Integrand size = 25, antiderivative size = 69 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {i c (1+i a x)^4}{3 a \left (c+a^2 c x^2\right )^{5/2}}+\frac {i c (1+i a x)^5}{15 a \left (c+a^2 c x^2\right )^{5/2}} \] Output:
-1/3*I*c*(1+I*a*x)^4/a/(a^2*c*x^2+c)^(5/2)+1/15*I*c*(1+I*a*x)^5/a/(a^2*c*x ^2+c)^(5/2)
Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.12 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {(1+i a x)^{3/2} (4 i+a x) \sqrt {1+a^2 x^2}}{15 a c \sqrt {1-i a x} (i+a x)^2 \sqrt {c+a^2 c x^2}} \] Input:
Integrate[E^((4*I)*ArcTan[a*x])/(c + a^2*c*x^2)^(3/2),x]
Output:
((1 + I*a*x)^(3/2)*(4*I + a*x)*Sqrt[1 + a^2*x^2])/(15*a*c*Sqrt[1 - I*a*x]* (I + a*x)^2*Sqrt[c + a^2*c*x^2])
Time = 0.44 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5598, 461, 460}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{4 i \arctan (a x)}}{\left (a^2 c x^2+c\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 5598 |
\(\displaystyle c^2 \int \frac {(i a x+1)^4}{\left (a^2 c x^2+c\right )^{7/2}}dx\) |
\(\Big \downarrow \) 461 |
\(\displaystyle c^2 \left (-\frac {1}{3} \int \frac {(i a x+1)^5}{\left (a^2 c x^2+c\right )^{7/2}}dx-\frac {i (1+i a x)^4}{3 a c \left (a^2 c x^2+c\right )^{5/2}}\right )\) |
\(\Big \downarrow \) 460 |
\(\displaystyle c^2 \left (\frac {i (1+i a x)^5}{15 a c \left (a^2 c x^2+c\right )^{5/2}}-\frac {i (1+i a x)^4}{3 a c \left (a^2 c x^2+c\right )^{5/2}}\right )\) |
Input:
Int[E^((4*I)*ArcTan[a*x])/(c + a^2*c*x^2)^(3/2),x]
Output:
c^2*(((-1/3*I)*(1 + I*a*x)^4)/(a*c*(c + a^2*c*x^2)^(5/2)) + ((I/15)*(1 + I *a*x)^5)/(a*c*(c + a^2*c*x^2)^(5/2)))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
Int[E^(ArcTan[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Si mp[1/c^(I*(n/2)) Int[(c + d*x^2)^(p + I*(n/2))/(1 + I*a*x)^(I*n), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[d, a^2*c] && !(IntegerQ[p] || GtQ[c, 0]) && ILtQ[I*(n/2), 0]
Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.64
method | result | size |
orering | \(-\frac {\left (a x +4 i\right ) \left (i a x +1\right )^{4}}{15 a \left (a^{2} x^{2}+1\right ) \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\) | \(44\) |
gosper | \(\frac {\left (-a x +i\right ) \left (a x +i\right ) \left (a x +4 i\right ) \left (i a x +1\right )^{4}}{15 a \left (a^{2} x^{2}+1\right )^{2} \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\) | \(57\) |
trager | \(\frac {\left (-a^{5} x^{5}-10 a^{3} x^{3}+20 i x^{2} a^{2}+15 a x -4 i\right ) \sqrt {a^{2} c \,x^{2}+c}}{15 c^{2} \left (a^{2} x^{2}+1\right )^{3} a}\) | \(64\) |
default | \(\frac {x}{c \sqrt {a^{2} c \,x^{2}+c}}+\frac {2 \left (i \sqrt {-a^{2}}-a \right ) \left (\frac {1}{5 c \sqrt {-a^{2}}\, {\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} \sqrt {{\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}+\frac {3 a^{2} \left (\frac {1}{3 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right ) \sqrt {{\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}+\frac {2 \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right ) a^{2} c -2 c \sqrt {-a^{2}}}{3 c^{2} \sqrt {-a^{2}}\, \sqrt {{\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}\right )}{5 \sqrt {-a^{2}}}\right )}{a^{3}}-\frac {2 \left (i \sqrt {-a^{2}}+a \right ) \left (-\frac {1}{5 c \sqrt {-a^{2}}\, {\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} \sqrt {{\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}-\frac {3 a^{2} \left (-\frac {1}{3 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right ) \sqrt {{\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}-\frac {2 \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right ) a^{2} c +2 c \sqrt {-a^{2}}}{3 c^{2} \sqrt {-a^{2}}\, \sqrt {{\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}\right )}{5 \sqrt {-a^{2}}}\right )}{a^{3}}-\frac {2 \left (i \sqrt {-a^{2}}-a \right ) \left (\frac {1}{3 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right ) \sqrt {{\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}+\frac {2 \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right ) a^{2} c -2 c \sqrt {-a^{2}}}{3 c^{2} \sqrt {-a^{2}}\, \sqrt {{\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}\right )}{a \sqrt {-a^{2}}}-\frac {2 \left (i \sqrt {-a^{2}}+a \right ) \left (-\frac {1}{3 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right ) \sqrt {{\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}-\frac {2 \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right ) a^{2} c +2 c \sqrt {-a^{2}}}{3 c^{2} \sqrt {-a^{2}}\, \sqrt {{\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}\right )}{a \sqrt {-a^{2}}}\) | \(940\) |
Input:
int((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/15*(a*x+4*I)/a/(a^2*x^2+1)*(1+I*a*x)^4/(a^2*c*x^2+c)^(3/2)
Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 3 i \, a x + 4\right )}}{15 \, {\left (a^{4} c^{2} x^{3} + 3 i \, a^{3} c^{2} x^{2} - 3 \, a^{2} c^{2} x - i \, a c^{2}\right )}} \] Input:
integrate((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x, algorithm="fric as")
Output:
-1/15*sqrt(a^2*c*x^2 + c)*(a^2*x^2 + 3*I*a*x + 4)/(a^4*c^2*x^3 + 3*I*a^3*c ^2*x^2 - 3*a^2*c^2*x - I*a*c^2)
\[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {\left (a x - i\right )^{4}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a^{2} x^{2} + 1\right )^{2}}\, dx \] Input:
integrate((1+I*a*x)**4/(a**2*x**2+1)**2/(a**2*c*x**2+c)**(3/2),x)
Output:
Integral((a*x - I)**4/((c*(a**2*x**2 + 1))**(3/2)*(a**2*x**2 + 1)**2), x)
\[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{4}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (a^{2} x^{2} + 1\right )}^{2}} \,d x } \] Input:
integrate((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x, algorithm="maxi ma")
Output:
integrate((I*a*x + 1)^4/((a^2*c*x^2 + c)^(3/2)*(a^2*x^2 + 1)^2), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (53) = 106\).
Time = 0.14 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.94 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (15 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )}^{3} \sqrt {c} - 5 i \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )}^{2} c - 5 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )} c^{\frac {3}{2}} - i \, c^{2}\right )}}{15 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c} + i \, \sqrt {c}\right )}^{5} a c} \] Input:
integrate((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x, algorithm="giac ")
Output:
2/15*(15*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))^3*sqrt(c) - 5*I*(sqrt(a^2*c )*x - sqrt(a^2*c*x^2 + c))^2*c - 5*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))*c ^(3/2) - I*c^2)/((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c) + I*sqrt(c))^5*a*c)
Time = 23.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.67 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c\,\left (a^2\,x^2+1\right )}\,\left (a^2\,x^2\,1{}\mathrm {i}-3\,a\,x+4{}\mathrm {i}\right )}{15\,a\,c^2\,{\left (-1+a\,x\,1{}\mathrm {i}\right )}^3} \] Input:
int((a*x*1i + 1)^4/((c + a^2*c*x^2)^(3/2)*(a^2*x^2 + 1)^2),x)
Output:
((c*(a^2*x^2 + 1))^(1/2)*(a^2*x^2*1i - 3*a*x + 4i))/(15*a*c^2*(a*x*1i - 1) ^3)
Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.10 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (-\sqrt {a^{2} x^{2}+1}\, a^{5} x^{5}-10 \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+20 \sqrt {a^{2} x^{2}+1}\, a^{2} i \,x^{2}+15 \sqrt {a^{2} x^{2}+1}\, a x -4 \sqrt {a^{2} x^{2}+1}\, i +7 a^{6} x^{6}+21 a^{4} x^{4}+21 a^{2} x^{2}+7\right )}{15 a \,c^{2} \left (a^{6} x^{6}+3 a^{4} x^{4}+3 a^{2} x^{2}+1\right )} \] Input:
int((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x)
Output:
(sqrt(c)*( - sqrt(a**2*x**2 + 1)*a**5*x**5 - 10*sqrt(a**2*x**2 + 1)*a**3*x **3 + 20*sqrt(a**2*x**2 + 1)*a**2*i*x**2 + 15*sqrt(a**2*x**2 + 1)*a*x - 4* sqrt(a**2*x**2 + 1)*i + 7*a**6*x**6 + 21*a**4*x**4 + 21*a**2*x**2 + 7))/(1 5*a*c**2*(a**6*x**6 + 3*a**4*x**4 + 3*a**2*x**2 + 1))