\(\int \frac {e^{4 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx\) [302]

Optimal result

Integrand size = 24, antiderivative size = 73 \[ \int \frac {e^{4 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {2 i \sqrt {1+i a x}}{a \sqrt {1-i a x}}-\frac {2 i (1+i a x)^{3/2}}{3 a (1-i a x)^{3/2}}+\frac {\text {arcsinh}(a x)}{a} \]

[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] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5181, 49, 41, 221} \[ \int \frac {e^{4 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {\text {arcsinh}(a x)}{a}-\frac {2 i (1+i a x)^{3/2}}{3 a (1-i a x)^{3/2}}+\frac {2 i \sqrt {1+i a x}}{a \sqrt {1-i a x}} \]

[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 49

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 221

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 5181

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*} \text {integral}& = \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 {\text {arcsinh}(a x)}{a} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.66 \[ \int \frac {e^{4 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {4 i \sqrt {2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1-i a x)\right )}{3 a (1-i a x)^{3/2}} \]

[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))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (57 ) = 114\).

Time = 0.33 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.44

method	result	size
meijerg	\(\frac {x \left (2 a^{2} x^{2}+3\right )}{3 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {8 i \left (\frac {\sqrt {\pi }}{2}-\frac {\sqrt {\pi }}{2 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}\right )}{3 a \sqrt {\pi }}-\frac {2 a^{2} x^{3}}{\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {8 i \left (\sqrt {\pi }-\frac {\sqrt {\pi }\, \left (12 a^{2} x^{2}+8\right )}{8 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}\right )}{3 a \sqrt {\pi }}+\frac {-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {5}{2}} \left (20 a^{2} x^{2}+15\right )}{15 a^{4} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {\sqrt {\pi }\, \left (a^{2}\right )^{\frac {5}{2}} \operatorname {arcsinh}\left (a x \right )}{a^{5}}}{\sqrt {\pi }\, \sqrt {a^{2}}}\)	\(178\)
default	\(\frac {x}{3 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {2 x}{3 \sqrt {a^{2} x^{2}+1}}+a^{4} \left (-\frac {x^{3}}{3 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}}{a^{2}}\right )-6 a^{2} \left (-\frac {x}{2 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {\frac {x}{3 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {2 x}{3 \sqrt {a^{2} x^{2}+1}}}{2 a^{2}}\right )-4 i a^{3} \left (-\frac {x^{2}}{a^{2} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {2}{3 a^{4} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}\right )-\frac {4 i}{3 a \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}\)	\(224\)

[In]

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

[Out]

1/3*x*(2*a^2*x^2+3)/(a^2*x^2+1)^(3/2)+8/3*I/a/Pi^(1/2)*(1/2*Pi^(1/2)-1/2*Pi^(1/2)/(a^2*x^2+1)^(3/2))-2*a^2*x^3
/(a^2*x^2+1)^(3/2)-8/3*I/a/Pi^(1/2)*(Pi^(1/2)-1/8*Pi^(1/2)*(12*a^2*x^2+8)/(a^2*x^2+1)^(3/2))+2/3/Pi^(1/2)/(a^2
)^(1/2)*(-1/10*Pi^(1/2)*x*(a^2)^(5/2)*(20*a^2*x^2+15)/a^4/(a^2*x^2+1)^(3/2)+3/2*Pi^(1/2)*(a^2)^(5/2)/a^5*arcsi
nh(a*x))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.18 \[ \int \frac {e^{4 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {8 \, a^{2} x^{2} + 16 i \, a x + 3 \, {\left (a^{2} x^{2} + 2 i \, a x - 1\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + 4 \, \sqrt {a^{2} x^{2} + 1} {\left (2 \, a x + i\right )} - 8}{3 \, {\left (a^{3} x^{2} + 2 i \, a^{2} x - a\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {e^{4 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\int \frac {\left (a x - i\right )^{4}}{\left (a^{2} x^{2} + 1\right )^{\frac {5}{2}}}\, dx \]

[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)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (51) = 102\).

Time = 0.18 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.53 \[ \int \frac {e^{4 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {1}{3} \, a^{4} x {\left (\frac {3 \, x^{2}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{2}} + \frac {2}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{4}}\right )} + \frac {4 i \, a x^{2}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} - \frac {5 \, x}{3 \, \sqrt {a^{2} x^{2} + 1}} + \frac {\operatorname {arsinh}\left (a x\right )}{a} + \frac {7 \, x}{3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} + \frac {4 i}{3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.33 \[ \int \frac {e^{4 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {\log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right )}{{\left | a \right |}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.26 \[ \int \frac {e^{4 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{\sqrt {a^2}}-\frac {8\,\sqrt {a^2\,x^2+1}}{3\,\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}}+\frac {a\,\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{3\,\left (a^4\,x^2+a^3\,x\,2{}\mathrm {i}-a^2\right )} \]

[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))