\(\int e^{n \arctan (a x)} x^2 (c+a^2 c x^2)^{3/2} \, dx\) [365]

Optimal result

Integrand size = 26, antiderivative size = 283 \[ \int e^{n \arctan (a x)} x^2 \left (c+a^2 c x^2\right )^{3/2} \, dx=-\frac {c n (1-i a x)^{\frac {1}{2} (5+i n)} (1+i a x)^{\frac {1}{2} (5-i n)} \sqrt {c+a^2 c x^2}}{30 a^3 \sqrt {1+a^2 x^2}}+\frac {c x (1-i a x)^{\frac {1}{2} (5+i n)} (1+i a x)^{\frac {1}{2} (5-i n)} \sqrt {c+a^2 c x^2}}{6 a^2 \sqrt {1+a^2 x^2}}+\frac {2^{\frac {3}{2}-\frac {i n}{2}} c \left (5-n^2\right ) (1-i a x)^{\frac {1}{2} (5+i n)} \sqrt {c+a^2 c x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-3+i n),\frac {1}{2} (5+i n),\frac {1}{2} (7+i n),\frac {1}{2} (1-i a x)\right )}{15 a^3 (5 i-n) \sqrt {1+a^2 x^2}} \] Output:

-1/30*c*n*(1-I*a*x)^(5/2+1/2*I*n)*(1+I*a*x)^(5/2-1/2*I*n)*(a^2*c*x^2+c)^(1 
/2)/a^3/(a^2*x^2+1)^(1/2)+1/6*c*x*(1-I*a*x)^(5/2+1/2*I*n)*(1+I*a*x)^(5/2-1 
/2*I*n)*(a^2*c*x^2+c)^(1/2)/a^2/(a^2*x^2+1)^(1/2)+1/15*2^(3/2-1/2*I*n)*c*( 
-n^2+5)*(1-I*a*x)^(5/2+1/2*I*n)*(a^2*c*x^2+c)^(1/2)*hypergeom([5/2+1/2*I*n 
, -3/2+1/2*I*n],[7/2+1/2*I*n],1/2-1/2*I*a*x)/a^3/(5*I-n)/(a^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.77 \[ \int e^{n \arctan (a x)} x^2 \left (c+a^2 c x^2\right )^{3/2} \, dx=\frac {2^{-1-\frac {i n}{2}} c (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} (i+a x)^2 \sqrt {c+a^2 c x^2} \left (2^{\frac {i n}{2}} (-5 i+n) \sqrt {1+i a x} (-i+a x)^2 (-n+5 a x)-4 \sqrt {2} \left (-5+n^2\right ) (1+i a x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (5+i n),\frac {1}{2} i (3 i+n),\frac {1}{2} (7+i n),\frac {1}{2} (1-i a x)\right )\right )}{15 a^3 (-5 i+n) \sqrt {1+a^2 x^2}} \] Input:

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

Output:

(2^(-1 - (I/2)*n)*c*(1 - I*a*x)^(1/2 + (I/2)*n)*(I + a*x)^2*Sqrt[c + a^2*c 
*x^2]*(2^((I/2)*n)*(-5*I + n)*Sqrt[1 + I*a*x]*(-I + a*x)^2*(-n + 5*a*x) - 
4*Sqrt[2]*(-5 + n^2)*(1 + I*a*x)^((I/2)*n)*Hypergeometric2F1[(5 + I*n)/2, 
(I/2)*(3*I + n), (7 + I*n)/2, (1 - I*a*x)/2]))/(15*a^3*(-5*I + n)*(1 + I*a 
*x)^((I/2)*n)*Sqrt[1 + a^2*x^2])
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5608, 5605, 101, 25, 90, 79}

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.

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

\(\Big \downarrow \) 5608

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

\(\Big \downarrow \) 5605

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

\(\Big \downarrow \) 101

\(\displaystyle \frac {c \sqrt {a^2 c x^2+c} \left (\frac {\int -(1-i a x)^{\frac {1}{2} (i n+3)} (i a x+1)^{\frac {1}{2} (3-i n)} (a n x+1)dx}{6 a^2}+\frac {x (1-i a x)^{\frac {1}{2} (5+i n)} (1+i a x)^{\frac {1}{2} (5-i n)}}{6 a^2}\right )}{\sqrt {a^2 x^2+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {c \sqrt {a^2 c x^2+c} \left (\frac {x (1-i a x)^{\frac {1}{2} (5+i n)} (1+i a x)^{\frac {1}{2} (5-i n)}}{6 a^2}-\frac {\int (1-i a x)^{\frac {1}{2} (i n+3)} (i a x+1)^{\frac {1}{2} (3-i n)} (a n x+1)dx}{6 a^2}\right )}{\sqrt {a^2 x^2+1}}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {c \sqrt {a^2 c x^2+c} \left (\frac {x (1-i a x)^{\frac {1}{2} (5+i n)} (1+i a x)^{\frac {1}{2} (5-i n)}}{6 a^2}-\frac {\frac {1}{5} \left (5-n^2\right ) \int (1-i a x)^{\frac {1}{2} (i n+3)} (i a x+1)^{\frac {1}{2} (3-i n)}dx+\frac {n (1-i a x)^{\frac {1}{2} (5+i n)} (1+i a x)^{\frac {1}{2} (5-i n)}}{5 a}}{6 a^2}\right )}{\sqrt {a^2 x^2+1}}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {c \sqrt {a^2 c x^2+c} \left (\frac {x (1-i a x)^{\frac {1}{2} (5+i n)} (1+i a x)^{\frac {1}{2} (5-i n)}}{6 a^2}-\frac {\frac {n (1-i a x)^{\frac {1}{2} (5+i n)} (1+i a x)^{\frac {1}{2} (5-i n)}}{5 a}-\frac {2^{\frac {5}{2}-\frac {i n}{2}} \left (5-n^2\right ) (1-i a x)^{\frac {1}{2} (5+i n)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (i n-3),\frac {1}{2} (i n+5),\frac {1}{2} (i n+7),\frac {1}{2} (1-i a x)\right )}{5 a (-n+5 i)}}{6 a^2}\right )}{\sqrt {a^2 x^2+1}}\)

Input:

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

Output:

(c*Sqrt[c + a^2*c*x^2]*((x*(1 - I*a*x)^((5 + I*n)/2)*(1 + I*a*x)^((5 - I*n 
)/2))/(6*a^2) - ((n*(1 - I*a*x)^((5 + I*n)/2)*(1 + I*a*x)^((5 - I*n)/2))/( 
5*a) - (2^(5/2 - (I/2)*n)*(5 - n^2)*(1 - I*a*x)^((5 + I*n)/2)*Hypergeometr 
ic2F1[(-3 + I*n)/2, (5 + I*n)/2, (7 + I*n)/2, (1 - I*a*x)/2])/(5*a*(5*I - 
n)))/(6*a^2)))/Sqrt[1 + a^2*x^2]
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 90

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

rule 101

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

\[\int {\mathrm e}^{n \arctan \left (a x \right )} x^{2} \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}d x\]

Input:

int(exp(n*arctan(a*x))*x^2*(a^2*c*x^2+c)^(3/2),x)
 

Output:

int(exp(n*arctan(a*x))*x^2*(a^2*c*x^2+c)^(3/2),x)
 

Fricas [F]

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

integrate(exp(n*arctan(a*x))*x^2*(a^2*c*x^2+c)^(3/2),x, algorithm="fricas" 
)
 

Output:

integral((a^2*c*x^4 + c*x^2)*sqrt(a^2*c*x^2 + c)*e^(n*arctan(a*x)), x)
 

Sympy [F(-1)]

Timed out. \[ \int e^{n \arctan (a x)} x^2 \left (c+a^2 c x^2\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(exp(n*atan(a*x))*x**2*(a**2*c*x**2+c)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(exp(n*arctan(a*x))*x^2*(a^2*c*x^2+c)^(3/2),x, algorithm="maxima" 
)
 

Output:

integrate((a^2*c*x^2 + c)^(3/2)*x^2*e^(n*arctan(a*x)), x)
 

Giac [F]

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

integrate(exp(n*arctan(a*x))*x^2*(a^2*c*x^2+c)^(3/2),x, algorithm="giac")
 

Output:

integrate((a^2*c*x^2 + c)^(3/2)*x^2*e^(n*arctan(a*x)), x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{n \arctan (a x)} x^2 \left (c+a^2 c x^2\right )^{3/2} \, dx=\int x^2\,{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}\,{\left (c\,a^2\,x^2+c\right )}^{3/2} \,d x \] Input:

int(x^2*exp(n*atan(a*x))*(c + a^2*c*x^2)^(3/2),x)
 

Output:

int(x^2*exp(n*atan(a*x))*(c + a^2*c*x^2)^(3/2), x)
 

Reduce [F]

\[ \int e^{n \arctan (a x)} x^2 \left (c+a^2 c x^2\right )^{3/2} \, dx=\sqrt {c}\, c \left (\left (\int e^{\mathit {atan} \left (a x \right ) n} \sqrt {a^{2} x^{2}+1}\, x^{4}d x \right ) a^{2}+\int e^{\mathit {atan} \left (a x \right ) n} \sqrt {a^{2} x^{2}+1}\, x^{2}d x \right ) \] Input:

int(exp(n*atan(a*x))*x^2*(a^2*c*x^2+c)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

sqrt(c)*c*(int(e**(atan(a*x)*n)*sqrt(a**2*x**2 + 1)*x**4,x)*a**2 + int(e** 
(atan(a*x)*n)*sqrt(a**2*x**2 + 1)*x**2,x))