3.2.69 \(\int \frac {e^{i \text {ArcTan}(a+b x)}}{x^3} \, dx\) [169]

Optimal. Leaf size=201 \[ -\frac {(1+2 i a) b \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 (i-a) (i+a)^2 x}-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 \left (1+a^2\right ) x^2}+\frac {(1+2 i a) b^2 \tanh ^{-1}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{(i-a)^{3/2} (i+a)^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5203, 98, 96, 95, 214} \begin {gather*} -\frac {\sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{2 \left (a^2+1\right ) x^2}+\frac {(1+2 i a) b^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} (a+i)^{5/2}}-\frac {(1+2 i a) b \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{2 (-a+i) (a+i)^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a + b*x])/x^3,x]

[Out]

-1/2*((1 + (2*I)*a)*b*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/((I - a)*(I + a)^2*x) - (Sqrt[1 - I*a - I*b
*x]*(1 + I*a + I*b*x)^(3/2))/(2*(1 + a^2)*x^2) + ((1 + (2*I)*a)*b^2*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x]
)/(Sqrt[I - a]*Sqrt[1 - I*a - I*b*x])])/((I - a)^(3/2)*(I + a)^(5/2))

Rule 95

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 96

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[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 98

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 214

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

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 \frac {e^{i \tan ^{-1}(a+b x)}}{x^3} \, dx &=\int \frac {\sqrt {1+i a+i b x}}{x^3 \sqrt {1-i a-i b x}} \, dx\\ &=-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 \left (1+a^2\right ) x^2}+\frac {((i-2 a) b) \int \frac {\sqrt {1+i a+i b x}}{x^2 \sqrt {1-i a-i b x}} \, dx}{2 \left (1+a^2\right )}\\ &=-\frac {(i-2 a) b \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 (1-i a) \left (1+a^2\right ) x}-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 \left (1+a^2\right ) x^2}-\frac {\left ((i-2 a) b^2\right ) \int \frac {1}{x \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{2 (i+a) \left (1+a^2\right )}\\ &=-\frac {(i-2 a) b \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 (1-i a) \left (1+a^2\right ) x}-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 \left (1+a^2\right ) x^2}-\frac {\left ((i-2 a) b^2\right ) \text {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) \left (1+a^2\right )}\\ &=-\frac {(i-2 a) b \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 (1-i a) \left (1+a^2\right ) x}-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 \left (1+a^2\right ) x^2}+\frac {(1+2 i a) b^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} (i+a)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 154, normalized size = 0.77 \begin {gather*} \frac {-\frac {i \left (1+a^2+2 i b x-a b x\right ) \sqrt {1+a^2+2 a b x+b^2 x^2}}{x^2}+\frac {2 (-i+2 a) b^2 \tanh ^{-1}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1-i a} \sqrt {-1+i a}}}{2 (-i+a) (i+a)^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 287, normalized size = 1.43

method	result	size
risch	\(-\frac {i \left (-a \,b^{3} x^{3}+2 i b^{3} x^{3}-a^{2} b^{2} x^{2}+4 i a \,b^{2} x^{2}+a^{3} b x +2 i x \,a^{2} b +a^{4}+b^{2} x^{2}+a b x +2 i b x +2 a^{2}+1\right )}{2 x^{2} \left (i+a \right )^{2} \left (a -i\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}+\frac {i b^{2} \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{2 \left (a^{2}+1\right )^{\frac {3}{2}} \left (i+a \right )}-\frac {b^{2} \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right ) a}{\left (a^{2}+1\right )^{\frac {3}{2}} \left (i+a \right )}\)	\(246\)
default	\(\left (i a +1\right ) \left (-\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 \left (a^{2}+1\right ) x^{2}}-\frac {3 a b \left (-\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (a^{2}+1\right ) x}+\frac {a b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (a^{2}+1\right )^{\frac {3}{2}}}\right )}{2 \left (a^{2}+1\right )}+\frac {b^{2} \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{2 \left (a^{2}+1\right )^{\frac {3}{2}}}\right )+i b \left (-\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (a^{2}+1\right ) x}+\frac {a b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (a^{2}+1\right )^{\frac {3}{2}}}\right )\)	\(287\)

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^3,x,method=_RETURNVERBOSE)

[Out]

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

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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. 424 vs. \(2 (135) = 270\).
time = 0.28, size = 424, normalized size = 2.11 \begin {gather*} -\frac {3 \, a^{2} {\left (i \, a + 1\right )} b^{2} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{2 \, {\left (a^{2} + 1\right )}^{\frac {5}{2}}} + \frac {i \, a b^{2} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{{\left (a^{2} + 1\right )}^{\frac {3}{2}}} - \frac {{\left (-i \, a - 1\right )} b^{2} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{2 \, {\left (a^{2} + 1\right )}^{\frac {3}{2}}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a {\left (i \, a + 1\right )} b}{2 \, {\left (a^{2} + 1\right )}^{2} x} - \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b}{{\left (a^{2} + 1\right )} x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (i \, a + 1\right )}}{2 \, {\left (a^{2} + 1\right )} x^{2}} \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^3,x, algorithm="maxima")

[Out]

-3/2*a^2*(I*a + 1)*b^2*arcsinh(2*a*b*x/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)) + 2*a^2/(sqrt(-4*a^2*b^2 +
4*(a^2 + 1)*b^2)*abs(x)) + 2/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)))/(a^2 + 1)^(5/2) + I*a*b^2*arcsinh(2*
a*b*x/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)) + 2*a^2/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)) + 2/(sqr
t(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)))/(a^2 + 1)^(3/2) - 1/2*(-I*a - 1)*b^2*arcsinh(2*a*b*x/(sqrt(-4*a^2*b^2
+ 4*(a^2 + 1)*b^2)*abs(x)) + 2*a^2/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)) + 2/(sqrt(-4*a^2*b^2 + 4*(a^2
+ 1)*b^2)*abs(x)))/(a^2 + 1)^(3/2) + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*(I*a + 1)*b/((a^2 + 1)^2*x) - I*s
qrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b/((a^2 + 1)*x) - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(I*a + 1)/((a^2 + 1)*
x^2)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (135) = 270\).
time = 2.09, size = 452, normalized size = 2.25 \begin {gather*} \frac {{\left (i \, a + 2\right )} b^{2} x^{2} + \sqrt {\frac {{\left (4 \, a^{2} - 4 i \, a - 1\right )} b^{4}}{a^{8} + 2 i \, a^{7} + 2 \, a^{6} + 6 i \, a^{5} + 6 i \, a^{3} - 2 \, a^{2} + 2 i \, a - 1}} {\left (a^{3} + i \, a^{2} + a + i\right )} x^{2} \log \left (-\frac {{\left (2 \, a - i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a - i\right )} b^{2} + {\left (a^{5} + i \, a^{4} + 2 \, a^{3} + 2 i \, a^{2} + a + i\right )} \sqrt {\frac {{\left (4 \, a^{2} - 4 i \, a - 1\right )} b^{4}}{a^{8} + 2 i \, a^{7} + 2 \, a^{6} + 6 i \, a^{5} + 6 i \, a^{3} - 2 \, a^{2} + 2 i \, a - 1}}}{{\left (2 \, a - i\right )} b^{2}}\right ) - \sqrt {\frac {{\left (4 \, a^{2} - 4 i \, a - 1\right )} b^{4}}{a^{8} + 2 i \, a^{7} + 2 \, a^{6} + 6 i \, a^{5} + 6 i \, a^{3} - 2 \, a^{2} + 2 i \, a - 1}} {\left (a^{3} + i \, a^{2} + a + i\right )} x^{2} \log \left (-\frac {{\left (2 \, a - i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a - i\right )} b^{2} - {\left (a^{5} + i \, a^{4} + 2 \, a^{3} + 2 i \, a^{2} + a + i\right )} \sqrt {\frac {{\left (4 \, a^{2} - 4 i \, a - 1\right )} b^{4}}{a^{8} + 2 i \, a^{7} + 2 \, a^{6} + 6 i \, a^{5} + 6 i \, a^{3} - 2 \, a^{2} + 2 i \, a - 1}}}{{\left (2 \, a - i\right )} b^{2}}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (i \, a + 2\right )} b x - i \, a^{2} - i\right )}}{2 \, {\left (a^{3} + i \, a^{2} + a + i\right )} x^{2}} \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^3,x, algorithm="fricas")

[Out]

1/2*((I*a + 2)*b^2*x^2 + sqrt((4*a^2 - 4*I*a - 1)*b^4/(a^8 + 2*I*a^7 + 2*a^6 + 6*I*a^5 + 6*I*a^3 - 2*a^2 + 2*I
*a - 1))*(a^3 + I*a^2 + a + I)*x^2*log(-((2*a - I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a - I)*b^2 + (
a^5 + I*a^4 + 2*a^3 + 2*I*a^2 + a + I)*sqrt((4*a^2 - 4*I*a - 1)*b^4/(a^8 + 2*I*a^7 + 2*a^6 + 6*I*a^5 + 6*I*a^3
- 2*a^2 + 2*I*a - 1)))/((2*a - I)*b^2)) - sqrt((4*a^2 - 4*I*a - 1)*b^4/(a^8 + 2*I*a^7 + 2*a^6 + 6*I*a^5 + 6*I
*a^3 - 2*a^2 + 2*I*a - 1))*(a^3 + I*a^2 + a + I)*x^2*log(-((2*a - I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)
*(2*a - I)*b^2 - (a^5 + I*a^4 + 2*a^3 + 2*I*a^2 + a + I)*sqrt((4*a^2 - 4*I*a - 1)*b^4/(a^8 + 2*I*a^7 + 2*a^6 +
6*I*a^5 + 6*I*a^3 - 2*a^2 + 2*I*a - 1)))/((2*a - I)*b^2)) + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*((I*a + 2)*b*x
- I*a^2 - I))/((a^3 + I*a^2 + a + I)*x^2)

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

[Out]

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

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (135) = 270\).
time = 0.53, size = 471, normalized size = 2.34 \begin {gather*} -\frac {{\left (2 \, a b^{2} - i \, b^{2}\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 )}{2 \, {\left (a^{3} + i \, a^{2} + a + i\right )} \sqrt {a^{2} + 1}} - \frac {4 \, {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{4} b^{2} - 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a^{3} b {\left | b \right |} - 2 i \, a^{5} b {\left | b \right |} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} a b^{2} - 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{3} b^{2} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a^{2} b {\left | b \right |} - 2 \, a^{4} b {\left | b \right |} - i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} b^{2} + 5 \, {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} b^{2} - 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b {\left | b \right |} - 4 i \, a^{3} b {\left | b \right |} - 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a b^{2} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b {\left | b \right |} - 4 \, a^{2} b {\left | b \right |} - {\left (i \, x {\left | b \right |} - i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} b^{2} - 2 i \, a b {\left | b \right |} - 2 \, b {\left | b \right |}}{{\left (a^{3} + i \, a^{2} + a + i\right )} {\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} - a^{2} - 1\right )}^{2}} \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^3,x, algorithm="giac")

[Out]

-1/2*(2*a*b^2 - I*b^2)*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)))/((a^3 + I*a^2 + a + I)*sqrt(a^2 + 1)) - (4*(-I*x*abs(b) + I*sqrt((b*x +
a)^2 + 1))*a^4*b^2 - 2*I*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a^3*b*abs(b) - 2*I*a^5*b*abs(b) + 2*(x*abs(b) -
sqrt((b*x + a)^2 + 1))^3*a*b^2 - 2*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a^3*b^2 + 2*(x*abs(b) - sqrt((b*x + a)^2
+ 1))^2*a^2*b*abs(b) - 2*a^4*b*abs(b) - I*(x*abs(b) - sqrt((b*x + a)^2 + 1))^3*b^2 + 5*(-I*x*abs(b) + I*sqrt(
(b*x + a)^2 + 1))*a^2*b^2 - 2*I*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a*b*abs(b) - 4*I*a^3*b*abs(b) - 2*(x*abs(
b) - sqrt((b*x + a)^2 + 1))*a*b^2 + 2*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*b*abs(b) - 4*a^2*b*abs(b) - (I*x*ab
s(b) - I*sqrt((b*x + a)^2 + 1))*b^2 - 2*I*a*b*abs(b) - 2*b*abs(b))/((a^3 + I*a^2 + a + I)*((x*abs(b) - sqrt((b
*x + a)^2 + 1))^2 - a^2 - 1)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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