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

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

[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)^(5/2)/(I+a)^(3/2)
-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)^2/(I+a)/x

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Rubi [A]  time = 0.12, 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 = {5095, 96, 94, 93, 208} \[ -\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{2 \left (a^2+1\right ) x^2}+\frac {(1-2 i a) b^2 \tanh ^{-1}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{5/2} (a+i)^{3/2}}+\frac {(1-2 i a) b \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{2 (-a+i)^2 (a+i) x} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(I*ArcTan[a + b*x])*x^3),x]

[Out]

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

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

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Mathematica [A]  time = 0.15, size = 154, normalized size = 0.77 \[ \frac {\frac {i \left (a^2-a b x-2 i b x+1\right ) \sqrt {a^2+2 a b x+b^2 x^2+1}}{x^2}+\frac {2 (2 a+i) b^2 \tanh ^{-1}\left (\frac {\sqrt {-1-i a} \sqrt {-i (a+b x+i)}}{\sqrt {-1+i a} \sqrt {i a+i b x+1}}\right )}{\sqrt {-1-i a} \sqrt {-1+i a}}}{2 (a-i)^2 (a+i)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(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)^2*(I + a))

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fricas [B]  time = 0.52, size = 455, normalized size = 2.26 \[ \frac {{\left (-i \, a + 2\right )} b^{2} x^{2} + \sqrt {\frac {{\left (4 \, a^{2} + 4 i \, a - 1\right )} b^{4}}{a^{8} - 2 i \, a^{7} + 2 \, a^{6} - 6 i \, a^{5} - 6 i \, a^{3} - 2 \, a^{2} - 2 i \, a - 1}} {\left (a^{3} - i \, a^{2} + a - i\right )} x^{2} \log \left (-\frac {{\left (2 \, a + i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a + i\right )} b^{2} + {\left (a^{5} - i \, a^{4} + 2 \, a^{3} - 2 i \, a^{2} + a - i\right )} \sqrt {\frac {{\left (4 \, a^{2} + 4 i \, a - 1\right )} b^{4}}{a^{8} - 2 i \, a^{7} + 2 \, a^{6} - 6 i \, a^{5} - 6 i \, a^{3} - 2 \, a^{2} - 2 i \, a - 1}}}{{\left (2 \, a + i\right )} b^{2}}\right ) - \sqrt {\frac {{\left (4 \, a^{2} + 4 i \, a - 1\right )} b^{4}}{a^{8} - 2 i \, a^{7} + 2 \, a^{6} - 6 i \, a^{5} - 6 i \, a^{3} - 2 \, a^{2} - 2 i \, a - 1}} {\left (a^{3} - i \, a^{2} + a - i\right )} x^{2} \log \left (-\frac {{\left (2 \, a + i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a + i\right )} b^{2} - {\left (a^{5} - i \, a^{4} + 2 \, a^{3} - 2 i \, a^{2} + a - i\right )} \sqrt {\frac {{\left (4 \, a^{2} + 4 i \, a - 1\right )} b^{4}}{a^{8} - 2 i \, a^{7} + 2 \, a^{6} - 6 i \, a^{5} - 6 i \, a^{3} - 2 \, a^{2} - 2 i \, a - 1}}}{{\left (2 \, a + i\right )} b^{2}}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (-i \, a + 2\right )} b x + i \, a^{2} + i\right )}}{{\left (2 \, a^{3} - 2 i \, a^{2} + 2 \, a - 2 i\right )} x^{2}} \]

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^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.26, size = 481, normalized size = 2.39 \[ -\frac {{\left (2 \, a b^{2} + b^{2} i\right )} \log \left (\frac {{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{2 \, {\left (a^{3} - a^{2} i + a - i\right )} \sqrt {a^{2} + 1}} - \frac {4 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{4} b^{2} i + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a^{3} b i {\left | b \right |} + 2 \, a^{5} b i {\left | b \right |} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} a b^{2} - 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{3} b^{2} + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} b^{2} i + 5 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} b^{2} i + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a^{2} b {\left | b \right |} - 2 \, a^{4} b {\left | b \right |} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b i {\left | b \right |} + 4 \, a^{3} b i {\left | b \right |} - 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a b^{2} + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} b^{2} i + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b {\left | b \right |} - 4 \, a^{2} b {\left | b \right |} + 2 \, a b i {\left | b \right |} - 2 \, b {\left | b \right |}}{{\left (a^{3} - a^{2} i + a - i\right )} {\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} - a^{2} - 1\right )}^{2}} \]

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^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.18, size = 1146, normalized size = 5.70 \[ -\frac {i b^{2} a \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 (i-a \right )^{2} \sqrt {a^{2}+1}}+\frac {i b^{3} 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 )}{2 \left (i-a \right ) \left (a^{2}+1\right ) \sqrt {b^{2}}}-\frac {i a^{2} b^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (i-a \right ) \left (a^{2}+1\right )^{2}}+\frac {i a^{2} 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 (i-a \right ) \left (a^{2}+1\right )^{\frac {3}{2}}}+\frac {i b^{3} 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 )}{\left (i-a \right )^{3} \sqrt {b^{2}}}-\frac {i b \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{\left (i-a \right )^{2} \left (a^{2}+1\right ) x}-\frac {i b^{2} \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{\left (i-a \right )^{3}}+\frac {i b^{3} a^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\left (i-a \right )^{2} \left (a^{2}+1\right ) \sqrt {b^{2}}}+\frac {2 i b^{2} a \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (i-a \right )^{2} \left (a^{2}+1\right )}-\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 (i-a \right ) \sqrt {a^{2}+1}}+\frac {b^{3} \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\left (i-a \right )^{3} \sqrt {b^{2}}}-\frac {i b^{2} \sqrt {a^{2}+1}\, \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 (i-a \right )^{3}}+\frac {i b^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x}{\left (i-a \right )^{2} \left (a^{2}+1\right )}-\frac {i \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{2 \left (i-a \right ) \left (a^{2}+1\right ) x^{2}}-\frac {i a \,b^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x}{2 \left (i-a \right ) \left (a^{2}+1\right )^{2}}-\frac {i a \,b^{3} \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 \left (i-a \right ) \left (a^{2}+1\right )^{2} \sqrt {b^{2}}}+\frac {i b^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 \left (i-a \right ) \left (a^{2}+1\right )}+\frac {i a b \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{2 \left (i-a \right ) \left (a^{2}+1\right )^{2} x}+\frac {i b^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (i-a \right )^{3}}+\frac {i b^{3} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\left (i-a \right )^{2} \left (a^{2}+1\right ) \sqrt {b^{2}}}-\frac {i a^{3} b^{3} \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 \left (i-a \right ) \left (a^{2}+1\right )^{2} \sqrt {b^{2}}} \]

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^3,x)

[Out]

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

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

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^3,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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**3,x)

[Out]

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

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