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

Optimal. Leaf size=94 \[ \frac {2 i (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}+\frac {3 i \sqrt {i a+i b x+1} \sqrt {-i a-i b x+1}}{b}-\frac {3 \sinh ^{-1}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5093, 47, 50, 53, 619, 215} \[ \frac {2 i (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}+\frac {3 i \sqrt {i a+i b x+1} \sqrt {-i a-i b x+1}}{b}-\frac {3 \sinh ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^((-3*I)*ArcTan[a + b*x]),x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 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 5093

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.)), x_Symbol] :> Int[(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, n}, x]

Rubi steps

\begin {align*} \int e^{-3 i \tan ^{-1}(a+b x)} \, dx &=\int \frac {(1-i a-i b x)^{3/2}}{(1+i a+i b x)^{3/2}} \, dx\\ &=\frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}-3 \int \frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx\\ &=\frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-3 \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx\\ &=\frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-3 \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx\\ &=\frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )}{2 b^2}\\ &=\frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-\frac {3 \sinh ^{-1}(a+b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 45, normalized size = 0.48 \[ -\frac {3 \sinh ^{-1}(a+b x)}{b}+\frac {\sqrt {(a+b x)^2+1} \left (\frac {4}{a+b x-i}+i\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((-3*I)*ArcTan[a + b*x]),x]

[Out]

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

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fricas [A]  time = 0.47, size = 102, normalized size = 1.09 \[ \frac {{\left (i \, a + 8\right )} b x + i \, a^{2} + {\left (6 \, b x + 6 \, a - 6 i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 i \, b x + 2 i \, a + 10\right )} + 9 \, a - 8 i}{2 \, b^{2} x + {\left (2 \, a - 2 i\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="fricas")

[Out]

((I*a + 8)*b*x + I*a^2 + (6*b*x + 6*a - 6*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)*(2*I*b*x + 2*I*a + 10) + 9*a - 8*I)/(2*b^2*x + (2*a - 2*I)*b)

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giac [B]  time = 0.17, size = 185, normalized size = 1.97 \[ \frac {\sqrt {{\left (b x + a\right )}^{2} + 1} i}{b} + \frac {\log \left (3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b i + a^{3} b i + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} i {\left | b \right |} + 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} i {\left | b \right |} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b + 2 \, a^{2} b - a b i + 4 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} i {\left | b \right |}\right )}{{\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="giac")

[Out]

sqrt((b*x + a)^2 + 1)*i/b + log(3*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a*b*i + a^3*b*i + (x*abs(b) - sqrt((b*x
+ a)^2 + 1))^3*i*abs(b) + 3*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a^2*i*abs(b) + 2*(x*abs(b) - sqrt((b*x + a)^2
+ 1))^2*b + 2*a^2*b - a*b*i + 4*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a*abs(b) - (x*abs(b) - sqrt((b*x + a)^2 + 1
))*i*abs(b))/abs(b)

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maple [B]  time = 0.12, size = 329, normalized size = 3.50 \[ -\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b^{4} \left (x -\frac {i}{b}+\frac {a}{b}\right )^{3}}-\frac {2 i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b^{3} \left (x -\frac {i}{b}+\frac {a}{b}\right )^{2}}+\frac {2 i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{b}-3 \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\, x -\frac {3 \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\, a}{b}-\frac {3 \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 103, normalized size = 1.10 \[ \frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b - 2 i \, b^{2} x - 2 i \, a b - b} - \frac {3 \, \operatorname {arsinh}\left (b x + a\right )}{b} + \frac {6 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{i \, b^{2} x + i \, a b + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="maxima")

[Out]

I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/(b^3*x^2 + 2*a*b^2*x + a^2*b - 2*I*b^2*x - 2*I*a*b - b) - 3*arcsinh(b*x
+ a)/b + 6*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(I*b^2*x + I*a*b + b)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left ({\left (a+b\,x\right )}^2+1\right )}^{3/2}}{{\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i \left (\int \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx + \int \frac {a^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx + \int \frac {b^{2} x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx + \int \frac {2 a b x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 + 3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3
*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x) + Integral(a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 + 3
*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x) + Integral
(b**2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 + 3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*
a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x) + Integral(2*a*b*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 +
3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x))

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