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

Optimal. Leaf size=113 \[ -\frac {\text {Si}\left (2 \tan ^{-1}(a x)\right )}{2 a^2 c^3}-\frac {\text {Si}\left (4 \tan ^{-1}(a x)\right )}{a^2 c^3}-\frac {x}{2 a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2}+\frac {3}{2 a^2 c^3 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}-\frac {2}{a^2 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.45, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {4968, 4964, 4902, 4970, 4406, 12, 3299} \[ -\frac {\text {Si}\left (2 \tan ^{-1}(a x)\right )}{2 a^2 c^3}-\frac {\text {Si}\left (4 \tan ^{-1}(a x)\right )}{a^2 c^3}-\frac {x}{2 a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2}+\frac {3}{2 a^2 c^3 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}-\frac {2}{a^2 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3299

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 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 4902

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 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}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^3} \, dx &=-\frac {x}{2 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}+\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^2} \, dx}{2 a}-\frac {1}{2} (3 a) \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^2} \, dx\\ &=-\frac {x}{2 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac {1}{2 a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}-2 \int \frac {x}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)} \, dx+\frac {3 \int \frac {1}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^2} \, dx}{2 a}-\frac {3 \int \frac {1}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^2} \, dx}{2 a c}\\ &=-\frac {x}{2 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac {2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}+\frac {3}{2 a^2 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-6 \int \frac {x}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)} \, dx-\frac {2 \operatorname {Subst}\left (\int \frac {\cos ^3(x) \sin (x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^3}+\frac {3 \int \frac {x}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)} \, dx}{c}\\ &=-\frac {x}{2 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac {2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}+\frac {3}{2 a^2 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac {2 \operatorname {Subst}\left (\int \left (\frac {\sin (2 x)}{4 x}+\frac {\sin (4 x)}{8 x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^3}+\frac {3 \operatorname {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^3}-\frac {6 \operatorname {Subst}\left (\int \frac {\cos ^3(x) \sin (x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^3}\\ &=-\frac {x}{2 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac {2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}+\frac {3}{2 a^2 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac {\operatorname {Subst}\left (\int \frac {\sin (4 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{4 a^2 c^3}-\frac {\operatorname {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{2 a^2 c^3}+\frac {3 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^3}-\frac {6 \operatorname {Subst}\left (\int \left (\frac {\sin (2 x)}{4 x}+\frac {\sin (4 x)}{8 x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^3}\\ &=-\frac {x}{2 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac {2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}+\frac {3}{2 a^2 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac {\text {Si}\left (2 \tan ^{-1}(a x)\right )}{2 a^2 c^3}-\frac {\text {Si}\left (4 \tan ^{-1}(a x)\right )}{4 a^2 c^3}-\frac {3 \operatorname {Subst}\left (\int \frac {\sin (4 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{4 a^2 c^3}\\ &=-\frac {x}{2 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac {2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}+\frac {3}{2 a^2 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac {\text {Si}\left (2 \tan ^{-1}(a x)\right )}{2 a^2 c^3}-\frac {\text {Si}\left (4 \tan ^{-1}(a x)\right )}{a^2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 98, normalized size = 0.87 \[ -\frac {\left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2 \text {Si}\left (2 \tan ^{-1}(a x)\right )+2 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2 \text {Si}\left (4 \tan ^{-1}(a x)\right )-3 a^2 x^2 \tan ^{-1}(a x)+a x+\tan ^{-1}(a x)}{2 a^2 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.43, size = 314, normalized size = 2.78 \[ \frac {{\left (-2 i \, a^{4} x^{4} - 4 i \, a^{2} x^{2} - 2 i\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (\frac {a^{4} x^{4} + 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) + {\left (2 i \, a^{4} x^{4} + 4 i \, a^{2} x^{2} + 2 i\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (\frac {a^{4} x^{4} - 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} + 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) + {\left (-i \, a^{4} x^{4} - 2 i \, a^{2} x^{2} - i\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) + {\left (i \, a^{4} x^{4} + 2 i \, a^{2} x^{2} + i\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - 2 \, a x + 2 \, {\left (3 \, a^{2} x^{2} - 1\right )} \arctan \left (a x\right )}{4 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )} \arctan \left (a x\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [A]  time = 0.25, size = 88, normalized size = 0.78 \[ -\frac {8 \Si \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}+16 \Si \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}+4 \cos \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+4 \cos \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+2 \sin \left (2 \arctan \left (a x \right )\right )+\sin \left (4 \arctan \left (a x \right )\right )}{16 a^{2} c^{3} \arctan \left (a x \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/a^2/c^3*(8*Si(2*arctan(a*x))*arctan(a*x)^2+16*Si(4*arctan(a*x))*arctan(a*x)^2+4*cos(2*arctan(a*x))*arcta
n(a*x)+4*cos(4*arctan(a*x))*arctan(a*x)+2*sin(2*arctan(a*x))+sin(4*arctan(a*x)))/arctan(a*x)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )} \mathit {sage}_{0} x \arctan \left (a x\right )^{2} - a x + {\left (3 \, a^{2} x^{2} - 1\right )} \arctan \left (a x\right )}{2 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )} \arctan \left (a x\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a^6*c^3*x^4 + 2*a^4*c^3*x^2 + a^2*c^3)*arctan(a*x)^2*integrate((3*a^2*x^3 - 5*x)/((a^6*c^3*x^6 + 3*a^4
*c^3*x^4 + 3*a^2*c^3*x^2 + c^3)*arctan(a*x)), x) - a*x + (3*a^2*x^2 - 1)*arctan(a*x))/((a^6*c^3*x^4 + 2*a^4*c^
3*x^2 + a^2*c^3)*arctan(a*x)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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