\(\int \frac {x^3 \arctan (a x)^2}{(c+a^2 c x^2)^{5/2}} \, dx\) [349]

Optimal result

Integrand size = 24, antiderivative size = 172 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {2}{27 a^4 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {14}{9 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {2 x^3 \arctan (a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {4 x \arctan (a x)}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^2}{3 a^4 c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5060, 5050, 5014, 272, 45} \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 x^3 \arctan (a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 \arctan (a x)^2}{3 a^4 c^2 \sqrt {a^2 c x^2+c}}+\frac {14}{9 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{27 a^4 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {4 x \arctan (a x)}{3 a^3 c^2 \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

-2/(27*a^4*c*(c + a^2*c*x^2)^(3/2)) + 14/(9*a^4*c^2*Sqrt[c + a^2*c*x^2]) + (2*x^3*ArcTan[a*x])/(9*a*c*(c + a^2
*c*x^2)^(3/2)) + (4*x*ArcTan[a*x])/(3*a^3*c^2*Sqrt[c + a^2*c*x^2]) - (x^2*ArcTan[a*x]^2)/(3*a^2*c*(c + a^2*c*x
^2)^(3/2)) - (2*ArcTan[a*x]^2)/(3*a^4*c^2*Sqrt[c + a^2*c*x^2])

Rule 45

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] && NeQ[b*c - a*d, 0] && 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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5014

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

Rule 5050

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] - Dist[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 5060

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*ArcTan[c*x])^(p - 1)/(c*d*m^2)), x] + (Dist[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] - Dist[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] - Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p
/(c^2*d*m)), x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[q, -1] && G
tQ[p, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x^3 \arctan (a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2}{9} \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx+\frac {2 \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^2 c} \\ & = \frac {2 x^3 \arctan (a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^2}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {1}{9} \text {Subst}\left (\int \frac {x}{\left (c+a^2 c x\right )^{5/2}} \, dx,x,x^2\right )+\frac {4 \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^3 c} \\ & = \frac {4}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {2 x^3 \arctan (a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {4 x \arctan (a x)}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^2}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {1}{9} \text {Subst}\left (\int \left (-\frac {1}{a^2 \left (c+a^2 c x\right )^{5/2}}+\frac {1}{a^2 c \left (c+a^2 c x\right )^{3/2}}\right ) \, dx,x,x^2\right ) \\ & = -\frac {2}{27 a^4 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {14}{9 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {2 x^3 \arctan (a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {4 x \arctan (a x)}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \arctan (a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^2}{3 a^4 c^2 \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.47 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c+a^2 c x^2} \left (40+42 a^2 x^2+6 a x \left (6+7 a^2 x^2\right ) \arctan (a x)-9 \left (2+3 a^2 x^2\right ) \arctan (a x)^2\right )}{27 a^4 c^3 \left (1+a^2 x^2\right )^2} \]

[In]

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

[Out]

(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))/
(27*a^4*c^3*(1 + a^2*x^2)^2)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.87 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.60

method	result	size
default	\(-\frac {\left (6 i \arctan \left (a x \right )+9 \arctan \left (a x \right )^{2}-2\right ) \left (i a^{3} x^{3}+3 a^{2} x^{2}-3 i a x -1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{216 \left (a^{2} x^{2}+1\right )^{2} a^{4} c^{3}}-\frac {3 \left (\arctan \left (a x \right )^{2}-2+2 i \arctan \left (a x \right )\right ) \left (i a x +1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 c^{3} a^{4} \left (a^{2} x^{2}+1\right )}+\frac {3 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a x -1\right ) \left (\arctan \left (a x \right )^{2}-2-2 i \arctan \left (a x \right )\right )}{8 c^{3} a^{4} \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a^{3} x^{3}-3 a^{2} x^{2}-3 i a x +1\right ) \left (-6 i \arctan \left (a x \right )+9 \arctan \left (a x \right )^{2}-2\right )}{216 c^{3} a^{4} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}\)	\(276\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.53 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {a^{2} c x^{2} + c} {\left (42 \, a^{2} x^{2} - 9 \, {\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )^{2} + 6 \, {\left (7 \, a^{3} x^{3} + 6 \, a x\right )} \arctan \left (a x\right ) + 40\right )}}{27 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \]

[In]

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

[Out]

1/27*sqrt(a^2*c*x^2 + c)*(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) + 4
0)/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

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