3.2.64 \(\int e^{i \text {ArcTan}(a+b x)} x^2 \, dx\) [164]

Optimal. Leaf size=171 \[ -\frac {\left (i+2 a-2 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}-\frac {(i+4 a) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{3 b^2}-\frac {\left (1-2 i a-2 a^2\right ) \sinh ^{-1}(a+b x)}{2 b^3} \]

[Out]

-1/2*(1-2*I*a-2*a^2)*arcsinh(b*x+a)/b^3-1/6*(I+4*a)*(1+I*a+I*b*x)^(3/2)*(1-I*a-I*b*x)^(1/2)/b^3+1/3*x*(1+I*a+I
*b*x)^(3/2)*(1-I*a-I*b*x)^(1/2)/b^2-1/2*(I+2*a-2*I*a^2)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/b^3

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Rubi [A]
time = 0.09, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5203, 92, 81, 52, 55, 633, 221} \begin {gather*} -\frac {\left (-2 i a^2+2 a+i\right ) \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{2 b^3}-\frac {\left (-2 a^2-2 i a+1\right ) \sinh ^{-1}(a+b x)}{2 b^3}-\frac {(4 a+i) \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{6 b^3}+\frac {x \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a + b*x])*x^2,x]

[Out]

-1/2*((I + 2*a - (2*I)*a^2)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/b^3 - ((I + 4*a)*Sqrt[1 - I*a - I*b*x
]*(1 + I*a + I*b*x)^(3/2))/(6*b^3) + (x*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2))/(3*b^2) - ((1 - (2*I)*a
- 2*a^2)*ArcSinh[a + b*x])/(2*b^3)

Rule 52

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 55

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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 221

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 633

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 5203

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 e^{i \tan ^{-1}(a+b x)} x^2 \, dx &=\int \frac {x^2 \sqrt {1+i a+i b x}}{\sqrt {1-i a-i b x}} \, dx\\ &=\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{3 b^2}+\frac {\int \frac {\sqrt {1+i a+i b x} \left (-1-a^2-(i+4 a) b x\right )}{\sqrt {1-i a-i b x}} \, dx}{3 b^2}\\ &=-\frac {(i+4 a) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{3 b^2}-\frac {\left (1-2 i a-2 a^2\right ) \int \frac {\sqrt {1+i a+i b x}}{\sqrt {1-i a-i b x}} \, dx}{2 b^2}\\ &=-\frac {\left (i+2 a-2 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}-\frac {(i+4 a) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{3 b^2}-\frac {\left (1-2 i a-2 a^2\right ) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{2 b^2}\\ &=-\frac {\left (i+2 a-2 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}-\frac {(i+4 a) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{3 b^2}-\frac {\left (1-2 i a-2 a^2\right ) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{2 b^2}\\ &=-\frac {\left (i+2 a-2 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}-\frac {(i+4 a) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{3 b^2}-\frac {\left (1-2 i a-2 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )}{4 b^4}\\ &=-\frac {\left (i+2 a-2 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}-\frac {(i+4 a) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{3 b^2}-\frac {\left (1-2 i a-2 a^2\right ) \sinh ^{-1}(a+b x)}{2 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 135, normalized size = 0.79 \begin {gather*} \frac {\sqrt {1+a^2+2 a b x+b^2 x^2} \left (-4 i+2 i a^2+3 b x+2 i b^2 x^2+a (-9-2 i b x)\right )}{6 b^3}+\frac {\sqrt [4]{-1} \left (-1+2 i a+2 a^2\right ) \sqrt {-i b} \sinh ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{b^{7/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(I*ArcTan[a + b*x])*x^2,x]

[Out]

(Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(-4*I + (2*I)*a^2 + 3*b*x + (2*I)*b^2*x^2 + a*(-9 - (2*I)*b*x)))/(6*b^3) +
((-1)^(1/4)*(-1 + (2*I)*a + 2*a^2)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-
I)*b]])/b^(7/2)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 435 vs. \(2 (135 ) = 270\).
time = 0.09, size = 436, normalized size = 2.55

method	result	size
risch	\(\frac {i \left (2 b^{2} x^{2}-2 a b x -3 i b x +2 a^{2}+9 i a -4\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{6 b^{3}}+\frac {i \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right ) a}{b^{2} \sqrt {b^{2}}}+\frac {\ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right ) a^{2}}{b^{2} \sqrt {b^{2}}}-\frac {\ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\)	\(198\)
default	\(i b \left (\frac {x^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{3 b^{2}}-\frac {5 a \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )}{3 b}-\frac {2 \left (a^{2}+1\right ) \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{3 b^{2}}\right )+\left (i a +1\right ) \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )\)	\(436\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2)*x^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (119) = 238\).
time = 0.26, size = 351, normalized size = 2.05 \begin {gather*} \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x^{2}}{3 \, b} - \frac {5 i \, a^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} + \frac {3 \, a^{2} {\left (i \, a + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} - \frac {5 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{6 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-i \, a - 1\right )} x}{2 \, b^{2}} + \frac {3 i \, {\left (a^{2} + 1\right )} a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} - \frac {{\left (a^{2} + 1\right )} {\left (i \, a + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} + \frac {5 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{2 \, b^{3}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a {\left (i \, a + 1\right )}}{2 \, b^{3}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (i \, a^{2} + i\right )}}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*x^2/b - 5/2*I*a^3*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1
)*b^2))/b^3 + 3/2*a^2*(I*a + 1)*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 - 5/6*I*sqrt(b
^2*x^2 + 2*a*b*x + a^2 + 1)*a*x/b^2 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(-I*a - 1)*x/b^2 + 3/2*I*(a^2 + 1)
*a*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 - 1/2*(a^2 + 1)*(I*a + 1)*arcsinh(2*(b^2*x
+ a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 + 5/2*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2/b^3 - 3/2*sqrt(b^
2*x^2 + 2*a*b*x + a^2 + 1)*a*(I*a + 1)/b^3 - 2/3*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(I*a^2 + I)/b^3

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Fricas [A]
time = 1.37, size = 106, normalized size = 0.62 \begin {gather*} \frac {7 i \, a^{3} - 21 \, a^{2} - 12 \, {\left (2 \, a^{2} + 2 i \, a - 1\right )} \log \left (-b x - a + \sqrt {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 (-2 i \, b^{2} x^{2} + {\left (2 i \, a - 3\right )} b x - 2 i \, a^{2} + 9 \, a + 4 i\right )} - 9 i \, a}{24 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(7*I*a^3 - 21*a^2 - 12*(2*a^2 + 2*I*a - 1)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - 4*sqrt(b^2
*x^2 + 2*a*b*x + a^2 + 1)*(-2*I*b^2*x^2 + (2*I*a - 3)*b*x - 2*I*a^2 + 9*a + 4*I) - 9*I*a)/b^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i \left (\int \left (- \frac {i x^{2}}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\right )\, dx + \int \frac {a x^{2}}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx + \int \frac {b x^{3}}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.44, size = 113, normalized size = 0.66 \begin {gather*} -\frac {1}{6} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (x {\left (-\frac {2 i \, x}{b} - \frac {-2 i \, a b^{3} + 3 \, b^{3}}{b^{5}}\right )} - \frac {2 i \, a^{2} b^{2} - 9 \, a b^{2} - 4 i \, b^{2}}{b^{5}}\right )} - \frac {{\left (2 \, a^{2} + 2 i \, a - 1\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b^{2} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt((b*x + a)^2 + 1)*(x*(-2*I*x/b - (-2*I*a*b^3 + 3*b^3)/b^5) - (2*I*a^2*b^2 - 9*a*b^2 - 4*I*b^2)/b^5) -
1/2*(2*a^2 + 2*I*a - 1)*log(-a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/(b^2*abs(b))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}{\sqrt {{\left (a+b\,x\right )}^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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