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

Optimal. Leaf size=112 \[ \frac{2 x}{3 a^3 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 \tan ^{-1}(a x)}{3 a^4 c^2 \sqrt{a^2 c x^2+c}}+\frac{x^3}{9 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac{x^2 \tan ^{-1}(a x)}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.137188, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4938, 4930, 191} \[ \frac{2 x}{3 a^3 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 \tan ^{-1}(a x)}{3 a^4 c^2 \sqrt{a^2 c x^2+c}}+\frac{x^3}{9 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac{x^2 \tan ^{-1}(a x)}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0792966, size = 65, normalized size = 0.58 \[ \frac{\sqrt{a^2 c x^2+c} \left (a x \left (7 a^2 x^2+6\right )-3 \left (3 a^2 x^2+2\right ) \tan ^{-1}(a x)\right )}{9 a^4 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.971, size = 244, normalized size = 2.2 \begin{align*} -{\frac{ \left ( i+3\,\arctan \left ( ax \right ) \right ) \left ( i{x}^{3}{a}^{3}+3\,{a}^{2}{x}^{2}-3\,iax-1 \right ) }{72\, \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\,\arctan \left ( ax \right ) +3\,i \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 ( \arctan \left ( ax \right ) -i \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 ( -i+3\,\arctan \left ( ax \right ) \right ) }{72\,{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)/(a^2*c*x^2+c)^(5/2),x)

[Out]

-1/72*(I+3*arctan(a*x))*(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*(a
rctan(a*x)+I)*(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)-I)/a^4/c^3/(a^2*x^2+1)+1/72*(c*(a*x-I)*(a*x+I))^(1/2)*(I*x^3*a^3-3*a^2*x^2-3*I*a*x+1)*(-I+3*arct
an(a*x))/a^4/c^3/(a^4*x^4+2*a^2*x^2+1)

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Maxima [A]  time = 1.39263, size = 88, normalized size = 0.79 \begin{align*} \frac{7 \, a^{3} x^{3} + 6 \, a x - 3 \,{\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )}{9 \,{\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^3*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

1/9*(7*a^3*x^3 + 6*a*x - 3*(3*a^2*x^2 + 2)*arctan(a*x))/((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 = 2.66274, size = 158, normalized size = 1.41 \begin{align*} \frac{{\left (7 \, a^{3} x^{3} + 6 \, a x - 3 \,{\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )\right )} \sqrt{a^{2} c x^{2} + c}}{9 \,{\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)/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.21678, size = 99, normalized size = 0.88 \begin{align*} \frac{x{\left (\frac{7 \, x^{2}}{a c} + \frac{6}{a^{3} c}\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 )}{3 \,{\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)/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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