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

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

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Rubi [A]  time = 0.0418089, 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]

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

Mathematica [A]  time = 0.0393388, 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 = 0.092, size = 362, normalized size = 3.9 \begin{align*}{\frac{-5\,i}{b}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}+3\,{\frac{a}{b\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}+3\,{\frac{{a}^{2}x}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}+3\,{\frac{{a}^{3}}{b\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}+2\,{\frac{ \left ( 1+ia \right ) ^{3} \left ( 2\,{b}^{2}x+2\,ab \right ) }{ \left ( 4\,{b}^{2} \left ({a}^{2}+1 \right ) -4\,{a}^{2}{b}^{2} \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}+{i{a}^{3}x{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}+{\frac{i{a}^{4}}{b}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}-{ib{x}^{2}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}-{5\,iax{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}+3\,{\frac{x}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}-{\frac{4\,i{a}^{2}}{b}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}-3\,{\frac{1}{\sqrt{{b}^{2}}}\ln \left ({\frac{{b}^{2}x+ab}{\sqrt{{b}^{2}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*I/b/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3/b*a/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3*a^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x
+3/b*a^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+2*(1+I*a)^3*(2*b^2*x+2*a*b)/(4*b^2*(a^2+1)-4*a^2*b^2)/(b^2*x^2+2*a*b*x+
a^2+1)^(1/2)+I*a^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+I/b*a^4/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-I*b*x^2/(b^2*x^2+2*a*
b*x+a^2+1)^(1/2)-5*I*a*x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3*x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-4*I/b*a^2/(b^2*x^2+2*
a*b*x+a^2+1)^(1/2)-3*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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.16026, size = 265, normalized size = 2.82 \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+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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (i a + i b x + 1\right )^{3}}{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.16944, size = 262, normalized size = 2.79 \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 + a^{3} b + 2 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}^{2} b i + 2 \, a^{2} b i +{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}^{3}{\left | b \right |} + 3 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )} a^{2}{\left | b \right |} + 4 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )} a i{\left | b \right |} - a b -{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}{\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+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 + a^3*b + 2*(x*abs(b) - sqrt((b*x
+ a)^2 + 1))^2*b*i + 2*a^2*b*i + (x*abs(b) - sqrt((b*x + a)^2 + 1))^3*abs(b) + 3*(x*abs(b) - sqrt((b*x + a)^2
+ 1))*a^2*abs(b) + 4*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a*i*abs(b) - a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*
abs(b))/abs(b) + 2*abs(b)*log(12*b)/b^2