3.302 \(\int \frac{e^{4 i \tan ^{-1}(a x)}}{\sqrt{1+a^2 x^2}} \, dx\)

Optimal. Leaf size=73 \[ -\frac{2 i (1+i a x)^{3/2}}{3 a (1-i a x)^{3/2}}+\frac{2 i \sqrt{1+i a x}}{a \sqrt{1-i a x}}+\frac{\sinh ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.0388457, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5073, 47, 41, 215} \[ -\frac{2 i (1+i a x)^{3/2}}{3 a (1-i a x)^{3/2}}+\frac{2 i \sqrt{1+i a x}}{a \sqrt{1-i a x}}+\frac{\sinh ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x])/Sqrt[1 + a^2*x^2],x]

[Out]

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

Rule 5073

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - I*a*x)^(p + (I*n
)/2)*(1 + I*a*x)^(p - (I*n)/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[d, a^2*c] && (IntegerQ[p] || GtQ[c,
0])

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 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[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 \frac{e^{4 i \tan ^{-1}(a x)}}{\sqrt{1+a^2 x^2}} \, dx &=\int \frac{(1+i a x)^{3/2}}{(1-i a x)^{5/2}} \, dx\\ &=-\frac{2 i (1+i a x)^{3/2}}{3 a (1-i a x)^{3/2}}-\int \frac{\sqrt{1+i a x}}{(1-i a x)^{3/2}} \, dx\\ &=\frac{2 i \sqrt{1+i a x}}{a \sqrt{1-i a x}}-\frac{2 i (1+i a x)^{3/2}}{3 a (1-i a x)^{3/2}}+\int \frac{1}{\sqrt{1-i a x} \sqrt{1+i a x}} \, dx\\ &=\frac{2 i \sqrt{1+i a x}}{a \sqrt{1-i a x}}-\frac{2 i (1+i a x)^{3/2}}{3 a (1-i a x)^{3/2}}+\int \frac{1}{\sqrt{1+a^2 x^2}} \, dx\\ &=\frac{2 i \sqrt{1+i a x}}{a \sqrt{1-i a x}}-\frac{2 i (1+i a x)^{3/2}}{3 a (1-i a x)^{3/2}}+\frac{\sinh ^{-1}(a x)}{a}\\ \end{align*}

Mathematica [C]  time = 0.0119682, size = 48, normalized size = 0.66 \[ -\frac{4 i \sqrt{2} \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};\frac{1}{2} (1-i a x)\right )}{3 a (1-i a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((4*I)*ArcTan[a*x])/Sqrt[1 + a^2*x^2],x]

[Out]

(((-4*I)/3)*Sqrt[2]*Hypergeometric2F1[-3/2, -3/2, -1/2, (1 - I*a*x)/2])/(a*(1 - I*a*x)^(3/2))

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Maple [A]  time = 0.067, size = 113, normalized size = 1.6 \begin{align*}{\frac{7\,x}{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{\frac{7\,x}{3}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{{a}^{2}{x}^{3}}{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{4\,ia{x}^{2} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{{\frac{4\,i}{3}}}{a} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^(5/2),x)

[Out]

7/3*x/(a^2*x^2+1)^(3/2)-7/3*x/(a^2*x^2+1)^(1/2)-1/3*a^2*x^3/(a^2*x^2+1)^(3/2)+ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)
^(1/2))/(a^2)^(1/2)+4*I*a*x^2/(a^2*x^2+1)^(3/2)+4/3*I/a/(a^2*x^2+1)^(3/2)

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Maxima [B]  time = 0.995436, size = 163, normalized size = 2.23 \begin{align*} -\frac{1}{3} \, a^{4} x{\left (\frac{3 \, x^{2}}{{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a^{2}} + \frac{2}{{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a^{4}}\right )} + \frac{4 i \, a x^{2}}{{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}} - \frac{5 \, x}{3 \, \sqrt{a^{2} x^{2} + 1}} + \frac{\operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}} + \frac{7 \, x}{3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}} + \frac{4 i}{3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^(5/2),x, algorithm="maxima")

[Out]

-1/3*a^4*x*(3*x^2/((a^2*x^2 + 1)^(3/2)*a^2) + 2/((a^2*x^2 + 1)^(3/2)*a^4)) + 4*I*a*x^2/(a^2*x^2 + 1)^(3/2) - 5
/3*x/sqrt(a^2*x^2 + 1) + arcsinh(a^2*x/sqrt(a^2))/sqrt(a^2) + 7/3*x/(a^2*x^2 + 1)^(3/2) + 4/3*I/((a^2*x^2 + 1)
^(3/2)*a)

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Fricas [A]  time = 1.88621, size = 204, normalized size = 2.79 \begin{align*} -\frac{8 \, a^{2} x^{2} + 16 i \, a x +{\left (3 \, a^{2} x^{2} + 6 i \, a x - 3\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right ) + \sqrt{a^{2} x^{2} + 1}{\left (8 \, a x + 4 i\right )} - 8}{3 \, a^{3} x^{2} + 6 i \, a^{2} x - 3 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^(5/2),x, algorithm="fricas")

[Out]

-(8*a^2*x^2 + 16*I*a*x + (3*a^2*x^2 + 6*I*a*x - 3)*log(-a*x + sqrt(a^2*x^2 + 1)) + sqrt(a^2*x^2 + 1)*(8*a*x +
4*I) - 8)/(3*a^3*x^2 + 6*I*a^2*x - 3*a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (i a x + 1\right )^{4}}{\left (a^{2} x^{2} + 1\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**4/(a**2*x**2+1)**(5/2),x)

[Out]

Integral((I*a*x + 1)**4/(a**2*x**2 + 1)**(5/2), x)

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Giac [A]  time = 1.13991, size = 32, normalized size = 0.44 \begin{align*} -\frac{\log \left (-x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1}\right )}{{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^(5/2),x, algorithm="giac")

[Out]

-log(-x*abs(a) + sqrt(a^2*x^2 + 1))/abs(a)