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

Optimal. Leaf size=77 \[ \frac {x^3 \tan ^{-1}(a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {1}{3 a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {1}{9 a^3 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4944, 266, 43} \[ \frac {1}{3 a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {1}{9 a^3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {x^3 \tan ^{-1}(a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

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)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=\frac {x^3 \tan ^{-1}(a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {1}{3} a \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx\\ &=\frac {x^3 \tan ^{-1}(a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {1}{6} a \operatorname {Subst}\left (\int \frac {x}{\left (c+a^2 c x\right )^{5/2}} \, dx,x,x^2\right )\\ &=\frac {x^3 \tan ^{-1}(a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {1}{6} a \operatorname {Subst}\left (\int \left (-\frac {1}{a^2 \left (c+a^2 c x\right )^{5/2}}+\frac {1}{a^2 c \left (c+a^2 c x\right )^{3/2}}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{9 a^3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {1}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}+\frac {x^3 \tan ^{-1}(a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 57, normalized size = 0.74 \[ \frac {\sqrt {a^2 c x^2+c} \left (3 a^3 x^3 \tan ^{-1}(a x)+3 a^2 x^2+2\right )}{9 a^3 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 67, normalized size = 0.87 \[ \frac {{\left (3 \, a^{3} x^{3} \arctan \left (a x\right ) + 3 \, a^{2} x^{2} + 2\right )} \sqrt {a^{2} c x^{2} + c}}{9 \, {\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*a^3*x^3*arctan(a*x) + 3*a^2*x^2 + 2)*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)/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 2.83, size = 240, normalized size = 3.12 \[ \frac {\left (i+3 \arctan \left (a x \right )\right ) \left (a^{3} x^{3}-3 i x^{2} a^{2}-3 a x +i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{72 \left (a^{2} x^{2}+1\right )^{2} c^{3} a^{3}}+\frac {\left (i+\arctan \left (a x \right )\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 a^{3} c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )-i\right )}{8 a^{3} c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\left (-i+3 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a^{3} x^{3}+3 i x^{2} a^{2}-3 a x -i\right )}{72 \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) c^{3} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(I+3*arctan(a*x))*(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*(I+ar
ctan(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)-I)/a^3/c^3/(a^2*x^2+1)+1/72*(-I+3*arctan(a*x))*(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 = 0.36, size = 93, normalized size = 1.21 \[ \frac {1}{9} \, a {\left (\frac {3}{\sqrt {a^{2} c x^{2} + c} a^{4} c^{2}} - \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{4} c}\right )} + \frac {1}{3} \, {\left (\frac {x}{\sqrt {a^{2} c x^{2} + c} a^{2} c^{2}} - \frac {x}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{2} c}\right )} \arctan \left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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