3.197 \(\int \frac{e^{-i \tan ^{-1}(a+b x)}}{x^4} \, dx\)
Optimal. Leaf size=283 \[ \frac{\left (-2 a^2-9 i a+4\right ) b^2 \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{6 (1+i a) \left (a^2+1\right )^2 x}+\frac{\left (2 a+i \left (1-2 a^2\right )\right ) b^3 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{7/2} (a+i)^{5/2}}+\frac{(3-2 i a) b \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{6 (-a+i)^2 (a+i) x^2}-\frac{\sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{3 (1+i a) x^3} \]
[Out]
-(Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(3*(1 + I*a)*x^3) + ((3 - (2*I)*a)*b*Sqrt[1 - I*a - I*b*x]*Sqrt
[1 + I*a + I*b*x])/(6*(I - a)^2*(I + a)*x^2) + ((4 - (9*I)*a - 2*a^2)*b^2*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a +
I*b*x])/(6*(1 + I*a)*(1 + a^2)^2*x) + ((2*a + I*(1 - 2*a^2))*b^3*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/
(Sqrt[I - a]*Sqrt[1 - I*a - I*b*x])])/((I - a)^(7/2)*(I + a)^(5/2))
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Rubi [A] time = 0.183009, antiderivative size = 282, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used =
{5095, 99, 151, 12, 93, 208} \[ \frac{\left (-2 a^2-9 i a+4\right ) b^2 \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{6 (1+i a) \left (a^2+1\right )^2 x}+\frac{\left (-2 i a^2+2 a+i\right ) b^3 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{7/2} (a+i)^{5/2}}+\frac{(3-2 i a) b \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{6 (-a+i)^2 (a+i) x^2}-\frac{\sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{3 (1+i a) x^3} \]
Antiderivative was successfully verified.
[In]
Int[1/(E^(I*ArcTan[a + b*x])*x^4),x]
[Out]
-(Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(3*(1 + I*a)*x^3) + ((3 - (2*I)*a)*b*Sqrt[1 - I*a - I*b*x]*Sqrt
[1 + I*a + I*b*x])/(6*(I - a)^2*(I + a)*x^2) + ((4 - (9*I)*a - 2*a^2)*b^2*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a +
I*b*x])/(6*(1 + I*a)*(1 + a^2)^2*x) + ((I + 2*a - (2*I)*a^2)*b^3*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/
(Sqrt[I - a]*Sqrt[1 - I*a - I*b*x])])/((I - a)^(7/2)*(I + a)^(5/2))
Rule 5095
Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1 -
I*a*c - I*b*c*x)^((I*n)/2))/(1 + I*a*c + I*b*c*x)^((I*n)/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]
Rule 99
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[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])
Rule 151
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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^{-i \tan ^{-1}(a+b x)}}{x^4} \, dx &=\int \frac{\sqrt{1-i a-i b x}}{x^4 \sqrt{1+i a+i b x}} \, dx\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1+i a) x^3}+\frac{\int \frac{-(3 i+2 a) b-2 b^2 x}{x^3 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{3 (1+i a)}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1+i a) x^3}+\frac{(3-2 i a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (i-a)^2 (i+a) x^2}-\frac{\int \frac{\left (4-9 i a-2 a^2\right ) b^2-(3 i+2 a) b^3 x}{x^2 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{6 (1+i a) \left (1+a^2\right )}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1+i a) x^3}+\frac{(3-2 i a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (i-a)^2 (i+a) x^2}+\frac{\left (4-9 i a-2 a^2\right ) b^2 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1+i a) \left (1+a^2\right )^2 x}+\frac{\int \frac{3 \left (i+2 a-2 i a^2\right ) b^3}{x \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{6 (1+i a) \left (1+a^2\right )^2}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1+i a) x^3}+\frac{(3-2 i a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (i-a)^2 (i+a) x^2}+\frac{\left (4-9 i a-2 a^2\right ) b^2 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1+i a) \left (1+a^2\right )^2 x}-\frac{\left (\left (1-2 i a-2 a^2\right ) b^3\right ) \int \frac{1}{x \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{2 (i-a)^3 (i+a)^2}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1+i a) x^3}+\frac{(3-2 i a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (i-a)^2 (i+a) x^2}+\frac{\left (4-9 i a-2 a^2\right ) b^2 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1+i a) \left (1+a^2\right )^2 x}-\frac{\left (\left (1-2 i a-2 a^2\right ) b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-1-i a-(-1+i a) x^2} \, dx,x,\frac{\sqrt{1+i a+i b x}}{\sqrt{1-i a-i b x}}\right )}{(i-a)^3 (i+a)^2}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1+i a) x^3}+\frac{(3-2 i a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (i-a)^2 (i+a) x^2}+\frac{\left (4-9 i a-2 a^2\right ) b^2 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1+i a) \left (1+a^2\right )^2 x}+\frac{\left (i+2 a-2 i a^2\right ) b^3 \tanh ^{-1}\left (\frac{\sqrt{i+a} \sqrt{1+i a+i b x}}{\sqrt{i-a} \sqrt{1-i a-i b x}}\right )}{(i-a)^{7/2} (i+a)^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.23213, size = 251, normalized size = 0.89 \[ \frac{(1-4 i a) b x (a+b x+i) \sqrt{a^2+2 a b x+b^2 x^2+1}+2 (1+i a) (a+i) (a+b x+i) \sqrt{a^2+2 a b x+b^2 x^2+1}+\frac{3 i \left (2 a^2+2 i a-1\right ) b^2 x^2 \left (\sqrt{-1+i a} \sqrt{1+i a} \sqrt{a^2+2 a b x+b^2 x^2+1}+2 i b x \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )\right )}{\sqrt{-1+i a} \sqrt{1+i a} (a-i)}}{6 \left (a^2+1\right )^2 x^3} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(E^(I*ArcTan[a + b*x])*x^4),x]
[Out]
(2*(1 + I*a)*(I + a)*(I + a + b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + (1 - (4*I)*a)*b*x*(I + a + b*x)*Sqrt[1
+ a^2 + 2*a*b*x + b^2*x^2] + ((3*I)*(-1 + (2*I)*a + 2*a^2)*b^2*x^2*(Sqrt[-1 + I*a]*Sqrt[1 + I*a]*Sqrt[1 + a^2
+ 2*a*b*x + b^2*x^2] + (2*I)*b*x*ArcTan[Sqrt[(-I)*(I + a + b*x)]/(Sqrt[(I + a)/(-I + a)]*Sqrt[1 + I*a + I*b*x]
)]))/(Sqrt[-1 + I*a]*Sqrt[1 + I*a]*(-I + a)))/(6*(1 + a^2)^2*x^3)
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Maple [B] time = 0.129, size = 1738, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2)/x^4,x)
[Out]
I*b^3/(I-a)^4*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+b^4/(I-a)^4*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((x-(I-a)/b)^2*b^
2+2*I*(x-(I-a)/b)*b)^(1/2))/(b^2)^(1/2)-I*b^3/(I-a)^4*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)-1/2*I*b^4/(I
-a)^2*a/(a^2+1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-1/2*I*b^4/(I-a)^2*a/(a^2+1)^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^
2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)+1/2*I*b^4/(I-a)^2/(a^2+1)*a*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*
x+a^2+1)^(1/2))/(b^2)^(1/2)+I*b^4/(I-a)^3*a^2/(a^2+1)*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2)
)/(b^2)^(1/2)+1/2*I/(I-a)*a*b/(a^2+1)^2/x^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-1/2*I/(I-a)*a^2*b^2/(a^2+1)^3/x*(b^2
*x^2+2*a*b*x+a^2+1)^(3/2)+1/2*I/(I-a)*a^4*b^4/(a^2+1)^3*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/
2))/(b^2)^(1/2)+1/2*I/(I-a)*a^2*b^4/(a^2+1)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+1/2*I/(I-a)*a^2*b^4/(a^2+1)^3*ln
((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)-1/2*I/(I-a)*a^2*b^4/(a^2+1)^2*ln((b^2*x+a*
b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)+1/2*I*b^2/(I-a)^2*a/(a^2+1)^2/x*(b^2*x^2+2*a*b*x+a^2
+1)^(3/2)-1/2*I*b^4/(I-a)^2*a^3/(a^2+1)^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2
)-1/3*I/(I-a)/(a^2+1)/x^3*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+1/2*I*b^3/(I-a)^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2
)-1/2*I*b^3/(I-a)^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)-I*b^3/
(I-a)^4*(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)+2*I*b^3/(I-a)^3*a/
(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-I*b^3/(I-a)^3*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)+I*b^4/(I-a)^3/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+I*b^4/(I-a)^3/(a^2+1)*ln((b^
2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)-I*b^2/(I-a)^3/(a^2+1)/x*(b^2*x^2+2*a*b*x+a^2+1
)^(3/2)-1/2*I/(I-a)*a^3*b^3/(a^2+1)^(5/2)*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
)-1/2*I/(I-a)*a*b^3/(a^2+1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/2*I/(I-a)*a*b^3/(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)+I*b^4/(I-a)^4*a*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*
x+a^2+1)^(1/2))/(b^2)^(1/2)+1/2*I*b^3/(I-a)^2*a^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)-1/2*I*b/(I-a)^2/(a^2+1)/x^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-I*b^3/(I-a)^2*a^2/(a^2+1)^2*
(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+I/(I-a)*a^3*b^3/(a^2+1)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{{\left (b x + a\right )}^{2} + 1}}{{\left (i \, b x + i \, a + 1\right )} x^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2)/x^4,x, algorithm="maxima")
[Out]
integrate(sqrt((b*x + a)^2 + 1)/((I*b*x + I*a + 1)*x^4), x)
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Fricas [B] time = 2.45153, size = 1766, normalized size = 6.24 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2)/x^4,x, algorithm="fricas")
[Out]
((2*I*a^2 - 9*a - 4*I)*b^3*x^3 - sqrt((4*a^4 + 8*I*a^3 - 8*a^2 - 4*I*a + 1)*b^6/(a^12 - 2*I*a^11 + 4*a^10 - 10
*I*a^9 + 5*a^8 - 20*I*a^7 - 20*I*a^5 - 5*a^4 - 10*I*a^3 - 4*a^2 - 2*I*a - 1))*(3*a^5 - 3*I*a^4 + 6*a^3 - 6*I*a
^2 + 3*a - 3*I)*x^3*log(-((2*a^2 + 2*I*a - 1)*b^4*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a^2 + 2*I*a - 1)*b^
3 + (a^7 - I*a^6 + 3*a^5 - 3*I*a^4 + 3*a^3 - 3*I*a^2 + a - I)*sqrt((4*a^4 + 8*I*a^3 - 8*a^2 - 4*I*a + 1)*b^6/(
a^12 - 2*I*a^11 + 4*a^10 - 10*I*a^9 + 5*a^8 - 20*I*a^7 - 20*I*a^5 - 5*a^4 - 10*I*a^3 - 4*a^2 - 2*I*a - 1)))/((
2*a^2 + 2*I*a - 1)*b^3)) + sqrt((4*a^4 + 8*I*a^3 - 8*a^2 - 4*I*a + 1)*b^6/(a^12 - 2*I*a^11 + 4*a^10 - 10*I*a^9
+ 5*a^8 - 20*I*a^7 - 20*I*a^5 - 5*a^4 - 10*I*a^3 - 4*a^2 - 2*I*a - 1))*(3*a^5 - 3*I*a^4 + 6*a^3 - 6*I*a^2 + 3
*a - 3*I)*x^3*log(-((2*a^2 + 2*I*a - 1)*b^4*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a^2 + 2*I*a - 1)*b^3 - (a
^7 - I*a^6 + 3*a^5 - 3*I*a^4 + 3*a^3 - 3*I*a^2 + a - I)*sqrt((4*a^4 + 8*I*a^3 - 8*a^2 - 4*I*a + 1)*b^6/(a^12 -
2*I*a^11 + 4*a^10 - 10*I*a^9 + 5*a^8 - 20*I*a^7 - 20*I*a^5 - 5*a^4 - 10*I*a^3 - 4*a^2 - 2*I*a - 1)))/((2*a^2
+ 2*I*a - 1)*b^3)) + ((2*I*a^2 - 9*a - 4*I)*b^2*x^2 + 2*I*a^4 + (-2*I*a^3 + 3*a^2 - 2*I*a + 3)*b*x + 4*I*a^2 +
2*I)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/((6*a^5 - 6*I*a^4 + 12*a^3 - 12*I*a^2 + 6*a - 6*I)*x^3)
________________________________________________________________________________________
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/(1+I*(b*x+a))*(1+(b*x+a)**2)**(1/2)/x**4,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2)/x^4,x, algorithm="giac")
[Out]
Exception raised: NotImplementedError