\(\int \frac {e^{3 i \arctan (a x)} x^2}{(c+a^2 c x^2)^{11/2}} \, dx\) [395]

Optimal result

Integrand size = 28, antiderivative size = 65 \[ \int \frac {e^{3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=-\frac {(i+3 a x) \sqrt {1+a^2 x^2}}{24 a^3 c^5 (1-i a x)^6 (1+i a x)^3 \sqrt {c+a^2 c x^2}} \] Output:

-1/24*(I+3*a*x)*(a^2*x^2+1)^(1/2)/a^3/c^5/(1-I*a*x)^6/(1+I*a*x)^3/(a^2*c*x 
^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\frac {i (i+3 a x) \sqrt {1+a^2 x^2}}{24 a^3 c^5 (-i+a x)^3 (i+a x)^6 \sqrt {c+a^2 c x^2}} \] Input:

Integrate[(E^((3*I)*ArcTan[a*x])*x^2)/(c + a^2*c*x^2)^(11/2),x]
 

Output:

((I/24)*(I + 3*a*x)*Sqrt[1 + a^2*x^2])/(a^3*c^5*(-I + a*x)^3*(I + a*x)^6*S 
qrt[c + a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {5608, 5605, 91}

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.

\(\displaystyle \int \frac {x^2 e^{3 i \arctan (a x)}}{\left (a^2 c x^2+c\right )^{11/2}} \, dx\)

\(\Big \downarrow \) 5608

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {e^{3 i \arctan (a x)} x^2}{\left (a^2 x^2+1\right )^{11/2}}dx}{c^5 \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 5605

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {x^2}{(1-i a x)^7 (i a x+1)^4}dx}{c^5 \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 91

\(\displaystyle -\frac {(3 a x+i) \sqrt {a^2 x^2+1}}{24 a^3 c^5 (1-i a x)^6 (1+i a x)^3 \sqrt {a^2 c x^2+c}}\)

Input:

Int[(E^((3*I)*ArcTan[a*x])*x^2)/(c + a^2*c*x^2)^(11/2),x]
 

Output:

-1/24*((I + 3*a*x)*Sqrt[1 + a^2*x^2])/(a^3*c^5*(1 - I*a*x)^6*(1 + I*a*x)^3 
*Sqrt[c + a^2*c*x^2])
 

Defintions of rubi rules used

rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(c + d*x)^(n + 1)*(e + f*x)^(p + 1)*((2*a*d*f*(n + p + 3 
) - b*(d*e*(n + 2) + c*f*(p + 2)) + b*d*f*(n + p + 2)*x)/(d^2*f^2*(n + p + 
2)*(n + p + 3))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2 
, 0] && NeQ[n + p + 3, 0] && EqQ[d*f*(n + p + 2)*(a^2*d*f*(n + p + 3) - b*( 
b*c*e + a*(d*e*(n + 1) + c*f*(p + 1)))) - b*(d*e*(n + 1) + c*f*(p + 1))*(a* 
d*f*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2))), 0]
 

rule 5605

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[c^p   Int[x^m*(1 - I*a*x)^(p + I*(n/2))*(1 + I*a*x)^(p - I* 
(n/2)), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[d, a^2*c] && (Integer 
Q[p] || GtQ[c, 0])
 

rule 5608

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 + a^2*x^2)^FracPart 
[p])   Int[x^m*(1 + a^2*x^2)^p*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, 
 m, n, p}, x] && EqQ[d, a^2*c] &&  !(IntegerQ[p] || GtQ[c, 0])
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (3 i a x -1\right )}{24 \sqrt {a^{2} x^{2}+1}\, c^{6} a^{3} \left (a x +i\right )^{6} \left (-a x +i\right )^{3}}\)	\(57\)
gosper	\(\frac {\left (-a x +i\right ) \left (a x +i\right ) \left (3 a x +i\right ) \left (i a x +1\right )^{3}}{24 a^{3} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}} \left (a^{2} c \,x^{2}+c \right )^{\frac {11}{2}}}\)	\(58\)
risch	\(\frac {\sqrt {a^{2} x^{2}+1}\, \left (\frac {i x}{8 a^{2}}-\frac {1}{24 a^{3}}\right )}{c^{5} \sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (a x +i\right )^{6} \left (a x -i\right )^{3}}\)	\(58\)
orering	\(-\frac {x^{3} \left (x^{6} a^{6}+3 i x^{5} a^{5}+8 i x^{3} a^{3}-6 a^{2} x^{2}+6 i a x -8\right ) \left (i a x +1\right )^{3}}{24 \sqrt {a^{2} x^{2}+1}\, \left (a^{2} c \,x^{2}+c \right )^{\frac {11}{2}}}\)	\(78\)

Input:

int((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^2/(a^2*c*x^2+c)^(11/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-1/24/(a^2*x^2+1)^(1/2)*(c*(a^2*x^2+1))^(1/2)*(3*I*a*x-1)/c^6/a^3/(I+a*x)^ 
6/(I-a*x)^3
 

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. 192 vs. \(2 (53) = 106\).

Time = 0.13 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.95 \[ \int \frac {e^{3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\frac {{\left (i \, a^{6} x^{9} - 3 \, a^{5} x^{8} - 8 \, a^{3} x^{6} - 6 i \, a^{2} x^{5} - 6 \, a x^{4} - 8 i \, x^{3}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1}}{24 \, {\left (a^{11} c^{6} x^{11} + 3 i \, a^{10} c^{6} x^{10} + a^{9} c^{6} x^{9} + 11 i \, a^{8} c^{6} x^{8} - 6 \, a^{7} c^{6} x^{7} + 14 i \, a^{6} c^{6} x^{6} - 14 \, a^{5} c^{6} x^{5} + 6 i \, a^{4} c^{6} x^{4} - 11 \, a^{3} c^{6} x^{3} - i \, a^{2} c^{6} x^{2} - 3 \, a c^{6} x - i \, c^{6}\right )}} \] Input:

integrate((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^2/(a^2*c*x^2+c)^(11/2),x, algori 
thm="fricas")
 

Output:

1/24*(I*a^6*x^9 - 3*a^5*x^8 - 8*a^3*x^6 - 6*I*a^2*x^5 - 6*a*x^4 - 8*I*x^3) 
*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)/(a^11*c^6*x^11 + 3*I*a^10*c^6*x^10 
+ a^9*c^6*x^9 + 11*I*a^8*c^6*x^8 - 6*a^7*c^6*x^7 + 14*I*a^6*c^6*x^6 - 14*a 
^5*c^6*x^5 + 6*I*a^4*c^6*x^4 - 11*a^3*c^6*x^3 - I*a^2*c^6*x^2 - 3*a*c^6*x 
- I*c^6)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {e^{3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx =\text {Too large to display} \] Input:

integrate((1+I*a*x)**3/(a**2*x**2+1)**(3/2)*x**2/(a**2*c*x**2+c)**(11/2),x 
)
 

Output:

-I*(Integral(I*x**2/(a**12*c**5*x**12*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 
 + c) + 6*a**10*c**5*x**10*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 15* 
a**8*c**5*x**8*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 20*a**6*c**5*x* 
*6*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 15*a**4*c**5*x**4*sqrt(a**2 
*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 6*a**2*c**5*x**2*sqrt(a**2*x**2 + 1)*sq 
rt(a**2*c*x**2 + c) + c**5*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c)), x) 
+ Integral(-3*a*x**3/(a**12*c**5*x**12*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x** 
2 + c) + 6*a**10*c**5*x**10*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 15 
*a**8*c**5*x**8*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 20*a**6*c**5*x 
**6*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 15*a**4*c**5*x**4*sqrt(a** 
2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 6*a**2*c**5*x**2*sqrt(a**2*x**2 + 1)*s 
qrt(a**2*c*x**2 + c) + c**5*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c)), x) 
 + Integral(a**3*x**5/(a**12*c**5*x**12*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x* 
*2 + c) + 6*a**10*c**5*x**10*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 1 
5*a**8*c**5*x**8*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 20*a**6*c**5* 
x**6*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 15*a**4*c**5*x**4*sqrt(a* 
*2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + 6*a**2*c**5*x**2*sqrt(a**2*x**2 + 1)* 
sqrt(a**2*c*x**2 + c) + c**5*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c)), x 
) + Integral(-3*I*a**2*x**4/(a**12*c**5*x**12*sqrt(a**2*x**2 + 1)*sqrt(a** 
2*c*x**2 + c) + 6*a**10*c**5*x**10*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2...
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^2/(a^2*c*x^2+c)^(11/2),x, algori 
thm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

\[ \int \frac {e^{3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3} x^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {11}{2}} {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^2/(a^2*c*x^2+c)^(11/2),x, algori 
thm="giac")
 

Output:

integrate((I*a*x + 1)^3*x^2/((a^2*c*x^2 + c)^(11/2)*(a^2*x^2 + 1)^(3/2)), 
x)
 

Mupad [B] (verification not implemented)

Time = 24.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {e^{3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\frac {\sqrt {c\,\left (a^2\,x^2+1\right )}\,{\left (a\,x-\mathrm {i}\right )}^3\,\left (3\,a\,x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{24\,a^3\,c^6\,{\left (a^2\,x^2+1\right )}^{13/2}} \] Input:

int((x^2*(a*x*1i + 1)^3)/((c + a^2*c*x^2)^(11/2)*(a^2*x^2 + 1)^(3/2)),x)
 

Output:

((c*(a^2*x^2 + 1))^(1/2)*(a*x - 1i)^3*(3*a*x + 1i)*1i)/(24*a^3*c^6*(a^2*x^ 
2 + 1)^(13/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.40 \[ \int \frac {e^{3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\frac {\sqrt {c}\, \left (3 a^{4} i \,x^{4}+8 a^{3} x^{3}-6 a^{2} i \,x^{2}-i \right )}{24 a^{3} c^{6} \left (a^{12} x^{12}+6 a^{10} x^{10}+15 a^{8} x^{8}+20 a^{6} x^{6}+15 a^{4} x^{4}+6 a^{2} x^{2}+1\right )} \] Input:

int((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^2/(a^2*c*x^2+c)^(11/2),x)
 

Output:

(sqrt(c)*(3*a**4*i*x**4 + 8*a**3*x**3 - 6*a**2*i*x**2 - i))/(24*a**3*c**6* 
(a**12*x**12 + 6*a**10*x**10 + 15*a**8*x**8 + 20*a**6*x**6 + 15*a**4*x**4 
+ 6*a**2*x**2 + 1))