3.5.53 \(\int \frac {x^2 \arctan (a x)^3}{(c+a^2 c x^2)^{5/2}} \, dx\) [453]

3.5.53.1 Optimal result

Integrand size = 24, antiderivative size = 199 \[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {2}{27 a^3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {14}{9 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {2 x^3 \arctan (a x)}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {4 x \arctan (a x)}{3 a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {x^2 \arctan (a x)^2}{3 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 \arctan (a x)^2}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}+\frac {x^3 \arctan (a x)^3}{3 c \left (c+a^2 c x^2\right )^{3/2}} \]

output

2/27/a^3/c/(a^2*c*x^2+c)^(3/2)-2/9*x^3*arctan(a*x)/c/(a^2*c*x^2+c)^(3/2)+1 
/3*x^2*arctan(a*x)^2/a/c/(a^2*c*x^2+c)^(3/2)+1/3*x^3*arctan(a*x)^3/c/(a^2* 
c*x^2+c)^(3/2)-14/9/a^3/c^2/(a^2*c*x^2+c)^(1/2)-4/3*x*arctan(a*x)/a^2/c^2/ 
(a^2*c*x^2+c)^(1/2)+2/3*arctan(a*x)^2/a^3/c^2/(a^2*c*x^2+c)^(1/2)
 

3.5.53.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.48 \[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c+a^2 c x^2} \left (-40-42 a^2 x^2-6 a x \left (6+7 a^2 x^2\right ) \arctan (a x)+9 \left (2+3 a^2 x^2\right ) \arctan (a x)^2+9 a^3 x^3 \arctan (a x)^3\right )}{27 a^3 c^3 \left (1+a^2 x^2\right )^2} \]

input

Integrate[(x^2*ArcTan[a*x]^3)/(c + a^2*c*x^2)^(5/2),x]
 

output

(Sqrt[c + a^2*c*x^2]*(-40 - 42*a^2*x^2 - 6*a*x*(6 + 7*a^2*x^2)*ArcTan[a*x] 
 + 9*(2 + 3*a^2*x^2)*ArcTan[a*x]^2 + 9*a^3*x^3*ArcTan[a*x]^3))/(27*a^3*c^3 
*(1 + a^2*x^2)^2)
 

3.5.53.3 Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.20, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {5479, 5475, 243, 53, 2009, 5465, 5429}

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.

\(\displaystyle \int \frac {x^2 \arctan (a x)^3}{\left (a^2 c x^2+c\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 5479

\(\displaystyle \frac {x^3 \arctan (a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}}-a \int \frac {x^3 \arctan (a x)^2}{\left (a^2 c x^2+c\right )^{5/2}}dx\)

\(\Big \downarrow \) 5475

\(\displaystyle \frac {x^3 \arctan (a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}}-a \left (\frac {2 \int \frac {x \arctan (a x)^2}{\left (a^2 c x^2+c\right )^{3/2}}dx}{3 a^2 c}-\frac {2}{9} \int \frac {x^3}{\left (a^2 c x^2+c\right )^{5/2}}dx-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 x^3 \arctan (a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {x^3 \arctan (a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}}-a \left (\frac {2 \int \frac {x \arctan (a x)^2}{\left (a^2 c x^2+c\right )^{3/2}}dx}{3 a^2 c}-\frac {1}{9} \int \frac {x^2}{\left (a^2 c x^2+c\right )^{5/2}}dx^2-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 x^3 \arctan (a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

\(\Big \downarrow \) 53

\(\displaystyle \frac {x^3 \arctan (a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}}-a \left (\frac {2 \int \frac {x \arctan (a x)^2}{\left (a^2 c x^2+c\right )^{3/2}}dx}{3 a^2 c}-\frac {1}{9} \int \left (\frac {1}{a^2 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {1}{a^2 \left (a^2 c x^2+c\right )^{5/2}}\right )dx^2-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 x^3 \arctan (a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^3 \arctan (a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}}-a \left (\frac {2 \int \frac {x \arctan (a x)^2}{\left (a^2 c x^2+c\right )^{3/2}}dx}{3 a^2 c}-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 x^3 \arctan (a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}}+\frac {1}{9} \left (\frac {2}{a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{3 a^4 c \left (a^2 c x^2+c\right )^{3/2}}\right )\right )\)

\(\Big \downarrow \) 5465

\(\displaystyle \frac {x^3 \arctan (a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}}-a \left (\frac {2 \left (\frac {2 \int \frac {\arctan (a x)}{\left (a^2 c x^2+c\right )^{3/2}}dx}{a}-\frac {\arctan (a x)^2}{a^2 c \sqrt {a^2 c x^2+c}}\right )}{3 a^2 c}-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 x^3 \arctan (a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}}+\frac {1}{9} \left (\frac {2}{a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{3 a^4 c \left (a^2 c x^2+c\right )^{3/2}}\right )\right )\)

\(\Big \downarrow \) 5429

\(\displaystyle \frac {x^3 \arctan (a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}}-a \left (-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 \left (\frac {2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{a}-\frac {\arctan (a x)^2}{a^2 c \sqrt {a^2 c x^2+c}}\right )}{3 a^2 c}+\frac {2 x^3 \arctan (a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}}+\frac {1}{9} \left (\frac {2}{a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{3 a^4 c \left (a^2 c x^2+c\right )^{3/2}}\right )\right )\)

input

Int[(x^2*ArcTan[a*x]^3)/(c + a^2*c*x^2)^(5/2),x]
 

output

(x^3*ArcTan[a*x]^3)/(3*c*(c + a^2*c*x^2)^(3/2)) - a*((-2/(3*a^4*c*(c + a^2 
*c*x^2)^(3/2)) + 2/(a^4*c^2*Sqrt[c + a^2*c*x^2]))/9 + (2*x^3*ArcTan[a*x])/ 
(9*a*c*(c + a^2*c*x^2)^(3/2)) - (x^2*ArcTan[a*x]^2)/(3*a^2*c*(c + a^2*c*x^ 
2)^(3/2)) + (2*(-(ArcTan[a*x]^2/(a^2*c*Sqrt[c + a^2*c*x^2])) + (2*(1/(a*c* 
Sqrt[c + a^2*c*x^2]) + (x*ArcTan[a*x])/(c*Sqrt[c + a^2*c*x^2])))/a))/(3*a^ 
2*c))
 

3.5.53.3.1 Defintions of rubi rules used

rule 53

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, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

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

rule 5429

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbo 
l] :> Simp[b/(c*d*Sqrt[d + e*x^2]), x] + Simp[x*((a + b*ArcTan[c*x])/(d*Sqr 
t[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]
 

rule 5465

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_ 
.), x_Symbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(2*e*(q + 
1))), x] - Simp[b*(p/(2*c*(q + 1)))   Int[(d + e*x^2)^q*(a + b*ArcTan[c*x]) 
^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 
 0] && NeQ[q, -1]
 

rule 5475

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_)*((d_) + (e_.) 
*(x_)^2)^(q_), x_Symbol] :> Simp[b*p*(f*x)^m*(d + e*x^2)^(q + 1)*((a + b*Ar 
cTan[c*x])^(p - 1)/(c*d*m^2)), x] + (-Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(q + 
 1)*((a + b*ArcTan[c*x])^p/(c^2*d*m)), x] + Simp[f^2*((m - 1)/(c^2*d*m)) 
Int[(f*x)^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Simp[ 
b^2*p*((p - 1)/m^2)   Int[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 2) 
, x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && EqQ[m + 2* 
q + 2, 0] && LtQ[q, -1] && GtQ[p, 1]
 

rule 5479

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

3.5.53.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.79 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.55

method	result	size
default	\(\frac {\left (9 i \arctan \left (a x \right )^{2}+9 \arctan \left (a x \right )^{3}-2 i-6 \arctan \left (a x \right )\right ) \left (a^{3} x^{3}-3 i a^{2} x^{2}-3 a x +i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{216 \left (a^{2} x^{2}+1\right )^{2} a^{3} c^{3}}+\frac {\left (\arctan \left (a x \right )^{3}-6 \arctan \left (a x \right )+3 i \arctan \left (a x \right )^{2}-6 i\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 c^{3} a^{3} \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )^{3}-6 \arctan \left (a x \right )-3 i \arctan \left (a x \right )^{2}+6 i\right )}{8 c^{3} a^{3} \left (a^{2} x^{2}+1\right )}+\frac {\left (-9 i \arctan \left (a x \right )^{2}+9 \arctan \left (a x \right )^{3}+2 i-6 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a^{3} x^{3}+3 i a^{2} x^{2}-3 a x -i\right )}{216 \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) a^{3} c^{3}}\)	\(308\)

input

int(x^2*arctan(a*x)^3/(a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

output

1/216*(9*I*arctan(a*x)^2+9*arctan(a*x)^3-2*I-6*arctan(a*x))*(a^3*x^3-3*I*a 
^2*x^2-3*a*x+I)*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^2/a^3/c^3+1/8*(arcta 
n(a*x)^3-6*arctan(a*x)+3*I*arctan(a*x)^2-6*I)*(a*x-I)*(c*(a*x-I)*(I+a*x))^ 
(1/2)/c^3/a^3/(a^2*x^2+1)+1/8*(c*(a*x-I)*(I+a*x))^(1/2)*(I+a*x)*(arctan(a* 
x)^3-6*arctan(a*x)-3*I*arctan(a*x)^2+6*I)/c^3/a^3/(a^2*x^2+1)+1/216*(-9*I* 
arctan(a*x)^2+9*arctan(a*x)^3+2*I-6*arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1/2) 
*(a^3*x^3+3*I*a^2*x^2-3*a*x-I)/(a^4*x^4+2*a^2*x^2+1)/a^3/c^3
 

3.5.53.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.53 \[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (9 \, a^{3} x^{3} \arctan \left (a x\right )^{3} - 42 \, a^{2} x^{2} + 9 \, {\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )^{2} - 6 \, {\left (7 \, a^{3} x^{3} + 6 \, a x\right )} \arctan \left (a x\right ) - 40\right )} \sqrt {a^{2} c x^{2} + c}}{27 \, {\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )}} \]

input

integrate(x^2*arctan(a*x)^3/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")
 

output

1/27*(9*a^3*x^3*arctan(a*x)^3 - 42*a^2*x^2 + 9*(3*a^2*x^2 + 2)*arctan(a*x) 
^2 - 6*(7*a^3*x^3 + 6*a*x)*arctan(a*x) - 40)*sqrt(a^2*c*x^2 + c)/(a^7*c^3* 
x^4 + 2*a^5*c^3*x^2 + a^3*c^3)
 

3.5.53.6 Sympy [F]

\[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^{2} \operatorname {atan}^{3}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

input

integrate(x**2*atan(a*x)**3/(a**2*c*x**2+c)**(5/2),x)
 

output

Integral(x**2*atan(a*x)**3/(c*(a**2*x**2 + 1))**(5/2), x)
 

3.5.53.7 Maxima [F]

\[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{2} \arctan \left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(x^2*arctan(a*x)^3/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")
 

output

integrate(x^2*arctan(a*x)^3/(a^2*c*x^2 + c)^(5/2), x)
 

3.5.53.8 Giac [F]

\[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{2} \arctan \left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(x^2*arctan(a*x)^3/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")
 

output

sage0*x
 

3.5.53.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^2\,{\mathrm {atan}\left (a\,x\right )}^3}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

input

int((x^2*atan(a*x)^3)/(c + a^2*c*x^2)^(5/2),x)
 

output

int((x^2*atan(a*x)^3)/(c + a^2*c*x^2)^(5/2), x)