3.349 \(\int \frac{x^3 \tan ^{-1}(a x)^2}{(c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=172 \[ \frac{14}{9 a^4 c^2 \sqrt{a^2 c x^2+c}}+\frac{4 x \tan ^{-1}(a x)}{3 a^3 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 \tan ^{-1}(a x)^2}{3 a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{2}{27 a^4 c \left (a^2 c x^2+c\right )^{3/2}}+\frac{2 x^3 \tan ^{-1}(a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac{x^2 \tan ^{-1}(a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

-2/(27*a^4*c*(c + a^2*c*x^2)^(3/2)) + 14/(9*a^4*c^2*Sqrt[c + a^2*c*x^2]) + (2*x^3*ArcTan[a*x])/(9*a*c*(c + a^2
*c*x^2)^(3/2)) + (4*x*ArcTan[a*x])/(3*a^3*c^2*Sqrt[c + a^2*c*x^2]) - (x^2*ArcTan[a*x]^2)/(3*a^2*c*(c + a^2*c*x
^2)^(3/2)) - (2*ArcTan[a*x]^2)/(3*a^4*c^2*Sqrt[c + a^2*c*x^2])

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Rubi [A]  time = 0.282745, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {4940, 4930, 4894, 266, 43} \[ \frac{14}{9 a^4 c^2 \sqrt{a^2 c x^2+c}}+\frac{4 x \tan ^{-1}(a x)}{3 a^3 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 \tan ^{-1}(a x)^2}{3 a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{2}{27 a^4 c \left (a^2 c x^2+c\right )^{3/2}}+\frac{2 x^3 \tan ^{-1}(a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac{x^2 \tan ^{-1}(a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[a*x]^2)/(c + a^2*c*x^2)^(5/2),x]

[Out]

-2/(27*a^4*c*(c + a^2*c*x^2)^(3/2)) + 14/(9*a^4*c^2*Sqrt[c + a^2*c*x^2]) + (2*x^3*ArcTan[a*x])/(9*a*c*(c + a^2
*c*x^2)^(3/2)) + (4*x*ArcTan[a*x])/(3*a^3*c^2*Sqrt[c + a^2*c*x^2]) - (x^2*ArcTan[a*x]^2)/(3*a^2*c*(c + a^2*c*x
^2)^(3/2)) - (2*ArcTan[a*x]^2)/(3*a^4*c^2*Sqrt[c + a^2*c*x^2])

Rule 4940

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(b
*p*(f*x)^m*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^(p - 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])^p, x], x] - Dist[(b^2*p*(p - 1))/m^2, Int[(f*x)^m*(d +
e*x^2)^q*(a + b*ArcTan[c*x])^(p - 2), x], x] - Simp[(f*(f*x)^(m - 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p
)/(c^2*d*m), x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[q, -1] && G
tQ[p, 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 4894

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[b/(c*d*Sqrt[d + e*x^2]),
x] + Simp[(x*(a + b*ArcTan[c*x]))/(d*Sqrt[d + e*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]

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 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])

Rubi steps

\begin{align*} \int \frac{x^3 \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=\frac{2 x^3 \tan ^{-1}(a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac{x^2 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{2}{9} \int \frac{x^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx+\frac{2 \int \frac{x \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^2 c}\\ &=\frac{2 x^3 \tan ^{-1}(a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac{x^2 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{2 \tan ^{-1}(a x)^2}{3 a^4 c^2 \sqrt{c+a^2 c x^2}}-\frac{1}{9} \operatorname{Subst}\left (\int \frac{x}{\left (c+a^2 c x\right )^{5/2}} \, dx,x,x^2\right )+\frac{4 \int \frac{\tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^3 c}\\ &=\frac{4}{3 a^4 c^2 \sqrt{c+a^2 c x^2}}+\frac{2 x^3 \tan ^{-1}(a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{4 x \tan ^{-1}(a x)}{3 a^3 c^2 \sqrt{c+a^2 c x^2}}-\frac{x^2 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{2 \tan ^{-1}(a x)^2}{3 a^4 c^2 \sqrt{c+a^2 c x^2}}-\frac{1}{9} \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{2}{27 a^4 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{14}{9 a^4 c^2 \sqrt{c+a^2 c x^2}}+\frac{2 x^3 \tan ^{-1}(a x)}{9 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{4 x \tan ^{-1}(a x)}{3 a^3 c^2 \sqrt{c+a^2 c x^2}}-\frac{x^2 \tan ^{-1}(a x)^2}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{2 \tan ^{-1}(a x)^2}{3 a^4 c^2 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.113212, size = 81, normalized size = 0.47 \[ \frac{\sqrt{a^2 c x^2+c} \left (42 a^2 x^2+6 a x \left (7 a^2 x^2+6\right ) \tan ^{-1}(a x)-9 \left (3 a^2 x^2+2\right ) \tan ^{-1}(a x)^2+40\right )}{27 a^4 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcTan[a*x]^2)/(c + a^2*c*x^2)^(5/2),x]

[Out]

(Sqrt[c + a^2*c*x^2]*(40 + 42*a^2*x^2 + 6*a*x*(6 + 7*a^2*x^2)*ArcTan[a*x] - 9*(2 + 3*a^2*x^2)*ArcTan[a*x]^2))/
(27*a^4*c^3*(1 + a^2*x^2)^2)

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Maple [C]  time = 0.982, size = 276, normalized size = 1.6 \begin{align*} -{\frac{ \left ( 6\,i\arctan \left ( ax \right ) +9\, \left ( \arctan \left ( ax \right ) \right ) ^{2}-2 \right ) \left ( i{x}^{3}{a}^{3}+3\,{a}^{2}{x}^{2}-3\,iax-1 \right ) }{216\, \left ({a}^{2}{x}^{2}+1 \right ) ^{2}{c}^{3}{a}^{4}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{ \left ( 3\, \left ( \arctan \left ( ax \right ) \right ) ^{2}-6+6\,i\arctan \left ( ax \right ) \right ) \left ( 1+iax \right ) }{8\,{c}^{3}{a}^{4} \left ({a}^{2}{x}^{2}+1 \right ) }\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{ \left ( -3+3\,iax \right ) \left ( \left ( \arctan \left ( ax \right ) \right ) ^{2}-2-2\,i\arctan \left ( ax \right ) \right ) }{8\,{c}^{3}{a}^{4} \left ({a}^{2}{x}^{2}+1 \right ) }\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{ \left ( i{x}^{3}{a}^{3}-3\,{a}^{2}{x}^{2}-3\,iax+1 \right ) \left ( -6\,i\arctan \left ( ax \right ) +9\, \left ( \arctan \left ( ax \right ) \right ) ^{2}-2 \right ) }{216\,{c}^{3}{a}^{4} \left ({a}^{4}{x}^{4}+2\,{a}^{2}{x}^{2}+1 \right ) }\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^3*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)*(I*x^3*a^3+3*a^2*x^2-3*I*a*x-1)*(c*(a*x-I)*(a*x+I))^(1/2)/(a^2*x^2+
1)^2/c^3/a^4-3/8*(arctan(a*x)^2-2+2*I*arctan(a*x))*(1+I*a*x)*(c*(a*x-I)*(a*x+I))^(1/2)/a^4/c^3/(a^2*x^2+1)+3/8
*(c*(a*x-I)*(a*x+I))^(1/2)*(-1+I*a*x)*(arctan(a*x)^2-2-2*I*arctan(a*x))/a^4/c^3/(a^2*x^2+1)+1/216*(c*(a*x-I)*(
a*x+I))^(1/2)*(I*x^3*a^3-3*a^2*x^2-3*I*a*x+1)*(-6*I*arctan(a*x)+9*arctan(a*x)^2-2)/a^4/c^3/(a^4*x^4+2*a^2*x^2+
1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^3*arctan(a*x)^2/(a^2*c*x^2 + c)^(5/2), x)

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Fricas [A]  time = 2.2528, size = 208, normalized size = 1.21 \begin{align*} \frac{\sqrt{a^{2} c x^{2} + c}{\left (42 \, a^{2} x^{2} - 9 \,{\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )^{2} + 6 \,{\left (7 \, a^{3} x^{3} + 6 \, a x\right )} \arctan \left (a x\right ) + 40\right )}}{27 \,{\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)^2/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/27*sqrt(a^2*c*x^2 + c)*(42*a^2*x^2 - 9*(3*a^2*x^2 + 2)*arctan(a*x)^2 + 6*(7*a^3*x^3 + 6*a*x)*arctan(a*x) + 4
0)/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \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**3*atan(a*x)**2/(a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(x**3*atan(a*x)**2/(c*(a**2*x**2 + 1))**(5/2), x)

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Giac [A]  time = 1.26272, size = 151, normalized size = 0.88 \begin{align*} \frac{2 \, x{\left (\frac{7 \, x^{2}}{a c} + \frac{6}{a^{3} c}\right )} \arctan \left (a x\right )}{9 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}} - \frac{{\left (3 \, a^{2} c x^{2} + 2 \, c\right )} \arctan \left (a x\right )^{2}}{3 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} a^{4} c^{2}} + \frac{2 \,{\left (21 \, a^{2} c x^{2} + 20 \, c\right )}}{27 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} a^{4} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*x*(7*x^2/(a*c) + 6/(a^3*c))*arctan(a*x)/(a^2*c*x^2 + c)^(3/2) - 1/3*(3*a^2*c*x^2 + 2*c)*arctan(a*x)^2/((a^
2*c*x^2 + c)^(3/2)*a^4*c^2) + 2/27*(21*a^2*c*x^2 + 20*c)/((a^2*c*x^2 + c)^(3/2)*a^4*c^2)