3.4.10 \(\int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx\) [310]

3.4.10.1 Optimal result
3.4.10.2 Mathematica [A] (verified)
3.4.10.3 Rubi [A] (verified)
3.4.10.4 Maple [A] (verified)
3.4.10.5 Fricas [B] (verification not implemented)
3.4.10.6 Sympy [F]
3.4.10.7 Maxima [F]
3.4.10.8 Giac [F]
3.4.10.9 Mupad [F(-1)]

3.4.10.1 Optimal result

Integrand size = 25, antiderivative size = 131 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {2 i \sqrt {1+a^2 x^2}}{a (1-i a x)^2 \sqrt {c+a^2 c x^2}}+\frac {4 i \sqrt {1+a^2 x^2}}{a (1-i a x) \sqrt {c+a^2 c x^2}}+\frac {i \sqrt {1+a^2 x^2} \log (i+a x)}{a \sqrt {c+a^2 c x^2}} \]

output
-2*I*(a^2*x^2+1)^(1/2)/a/(1-I*a*x)^2/(a^2*c*x^2+c)^(1/2)+4*I*(a^2*x^2+1)^( 
1/2)/a/(1-I*a*x)/(a^2*c*x^2+c)^(1/2)+I*ln(I+a*x)*(a^2*x^2+1)^(1/2)/a/(a^2* 
c*x^2+c)^(1/2)
 
3.4.10.2 Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.53 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {i \sqrt {1+a^2 x^2} \left (-2+4 i a x+(i+a x)^2 \log (i+a x)\right )}{a (i+a x)^2 \sqrt {c+a^2 c x^2}} \]

input
Integrate[E^((5*I)*ArcTan[a*x])/Sqrt[c + a^2*c*x^2],x]
 
output
(I*Sqrt[1 + a^2*x^2]*(-2 + (4*I)*a*x + (I + a*x)^2*Log[I + a*x]))/(a*(I + 
a*x)^2*Sqrt[c + a^2*c*x^2])
 
3.4.10.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.60, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {5599, 5596, 49, 2009}

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.

\(\displaystyle \int \frac {e^{5 i \arctan (a x)}}{\sqrt {a^2 c x^2+c}} \, dx\)

\(\Big \downarrow \) 5599

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {e^{5 i \arctan (a x)}}{\sqrt {a^2 x^2+1}}dx}{\sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 5596

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {(i a x+1)^2}{(1-i a x)^3}dx}{\sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \left (\frac {1}{1-i a x}-\frac {4}{(1-i a x)^2}+\frac {4}{(1-i a x)^3}\right )dx}{\sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (\frac {4 i}{a (1-i a x)}-\frac {2 i}{a (1-i a x)^2}+\frac {i \log (a x+i)}{a}\right )}{\sqrt {a^2 c x^2+c}}\)

input
Int[E^((5*I)*ArcTan[a*x])/Sqrt[c + a^2*c*x^2],x]
 
output
(Sqrt[1 + a^2*x^2]*((-2*I)/(a*(1 - I*a*x)^2) + (4*I)/(a*(1 - I*a*x)) + (I* 
Log[I + a*x])/a))/Sqrt[c + a^2*c*x^2]
 

3.4.10.3.1 Defintions of rubi rules used

rule 49
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5596
Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> 
Simp[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])
 

rule 5599
Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> S 
imp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 + a^2*x^2)^FracPart[p])   Int[ 
(1 + a^2*x^2)^p*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && E 
qQ[d, a^2*c] &&  !(IntegerQ[p] || GtQ[c, 0])
 
3.4.10.4 Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.63

method result size
risch \(\frac {\sqrt {a^{2} x^{2}+1}\, \left (-4 x -\frac {2 i}{a}\right )}{\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (a x +i\right )^{2}}+\frac {i \sqrt {a^{2} x^{2}+1}\, \ln \left (a x +i\right )}{\sqrt {c \left (a^{2} x^{2}+1\right )}\, a}\) \(82\)
default \(\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (i \ln \left (a x +i\right ) x^{2} a^{2}-2 \ln \left (a x +i\right ) a x -i \ln \left (a x +i\right )-4 a x -2 i\right )}{\sqrt {a^{2} x^{2}+1}\, c a \left (a x +i\right )^{2}}\) \(84\)

input
int((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(1/2),x,method=_RETURNVERB 
OSE)
 
output
(a^2*x^2+1)^(1/2)/(c*(a^2*x^2+1))^(1/2)*(-4*x-2*I/a)/(I+a*x)^2+I*(a^2*x^2+ 
1)^(1/2)/(c*(a^2*x^2+1))^(1/2)/a*ln(I+a*x)
 
3.4.10.5 Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (107) = 214\).

Time = 0.29 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.78 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {-4 i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a x^{2} + {\left (i \, a^{4} c x^{4} - 2 \, a^{3} c x^{3} - 2 \, a c x - i \, c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (i \, a^{6} x^{2} - 2 \, a^{5} x - 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (i \, a^{9} c x^{4} - 2 \, a^{8} c x^{3} + i \, a^{7} c x^{2} - 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2} + a x + i\right )}}\right ) + {\left (-i \, a^{4} c x^{4} + 2 \, a^{3} c x^{3} + 2 \, a c x + i \, c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (i \, a^{6} x^{2} - 2 \, a^{5} x - 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (-i \, a^{9} c x^{4} + 2 \, a^{8} c x^{3} - i \, a^{7} c x^{2} + 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2} + a x + i\right )}}\right )}{2 \, {\left (a^{4} c x^{4} + 2 i \, a^{3} c x^{3} + 2 i \, a c x - c\right )}} \]

input
integrate((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(1/2),x, algorithm=" 
fricas")
 
output
1/2*(-4*I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a*x^2 + (I*a^4*c*x^4 - 2*a 
^3*c*x^3 - 2*a*c*x - I*c)*sqrt(1/(a^2*c))*log(1/8*((I*a^6*x^2 - 2*a^5*x - 
2*I*a^4)*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1) + (I*a^9*c*x^4 - 2*a^8*c*x^ 
3 + I*a^7*c*x^2 - 2*a^6*c*x)*sqrt(1/(a^2*c)))/(a^3*x^3 + I*a^2*x^2 + a*x + 
 I)) + (-I*a^4*c*x^4 + 2*a^3*c*x^3 + 2*a*c*x + I*c)*sqrt(1/(a^2*c))*log(1/ 
8*((I*a^6*x^2 - 2*a^5*x - 2*I*a^4)*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1) + 
 (-I*a^9*c*x^4 + 2*a^8*c*x^3 - I*a^7*c*x^2 + 2*a^6*c*x)*sqrt(1/(a^2*c)))/( 
a^3*x^3 + I*a^2*x^2 + a*x + I)))/(a^4*c*x^4 + 2*I*a^3*c*x^3 + 2*I*a*c*x - 
c)
 
3.4.10.6 Sympy [F]

\[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=i \left (\int \left (- \frac {i}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx + \int \frac {5 a x}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \left (- \frac {10 a^{3} x^{3}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx + \int \frac {a^{5} x^{5}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \frac {10 i a^{2} x^{2}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx + \int \left (- \frac {5 i a^{4} x^{4}}{a^{4} x^{4} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + 2 a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx\right ) \]

input
integrate((1+I*a*x)**5/(a**2*x**2+1)**(5/2)/(a**2*c*x**2+c)**(1/2),x)
 
output
I*(Integral(-I/(a**4*x**4*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 2*a* 
*2*x**2*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + sqrt(a**2*x**2 + 1)*sq 
rt(a**2*c*x**2 + c)), x) + Integral(5*a*x/(a**4*x**4*sqrt(a**2*x**2 + 1)*s 
qrt(a**2*c*x**2 + c) + 2*a**2*x**2*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + 
c) + sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c)), x) + Integral(-10*a**3*x* 
*3/(a**4*x**4*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 2*a**2*x**2*sqrt 
(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x* 
*2 + c)), x) + Integral(a**5*x**5/(a**4*x**4*sqrt(a**2*x**2 + 1)*sqrt(a**2 
*c*x**2 + c) + 2*a**2*x**2*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + sqr 
t(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c)), x) + Integral(10*I*a**2*x**2/(a** 
4*x**4*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 2*a**2*x**2*sqrt(a**2*x 
**2 + 1)*sqrt(a**2*c*x**2 + c) + sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) 
), x) + Integral(-5*I*a**4*x**4/(a**4*x**4*sqrt(a**2*x**2 + 1)*sqrt(a**2*c 
*x**2 + c) + 2*a**2*x**2*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + sqrt( 
a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c)), x))
 
3.4.10.7 Maxima [F]

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

input
integrate((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(1/2),x, algorithm=" 
maxima")
 
output
integrate((I*a*x + 1)^5/(sqrt(a^2*c*x^2 + c)*(a^2*x^2 + 1)^(5/2)), x)
 
3.4.10.8 Giac [F]

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

input
integrate((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(1/2),x, algorithm=" 
giac")
 
output
integrate((I*a*x + 1)^5/(sqrt(a^2*c*x^2 + c)*(a^2*x^2 + 1)^(5/2)), x)
 
3.4.10.9 Mupad [F(-1)]

Timed out. \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {{\left (1+a\,x\,1{}\mathrm {i}\right )}^5}{\sqrt {c\,a^2\,x^2+c}\,{\left (a^2\,x^2+1\right )}^{5/2}} \,d x \]

input
int((a*x*1i + 1)^5/((c + a^2*c*x^2)^(1/2)*(a^2*x^2 + 1)^(5/2)),x)
 
output
int((a*x*1i + 1)^5/((c + a^2*c*x^2)^(1/2)*(a^2*x^2 + 1)^(5/2)), x)