\(\int \frac {x \arctan (a x)}{(c+a^2 c x^2)^{5/2}} \, dx\) [244]

Optimal result

Integrand size = 20, antiderivative size = 79 \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {x}{9 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 x}{9 a c^2 \sqrt {c+a^2 c x^2}}-\frac {\arctan (a x)}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5050, 198, 197} \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {\arctan (a x)}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 x}{9 a c^2 \sqrt {a^2 c x^2+c}}+\frac {x}{9 a c \left (a^2 c x^2+c\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

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]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c+a^2 c x^2} \left (3 a x+2 a^3 x^3-3 \arctan (a x)\right )}{9 c^3 \left (a+a^3 x^2\right )^2} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.40 (sec) , antiderivative size = 244, normalized size of antiderivative = 3.09

method	result	size
default	\(\frac {\left (i+3 \arctan \left (a x \right )\right ) \left (i a^{3} x^{3}+3 a^{2} x^{2}-3 i a x -1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{72 \left (a^{2} x^{2}+1\right )^{2} a^{2} c^{3}}-\frac {\left (\arctan \left (a x \right )+i\right ) \left (i a x +1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 c^{3} a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a x -1\right ) \left (\arctan \left (a x \right )-i\right )}{8 c^{3} a^{2} \left (a^{2} x^{2}+1\right )}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a^{3} x^{3}-3 a^{2} x^{2}-3 i a x +1\right ) \left (-i+3 \arctan \left (a x \right )\right )}{72 c^{3} a^{2} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}\)	\(244\)

[In]

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

[Out]

1/72*(I+3*arctan(a*x))*(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^2/c^3-1/8*(ar
ctan(a*x)+I)*(1+I*a*x)*(c*(a*x-I)*(I+a*x))^(1/2)/c^3/a^2/(a^2*x^2+1)+1/8*(c*(a*x-I)*(I+a*x))^(1/2)*(I*a*x-1)*(
arctan(a*x)-I)/c^3/a^2/(a^2*x^2+1)-1/72*(c*(a*x-I)*(I+a*x))^(1/2)*(I*a^3*x^3-3*a^2*x^2-3*I*a*x+1)*(-I+3*arctan
(a*x))/c^3/a^2/(a^4*x^4+2*a^2*x^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.81 \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, a^{3} x^{3} + 3 \, a x - 3 \, \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c}}{9 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> Invalid comparison of non-real zoo

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, a^{3} x^{3} + 3 \, a x - 3 \, \arctan \left (a x\right )\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {c}}{9 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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