3.559 \(\int \frac {x^3}{(c+a^2 c x^2)^3 \tan ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=86 \[ \frac {\text {Ci}\left (2 \tan ^{-1}(a x)\right )}{2 a^4 c^3}-\frac {\text {Ci}\left (4 \tan ^{-1}(a x)\right )}{2 a^4 c^3}-\frac {x}{a^3 c^3 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}+\frac {x}{a^3 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.52, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {4964, 4968, 4970, 3312, 3302, 4904, 4406} \[ \frac {\text {CosIntegral}\left (2 \tan ^{-1}(a x)\right )}{2 a^4 c^3}-\frac {\text {CosIntegral}\left (4 \tan ^{-1}(a x)\right )}{2 a^4 c^3}-\frac {x}{a^3 c^3 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}+\frac {x}{a^3 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3302

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 3312

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 4406

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 4904

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

Rule 4964

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

Rule 4968

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

Rule 4970

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*} \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^2} \, dx &=-\frac {\int \frac {x}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^2} \, dx}{a^2}+\frac {\int \frac {x}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^2} \, dx}{a^2 c}\\ &=\frac {x}{a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}-\frac {x}{a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)} \, dx}{a^3}+\frac {3 \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)} \, dx}{a}+\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)} \, dx}{a^3 c}-\frac {\int \frac {x^2}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)} \, dx}{a c}\\ &=\frac {x}{a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}-\frac {x}{a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}+\frac {\operatorname {Subst}\left (\int \frac {\cos ^2(x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}-\frac {\operatorname {Subst}\left (\int \frac {\cos ^4(x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}-\frac {\operatorname {Subst}\left (\int \frac {\sin ^2(x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}+\frac {3 \operatorname {Subst}\left (\int \frac {\cos ^2(x) \sin ^2(x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}\\ &=\frac {x}{a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}-\frac {x}{a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{2 x}-\frac {\cos (2 x)}{2 x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}+\frac {\operatorname {Subst}\left (\int \left (\frac {1}{2 x}+\frac {\cos (2 x)}{2 x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}-\frac {\operatorname {Subst}\left (\int \left (\frac {3}{8 x}+\frac {\cos (2 x)}{2 x}+\frac {\cos (4 x)}{8 x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}+\frac {3 \operatorname {Subst}\left (\int \left (\frac {1}{8 x}-\frac {\cos (4 x)}{8 x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}\\ &=\frac {x}{a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}-\frac {x}{a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac {\operatorname {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^4 c^3}-\frac {3 \operatorname {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^4 c^3}+\frac {\operatorname {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{2 a^4 c^3}\\ &=\frac {x}{a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}-\frac {x}{a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}+\frac {\text {Ci}\left (2 \tan ^{-1}(a x)\right )}{2 a^4 c^3}-\frac {\text {Ci}\left (4 \tan ^{-1}(a x)\right )}{2 a^4 c^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 83, normalized size = 0.97 \[ \frac {-2 a^3 x^3+\left (a^2 x^2+1\right )^2 \tan ^{-1}(a x) \text {Ci}\left (2 \tan ^{-1}(a x)\right )-\left (a^2 x^2+1\right )^2 \tan ^{-1}(a x) \text {Ci}\left (4 \tan ^{-1}(a x)\right )}{2 a^4 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 0.47, size = 292, normalized size = 3.40 \[ -\frac {4 \, a^{3} x^{3} + {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \operatorname {log\_integral}\left (\frac {a^{4} x^{4} + 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) + {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \operatorname {log\_integral}\left (\frac {a^{4} x^{4} - 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} + 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) - {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right )}{4 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )} \arctan \left (a x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.24, size = 58, normalized size = 0.67 \[ \frac {4 \Ci \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-4 \Ci \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-2 \sin \left (2 \arctan \left (a x \right )\right )+\sin \left (4 \arctan \left (a x \right )\right )}{8 a^{4} c^{3} \arctan \left (a x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )} \mathit {sage}_{0} x \arctan \left (a x\right ) + x^{3}}{{\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )} \arctan \left (a x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^3 + (a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c^3)*arctan(a*x)*integrate((a^2*x^4 - 3*x^2)/((a^7*c^3*x^6 + 3*a^5*c^
3*x^4 + 3*a^3*c^3*x^2 + a*c^3)*arctan(a*x)), x))/((a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c^3)*arctan(a*x))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3}{{\mathrm {atan}\left (a\,x\right )}^2\,{\left (c\,a^2\,x^2+c\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{3}}{a^{6} x^{6} \operatorname {atan}^{2}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname {atan}^{2}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atan}^{2}{\left (a x \right )} + \operatorname {atan}^{2}{\left (a x \right )}}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________