Integrand size = 25, antiderivative size = 96 \[ \int \frac {e^{-4 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {2 i c (1-i a x)^3}{3 a \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 i (1-i a x)}{a \sqrt {c+a^2 c x^2}}+\frac {\text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a \sqrt {c}} \] Output:
2/3*I*c*(1-I*a*x)^3/a/(a^2*c*x^2+c)^(3/2)-2*I*(1-I*a*x)/a/(a^2*c*x^2+c)^(1 /2)+arctanh(a*c^(1/2)*x/(a^2*c*x^2+c)^(1/2))/a/c^(1/2)
Time = 0.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-4 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {2 \sqrt {1+a^2 x^2} \left (2 i \sqrt {1+i a x} \left (1+i a x+2 a^2 x^2\right )+3 i \sqrt {1-i a x} (-i+a x)^2 \arcsin \left (\frac {\sqrt {1-i a x}}{\sqrt {2}}\right )\right )}{3 a \sqrt {1-i a x} (-i+a x)^2 \sqrt {c+a^2 c x^2}} \] Input:
Integrate[1/(E^((4*I)*ArcTan[a*x])*Sqrt[c + a^2*c*x^2]),x]
Output:
(2*Sqrt[1 + a^2*x^2]*((2*I)*Sqrt[1 + I*a*x]*(1 + I*a*x + 2*a^2*x^2) + (3*I )*Sqrt[1 - I*a*x]*(-I + a*x)^2*ArcSin[Sqrt[1 - I*a*x]/Sqrt[2]]))/(3*a*Sqrt [1 - I*a*x]*(-I + a*x)^2*Sqrt[c + a^2*c*x^2])
Time = 0.51 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5597, 468, 457, 224, 219}
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.
\(\displaystyle \int \frac {e^{-4 i \arctan (a x)}}{\sqrt {a^2 c x^2+c}} \, dx\) |
\(\Big \downarrow \) 5597 |
\(\displaystyle c^2 \int \frac {(1-i a x)^4}{\left (a^2 c x^2+c\right )^{5/2}}dx\) |
\(\Big \downarrow \) 468 |
\(\displaystyle c^2 \left (\frac {2 i (1-i a x)^3}{3 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\int \frac {(1-i a x)^2}{\left (a^2 c x^2+c\right )^{3/2}}dx}{c}\right )\) |
\(\Big \downarrow \) 457 |
\(\displaystyle c^2 \left (\frac {2 i (1-i a x)^3}{3 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac {-\frac {\int \frac {1}{\sqrt {a^2 c x^2+c}}dx}{c}+\frac {2 i (1-i a x)}{a c \sqrt {a^2 c x^2+c}}}{c}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle c^2 \left (\frac {2 i (1-i a x)^3}{3 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac {-\frac {\int \frac {1}{1-\frac {a^2 c x^2}{a^2 c x^2+c}}d\frac {x}{\sqrt {a^2 c x^2+c}}}{c}+\frac {2 i (1-i a x)}{a c \sqrt {a^2 c x^2+c}}}{c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle c^2 \left (\frac {2 i (1-i a x)^3}{3 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac {-\frac {\text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a c^{3/2}}+\frac {2 i (1-i a x)}{a c \sqrt {a^2 c x^2+c}}}{c}\right )\) |
Input:
Int[1/(E^((4*I)*ArcTan[a*x])*Sqrt[c + a^2*c*x^2]),x]
Output:
c^2*((((2*I)/3)*(1 - I*a*x)^3)/(a*c*(c + a^2*c*x^2)^(3/2)) - (((2*I)*(1 - I*a*x))/(a*c*Sqrt[c + a^2*c*x^2]) - ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x ^2]]/(a*c^(3/2)))/c)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_) + (d_.)*(x_))^2*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*( c + d*x)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] - Simp[d^2*((p + 2)/(b*(p + 1))) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[ b*c^2 + a*d^2, 0] && LtQ[p, -1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] - Simp[d^2*((n + p)/(b*(p + 1))) Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1), x], x] /; Free Q[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[p, -1] && GtQ[n, 1] && I ntegerQ[2*p]
Int[E^(ArcTan[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Si mp[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]) && IGtQ[I*(n/2), 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (80 ) = 160\).
Time = 0.24 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.96
method | result | size |
default | \(\frac {\ln \left (\frac {a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}+c}\right )}{\sqrt {a^{2} c}}-\frac {4 \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}{a^{2} c \left (x -\frac {i}{a}\right )}-\frac {4 \left (\frac {i \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}{3 a c \left (x -\frac {i}{a}\right )^{2}}-\frac {\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}{3 c \left (x -\frac {i}{a}\right )}\right )}{a^{2}}\) | \(188\) |
Input:
int(1/(1+I*a*x)^4*(a^2*x^2+1)^2/(a^2*c*x^2+c)^(1/2),x,method=_RETURNVERBOS E)
Output:
ln(a^2*c*x/(a^2*c)^(1/2)+(a^2*c*x^2+c)^(1/2))/(a^2*c)^(1/2)-4/a^2/c/(x-I/a )*((x-I/a)^2*a^2*c+2*I*a*c*(x-I/a))^(1/2)-4/a^2*(1/3*I/a/c/(x-I/a)^2*((x-I /a)^2*a^2*c+2*I*a*c*(x-I/a))^(1/2)-1/3/c/(x-I/a)*((x-I/a)^2*a^2*c+2*I*a*c* (x-I/a))^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (75) = 150\).
Time = 0.09 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.94 \[ \int \frac {e^{-4 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {3 \, {\left (a^{3} c x^{2} - 2 i \, a^{2} c x - a c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {2 \, {\left (a^{2} c x + \sqrt {a^{2} c x^{2} + c} a^{2} c \sqrt {\frac {1}{a^{2} c}}\right )}}{x}\right ) - 3 \, {\left (a^{3} c x^{2} - 2 i \, a^{2} c x - a c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {2 \, {\left (a^{2} c x - \sqrt {a^{2} c x^{2} + c} a^{2} c \sqrt {\frac {1}{a^{2} c}}\right )}}{x}\right ) - 8 \, \sqrt {a^{2} c x^{2} + c} {\left (2 \, a x - i\right )}}{6 \, {\left (a^{3} c x^{2} - 2 i \, a^{2} c x - a c\right )}} \] Input:
integrate(1/(1+I*a*x)^4*(a^2*x^2+1)^2/(a^2*c*x^2+c)^(1/2),x, algorithm="fr icas")
Output:
1/6*(3*(a^3*c*x^2 - 2*I*a^2*c*x - a*c)*sqrt(1/(a^2*c))*log(2*(a^2*c*x + sq rt(a^2*c*x^2 + c)*a^2*c*sqrt(1/(a^2*c)))/x) - 3*(a^3*c*x^2 - 2*I*a^2*c*x - a*c)*sqrt(1/(a^2*c))*log(2*(a^2*c*x - sqrt(a^2*c*x^2 + c)*a^2*c*sqrt(1/(a ^2*c)))/x) - 8*sqrt(a^2*c*x^2 + c)*(2*a*x - I))/(a^3*c*x^2 - 2*I*a^2*c*x - a*c)
\[ \int \frac {e^{-4 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {\left (a^{2} x^{2} + 1\right )^{2}}{\sqrt {c \left (a^{2} x^{2} + 1\right )} \left (a x - i\right )^{4}}\, dx \] Input:
integrate(1/(1+I*a*x)**4*(a**2*x**2+1)**2/(a**2*c*x**2+c)**(1/2),x)
Output:
Integral((a**2*x**2 + 1)**2/(sqrt(c*(a**2*x**2 + 1))*(a*x - I)**4), x)
Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-4 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {4 i \, \sqrt {a^{2} c x^{2} + c}}{3 \, {\left (a^{3} c x^{2} - 2 i \, a^{2} c x - a c\right )}} - \frac {8 i \, \sqrt {a^{2} c x^{2} + c}}{3 i \, a^{2} c x + 3 \, a c} + \frac {\operatorname {arsinh}\left (a x\right )}{a \sqrt {c}} \] Input:
integrate(1/(1+I*a*x)^4*(a^2*x^2+1)^2/(a^2*c*x^2+c)^(1/2),x, algorithm="ma xima")
Output:
-4/3*I*sqrt(a^2*c*x^2 + c)/(a^3*c*x^2 - 2*I*a^2*c*x - a*c) - 8*I*sqrt(a^2* c*x^2 + c)/(3*I*a^2*c*x + 3*a*c) + arcsinh(a*x)/(a*sqrt(c))
Time = 0.19 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-4 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {\log \left ({\left | -\sqrt {a^{2} c} x + \sqrt {a^{2} c x^{2} + c} \right |}\right )}{a \sqrt {c}} - \frac {8 \, {\left (3 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )}^{2} - 3 i \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )} \sqrt {c} - 2 \, c\right )}}{3 \, {\left (-i \, \sqrt {a^{2} c} x + i \, \sqrt {a^{2} c x^{2} + c} - \sqrt {c}\right )}^{3} a} \] Input:
integrate(1/(1+I*a*x)^4*(a^2*x^2+1)^2/(a^2*c*x^2+c)^(1/2),x, algorithm="gi ac")
Output:
-log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 + c)))/(a*sqrt(c)) - 8/3*(3*(sqrt (a^2*c)*x - sqrt(a^2*c*x^2 + c))^2 - 3*I*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))*sqrt(c) - 2*c)/((-I*sqrt(a^2*c)*x + I*sqrt(a^2*c*x^2 + c) - sqrt(c))^ 3*a)
Timed out. \[ \int \frac {e^{-4 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {{\left (a^2\,x^2+1\right )}^2}{\sqrt {c\,a^2\,x^2+c}\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^4} \,d x \] Input:
int((a^2*x^2 + 1)^2/((c + a^2*c*x^2)^(1/2)*(a*x*1i + 1)^4),x)
Output:
int((a^2*x^2 + 1)^2/((c + a^2*c*x^2)^(1/2)*(a*x*1i + 1)^4), x)
\[ \int \frac {e^{-4 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {\sqrt {c}\, \left (-16 \left (\int -\frac {\sqrt {a^{2} x^{2}+1}\, x^{2}}{a^{6} x^{6}-4 a^{5} i \,x^{5}-5 a^{4} x^{4}-5 a^{2} x^{2}+4 a i x +1}d x \right ) a^{3}+8 \left (\int \frac {\sqrt {a^{2} x^{2}+1}\, x^{3}}{a^{6} x^{6}-4 a^{5} i \,x^{5}-5 a^{4} x^{4}-5 a^{2} x^{2}+4 a i x +1}d x \right ) a^{4} i -8 \left (\int \frac {\sqrt {a^{2} x^{2}+1}\, x}{a^{6} x^{6}-4 a^{5} i \,x^{5}-5 a^{4} x^{4}-5 a^{2} x^{2}+4 a i x +1}d x \right ) a^{2} i -\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}-a x \right )+\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x \right )\right )}{2 a c} \] Input:
int(1/(1+I*a*x)^4*(a^2*x^2+1)^2/(a^2*c*x^2+c)^(1/2),x)
Output:
(sqrt(c)*( - 16*int(( - sqrt(a**2*x**2 + 1)*x**2)/(a**6*x**6 - 4*a**5*i*x* *5 - 5*a**4*x**4 - 5*a**2*x**2 + 4*a*i*x + 1),x)*a**3 + 8*int((sqrt(a**2*x **2 + 1)*x**3)/(a**6*x**6 - 4*a**5*i*x**5 - 5*a**4*x**4 - 5*a**2*x**2 + 4* a*i*x + 1),x)*a**4*i - 8*int((sqrt(a**2*x**2 + 1)*x)/(a**6*x**6 - 4*a**5*i *x**5 - 5*a**4*x**4 - 5*a**2*x**2 + 4*a*i*x + 1),x)*a**2*i - log(sqrt(a**2 *x**2 + 1) - a*x) + log(sqrt(a**2*x**2 + 1) + a*x)))/(2*a*c)