\(\int e^{i \arctan (a+b x)} x \, dx\) [165]

Optimal result

Integrand size = 14, antiderivative size = 110 \[ \int e^{i \arctan (a+b x)} x \, dx=\frac {(1-2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 b^2}-\frac {(i+2 a) \text {arcsinh}(a+b x)}{2 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5203, 81, 52, 55, 633, 221} \[ \int e^{i \arctan (a+b x)} x \, dx=-\frac {(2 a+i) \text {arcsinh}(a+b x)}{2 b^2}+\frac {\sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{2 b^2}+\frac {(1-2 i a) \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{2 b^2} \]

[In]

Int[E^(I*ArcTan[a + b*x])*x,x]

[Out]

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

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 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*} \text {integral}& = \int \frac {x \sqrt {1+i a+i b x}}{\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 b^2}-\frac {(i+2 a) \int \frac {\sqrt {1+i a+i b x}}{\sqrt {1-i a-i b x}} \, dx}{2 b} \\ & = \frac {(1-2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 b^2}-\frac {(i+2 a) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{2 b} \\ & = \frac {(1-2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 b^2}-\frac {(i+2 a) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{2 b} \\ & = \frac {(1-2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 b^2}-\frac {(i+2 a) \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^3} \\ & = \frac {(1-2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 b^2}-\frac {(i+2 a) \text {arcsinh}(a+b x)}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.98 \[ \int e^{i \arctan (a+b x)} x \, dx=\frac {(2-i a+i b x) \sqrt {1+a^2+2 a b x+b^2 x^2}}{2 b^2}+\frac {(-1)^{3/4} (i+2 a) \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{\sqrt {-i b} b^{3/2}} \]

[In]

Integrate[E^(I*ArcTan[a + b*x])*x,x]

[Out]

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

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {i \left (-b x +a +2 i\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {\left (i+2 a \right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b \sqrt {b^{2}}}\)	\(87\)
default	\(i b \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )+\left (i a +1\right ) \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )\)	\(238\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72 \[ \int e^{i \arctan (a+b x)} x \, dx=\frac {-3 i \, a^{2} + 4 \, {\left (2 \, a + i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-i \, b x + i \, a - 2\right )} + 4 \, a}{8 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (85) = 170\).

Time = 0.92 (sec) , antiderivative size = 362, normalized size of antiderivative = 3.29 \[ \int e^{i \arctan (a+b x)} x \, dx=\begin {cases} \left (\frac {i x}{2 b} + \frac {- \frac {i a}{2} + 1}{b^{2}}\right ) \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} + \frac {\left (- \frac {a \left (- \frac {i a}{2} + 1\right )}{b} - \frac {i \left (a^{2} + 1\right )}{2 b}\right ) \log {\left (2 a b + 2 b^{2} x + 2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \sqrt {b^{2}} \right )}}{\sqrt {b^{2}}} & \text {for}\: b^{2} \neq 0 \\\frac {\frac {i \left (- a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}}}{3} - \sqrt {a^{2} + 2 a b x + 1}\right )}{b} + \frac {- a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}}}{3} - \sqrt {a^{2} + 2 a b x + 1}}{a b} + \frac {i \left (a^{4} \sqrt {a^{2} + 2 a b x + 1} + 2 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 2 a^{2} - 2\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \sqrt {a^{2} + 2 a b x + 1}\right )}{2 a^{2} b}}{2 a b} & \text {for}\: a b \neq 0 \\\frac {\frac {i a x^{2}}{2} + \frac {i b x^{3}}{3} + \frac {x^{2}}{2}}{\sqrt {a^{2} + 1}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((I*x/(2*b) + (-I*a/2 + 1)/b**2)*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + (-a*(-I*a/2 + 1)/b - I*(a**2
+ 1)/(2*b))*log(2*a*b + 2*b**2*x + 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*sqrt(b**2))/sqrt(b**2), Ne(b**2, 0)
), ((I*(-a**2*sqrt(a**2 + 2*a*b*x + 1) + (a**2 + 2*a*b*x + 1)**(3/2)/3 - sqrt(a**2 + 2*a*b*x + 1))/b + (-a**2*
sqrt(a**2 + 2*a*b*x + 1) + (a**2 + 2*a*b*x + 1)**(3/2)/3 - sqrt(a**2 + 2*a*b*x + 1))/(a*b) + I*(a**4*sqrt(a**2
+ 2*a*b*x + 1) + 2*a**2*sqrt(a**2 + 2*a*b*x + 1) + (-2*a**2 - 2)*(a**2 + 2*a*b*x + 1)**(3/2)/3 + (a**2 + 2*a*
b*x + 1)**(5/2)/5 + sqrt(a**2 + 2*a*b*x + 1))/(2*a**2*b))/(2*a*b), Ne(a*b, 0)), ((I*a*x**2/2 + I*b*x**3/3 + x*
*2/2)/sqrt(a**2 + 1), True))

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (76) = 152\).

Time = 0.18 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.90 \[ \int e^{i \arctan (a+b x)} x \, dx=\frac {3 i \, a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{2}} - \frac {a {\left (i \, a + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{2}} + \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{2 \, b} - \frac {{\left (i \, a^{2} + i\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{2}} - \frac {3 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{2 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (i \, a + 1\right )}}{b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.68 \[ \int e^{i \arctan (a+b x)} x \, dx=-\frac {1}{2} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (-\frac {i \, x}{b} + \frac {i \, a b - 2 \, b}{b^{3}}\right )} + \frac {{\left (2 \, a + i\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int e^{i \arctan (a+b x)} x \, dx=\int \frac {x\,\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}{\sqrt {{\left (a+b\,x\right )}^2+1}} \,d x \]

[In]

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

[Out]

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