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

Optimal. Leaf size=169 \[ -\frac{15 x}{64 c^3 \left (a^2 x^2+1\right )}-\frac{x}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac{3 x \tan ^{-1}(a x)^2}{8 c^3 \left (a^2 x^2+1\right )}+\frac{x \tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac{3 \tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac{\tan ^{-1}(a x)^3}{8 a c^3}-\frac{15 \tan ^{-1}(a x)}{64 a c^3} \]

[Out]

-x/(32*c^3*(1 + a^2*x^2)^2) - (15*x)/(64*c^3*(1 + a^2*x^2)) - (15*ArcTan[a*x])/(64*a*c^3) + ArcTan[a*x]/(8*a*c
^3*(1 + a^2*x^2)^2) + (3*ArcTan[a*x])/(8*a*c^3*(1 + a^2*x^2)) + (x*ArcTan[a*x]^2)/(4*c^3*(1 + a^2*x^2)^2) + (3
*x*ArcTan[a*x]^2)/(8*c^3*(1 + a^2*x^2)) + ArcTan[a*x]^3/(8*a*c^3)

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Rubi [A]  time = 0.118644, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {4900, 4892, 4930, 199, 205} \[ -\frac{15 x}{64 c^3 \left (a^2 x^2+1\right )}-\frac{x}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac{3 x \tan ^{-1}(a x)^2}{8 c^3 \left (a^2 x^2+1\right )}+\frac{x \tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac{3 \tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac{\tan ^{-1}(a x)^3}{8 a c^3}-\frac{15 \tan ^{-1}(a x)}{64 a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(32*c^3*(1 + a^2*x^2)^2) - (15*x)/(64*c^3*(1 + a^2*x^2)) - (15*ArcTan[a*x])/(64*a*c^3) + ArcTan[a*x]/(8*a*c
^3*(1 + a^2*x^2)^2) + (3*ArcTan[a*x])/(8*a*c^3*(1 + a^2*x^2)) + (x*ArcTan[a*x]^2)/(4*c^3*(1 + a^2*x^2)^2) + (3
*x*ArcTan[a*x]^2)/(8*c^3*(1 + a^2*x^2)) + ArcTan[a*x]^3/(8*a*c^3)

Rule 4900

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

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 4930

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx &=\frac{\tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac{x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac{1}{8} \int \frac{1}{\left (c+a^2 c x^2\right )^3} \, dx+\frac{3 \int \frac{\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c}\\ &=-\frac{x}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac{x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^3}{8 a c^3}-\frac{3 \int \frac{1}{\left (c+a^2 c x^2\right )^2} \, dx}{32 c}-\frac{(3 a) \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c}\\ &=-\frac{x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac{3 x}{64 c^3 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac{3 \tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^3}{8 a c^3}-\frac{3 \int \frac{1}{c+a^2 c x^2} \, dx}{64 c^2}-\frac{3 \int \frac{1}{\left (c+a^2 c x^2\right )^2} \, dx}{8 c}\\ &=-\frac{x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac{15 x}{64 c^3 \left (1+a^2 x^2\right )}-\frac{3 \tan ^{-1}(a x)}{64 a c^3}+\frac{\tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac{3 \tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^3}{8 a c^3}-\frac{3 \int \frac{1}{c+a^2 c x^2} \, dx}{16 c^2}\\ &=-\frac{x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac{15 x}{64 c^3 \left (1+a^2 x^2\right )}-\frac{15 \tan ^{-1}(a x)}{64 a c^3}+\frac{\tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac{3 \tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^3}{8 a c^3}\\ \end{align*}

Mathematica [A]  time = 0.0447453, size = 98, normalized size = 0.58 \[ \frac{-a x \left (15 a^2 x^2+17\right )+8 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^3+8 a x \left (3 a^2 x^2+5\right ) \tan ^{-1}(a x)^2+\left (-15 a^4 x^4-6 a^2 x^2+17\right ) \tan ^{-1}(a x)}{64 a c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*x*(17 + 15*a^2*x^2)) + (17 - 6*a^2*x^2 - 15*a^4*x^4)*ArcTan[a*x] + 8*a*x*(5 + 3*a^2*x^2)*ArcTan[a*x]^2 +
8*(1 + a^2*x^2)^2*ArcTan[a*x]^3)/(64*a*c^3*(1 + a^2*x^2)^2)

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Maple [A]  time = 0.041, size = 159, normalized size = 0.9 \begin{align*}{\frac{x \left ( \arctan \left ( ax \right ) \right ) ^{2}}{4\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3\,x \left ( \arctan \left ( ax \right ) \right ) ^{2}}{8\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{3}}{8\,a{c}^{3}}}+{\frac{\arctan \left ( ax \right ) }{8\,a{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3\,\arctan \left ( ax \right ) }{8\,a{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }}-{\frac{15\,{a}^{2}{x}^{3}}{64\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{17\,x}{64\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{15\,\arctan \left ( ax \right ) }{64\,a{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x*arctan(a*x)^2/c^3/(a^2*x^2+1)^2+3/8*x*arctan(a*x)^2/c^3/(a^2*x^2+1)+1/8*arctan(a*x)^3/a/c^3+1/8*arctan(a
*x)/a/c^3/(a^2*x^2+1)^2+3/8*arctan(a*x)/a/c^3/(a^2*x^2+1)-15/64*a^2/c^3/(a^2*x^2+1)^2*x^3-17/64*x/c^3/(a^2*x^2
+1)^2-15/64*arctan(a*x)/a/c^3

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Maxima [A]  time = 1.67147, size = 313, normalized size = 1.85 \begin{align*} \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 )^{2} - \frac{{\left (15 \, a^{3} x^{3} - 8 \,{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} + 17 \, a x + 15 \,{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )} a^{2}}{64 \,{\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\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 \arctan \left (a x\right )}{8 \,{\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/(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)^2 - 1/64*(15*a
^3*x^3 - 8*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^3 + 17*a*x + 15*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x))*a^2/(a
^7*c^3*x^4 + 2*a^5*c^3*x^2 + a^3*c^3) + 1/8*(3*a^2*x^2 - 3*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 4)*a*arct
an(a*x)/(a^6*c^3*x^4 + 2*a^4*c^3*x^2 + a^2*c^3)

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Fricas [A]  time = 2.15604, size = 261, normalized size = 1.54 \begin{align*} -\frac{15 \, a^{3} x^{3} - 8 \,{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} - 8 \,{\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \arctan \left (a x\right )^{2} + 17 \, a x +{\left (15 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 17\right )} \arctan \left (a x\right )}{64 \,{\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(15*a^3*x^3 - 8*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^3 - 8*(3*a^3*x^3 + 5*a*x)*arctan(a*x)^2 + 17*a*x +
(15*a^4*x^4 + 6*a^2*x^2 - 17)*arctan(a*x))/(a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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