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

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+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 = 0.08, 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 = {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[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 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 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 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]

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 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 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 {\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) \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 {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{2 b^2}-\frac {(i+2 a) \sinh ^{-1}(a+b x)}{2 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 108, normalized size = 0.98 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2+1} (-i a+i b x+2)}{2 b^2}+\frac {(-1)^{3/4} (2 a+i) \sinh ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (a+b x+i)}}{\sqrt {-i b}}\right )}{\sqrt {-i b} b^{3/2}} \]

Warning: Unable to verify antiderivative.

[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))

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fricas [A]  time = 0.43, size = 77, normalized size = 0.70 \[ \frac {-3 i \, a^{2} + {\left (8 \, a + 4 i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (4 i \, b x - 4 i \, a + 8\right )} + 4 \, a}{8 \, b^{2}} \]

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,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

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giac [A]  time = 0.19, size = 76, normalized size = 0.69 \[ \frac {1}{2} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (\frac {i x}{b} - \frac {a b i - 2 \, b}{b^{3}}\right )} + \frac {{\left (2 \, a + i\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b {\left | b \right |}} \]

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,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))

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maple [A]  time = 0.17, size = 171, normalized size = 1.55 \[ \frac {i x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b}-\frac {i a \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {i \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b \sqrt {b^{2}}}+\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}} \]

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

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maxima [B]  time = 0.33, size = 209, normalized size = 1.90 \[ \frac {3 i \, a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{2}} - \frac {a {\left (i \, a + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{2}} + \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{2 \, b} - \frac {{\left (i \, a^{2} + i\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{2}} - \frac {3 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{2 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (i \, a + 1\right )}}{b^{2}} \]

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,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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

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,x)

[Out]

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