\(\int \frac {x^3}{(c+a^2 c x^2)^3 \arctan (a x)} \, dx\) [490]

Optimal result

Integrand size = 22, antiderivative size = 35 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx=\frac {\text {Si}(2 \arctan (a x))}{4 a^4 c^3}-\frac {\text {Si}(4 \arctan (a x))}{8 a^4 c^3} \]

[Out]

1/4*Si(2*arctan(a*x))/a^4/c^3-1/8*Si(4*arctan(a*x))/a^4/c^3

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {5090, 4491, 3380} \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx=\frac {\text {Si}(2 \arctan (a x))}{4 a^4 c^3}-\frac {\text {Si}(4 \arctan (a x))}{8 a^4 c^3} \]

[In]

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

[Out]

SinIntegral[2*ArcTan[a*x]]/(4*a^4*c^3) - SinIntegral[4*ArcTan[a*x]]/(8*a^4*c^3)

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 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

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])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\cos (x) \sin ^3(x)}{x} \, dx,x,\arctan (a x)\right )}{a^4 c^3} \\ & = \frac {\text {Subst}\left (\int \left (\frac {\sin (2 x)}{4 x}-\frac {\sin (4 x)}{8 x}\right ) \, dx,x,\arctan (a x)\right )}{a^4 c^3} \\ & = -\frac {\text {Subst}\left (\int \frac {\sin (4 x)}{x} \, dx,x,\arctan (a x)\right )}{8 a^4 c^3}+\frac {\text {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\arctan (a x)\right )}{4 a^4 c^3} \\ & = \frac {\text {Si}(2 \arctan (a x))}{4 a^4 c^3}-\frac {\text {Si}(4 \arctan (a x))}{8 a^4 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx=-\frac {-2 \text {Si}(2 \arctan (a x))+\text {Si}(4 \arctan (a x))}{8 a^4 c^3} \]

[In]

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

[Out]

-1/8*(-2*SinIntegral[2*ArcTan[a*x]] + SinIntegral[4*ArcTan[a*x]])/(a^4*c^3)

Maple [A] (verified)

Time = 2.89 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(-\frac {\operatorname {Si}\left (4 \arctan \left (a x \right )\right )-2 \,\operatorname {Si}\left (2 \arctan \left (a x \right )\right )}{8 a^{4} c^{3}}\)	\(26\)
default	\(-\frac {\operatorname {Si}\left (4 \arctan \left (a x \right )\right )-2 \,\operatorname {Si}\left (2 \arctan \left (a x \right )\right )}{8 a^{4} c^{3}}\)	\(26\)

[In]

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

[Out]

-1/8/a^4*(Si(4*arctan(a*x))-2*Si(2*arctan(a*x)))/c^3

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 171, normalized size of antiderivative = 4.89 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx=\frac {-i \, \operatorname {log\_integral}\left (\frac {a^{4} x^{4} + 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) + i \, \operatorname {log\_integral}\left (\frac {a^{4} x^{4} - 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} + 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) + 2 i \, \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - 2 i \, \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right )}{16 \, a^{4} c^{3}} \]

[In]

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

[Out]

1/16*(-I*log_integral((a^4*x^4 + 4*I*a^3*x^3 - 6*a^2*x^2 - 4*I*a*x + 1)/(a^4*x^4 + 2*a^2*x^2 + 1)) + I*log_int
egral((a^4*x^4 - 4*I*a^3*x^3 - 6*a^2*x^2 + 4*I*a*x + 1)/(a^4*x^4 + 2*a^2*x^2 + 1)) + 2*I*log_integral(-(a^2*x^
2 + 2*I*a*x - 1)/(a^2*x^2 + 1)) - 2*I*log_integral(-(a^2*x^2 - 2*I*a*x - 1)/(a^2*x^2 + 1)))/(a^4*c^3)

Sympy [F]

\[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx=\frac {\int \frac {x^{3}}{a^{6} x^{6} \operatorname {atan}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname {atan}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atan}{\left (a x \right )} + \operatorname {atan}{\left (a x \right )}}\, dx}{c^{3}} \]

[In]

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

[Out]

Integral(x**3/(a**6*x**6*atan(a*x) + 3*a**4*x**4*atan(a*x) + 3*a**2*x**2*atan(a*x) + atan(a*x)), x)/c**3

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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