3.350 \(\int \frac{x^2 \tan ^{-1}(a x)^2}{(c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=139 \[ -\frac{4 x}{9 a^2 c^2 \sqrt{a^2 c x^2+c}}+\frac{4 \tan ^{-1}(a x)}{9 a^3 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 x^3}{27 c \left (a^2 c x^2+c\right )^{3/2}}+\frac{x^3 \tan ^{-1}(a x)^2}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac{2 x^2 \tan ^{-1}(a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

(-2*x^3)/(27*c*(c + a^2*c*x^2)^(3/2)) - (4*x)/(9*a^2*c^2*Sqrt[c + a^2*c*x^2]) + (2*x^2*ArcTan[a*x])/(9*a*c*(c
+ a^2*c*x^2)^(3/2)) + (4*ArcTan[a*x])/(9*a^3*c^2*Sqrt[c + a^2*c*x^2]) + (x^3*ArcTan[a*x]^2)/(3*c*(c + a^2*c*x^
2)^(3/2))

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Rubi [A]  time = 0.267383, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4944, 4938, 4930, 191} \[ -\frac{4 x}{9 a^2 c^2 \sqrt{a^2 c x^2+c}}+\frac{4 \tan ^{-1}(a x)}{9 a^3 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 x^3}{27 c \left (a^2 c x^2+c\right )^{3/2}}+\frac{x^3 \tan ^{-1}(a x)^2}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac{2 x^2 \tan ^{-1}(a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcTan[a*x]^2)/(c + a^2*c*x^2)^(5/2),x]

[Out]

(-2*x^3)/(27*c*(c + a^2*c*x^2)^(3/2)) - (4*x)/(9*a^2*c^2*Sqrt[c + a^2*c*x^2]) + (2*x^2*ArcTan[a*x])/(9*a*c*(c
+ a^2*c*x^2)^(3/2)) + (4*ArcTan[a*x])/(9*a^3*c^2*Sqrt[c + a^2*c*x^2]) + (x^3*ArcTan[a*x]^2)/(3*c*(c + a^2*c*x^
2)^(3/2))

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]

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 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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{x^2 \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=\frac{x^3 \tan ^{-1}(a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{1}{3} (2 a) \int \frac{x^3 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx\\ &=-\frac{2 x^3}{27 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{2 x^2 \tan ^{-1}(a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{x^3 \tan ^{-1}(a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{4 \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{9 a c}\\ &=-\frac{2 x^3}{27 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{2 x^2 \tan ^{-1}(a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{4 \tan ^{-1}(a x)}{9 a^3 c^2 \sqrt{c+a^2 c x^2}}+\frac{x^3 \tan ^{-1}(a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{4 \int \frac{1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{9 a^2 c}\\ &=-\frac{2 x^3}{27 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{4 x}{9 a^2 c^2 \sqrt{c+a^2 c x^2}}+\frac{2 x^2 \tan ^{-1}(a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{4 \tan ^{-1}(a x)}{9 a^3 c^2 \sqrt{c+a^2 c x^2}}+\frac{x^3 \tan ^{-1}(a x)^2}{3 c \left (c+a^2 c x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0866473, size = 80, normalized size = 0.58 \[ \frac{\sqrt{a^2 c x^2+c} \left (-2 a x \left (7 a^2 x^2+6\right )+9 a^3 x^3 \tan ^{-1}(a x)^2+6 \left (3 a^2 x^2+2\right ) \tan ^{-1}(a x)\right )}{27 a^3 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*ArcTan[a*x]^2)/(c + a^2*c*x^2)^(5/2),x]

[Out]

(Sqrt[c + a^2*c*x^2]*(-2*a*x*(6 + 7*a^2*x^2) + 6*(2 + 3*a^2*x^2)*ArcTan[a*x] + 9*a^3*x^3*ArcTan[a*x]^2))/(27*a
^3*c^3*(1 + a^2*x^2)^2)

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Maple [C]  time = 0.733, size = 272, normalized size = 2. \begin{align*}{\frac{ \left ( 6\,i\arctan \left ( ax \right ) +9\, \left ( \arctan \left ( ax \right ) \right ) ^{2}-2 \right ) \left ({a}^{3}{x}^{3}-3\,i{a}^{2}{x}^{2}-3\,ax+i \right ) }{216\, \left ({a}^{2}{x}^{2}+1 \right ) ^{2}{c}^{3}{a}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{ \left ( \left ( \arctan \left ( ax \right ) \right ) ^{2}-2+2\,i\arctan \left ( ax \right ) \right ) \left ( ax-i \right ) }{8\,{c}^{3}{a}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{ \left ( ax+i \right ) \left ( \left ( \arctan \left ( ax \right ) \right ) ^{2}-2-2\,i\arctan \left ( ax \right ) \right ) }{8\,{c}^{3}{a}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{ \left ( -6\,i\arctan \left ( ax \right ) +9\, \left ( \arctan \left ( ax \right ) \right ) ^{2}-2 \right ) \left ({a}^{3}{x}^{3}+3\,i{a}^{2}{x}^{2}-3\,ax-i \right ) }{ \left ( 216\,{a}^{4}{x}^{4}+432\,{a}^{2}{x}^{2}+216 \right ){c}^{3}{a}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x)

[Out]

1/216*(6*I*arctan(a*x)+9*arctan(a*x)^2-2)*(a^3*x^3-3*I*a^2*x^2-3*a*x+I)*(c*(a*x-I)*(a*x+I))^(1/2)/(a^2*x^2+1)^
2/c^3/a^3+1/8*(arctan(a*x)^2-2+2*I*arctan(a*x))*(a*x-I)*(c*(a*x-I)*(a*x+I))^(1/2)/a^3/c^3/(a^2*x^2+1)+1/8*(c*(
a*x-I)*(a*x+I))^(1/2)*(a*x+I)*(arctan(a*x)^2-2-2*I*arctan(a*x))/a^3/c^3/(a^2*x^2+1)+1/216*(-6*I*arctan(a*x)+9*
arctan(a*x)^2-2)*(c*(a*x-I)*(a*x+I))^(1/2)*(a^3*x^3+3*I*a^2*x^2-3*a*x-I)/(a^4*x^4+2*a^2*x^2+1)/c^3/a^3

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Maxima [A]  time = 1.39334, size = 158, normalized size = 1.14 \begin{align*} \frac{1}{3} \,{\left (\frac{x}{\sqrt{a^{2} c x^{2} + c} a^{2} c^{2}} - \frac{x}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} a^{2} c}\right )} \arctan \left (a x\right )^{2} - \frac{2 \,{\left (7 \, a^{3} x^{3} + 6 \, a x - 3 \,{\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )\right )} a}{27 \,{\left (a^{6} c^{2} x^{2} + a^{4} c^{2}\right )} \sqrt{a^{2} x^{2} + 1} \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

1/3*(x/(sqrt(a^2*c*x^2 + c)*a^2*c^2) - x/((a^2*c*x^2 + c)^(3/2)*a^2*c))*arctan(a*x)^2 - 2/27*(7*a^3*x^3 + 6*a*
x - 3*(3*a^2*x^2 + 2)*arctan(a*x))*a/((a^6*c^2*x^2 + a^4*c^2)*sqrt(a^2*x^2 + 1)*sqrt(c))

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Fricas [A]  time = 2.32938, size = 197, normalized size = 1.42 \begin{align*} \frac{{\left (9 \, a^{3} x^{3} \arctan \left (a x\right )^{2} - 14 \, a^{3} x^{3} - 12 \, a x + 6 \,{\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )\right )} \sqrt{a^{2} c x^{2} + c}}{27 \,{\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/27*(9*a^3*x^3*arctan(a*x)^2 - 14*a^3*x^3 - 12*a*x + 6*(3*a^2*x^2 + 2)*arctan(a*x))*sqrt(a^2*c*x^2 + c)/(a^7*
c^3*x^4 + 2*a^5*c^3*x^2 + a^3*c^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \operatorname{atan}^{2}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atan(a*x)**2/(a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(x**2*atan(a*x)**2/(c*(a**2*x**2 + 1))**(5/2), x)

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Giac [A]  time = 1.24564, size = 138, normalized size = 0.99 \begin{align*} \frac{x^{3} \arctan \left (a x\right )^{2}}{3 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} c} - \frac{2}{27} \, a{\left (\frac{x{\left (\frac{7 \, x^{2}}{a c} + \frac{6}{a^{3} c}\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}} - \frac{3 \,{\left (3 \, a^{2} c x^{2} + 2 \, c\right )} \arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} a^{4} c^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/3*x^3*arctan(a*x)^2/((a^2*c*x^2 + c)^(3/2)*c) - 2/27*a*(x*(7*x^2/(a*c) + 6/(a^3*c))/(a^2*c*x^2 + c)^(3/2) -
3*(3*a^2*c*x^2 + 2*c)*arctan(a*x)/((a^2*c*x^2 + c)^(3/2)*a^4*c^2))