\(\int \frac {x}{(c+a^2 c x^2)^{3/2} \arctan (a x)^3} \, dx\) [660]

Optimal result

Integrand size = 22, antiderivative size = 104 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^3} \, dx=-\frac {x}{2 a c \sqrt {c+a^2 c x^2} \arctan (a x)^2}-\frac {1}{2 a^2 c \sqrt {c+a^2 c x^2} \arctan (a x)}-\frac {\sqrt {1+a^2 x^2} \text {Si}(\arctan (a x))}{2 a^2 c \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5062, 5022, 5091, 5090, 3380} \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^3} \, dx=-\frac {\sqrt {a^2 x^2+1} \text {Si}(\arctan (a x))}{2 a^2 c \sqrt {a^2 c x^2+c}}-\frac {x}{2 a c \arctan (a x)^2 \sqrt {a^2 c x^2+c}}-\frac {1}{2 a^2 c \arctan (a x) \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

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

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 5022

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

Rule 5062

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

Rule 5090

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c^(m
+ 1), Subst[Int[(a + b*x)^p*(Sin[x]^m/Cos[x]^(m + 2*(q + 1))), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d,
e, p}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q] || GtQ[d, 0])

Rule 5091

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^(q + 1
/2)*(Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]), Int[x^m*(1 + c^2*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] &&  !(IntegerQ[q] || GtQ[d, 0])

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.61 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^3} \, dx=-\frac {a x+\arctan (a x)+\sqrt {1+a^2 x^2} \arctan (a x)^2 \text {Si}(\arctan (a x))}{2 a^2 c \sqrt {c+a^2 c x^2} \arctan (a x)^2} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 9.25 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.50

method	result	size
default	\(-\frac {i \left (\arctan \left (a x \right )^{2} \operatorname {Ei}_{1}\left (-i \arctan \left (a x \right )\right ) a^{2} x^{2}-\arctan \left (a x \right )^{2} \operatorname {Ei}_{1}\left (i \arctan \left (a x \right )\right ) a^{2} x^{2}+\operatorname {Ei}_{1}\left (-i \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}-\operatorname {Ei}_{1}\left (i \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}-2 i \sqrt {a^{2} x^{2}+1}\, a x -2 i \sqrt {a^{2} x^{2}+1}\, \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}} \arctan \left (a x \right )^{2} a^{2} c^{2}}\)	\(156\)

[In]

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

[Out]

-1/4*I*(arctan(a*x)^2*Ei(1,-I*arctan(a*x))*a^2*x^2-arctan(a*x)^2*Ei(1,I*arctan(a*x))*a^2*x^2+Ei(1,-I*arctan(a*
x))*arctan(a*x)^2-Ei(1,I*arctan(a*x))*arctan(a*x)^2-2*I*(a^2*x^2+1)^(1/2)*a*x-2*I*(a^2*x^2+1)^(1/2)*arctan(a*x
))*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^(3/2)/arctan(a*x)^2/a^2/c^2

Fricas [F]

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

[In]

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

[Out]

integral(sqrt(a^2*c*x^2 + c)*x/((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*arctan(a*x)^3), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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