3.866 \(\int \frac{e^{-3 \tanh ^{-1}(a+b x)}}{x^3} \, dx\)
Optimal. Leaf size=200 \[ -\frac{3 (3-2 a) b^2 \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{(a+1)^3 \sqrt{1-a^2}}-\frac{(-a-b x+1)^{5/2}}{2 \left (1-a^2\right ) x^2 \sqrt{a+b x+1}}+\frac{3 (3-2 a) b^2 \sqrt{-a-b x+1}}{(1-a) (a+1)^3 \sqrt{a+b x+1}}+\frac{(3-2 a) b (-a-b x+1)^{3/2}}{2 (1-a) (a+1)^2 x \sqrt{a+b x+1}} \]
[Out]
(3*(3 - 2*a)*b^2*Sqrt[1 - a - b*x])/((1 - a)*(1 + a)^3*Sqrt[1 + a + b*x]) + ((3 - 2*a)*b*(1 - a - b*x)^(3/2))/
(2*(1 - a)*(1 + a)^2*x*Sqrt[1 + a + b*x]) - (1 - a - b*x)^(5/2)/(2*(1 - a^2)*x^2*Sqrt[1 + a + b*x]) - (3*(3 -
2*a)*b^2*ArcTanh[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[1 - a - b*x])])/((1 + a)^3*Sqrt[1 - a^2])
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Rubi [A] time = 0.129499, antiderivative size = 200, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used =
{6163, 96, 94, 93, 208} \[ -\frac{3 (3-2 a) b^2 \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{(a+1)^3 \sqrt{1-a^2}}-\frac{(-a-b x+1)^{5/2}}{2 \left (1-a^2\right ) x^2 \sqrt{a+b x+1}}+\frac{3 (3-2 a) b^2 \sqrt{-a-b x+1}}{(1-a) (a+1)^3 \sqrt{a+b x+1}}+\frac{(3-2 a) b (-a-b x+1)^{3/2}}{2 (1-a) (a+1)^2 x \sqrt{a+b x+1}} \]
Antiderivative was successfully verified.
[In]
Int[1/(E^(3*ArcTanh[a + b*x])*x^3),x]
[Out]
(3*(3 - 2*a)*b^2*Sqrt[1 - a - b*x])/((1 - a)*(1 + a)^3*Sqrt[1 + a + b*x]) + ((3 - 2*a)*b*(1 - a - b*x)^(3/2))/
(2*(1 - a)*(1 + a)^2*x*Sqrt[1 + a + b*x]) - (1 - a - b*x)^(5/2)/(2*(1 - a^2)*x^2*Sqrt[1 + a + b*x]) - (3*(3 -
2*a)*b^2*ArcTanh[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[1 - a - b*x])])/((1 + a)^3*Sqrt[1 - a^2])
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 96
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))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])
Rule 94
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 + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && !(SumSimplerQ[p, 1] && !SumSimplerQ[m, 1])
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 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]
Rubi steps
\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a+b x)}}{x^3} \, dx &=\int \frac{(1-a-b x)^{3/2}}{x^3 (1+a+b x)^{3/2}} \, dx\\ &=-\frac{(1-a-b x)^{5/2}}{2 \left (1-a^2\right ) x^2 \sqrt{1+a+b x}}-\frac{((3-2 a) b) \int \frac{(1-a-b x)^{3/2}}{x^2 (1+a+b x)^{3/2}} \, dx}{2 \left (1-a^2\right )}\\ &=\frac{(3-2 a) b (1-a-b x)^{3/2}}{2 (1-a) (1+a)^2 x \sqrt{1+a+b x}}-\frac{(1-a-b x)^{5/2}}{2 \left (1-a^2\right ) x^2 \sqrt{1+a+b x}}+\frac{\left (3 (3-2 a) b^2\right ) \int \frac{\sqrt{1-a-b x}}{x (1+a+b x)^{3/2}} \, dx}{2 (1-a) (1+a)^2}\\ &=\frac{3 (3-2 a) b^2 \sqrt{1-a-b x}}{(1-a) (1+a)^3 \sqrt{1+a+b x}}+\frac{(3-2 a) b (1-a-b x)^{3/2}}{2 (1-a) (1+a)^2 x \sqrt{1+a+b x}}-\frac{(1-a-b x)^{5/2}}{2 \left (1-a^2\right ) x^2 \sqrt{1+a+b x}}+\frac{\left (3 (3-2 a) b^2\right ) \int \frac{1}{x \sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx}{2 (1+a)^3}\\ &=\frac{3 (3-2 a) b^2 \sqrt{1-a-b x}}{(1-a) (1+a)^3 \sqrt{1+a+b x}}+\frac{(3-2 a) b (1-a-b x)^{3/2}}{2 (1-a) (1+a)^2 x \sqrt{1+a+b x}}-\frac{(1-a-b x)^{5/2}}{2 \left (1-a^2\right ) x^2 \sqrt{1+a+b x}}+\frac{\left (3 (3-2 a) b^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)^3}\\ &=\frac{3 (3-2 a) b^2 \sqrt{1-a-b x}}{(1-a) (1+a)^3 \sqrt{1+a+b x}}+\frac{(3-2 a) b (1-a-b x)^{3/2}}{2 (1-a) (1+a)^2 x \sqrt{1+a+b x}}-\frac{(1-a-b x)^{5/2}}{2 \left (1-a^2\right ) x^2 \sqrt{1+a+b x}}-\frac{3 (3-2 a) b^2 \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{1+a+b x}}{\sqrt{1+a} \sqrt{1-a-b x}}\right )}{(1+a)^3 \sqrt{1-a^2}}\\ \end{align*}
Mathematica [A] time = 0.153857, size = 134, normalized size = 0.67 \[ \frac{\sqrt{-a-b x+1} \left (a^3+a^2-a \left (b^2 x^2-5 b x+1\right )+14 b^2 x^2+5 b x-1\right )}{2 (a+1)^3 x^2 \sqrt{a+b x+1}}-\frac{3 (2 a-3) b^2 \tan ^{-1}\left (\frac{\sqrt{-a-b x+1}}{\sqrt{\frac{a-1}{a+1}} \sqrt{a+b x+1}}\right )}{\sqrt{a-1} (a+1)^{7/2}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(E^(3*ArcTanh[a + b*x])*x^3),x]
[Out]
(Sqrt[1 - a - b*x]*(-1 + a^2 + a^3 + 5*b*x + 14*b^2*x^2 - a*(1 - 5*b*x + b^2*x^2)))/(2*(1 + a)^3*x^2*Sqrt[1 +
a + b*x]) - (3*(-3 + 2*a)*b^2*ArcTan[Sqrt[1 - a - b*x]/(Sqrt[(-1 + a)/(1 + a)]*Sqrt[1 + a + b*x])])/(Sqrt[-1 +
a]*(1 + a)^(7/2))
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Maple [B] time = 0.057, size = 2848, normalized size = 14.2 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(b*x+a+1)^3*(1-(b*x+a)^2)^(3/2)/x^3,x)
[Out]
2/(1+a)^3/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)+2/(1+a)^3*b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1
+a)/b))^(3/2)-3/(1+a)^4/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)-3/(1+a)^4*b^2*(-(x+(1+a)/b)^2
*b^2+2*b*(x+(1+a)/b))^(3/2)-2/(1+a)^5*b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)+2/(1+a)^5*b^2*(-b^2*x^2-2
*a*b*x-a^2+1)^(3/2)+6/(1+a)^5*b^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/2/(1+a)^3/(-a^2+1)/x^2*(-b^2*x^2-2*a*b*x-a^
2+1)^(5/2)-1/2/(1+a)^3*b^2/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)-3/2/(1+a)^3*b^2/(-a^2+1)*(-b^2*x^2-2*a*b*x-
a^2+1)^(1/2)+3/2/(1+a)^3*b^2/(-a^2+1)^(1/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/
2))/x)-9/(1+a)^5*b^2*a^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-6/(1+a)^5*b^2*(-a^2+1)^(1/2)*ln((-2*a^2+2-2*x*a*b+2*(-
a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-9/2/(1+a)^4*b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x-9
/2/(1+a)^4*b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a-9/2/(1+a)^4*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))+3/(1+a)^3*b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1
/2)*a-3/(1+a)^5*b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x-3/(1+a)^5*b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a
)/b))^(1/2)*a-3/(1+a)^5*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/(1+a)^3/b/(x+1/b+a/b)^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)+3/(1+a)^3*b^3*(-(x+(1+a)/b)^2*b^
2+2*b*(x+(1+a)/b))^(1/2)*x+1/2/(1+a)^3*a*b/(-a^2+1)^2/x*(-b^2*x^2-2*a*b*x-a^2+1)^(5/2)-3/4/(1+a)^3*a^3*b^3/(-a
^2+1)^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x-9/4/(1+a)^3*a^3*b^3/(-a^2+1)^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))+3/2/(1+a)^3*a^5*b^3/(-a^2+1)^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))-9/2/(1+a)^4*b^3*a^2/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x-27/2/(1+a)^4*b^3*a^2/(-
a^2+1)/(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))+9/(1+a)^4*b^3*a^4/(-a^2+1)/(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/(1+a)^3*a*b^3/(-a^2+1)^2*(-b^2*x^2-2*a*b
*x-a^2+1)^(3/2)*x+3/4/(1+a)^3*a*b^3/(-a^2+1)^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+3/4/(1+a)^3*a*b^3/(-a^2+1)^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))+3/4/(1+a)^3*b^3/(-a^2+1)*a*(-b^2*x^2-2*a
*b*x-a^2+1)^(1/2)*x+9/4/(1+a)^3*b^3/(-a^2+1)*a/(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/(1+a)^3*b^3/(-a^2+1)*a^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/
(1+a)^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))+6/(1+a)
^5*b^3*a^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))+6/(1+a)^5*b^2*(-a^2+1)^(1/2)
*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)*a^2-3/(1+a)^5*b^3*a*(-b^2*x^2-2*a*b*
x-a^2+1)^(1/2)*x-9/(1+a)^5*b^3*a/(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/(1+a
)^4*b/(-a^2+1)/x*(-b^2*x^2-2*a*b*x-a^2+1)^(5/2)+6/(1+a)^4*b^2*a/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)-27/2/(
1+a)^4*b^2*a^3/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+27/2/(1+a)^4*b^2*a/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1
/2)+9/(1+a)^4*b^2*a^3/(-a^2+1)^(1/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-
9/(1+a)^4*b^2*a/(-a^2+1)^(1/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)+3/(1+a
)^4*b^3/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)*x+9/2/(1+a)^4*b^3/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+9/
2/(1+a)^4*b^3/(-a^2+1)/(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/(1+a)^3*a^2*b^
2/(-a^2+1)^2*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)-9/4/(1+a)^3*a^4*b^2/(-a^2+1)^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+9/4/
(1+a)^3*a^2*b^2/(-a^2+1)^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+3/2/(1+a)^3*a^4*b^2/(-a^2+1)^(3/2)*ln((-2*a^2+2-2*x*
a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-3/2/(1+a)^3*a^2*b^2/(-a^2+1)^(3/2)*ln((-2*a^2+2-2*x*a*
b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)+9/4/(1+a)^3*b^2/(-a^2+1)*a^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1
/2)-3/2/(1+a)^3*b^2/(-a^2+1)^(1/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)*a^
2
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}}}{{\left (b x + a + 1\right )}^{3} x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*x+a+1)^3*(1-(b*x+a)^2)^(3/2)/x^3,x, algorithm="maxima")
[Out]
integrate((-(b*x + a)^2 + 1)^(3/2)/((b*x + a + 1)^3*x^3), x)
________________________________________________________________________________________
Fricas [A] time = 2.07525, size = 1184, normalized size = 5.92 \begin{align*} \left [-\frac{3 \,{\left ({\left (2 \, a - 3\right )} b^{3} x^{3} +{\left (2 \, a^{2} - a - 3\right )} b^{2} x^{2}\right )} \sqrt{-a^{2} + 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 - 2 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) - 2 \,{\left (a^{5} -{\left (a^{3} - 14 \, a^{2} - a + 14\right )} b^{2} x^{2} + a^{4} - 2 \, a^{3} + 5 \,{\left (a^{3} + a^{2} - a - 1\right )} b x - 2 \, a^{2} + a + 1\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{4 \,{\left ({\left (a^{5} + 3 \, a^{4} + 2 \, a^{3} - 2 \, a^{2} - 3 \, a - 1\right )} b x^{3} +{\left (a^{6} + 4 \, a^{5} + 5 \, a^{4} - 5 \, a^{2} - 4 \, a - 1\right )} x^{2}\right )}}, -\frac{3 \,{\left ({\left (2 \, a - 3\right )} b^{3} x^{3} +{\left (2 \, a^{2} - a - 3\right )} b^{2} x^{2}\right )} \sqrt{a^{2} - 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{a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \,{\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) -{\left (a^{5} -{\left (a^{3} - 14 \, a^{2} - a + 14\right )} b^{2} x^{2} + a^{4} - 2 \, a^{3} + 5 \,{\left (a^{3} + a^{2} - a - 1\right )} b x - 2 \, a^{2} + a + 1\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \,{\left ({\left (a^{5} + 3 \, a^{4} + 2 \, a^{3} - 2 \, a^{2} - 3 \, a - 1\right )} b x^{3} +{\left (a^{6} + 4 \, a^{5} + 5 \, a^{4} - 5 \, a^{2} - 4 \, a - 1\right )} x^{2}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*x+a+1)^3*(1-(b*x+a)^2)^(3/2)/x^3,x, algorithm="fricas")
[Out]
[-1/4*(3*((2*a - 3)*b^3*x^3 + (2*a^2 - a - 3)*b^2*x^2)*sqrt(-a^2 + 1)*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^
3 - a)*b*x - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(-a^2 + 1) - 4*a^2 + 2)/x^2) - 2*(a^5
- (a^3 - 14*a^2 - a + 14)*b^2*x^2 + a^4 - 2*a^3 + 5*(a^3 + a^2 - a - 1)*b*x - 2*a^2 + a + 1)*sqrt(-b^2*x^2 - 2
*a*b*x - a^2 + 1))/((a^5 + 3*a^4 + 2*a^3 - 2*a^2 - 3*a - 1)*b*x^3 + (a^6 + 4*a^5 + 5*a^4 - 5*a^2 - 4*a - 1)*x^
2), -1/2*(3*((2*a - 3)*b^3*x^3 + (2*a^2 - a - 3)*b^2*x^2)*sqrt(a^2 - 1)*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 +
1)*(a*b*x + a^2 - 1)*sqrt(a^2 - 1)/((a^2 - 1)*b^2*x^2 + a^4 + 2*(a^3 - a)*b*x - 2*a^2 + 1)) - (a^5 - (a^3 - 1
4*a^2 - a + 14)*b^2*x^2 + a^4 - 2*a^3 + 5*(a^3 + a^2 - a - 1)*b*x - 2*a^2 + a + 1)*sqrt(-b^2*x^2 - 2*a*b*x - a
^2 + 1))/((a^5 + 3*a^4 + 2*a^3 - 2*a^2 - 3*a - 1)*b*x^3 + (a^6 + 4*a^5 + 5*a^4 - 5*a^2 - 4*a - 1)*x^2)]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*x+a+1)**3*(1-(b*x+a)**2)**(3/2)/x**3,x)
[Out]
Timed out
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Giac [B] time = 1.21796, size = 1110, normalized size = 5.55 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*x+a+1)^3*(1-(b*x+a)^2)^(3/2)/x^3,x, algorithm="giac")
[Out]
-8*b^3/((a^3*abs(b) + 3*a^2*abs(b) + 3*a*abs(b) + abs(b))*((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^
2*x + a*b) + 1)) - 3*(2*a*b^3 - 3*b^3)*arctan(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/(b^2*x + a*b)
- 1)/sqrt(a^2 - 1))/((a^3*abs(b) + 3*a^2*abs(b) + 3*a*abs(b) + abs(b))*sqrt(a^2 - 1)) - (2*(sqrt(-b^2*x^2 - 2
*a*b*x - a^2 + 1)*abs(b) + b)^2*a^4*b^3/(b^2*x + a*b)^2 + 2*a^4*b^3 - 5*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*ab
s(b) + b)*a^3*b^3/(b^2*x + a*b) - 6*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^3*b^3/(b^2*x + a*b)^2
- 3*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a^3*b^3/(b^2*x + a*b)^3 - 6*a^3*b^3 + 18*(sqrt(-b^2*x^2
- 2*a*b*x - a^2 + 1)*abs(b) + b)*a^2*b^3/(b^2*x + a*b) + 3*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a
^2*b^3/(b^2*x + a*b)^2 + 6*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a^2*b^3/(b^2*x + a*b)^3 - a^2*b^3
+ 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a*b^3/(b^2*x + a*b) - 12*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 +
1)*abs(b) + b)^2*a*b^3/(b^2*x + a*b)^2 + 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a*b^3/(b^2*x + a
*b)^3 - 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*b^3/(b^2*x + a*b)^2)/((a^5*abs(b) + 3*a^4*abs(b) +
3*a^3*abs(b) + a^2*abs(b))*((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a/(b^2*x + a*b)^2 + a - 2*(sqrt
(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b))^2)