1}(a+b x) \]

[Out]

-arcsin(b*x+a)-2*(1+a)^2*arctanh((1-a)^(1/2)*(b*x+a+1)^(1/2)/(1+a)^(1/2)/(-b*x-a+1)^(1/2))/(1-a)/(-a^2+1)^(1/2
)+4*(b*x+a+1)^(1/2)/(1-a)/(-b*x-a+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 107, 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, 98, 157, 53, 619, 216, 93, 208} \[ -\frac {2 (a+1)^2 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(1-a) \sqrt {1-a^2}}+\frac {4 \sqrt {a+b x+1}}{(1-a) \sqrt {-a-b x+1}}-\sin ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a + b*x])/x,x]

[Out]

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

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 98

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

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.59, size = 153, normalized size = 1.43 \[ \frac {2 \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {-b} \sqrt {-a-b x+1}}{\sqrt {2} \sqrt {b}}\right )}{\sqrt {-b}}-\frac {2 \left (2 \sqrt {a-1} \sqrt {a+b x+1}+(-a-1)^{3/2} \sqrt {-a-b x+1} \tanh ^{-1}\left (\frac {\sqrt {-a-1} \sqrt {-a-b x+1}}{\sqrt {a-1} \sqrt {a+b x+1}}\right )\right )}{(a-1)^{3/2} \sqrt {-a-b x+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcTanh[a + b*x])/x,x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.70, size = 439, normalized size = 4.10 \[ \left [\frac {{\left ({\left (a + 1\right )} b x + a^{2} - 1\right )} \sqrt {-\frac {a + 1}{a - 1}} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 4 \, a^{2} + 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{3} + {\left (a^{2} - a\right )} b x - a^{2} - a + 1\right )} \sqrt {-\frac {a + 1}{a - 1}} + 2}{x^{2}}\right ) + 2 \, {\left ({\left (a - 1\right )} b x + a^{2} - 2 \, a + 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + 8 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \, {\left ({\left (a - 1\right )} b x + a^{2} - 2 \, a + 1\right )}}, -\frac {{\left ({\left (a + 1\right )} b x + a^{2} - 1\right )} \sqrt {\frac {a + 1}{a - 1}} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {\frac {a + 1}{a - 1}}}{{\left (a + 1\right )} b^{2} x^{2} + a^{3} + 2 \, {\left (a^{2} + a\right )} b x + a^{2} - a - 1}\right ) - {\left ({\left (a - 1\right )} b x + a^{2} - 2 \, a + 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 4 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{{\left (a - 1\right )} b x + a^{2} - 2 \, a + 1}\right ] \]

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*(((a + 1)*b*x + a^2 - 1)*sqrt(-(a + 1)/(a - 1))*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x - 4*a^
2 + 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^3 + (a^2 - a)*b*x - a^2 - a + 1)*sqrt(-(a + 1)/(a - 1)) + 2)/x^2)
+ 2*((a - 1)*b*x + a^2 - 2*a + 1)*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)) + 8*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/((a - 1)*b*x + a^2 - 2*a + 1), -(((a + 1)*b*x + a^2 - 1)*sqrt((
a + 1)/(a - 1))*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt((a + 1)/(a - 1))/((a + 1)*b^2
*x^2 + a^3 + 2*(a^2 + a)*b*x + a^2 - a - 1)) - ((a - 1)*b*x + a^2 - 2*a + 1)*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)) - 4*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/((a - 1)*b*x + a^2 -
2*a + 1)]

________________________________________________________________________________________

giac [A]  time = 0.32, size = 139, normalized size = 1.30 \[ \frac {b \arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{{\left | b \right |}} - \frac {2 \, {\left (a^{2} b + 2 \, a b + b\right )} \arctan \left (\frac {\frac {{\left (\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{\sqrt {a^{2} - 1} {\left (a {\left | b \right |} - {\left | b \right |}\right )}} - \frac {8 \, b}{{\left (a {\left | b \right |} - {\left | b \right |}\right )} {\left (\frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} - 1\right )}} \]

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]

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

________________________________________________________________________________________

maple [B]  time = 0.04, size = 1019, normalized size = 9.52 \[ -\frac {3 b a x}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {2 b \,a^{2} x}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {6 b \left (-2 b^{2} x -2 a b \right )}{\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {3}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {1}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {2 a}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {2 a^{3}}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {3 a^{2}}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}+\frac {a b x}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {a^{4} b x}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {6 a^{2} b \left (-2 b^{2} x -2 a b \right )}{\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {12 a b \left (-2 b^{2} x -2 a b \right )}{\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {3 a^{3} b x}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {3 a^{2} b x}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {b \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 )}{\sqrt {b^{2}}}+\frac {b x}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {4 a^{3}}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {4 a^{2}}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {3 a}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {a^{5}}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {3 a^{4}}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right ) a^{3}}{\left (-a^{2}+1\right )^{\frac {3}{2}}}-\frac {3 \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right ) a^{2}}{\left (-a^{2}+1\right )^{\frac {3}{2}}}-\frac {3 \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right ) a}{\left (-a^{2}+1\right )^{\frac {3}{2}}} \]

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]

-3*b*a/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x-2*b*a^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+6*b*(-2*b^2*x-2*a*b)/(-4*b^2*
(-a^2+1)-4*a^2*b^2)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/(-a^2+1)/(-b^2*x^2-2*a*b
*x-a^2+1)^(1/2)+2*a/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-2*a^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-3*a^2/(-b^2*x^2-2*a*b*
x-a^2+1)^(1/2)-1/(-a^2+1)^(3/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)+a*b/(
-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+a^4*b/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+6*a^2*b*(-2*b^2*x-2*a
*b)/(-4*b^2*(-a^2+1)-4*a^2*b^2)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+12*a*b*(-2*b^2*x-2*a*b)/(-4*b^2*(-a^2+1)-4*a^2*
b^2)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+3*a^3*b/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+3*a^2*b/(-a^2+1)/(-b^2*x
^2-2*a*b*x-a^2+1)^(1/2)*x-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))+b*x/(-b^2*x
^2-2*a*b*x-a^2+1)^(1/2)+4/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a^3+4*a^2/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^
(1/2)+3/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a+a^5/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+3*a^4/(-a^2+1)/(
-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/(-a^2+1)^(3/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^
(1/2))/x)*a^3-3/(-a^2+1)^(3/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)*a^2-3/
(-a^2+1)^(3/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)*a

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is a-1 positive, negative or zero?

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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

________________________________________________________________________________________