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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.0759064, antiderivative size = 110, normalized size of antiderivative = 1., 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 = {5095, 80, 50, 53, 619, 215} \[ \frac{\sqrt{i a+i b x+1} (-i a-i b x+1)^{3/2}}{2 b^2}+\frac{(1+2 i a) \sqrt{i a+i b x+1} \sqrt{-i a-i b x+1}}{2 b^2}+\frac{(-2 a+i) \sinh ^{-1}(a+b x)}{2 b^2} \]

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 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]

Rule 80

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 50

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 53

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 619

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 215

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]

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) \operatorname{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*}

Mathematica [A]  time = 0.129312, size = 131, normalized size = 1.19 \[ \frac{\sqrt{i a+i b x+1} \left (a^2-i a-b^2 x^2-3 i b x+2\right )}{2 b^2 \sqrt{-i (a+b x+i)}}+\frac{(-1)^{3/4} (1+2 i a) \sqrt{-i b} \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{b^{5/2}} \]

Warning: Unable to verify antiderivative.

[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)

________________________________________________________________________________________

Maple [B]  time = 0.08, size = 350, normalized size = 3.2 \begin{align*}{\frac{-{\frac{i}{2}}x}{b}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{i}{2}}a}{{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{i}{2}}}{b}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{ia}{{b}^{2}}\sqrt{ \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b}}+{\frac{1}{{b}^{2}}\sqrt{ \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b}}-{\frac{a}{b}\ln \left ({ \left ( ib+ \left ( x-{\frac{i-a}{b}} \right ){b}^{2} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{ \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{i}{b}\ln \left ({ \left ( ib+ \left ( x-{\frac{i-a}{b}} \right ){b}^{2} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{ \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

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)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.51266, size = 131, normalized size = 1.19 \begin{align*} -\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{align*}

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

________________________________________________________________________________________

Fricas [A]  time = 2.29105, size = 200, normalized size = 1.82 \begin{align*} \frac{3 i \, a^{2} +{\left (8 \, a - 4 i\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (-4 i \, b x + 4 i \, a + 8\right )} + 4 \, a}{8 \, b^{2}} \end{align*}

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

________________________________________________________________________________________

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

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]

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

________________________________________________________________________________________

Giac [A]  time = 1.13039, size = 105, normalized size = 0.95 \begin{align*} -\frac{1}{2} \, \sqrt{{\left (b x + a\right )}^{2} + 1}{\left (\frac{i x}{b} - \frac{a b i + 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{align*}

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 - (a*b*i + 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))