\(\int \frac {e^{4 i \arctan (a x)}}{(1+a^2 x^2)^{3/2}} \, dx\) [320]

Optimal result

Integrand size = 24, antiderivative size = 67 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {i (1+i a x)^{3/2}}{5 a (1-i a x)^{5/2}}-\frac {i (1+i a x)^{3/2}}{15 a (1-i a x)^{3/2}} \]

[Out]

-1/5*I*(1+I*a*x)^(3/2)/a/(1-I*a*x)^(5/2)-1/15*I*(1+I*a*x)^(3/2)/a/(1-I*a*x)^(3/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5181, 47, 37} \[ \int \frac {e^{4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {i (1+i a x)^{3/2}}{15 a (1-i a x)^{3/2}}-\frac {i (1+i a x)^{3/2}}{5 a (1-i a x)^{5/2}} \]

[In]

Int[E^((4*I)*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

((-1/5*I)*(1 + I*a*x)^(3/2))/(a*(1 - I*a*x)^(5/2)) - ((I/15)*(1 + I*a*x)^(3/2))/(a*(1 - I*a*x)^(3/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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 {\sqrt {1+i a x}}{(1-i a x)^{7/2}} \, dx \\ & = -\frac {i (1+i a x)^{3/2}}{5 a (1-i a x)^{5/2}}+\frac {1}{5} \int \frac {\sqrt {1+i a x}}{(1-i a x)^{5/2}} \, dx \\ & = -\frac {i (1+i a x)^{3/2}}{5 a (1-i a x)^{5/2}}-\frac {i (1+i a x)^{3/2}}{15 a (1-i a x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.70 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {(1+i a x)^{3/2} (4 i+a x)}{15 a \sqrt {1-i a x} (i+a x)^2} \]

[In]

Integrate[E^((4*I)*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

((1 + I*a*x)^(3/2)*(4*I + a*x))/(15*a*Sqrt[1 - I*a*x]*(I + a*x)^2)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {\left (-a x +i\right ) \left (a x +i\right ) \left (a x +4 i\right ) \left (i a x +1\right )^{4}}{15 a \left (a^{2} x^{2}+1\right )^{\frac {7}{2}}}\)	\(45\)
trager	\(\frac {-a^{5} x^{5}-10 a^{3} x^{3}+20 i x^{2} a^{2}+15 a x -4 i}{15 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}} a}\)	\(49\)
meijerg	\(\frac {x \left (8 a^{4} x^{4}+20 a^{2} x^{2}+15\right )}{15 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {16 i \left (\frac {3 \sqrt {\pi }}{4}-\frac {3 \sqrt {\pi }}{4 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}\right )}{15 a \sqrt {\pi }}-\frac {2 a^{2} x^{3} \left (2 a^{2} x^{2}+5\right )}{5 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}-\frac {16 i \left (\frac {\sqrt {\pi }}{2}-\frac {\sqrt {\pi }\, \left (20 a^{2} x^{2}+8\right )}{16 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}\right )}{15 a \sqrt {\pi }}+\frac {a^{4} x^{5}}{5 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}\)	\(154\)
default	\(\frac {x}{5 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {4 x}{15 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {8 x}{15 \sqrt {a^{2} x^{2}+1}}+a^{4} \left (-\frac {x^{3}}{2 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {-\frac {3 x}{8 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {3 \left (\frac {x}{5 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {4 x}{15 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {8 x}{15 \sqrt {a^{2} x^{2}+1}}\right )}{8 a^{2}}}{a^{2}}\right )-6 a^{2} \left (-\frac {x}{4 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {\frac {x}{5 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {4 x}{15 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {8 x}{15 \sqrt {a^{2} x^{2}+1}}}{4 a^{2}}\right )-4 i a^{3} \left (-\frac {x^{2}}{3 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}-\frac {2}{15 a^{4} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}\right )-\frac {4 i}{5 a \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}\)	\(269\)

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/15*(I-a*x)*(I+a*x)*(a*x+4*I)*(1+I*a*x)^4/a/(a^2*x^2+1)^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {a^{3} x^{3} + 3 i \, a^{2} x^{2} - 3 \, a x + {\left (a^{2} x^{2} + 3 i \, a x + 4\right )} \sqrt {a^{2} x^{2} + 1} - i}{15 \, {\left (a^{4} x^{3} + 3 i \, a^{3} x^{2} - 3 \, a^{2} x - i \, a\right )}} \]

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^(7/2),x, algorithm="fricas")

[Out]

-1/15*(a^3*x^3 + 3*I*a^2*x^2 - 3*a*x + (a^2*x^2 + 3*I*a*x + 4)*sqrt(a^2*x^2 + 1) - I)/(a^4*x^3 + 3*I*a^3*x^2 -
3*a^2*x - I*a)

Sympy [F]

\[ \int \frac {e^{4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (a x - i\right )^{4}}{\left (a^{2} x^{2} + 1\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate((1+I*a*x)**4/(a**2*x**2+1)**(7/2),x)

[Out]

Integral((a*x - I)**4/(a**2*x**2 + 1)**(7/2), 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. 95 vs. \(2 (43) = 86\).

Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.42 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {a^{2} x^{3}}{2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} - \frac {x}{15 \, \sqrt {a^{2} x^{2} + 1}} + \frac {4 i \, a x^{2}}{3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} - \frac {x}{30 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} + \frac {11 \, x}{10 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} - \frac {4 i}{15 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}} a} \]

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^(7/2),x, algorithm="maxima")

[Out]

-1/2*a^2*x^3/(a^2*x^2 + 1)^(5/2) - 1/15*x/sqrt(a^2*x^2 + 1) + 4/3*I*a*x^2/(a^2*x^2 + 1)^(5/2) - 1/30*x/(a^2*x^
2 + 1)^(3/2) + 11/10*x/(a^2*x^2 + 1)^(5/2) - 4/15*I/((a^2*x^2 + 1)^(5/2)*a)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (43) = 86\).

Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.66 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (4 \, a^{4} - 25 \, a^{2} {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{2} + 15 i \, a {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{3} + 15 \, {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{4} - 5 \, a^{3} {\left (i \, \sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {i}{x}\right )}\right )}}{15 \, {\left (i \, a + \sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{5}} \]

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^(7/2),x, algorithm="giac")

[Out]

2/15*(4*a^4 - 25*a^2*(sqrt(a^2 + 1/x^2) - 1/x)^2 + 15*I*a*(sqrt(a^2 + 1/x^2) - 1/x)^3 + 15*(sqrt(a^2 + 1/x^2)
- 1/x)^4 - 5*a^3*(I*sqrt(a^2 + 1/x^2) - I/x))/(I*a + sqrt(a^2 + 1/x^2) - 1/x)^5

Mupad [B] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.61 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {a^2\,x^2+1}\,\left (a^2\,x^2\,1{}\mathrm {i}-3\,a\,x+4{}\mathrm {i}\right )}{15\,a\,{\left (-1+a\,x\,1{}\mathrm {i}\right )}^3} \]

[In]

int((a*x*1i + 1)^4/(a^2*x^2 + 1)^(7/2),x)

[Out]

((a^2*x^2 + 1)^(1/2)*(a^2*x^2*1i - 3*a*x + 4i))/(15*a*(a*x*1i - 1)^3)