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]

((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

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Rubi [A]  time = 0.0440134, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, 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 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]

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 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^{-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*}

Mathematica [A]  time = 0.0346367, 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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Maple [B]  time = 0.062, size = 329, normalized size = 3.5 \begin{align*} -{\frac{1}{{b}^{4}} \left ( \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{b}}+{\frac{a}{b}} \right ) ^{-3}}-{\frac{2\,i}{{b}^{3}} \left ( \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{b}}+{\frac{a}{b}} \right ) ^{-2}}+{\frac{2\,i}{b} \left ( \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b \right ) ^{{\frac{3}{2}}}}-3\,\sqrt{ \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b}x-3\,{\frac{a}{b}\sqrt{ \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b}}-3\,{\frac{1}{\sqrt{{b}^{2}}}\ln \left ({\frac{1}{\sqrt{{b}^{2}}} \left ( ib+ \left ( x-{\frac{i-a}{b}} \right ){b}^{2} \right ) }+\sqrt{ \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b} \right ) } \end{align*}

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

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Maxima [A]  time = 1.48851, size = 139, normalized size = 1.48 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 2.22626, size = 262, normalized size = 2.79 \begin{align*} \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} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [B]  time = 1.14208, size = 265, normalized size = 2.82 \begin{align*} \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 |}} + \frac{2 \,{\left | b \right |} \log \left (12 \, b\right )}{b^{2}} \end{align*}

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) + 2*abs(b)*log(12*b)/b^2