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

Optimal. Leaf size=140 \[ -\frac {3 \tan ^{-1}(a x)^2}{32 a^4 c^3}-\frac {x^4}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac {x^4 \tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {x^3 \tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac {3}{32 a^4 c^3 \left (a^2 x^2+1\right )}+\frac {3 x \tan ^{-1}(a x)}{16 a^3 c^3 \left (a^2 x^2+1\right )} \]

[Out]

-1/32*x^4/c^3/(a^2*x^2+1)^2+3/32/a^4/c^3/(a^2*x^2+1)+1/8*x^3*arctan(a*x)/a/c^3/(a^2*x^2+1)^2+3/16*x*arctan(a*x
)/a^3/c^3/(a^2*x^2+1)-3/32*arctan(a*x)^2/a^4/c^3+1/4*x^4*arctan(a*x)^2/c^3/(a^2*x^2+1)^2

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Rubi [A]  time = 0.19, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4944, 4938, 4934, 4884} \[ -\frac {x^4}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac {3}{32 a^4 c^3 \left (a^2 x^2+1\right )}+\frac {x^4 \tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {x^3 \tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac {3 x \tan ^{-1}(a x)}{16 a^3 c^3 \left (a^2 x^2+1\right )}-\frac {3 \tan ^{-1}(a x)^2}{32 a^4 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^4/(32*c^3*(1 + a^2*x^2)^2) + 3/(32*a^4*c^3*(1 + a^2*x^2)) + (x^3*ArcTan[a*x])/(8*a*c^3*(1 + a^2*x^2)^2) + (
3*x*ArcTan[a*x])/(16*a^3*c^3*(1 + a^2*x^2)) - (3*ArcTan[a*x]^2)/(32*a^4*c^3) + (x^4*ArcTan[a*x]^2)/(4*c^3*(1 +
a^2*x^2)^2)

Rule 4884

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

Rule 4934

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q
+ 1))/(4*c^3*d*(q + 1)^2), x] + (-Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x],
x] + Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]))/(2*c^2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && E
qQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]

Rule 4938

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

Rule 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 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 + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {x^3 \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx &=\frac {x^4 \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {1}{2} a \int \frac {x^4 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^3} \, dx\\ &=-\frac {x^4}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {x^3 \tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {x^4 \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {3 \int \frac {x^2 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{8 a c}\\ &=-\frac {x^4}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {3}{32 a^4 c^3 \left (1+a^2 x^2\right )}+\frac {x^3 \tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \tan ^{-1}(a x)}{16 a^3 c^3 \left (1+a^2 x^2\right )}+\frac {x^4 \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {3 \int \frac {\tan ^{-1}(a x)}{c+a^2 c x^2} \, dx}{16 a^3 c^2}\\ &=-\frac {x^4}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {3}{32 a^4 c^3 \left (1+a^2 x^2\right )}+\frac {x^3 \tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \tan ^{-1}(a x)}{16 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {3 \tan ^{-1}(a x)^2}{32 a^4 c^3}+\frac {x^4 \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 74, normalized size = 0.53 \[ \frac {5 a^2 x^2+2 a x \left (5 a^2 x^2+3\right ) \tan ^{-1}(a x)+\left (5 a^4 x^4-6 a^2 x^2-3\right ) \tan ^{-1}(a x)^2+4}{32 a^4 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4 + 5*a^2*x^2 + 2*a*x*(3 + 5*a^2*x^2)*ArcTan[a*x] + (-3 - 6*a^2*x^2 + 5*a^4*x^4)*ArcTan[a*x]^2)/(32*a^4*c^3*(
1 + a^2*x^2)^2)

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fricas [A]  time = 0.65, size = 87, normalized size = 0.62 \[ \frac {5 \, a^{2} x^{2} + {\left (5 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 3\right )} \arctan \left (a x\right )^{2} + 2 \, {\left (5 \, a^{3} x^{3} + 3 \, a x\right )} \arctan \left (a x\right ) + 4}{32 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(5*a^2*x^2 + (5*a^4*x^4 - 6*a^2*x^2 - 3)*arctan(a*x)^2 + 2*(5*a^3*x^3 + 3*a*x)*arctan(a*x) + 4)/(a^8*c^3*
x^4 + 2*a^6*c^3*x^2 + a^4*c^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.07, size = 154, normalized size = 1.10 \[ \frac {\arctan \left (a x \right )^{2}}{4 a^{4} c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {\arctan \left (a x \right )^{2}}{2 a^{4} c^{3} \left (a^{2} x^{2}+1\right )}+\frac {5 x^{3} \arctan \left (a x \right )}{16 a \,c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 \arctan \left (a x \right ) x}{16 a^{3} c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {5 \arctan \left (a x \right )^{2}}{32 a^{4} c^{3}}-\frac {1}{32 a^{4} c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {5}{32 a^{4} c^{3} \left (a^{2} x^{2}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a^4/c^3*arctan(a*x)^2/(a^2*x^2+1)^2-1/2/a^4/c^3*arctan(a*x)^2/(a^2*x^2+1)+5/16*x^3*arctan(a*x)/a/c^3/(a^2*
x^2+1)^2+3/16/a^3/c^3*arctan(a*x)*x/(a^2*x^2+1)^2+5/32*arctan(a*x)^2/a^4/c^3-1/32/a^4/c^3/(a^2*x^2+1)^2+5/32/a
^4/c^3/(a^2*x^2+1)

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maxima [A]  time = 0.43, size = 185, normalized size = 1.32 \[ \frac {1}{16} \, a {\left (\frac {5 \, a^{2} x^{3} + 3 \, x}{a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}} + \frac {5 \, \arctan \left (a x\right )}{a^{5} c^{3}}\right )} \arctan \left (a x\right ) + \frac {{\left (5 \, a^{2} x^{2} - 5 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4\right )} a^{2}}{32 \, {\left (a^{10} c^{3} x^{4} + 2 \, a^{8} c^{3} x^{2} + a^{6} c^{3}\right )}} - \frac {{\left (2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2}}{4 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*a*((5*a^2*x^3 + 3*x)/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3) + 5*arctan(a*x)/(a^5*c^3))*arctan(a*x) + 1/3
2*(5*a^2*x^2 - 5*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 4)*a^2/(a^10*c^3*x^4 + 2*a^8*c^3*x^2 + a^6*c^3) - 1
/4*(2*a^2*x^2 + 1)*arctan(a*x)^2/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3)

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mupad [B]  time = 0.57, size = 85, normalized size = 0.61 \[ \frac {5\,a^4\,x^4\,{\mathrm {atan}\left (a\,x\right )}^2+10\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )-6\,a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+5\,a^2\,x^2+6\,a\,x\,\mathrm {atan}\left (a\,x\right )-3\,{\mathrm {atan}\left (a\,x\right )}^2+4}{32\,a^4\,c^3\,{\left (a^2\,x^2+1\right )}^2} \]

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]

(5*a^2*x^2 - 3*atan(a*x)^2 + 10*a^3*x^3*atan(a*x) + 6*a*x*atan(a*x) - 6*a^2*x^2*atan(a*x)^2 + 5*a^4*x^4*atan(a
*x)^2 + 4)/(32*a^4*c^3*(a^2*x^2 + 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{3} \operatorname {atan}^{2}{\left (a x \right )}}{a^{6} x^{6} + 3 a^{4} x^{4} + 3 a^{2} x^{2} + 1}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*atan(a*x)**2/(a**6*x**6 + 3*a**4*x**4 + 3*a**2*x**2 + 1), x)/c**3

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