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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 73, normalized size of antiderivative = 1.00, 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 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 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 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 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])

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 = 0.01, 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))

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 86, normalized size = 1.18 \[ -\frac {8 \, a^{2} x^{2} + 16 i \, a x + {\left (3 \, a^{2} x^{2} + 6 i \, a x - 3\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + \sqrt {a^{2} x^{2} + 1} {\left (8 \, a x + 4 i\right )} - 8}{3 \, a^{3} x^{2} + 6 i \, a^{2} x - 3 \, a} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.15, size = 24, normalized size = 0.33 \[ -\frac {\log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right )}{{\left | a \right |}} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.18, size = 113, normalized size = 1.55 \[ \frac {7 x}{3 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {7 x}{3 \sqrt {a^{2} x^{2}+1}}-\frac {a^{2} x^{3}}{3 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {\ln \left (\frac {x \,a^{2}}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}+\frac {4 i a \,x^{2}}{\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {4 i}{3 a \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}} \]

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(x*a^2/(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)

________________________________________________________________________________________

maxima [B]  time = 0.32, size = 112, normalized size = 1.53 \[ -\frac {1}{3} \, a^{4} x {\left (\frac {3 \, x^{2}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{2}} + \frac {2}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{4}}\right )} + \frac {4 i \, a x^{2}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} - \frac {5 \, x}{3 \, \sqrt {a^{2} x^{2} + 1}} + \frac {\operatorname {arsinh}\left (a x\right )}{a} + \frac {7 \, x}{3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} + \frac {4 i}{3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a} \]

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

________________________________________________________________________________________

mupad [B]  time = 0.53, size = 92, normalized size = 1.26 \[ \frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{\sqrt {a^2}}-\frac {8\,\sqrt {a^2\,x^2+1}}{3\,\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}}+\frac {a\,\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{3\,\left (a^4\,x^2+a^3\,x\,2{}\mathrm {i}-a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

asinh(x*(a^2)^(1/2))/(a^2)^(1/2) - (8*(a^2*x^2 + 1)^(1/2))/(3*(((a^2)^(1/2)*1i)/a + x*(a^2)^(1/2))*(a^2)^(1/2)
) + (a*(a^2*x^2 + 1)^(1/2)*4i)/(3*(a^3*x*2i - a^2 + a^4*x^2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x - i\right )^{4}}{\left (a^{2} x^{2} + 1\right )^{\frac {5}{2}}}\, dx \]

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

________________________________________________________________________________________