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

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

[Out]

-2/27*x^3/c/(a^2*c*x^2+c)^(3/2)+2/9*x^2*arctan(a*x)/a/c/(a^2*c*x^2+c)^(3/2)+1/3*x^3*arctan(a*x)^2/c/(a^2*c*x^2
+c)^(3/2)-4/9*x/a^2/c^2/(a^2*c*x^2+c)^(1/2)+4/9*arctan(a*x)/a^3/c^2/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 139, normalized size of antiderivative = 1.00, 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 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]

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 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^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*}

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Mathematica [A]  time = 0.09, size = 80, normalized size = 0.58 \[ \frac {\sqrt {a^2 c x^2+c} \left (9 a^3 x^3 \tan ^{-1}(a x)^2-2 a x \left (7 a^2 x^2+6\right )+6 \left (3 a^2 x^2+2\right ) \tan ^{-1}(a x)\right )}{27 a^3 c^3 \left (a^2 x^2+1\right )^2} \]

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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fricas [A]  time = 0.48, size = 88, normalized size = 0.63 \[ \frac {{\left (9 \, a^{3} x^{3} \arctan \left (a x\right )^{2} - 14 \, a^{3} x^{3} - 12 \, a x + 6 \, {\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c}}{27 \, {\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )}} \]

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

[Out]

sage0*x

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maple [C]  time = 2.80, size = 272, normalized size = 1.96 \[ \frac {\left (6 i \arctan \left (a x \right )+9 \arctan \left (a x \right )^{2}-2\right ) \left (a^{3} x^{3}-3 i x^{2} a^{2}-3 a x +i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{216 \left (a^{2} x^{2}+1\right )^{2} c^{3} a^{3}}+\frac {\left (\arctan \left (a x \right )^{2}-2+2 i \arctan \left (a x \right )\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 a^{3} c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )^{2}-2-2 i \arctan \left (a x \right )\right )}{8 a^{3} c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\left (-6 i \arctan \left (a x \right )+9 \arctan \left (a x \right )^{2}-2\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a^{3} x^{3}+3 i x^{2} a^{2}-3 a x -i\right )}{216 \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) c^{3} a^{3}} \]

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*x^2*a^2-3*a*x+I)*(c*(a*x-I)*(I+a*x))^(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)*(I+a*x))^(1/2)/a^3/c^3/(a^2*x^2+1)+1/8*(c*(
a*x-I)*(I+a*x))^(1/2)*(I+a*x)*(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)*(I+a*x))^(1/2)*(a^3*x^3+3*I*x^2*a^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 = 0.43, size = 117, normalized size = 0.84 \[ \frac {1}{3} \, {\left (\frac {x}{\sqrt {a^{2} c x^{2} + c} a^{2} c^{2}} - \frac {x}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{2} c}\right )} \arctan \left (a x\right )^{2} - \frac {2 \, {\left (7 \, a^{3} x^{3} + 6 \, a x - 3 \, {\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )\right )} a}{27 \, {\left (a^{6} c^{2} x^{2} + a^{4} c^{2}\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {c}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\mathrm {atan}\left (a\,x\right )}^2}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \operatorname {atan}^{2}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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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