\(\int \frac {x^4}{(c+a^2 c x^2)^3 \arctan (a x)} \, dx\) [489]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 50 \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx=-\frac {\operatorname {CosIntegral}(2 \arctan (a x))}{2 a^5 c^3}+\frac {\operatorname {CosIntegral}(4 \arctan (a x))}{8 a^5 c^3}+\frac {3 \log (\arctan (a x))}{8 a^5 c^3} \]

[Out]

-1/2*Ci(2*arctan(a*x))/a^5/c^3+1/8*Ci(4*arctan(a*x))/a^5/c^3+3/8*ln(arctan(a*x))/a^5/c^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 50, 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, 3393, 3383} \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx=-\frac {\operatorname {CosIntegral}(2 \arctan (a x))}{2 a^5 c^3}+\frac {\operatorname {CosIntegral}(4 \arctan (a x))}{8 a^5 c^3}+\frac {3 \log (\arctan (a x))}{8 a^5 c^3} \]

[In]

Int[x^4/((c + a^2*c*x^2)^3*ArcTan[a*x]),x]

[Out]

-1/2*CosIntegral[2*ArcTan[a*x]]/(a^5*c^3) + CosIntegral[4*ArcTan[a*x]]/(8*a^5*c^3) + (3*Log[ArcTan[a*x]])/(8*a
^5*c^3)

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 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])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sin ^4(x)}{x} \, dx,x,\arctan (a x)\right )}{a^5 c^3} \\ & = \frac {\text {Subst}\left (\int \left (\frac {3}{8 x}-\frac {\cos (2 x)}{2 x}+\frac {\cos (4 x)}{8 x}\right ) \, dx,x,\arctan (a x)\right )}{a^5 c^3} \\ & = \frac {3 \log (\arctan (a x))}{8 a^5 c^3}+\frac {\text {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\arctan (a x)\right )}{8 a^5 c^3}-\frac {\text {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\arctan (a x)\right )}{2 a^5 c^3} \\ & = -\frac {\operatorname {CosIntegral}(2 \arctan (a x))}{2 a^5 c^3}+\frac {\operatorname {CosIntegral}(4 \arctan (a x))}{8 a^5 c^3}+\frac {3 \log (\arctan (a x))}{8 a^5 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.68 \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx=\frac {-4 \operatorname {CosIntegral}(2 \arctan (a x))+\operatorname {CosIntegral}(4 \arctan (a x))+3 \log (\arctan (a x))}{8 a^5 c^3} \]

[In]

Integrate[x^4/((c + a^2*c*x^2)^3*ArcTan[a*x]),x]

[Out]

(-4*CosIntegral[2*ArcTan[a*x]] + CosIntegral[4*ArcTan[a*x]] + 3*Log[ArcTan[a*x]])/(8*a^5*c^3)

Maple [A] (verified)

Time = 3.51 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.70

method result size
derivativedivides \(-\frac {-3 \ln \left (\arctan \left (a x \right )\right )+4 \,\operatorname {Ci}\left (2 \arctan \left (a x \right )\right )-\operatorname {Ci}\left (4 \arctan \left (a x \right )\right )}{8 a^{5} c^{3}}\) \(35\)
default \(-\frac {-3 \ln \left (\arctan \left (a x \right )\right )+4 \,\operatorname {Ci}\left (2 \arctan \left (a x \right )\right )-\operatorname {Ci}\left (4 \arctan \left (a x \right )\right )}{8 a^{5} c^{3}}\) \(35\)

[In]

int(x^4/(a^2*c*x^2+c)^3/arctan(a*x),x,method=_RETURNVERBOSE)

[Out]

-1/8/a^5*(-3*ln(arctan(a*x))+4*Ci(2*arctan(a*x))-Ci(4*arctan(a*x)))/c^3

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 174, normalized size of antiderivative = 3.48 \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx=\frac {6 \, \log \left (\arctan \left (a x\right )\right ) + \operatorname {log\_integral}\left (\frac {a^{4} x^{4} + 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) + \operatorname {log\_integral}\left (\frac {a^{4} x^{4} - 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} + 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) - 4 \, \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - 4 \, \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right )}{16 \, a^{5} c^{3}} \]

[In]

integrate(x^4/(a^2*c*x^2+c)^3/arctan(a*x),x, algorithm="fricas")

[Out]

1/16*(6*log(arctan(a*x)) + 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)) + 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)) - 4*log_int
egral(-(a^2*x^2 + 2*I*a*x - 1)/(a^2*x^2 + 1)) - 4*log_integral(-(a^2*x^2 - 2*I*a*x - 1)/(a^2*x^2 + 1)))/(a^5*c
^3)

Sympy [F]

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

[In]

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

[Out]

Integral(x**4/(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]

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

[In]

integrate(x^4/(a^2*c*x^2+c)^3/arctan(a*x),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(x^4/(a^2*c*x^2+c)^3/arctan(a*x),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx=\int \frac {x^4}{\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^3} \,d x \]

[In]

int(x^4/(atan(a*x)*(c + a^2*c*x^2)^3),x)

[Out]

int(x^4/(atan(a*x)*(c + a^2*c*x^2)^3), x)