3.2.92 \(\int e^{-i \text {ArcTan}(a+b x)} x \, dx\) [192]

Optimal. Leaf size=110 \[ \frac {(1+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(i-2 a) \sinh ^{-1}(a+b x)}{2 b^2} \]

[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

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Rubi [A]
time = 0.06, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5203, 81, 52, 55, 633, 221} \begin {gather*} \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}+\frac {(-2 a+i) \sinh ^{-1}(a+b x)}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/E^(I*ArcTan[a + b*x]),x]

[Out]

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

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Mathematica [A]
time = 0.09, size = 131, normalized size = 1.19 \begin {gather*} \frac {\sqrt {1+i a+i b x} \left (2-i a+a^2-3 i b x-b^2 x^2\right )}{2 b^2 \sqrt {-i (i+a+b x)}}+\frac {(-1)^{3/4} (1+2 i a) \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^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/E^(I*ArcTan[a + b*x]),x]

[Out]

(Sqrt[1 + I*a + I*b*x]*(2 - I*a + a^2 - (3*I)*b*x - b^2*x^2))/(2*b^2*Sqrt[(-I)*(I + a + b*x)]) + ((-1)^(3/4)*(
1 + (2*I)*a)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/b^(5/2)

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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. 236 vs. \(2 (86 ) = 172\).
time = 0.11, size = 237, normalized size = 2.15

method	result	size
risch	\(\frac {i \left (-b x +a -2 i\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}+\frac {i \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b \sqrt {b^{2}}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\)	\(129\)
default	\(-\frac {i \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}+\frac {\left (i a +1\right ) \left (\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}+\frac {i b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}}\right )}{b^{2}}\)	\(237\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I/b*(1/4*(2*b^2*x+2*a*b)/b^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/8*(4*b^2*(a^2+1)-4*a^2*b^2)/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))+(1+I*a)/b^2*(((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2
)+I*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 [A]
time = 0.48, size = 97, normalized size = 0.88 \begin {gather*} -\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{2 \, b} - \frac {a \operatorname {arsinh}\left (b x + a\right )}{b^{2}} + \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{2 \, b^{2}} + \frac {i \, \operatorname {arsinh}\left (b x + a\right )}{2 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.73, size = 79, normalized size = 0.72 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.45, size = 75, normalized size = 0.68 \begin {gather*} -\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 |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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