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

Optimal. Leaf size=105 \[ \frac {3}{16 a c^3 \left (a^2 x^2+1\right )}+\frac {1}{16 a c^3 \left (a^2 x^2+1\right )^2}+\frac {3 x \tan ^{-1}(a x)}{8 c^3 \left (a^2 x^2+1\right )}+\frac {x \tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 \tan ^{-1}(a x)^2}{16 a c^3} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4896, 4892, 261} \[ \frac {3}{16 a c^3 \left (a^2 x^2+1\right )}+\frac {1}{16 a c^3 \left (a^2 x^2+1\right )^2}+\frac {3 x \tan ^{-1}(a x)}{8 c^3 \left (a^2 x^2+1\right )}+\frac {x \tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 \tan ^{-1}(a x)^2}{16 a c^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]/(c + a^2*c*x^2)^3,x]

[Out]

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 4892

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

Rule 4896

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 68, normalized size = 0.65 \[ \frac {3 a^2 x^2+2 a x \left (3 a^2 x^2+5\right ) \tan ^{-1}(a x)+3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2+4}{16 a c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4 + 3*a^2*x^2 + 2*a*x*(5 + 3*a^2*x^2)*ArcTan[a*x] + 3*(1 + a^2*x^2)^2*ArcTan[a*x]^2)/(16*a*c^3*(1 + a^2*x^2)^
2)

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fricas [A]  time = 0.46, size = 85, normalized size = 0.81 \[ \frac {3 \, a^{2} x^{2} + 3 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 2 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \arctan \left (a x\right ) + 4}{16 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*a^2*x^2 + 3*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 2*(3*a^3*x^3 + 5*a*x)*arctan(a*x) + 4)/(a^5*c^3*
x^4 + 2*a^3*c^3*x^2 + a*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(arctan(a*x)/(a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.04, size = 96, normalized size = 0.91 \[ \frac {1}{16 a \,c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3}{16 a \,c^{3} \left (a^{2} x^{2}+1\right )}+\frac {x \arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{16 a \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 129, normalized size = 1.23 \[ \frac {1}{8} \, {\left (\frac {3 \, a^{2} x^{3} + 5 \, x}{a^{4} c^{3} x^{4} + 2 \, a^{2} c^{3} x^{2} + c^{3}} + \frac {3 \, \arctan \left (a x\right )}{a c^{3}}\right )} \arctan \left (a x\right ) + \frac {{\left (3 \, a^{2} x^{2} - 3 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4\right )} a}{16 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.48, size = 85, normalized size = 0.81 \[ \frac {3\,a^4\,x^4\,{\mathrm {atan}\left (a\,x\right )}^2+6\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )+6\,a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+3\,a^2\,x^2+10\,a\,x\,\mathrm {atan}\left (a\,x\right )+3\,{\mathrm {atan}\left (a\,x\right )}^2+4}{16\,a\,c^3\,{\left (a^2\,x^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)/(c + a^2*c*x^2)^3,x)

[Out]

(3*a^2*x^2 + 3*atan(a*x)^2 + 6*a^3*x^3*atan(a*x) + 10*a*x*atan(a*x) + 6*a^2*x^2*atan(a*x)^2 + 3*a^4*x^4*atan(a
*x)^2 + 4)/(16*a*c^3*(a^2*x^2 + 1)^2)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RecursionError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RecursionError

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