3.212 \(\int \frac {e^{-3 i \tan ^{-1}(a+b x)}}{x} \, dx\)

Optimal. Leaf size=134 \[ \frac {4 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}+i \sinh ^{-1}(a+b x)-\frac {2 (a+i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{3/2}} \]

[Out]

I*arcsinh(b*x+a)-2*(I+a)^(3/2)*arctanh((I+a)^(1/2)*(1+I*a+I*b*x)^(1/2)/(I-a)^(1/2)/(1-I*a-I*b*x)^(1/2))/(I-a)^
(3/2)+4*(1-I*a-I*b*x)^(1/2)/(1+I*a)/(1+I*a+I*b*x)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5095, 98, 157, 53, 619, 215, 93, 208} \[ \frac {4 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}+i \sinh ^{-1}(a+b x)-\frac {2 (a+i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((3*I)*ArcTan[a + b*x])*x),x]

[Out]

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

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 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[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 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]

Rubi steps

\begin {align*} \int \frac {e^{-3 i \tan ^{-1}(a+b x)}}{x} \, dx &=\int \frac {(1-i a-i b x)^{3/2}}{x (1+i a+i b x)^{3/2}} \, dx\\ &=\frac {4 \sqrt {1-i a-i b x}}{(1+i a) \sqrt {1+i a+i b x}}+\frac {2 \int \frac {-\frac {1}{2} i (i+a)^2 b-\frac {1}{2} (1+i a) b^2 x}{x \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{(i-a) b}\\ &=\frac {4 \sqrt {1-i a-i b x}}{(1+i a) \sqrt {1+i a+i b x}}-\frac {(i+a)^2 \int \frac {1}{x \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{1+i a}+(i b) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx\\ &=\frac {4 \sqrt {1-i a-i b x}}{(1+i a) \sqrt {1+i a+i b x}}-\frac {\left (2 (i+a)^2\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 )}{1+i a}+(i b) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx\\ &=\frac {4 \sqrt {1-i a-i b x}}{(1+i a) \sqrt {1+i a+i b x}}-\frac {2 (i+a)^{3/2} \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)^{3/2}}+\frac {i \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-i a-i b x}}{(1+i a) \sqrt {1+i a+i b x}}+i \sinh ^{-1}(a+b x)-\frac {2 (i+a)^{3/2} \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)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 189, normalized size = 1.41 \[ \frac {2 \left (\frac {\sqrt {-1+i a} (a+i) \tanh ^{-1}\left (\frac {\sqrt {-1-i a} \sqrt {-i (a+b x+i)}}{\sqrt {-1+i a} \sqrt {i a+i b x+1}}\right )}{\sqrt {-1-i a}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2+1}}{a+b x-i}\right )}{a-i}+\frac {2 (-1)^{3/4} \sqrt {-i b} \sinh ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (a+b x+i)}}{\sqrt {-i b}}\right )}{\sqrt {b}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^((3*I)*ArcTan[a + b*x])*x),x]

[Out]

(2*(-1)^(3/4)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/Sqrt[b] + (2*
((-2*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])/(-I + a + b*x) + (Sqrt[-1 + I*a]*(I + a)*ArcTanh[(Sqrt[-1 - I*a]*Sqrt[
(-I)*(I + a + b*x)])/(Sqrt[-1 + I*a]*Sqrt[1 + I*a + I*b*x])])/Sqrt[-1 - I*a]))/(-I + a)

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fricas [B]  time = 0.49, size = 381, normalized size = 2.84 \[ \frac {{\left ({\left (a - i\right )} b x + a^{2} - 2 i \, a - 1\right )} \sqrt {-\frac {4 \, a^{3} + 12 i \, a^{2} - 12 \, a - 4 i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}} \log \left (-\frac {{\left (2 \, a + 2 i\right )} b x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a + 2 i\right )} - {\left (i \, a^{2} + 2 \, a - i\right )} \sqrt {-\frac {4 \, a^{3} + 12 i \, a^{2} - 12 \, a - 4 i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}}}{2 \, a + 2 i}\right ) - {\left ({\left (a - i\right )} b x + a^{2} - 2 i \, a - 1\right )} \sqrt {-\frac {4 \, a^{3} + 12 i \, a^{2} - 12 \, a - 4 i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}} \log \left (-\frac {{\left (2 \, a + 2 i\right )} b x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a + 2 i\right )} - {\left (-i \, a^{2} - 2 \, a + i\right )} \sqrt {-\frac {4 \, a^{3} + 12 i \, a^{2} - 12 \, a - 4 i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}}}{2 \, a + 2 i}\right ) - 8 \, b x - {\left (2 \, {\left (i \, a + 1\right )} b x + 2 i \, a^{2} + 4 \, a - 2 i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 8 \, a - 8 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 8 i}{{\left (2 \, a - 2 i\right )} b x + 2 \, a^{2} - 4 i \, a - 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a - I)*b*x + a^2 - 2*I*a - 1)*sqrt(-(4*a^3 + 12*I*a^2 - 12*a - 4*I)/(a^3 - 3*I*a^2 - 3*a + I))*log(-((2*a +
2*I)*b*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a + 2*I) - (I*a^2 + 2*a - I)*sqrt(-(4*a^3 + 12*I*a^2 - 12*a -
4*I)/(a^3 - 3*I*a^2 - 3*a + I)))/(2*a + 2*I)) - ((a - I)*b*x + a^2 - 2*I*a - 1)*sqrt(-(4*a^3 + 12*I*a^2 - 12*
a - 4*I)/(a^3 - 3*I*a^2 - 3*a + I))*log(-((2*a + 2*I)*b*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a + 2*I) - (-
I*a^2 - 2*a + I)*sqrt(-(4*a^3 + 12*I*a^2 - 12*a - 4*I)/(a^3 - 3*I*a^2 - 3*a + I)))/(2*a + 2*I)) - 8*b*x - (2*(
I*a + 1)*b*x + 2*I*a^2 + 4*a - 2*I)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - 8*a - 8*sqrt(b^2*x^2 +
2*a*b*x + a^2 + 1) + 8*I)/((2*a - 2*I)*b*x + 2*a^2 - 4*I*a - 2)

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giac [B]  time = 0.33, size = 263, normalized size = 1.96 \[ \frac {{\left (a^{2} i - 2 \, a - i\right )} \log \left (\frac {{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1} {\left (a - i\right )}} + \frac {b \log \left (-3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b i - a^{3} b i - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} i {\left | b \right |} - 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} i {\left | b \right |} - 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b - 2 \, a^{2} b + a b i - 4 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} i {\left | b \right |}\right )}{3 \, i {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*i - 2*a - i)*log(abs(-2*x*abs(b) + 2*sqrt((b*x + a)^2 + 1) - 2*sqrt(a^2 + 1))/abs(-2*x*abs(b) + 2*sqrt((b
*x + a)^2 + 1) + 2*sqrt(a^2 + 1)))/(sqrt(a^2 + 1)*(a - i)) + 1/3*b*log(-3*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2
*a*b*i - a^3*b*i - (x*abs(b) - sqrt((b*x + a)^2 + 1))^3*i*abs(b) - 3*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a^2*i*
abs(b) - 2*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*b - 2*a^2*b + a*b*i - 4*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a*a
bs(b) + (x*abs(b) - sqrt((b*x + a)^2 + 1))*i*abs(b))/(i*abs(b))

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maple [B]  time = 0.14, size = 1278, normalized size = 9.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(I-a)^2/b^2/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)+1/(I-a)^2*((x-(I-a)/b)^2*b^2+2*I*b*(x
-(I-a)/b))^(3/2)-3/2*I/(I-a)^3*a*b*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/(I-
a)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/3*I/(I-a)^3*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-1/2*I/(I-a)^3*a*b*(b^2*x^2+2*a*
b*x+a^2+1)^(1/2)*x+I/(I-a)^3*(a^2+1)^(1/2)*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)+3/2*I/(I-a)^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*a-I/(I-a)^3*a^3*b*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)+2*I/(I-a)*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(3/2)+1/3*I/(I-a)^3*((x
-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(3/2)-3/2*I/(I-a)^3*a^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3/2*I/(I-a)^2*b*((x-(
I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*x-1/(I-a)/b^3/(x-I/b+a/b)^3*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)
-2*I/(I-a)/b^2/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)+I/(I-a)^3*(a^2+1)^(1/2)*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/(I-a)*b*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(
1/2)*x-3/(I-a)*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*a-3/(I-a)*b*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((
x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2))/(b^2)^(1/2)+3/2*I/(I-a)^2*b*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((
x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2))/(b^2)^(1/2)-1/2/(I-a)^3*b*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/
2)*x-1/2/(I-a)^3*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*a-1/2/(I-a)^3*b*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1
/2)+((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2))/(b^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i \left (\int \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} x + 3 a^{2} b x^{2} - 3 i a^{2} x + 3 a b^{2} x^{3} - 6 i a b x^{2} - 3 a x + b^{3} x^{4} - 3 i b^{2} x^{3} - 3 b x^{2} + i x}\, dx + \int \frac {a^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} x + 3 a^{2} b x^{2} - 3 i a^{2} x + 3 a b^{2} x^{3} - 6 i a b x^{2} - 3 a x + b^{3} x^{4} - 3 i b^{2} x^{3} - 3 b x^{2} + i x}\, dx + \int \frac {b^{2} x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} x + 3 a^{2} b x^{2} - 3 i a^{2} x + 3 a b^{2} x^{3} - 6 i a b x^{2} - 3 a x + b^{3} x^{4} - 3 i b^{2} x^{3} - 3 b x^{2} + i x}\, dx + \int \frac {2 a b x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} x + 3 a^{2} b x^{2} - 3 i a^{2} x + 3 a b^{2} x^{3} - 6 i a b x^{2} - 3 a x + b^{3} x^{4} - 3 i b^{2} x^{3} - 3 b x^{2} + i x}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3*x + 3*a**2*b*x**2 - 3*I*a**2*x + 3*a*b**2*x**3 - 6*I*a*
b*x**2 - 3*a*x + b**3*x**4 - 3*I*b**2*x**3 - 3*b*x**2 + I*x), x) + Integral(a**2*sqrt(a**2 + 2*a*b*x + b**2*x*
*2 + 1)/(a**3*x + 3*a**2*b*x**2 - 3*I*a**2*x + 3*a*b**2*x**3 - 6*I*a*b*x**2 - 3*a*x + b**3*x**4 - 3*I*b**2*x**
3 - 3*b*x**2 + I*x), x) + Integral(b**2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3*x + 3*a**2*b*x**2 - 3*
I*a**2*x + 3*a*b**2*x**3 - 6*I*a*b*x**2 - 3*a*x + b**3*x**4 - 3*I*b**2*x**3 - 3*b*x**2 + I*x), x) + Integral(2
*a*b*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3*x + 3*a**2*b*x**2 - 3*I*a**2*x + 3*a*b**2*x**3 - 6*I*a*b*x**
2 - 3*a*x + b**3*x**4 - 3*I*b**2*x**3 - 3*b*x**2 + I*x), x))

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