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

Optimal. Leaf size=237 \[ -\frac{40 x}{9 a^3 c^2 \sqrt{a^2 c x^2+c}}+\frac{2 x \tan ^{-1}(a x)^2}{a^3 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 \tan ^{-1}(a x)^3}{3 a^4 c^2 \sqrt{a^2 c x^2+c}}+\frac{40 \tan ^{-1}(a x)}{9 a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 x^3}{27 a c \left (a^2 c x^2+c\right )^{3/2}}+\frac{x^3 \tan ^{-1}(a x)^2}{3 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac{x^2 \tan ^{-1}(a x)^3}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac{2 x^2 \tan ^{-1}(a x)}{9 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.412489, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {4940, 4930, 4898, 191, 4938} \[ -\frac{40 x}{9 a^3 c^2 \sqrt{a^2 c x^2+c}}+\frac{2 x \tan ^{-1}(a x)^2}{a^3 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 \tan ^{-1}(a x)^3}{3 a^4 c^2 \sqrt{a^2 c x^2+c}}+\frac{40 \tan ^{-1}(a x)}{9 a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 x^3}{27 a c \left (a^2 c x^2+c\right )^{3/2}}+\frac{x^3 \tan ^{-1}(a x)^2}{3 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac{x^2 \tan ^{-1}(a x)^3}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac{2 x^2 \tan ^{-1}(a x)}{9 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^3)/(27*a*c*(c + a^2*c*x^2)^(3/2)) - (40*x)/(9*a^3*c^2*Sqrt[c + a^2*c*x^2]) + (2*x^2*ArcTan[a*x])/(9*a^2*
c*(c + a^2*c*x^2)^(3/2)) + (40*ArcTan[a*x])/(9*a^4*c^2*Sqrt[c + a^2*c*x^2]) + (x^3*ArcTan[a*x]^2)/(3*a*c*(c +
a^2*c*x^2)^(3/2)) + (2*x*ArcTan[a*x]^2)/(a^3*c^2*Sqrt[c + a^2*c*x^2]) - (x^2*ArcTan[a*x]^3)/(3*a^2*c*(c + a^2*
c*x^2)^(3/2)) - (2*ArcTan[a*x]^3)/(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 4898

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(b*p*(a + b*ArcTan[
c*x])^(p - 1))/(c*d*Sqrt[d + e*x^2]), x] + (-Dist[b^2*p*(p - 1), Int[(a + b*ArcTan[c*x])^(p - 2)/(d + e*x^2)^(
3/2), x], x] + Simp[(x*(a + b*ArcTan[c*x])^p)/(d*Sqrt[d + e*x^2]), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e,
c^2*d] && GtQ[p, 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]

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]

Rubi steps

\begin{align*} \int \frac{x^3 \tan ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=\frac{x^3 \tan ^{-1}(a x)^2}{3 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac{x^2 \tan ^{-1}(a x)^3}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{2}{3} \int \frac{x^3 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx+\frac{2 \int \frac{x \tan ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^2 c}\\ &=-\frac{2 x^3}{27 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{2 x^2 \tan ^{-1}(a x)}{9 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{x^3 \tan ^{-1}(a x)^2}{3 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac{x^2 \tan ^{-1}(a x)^3}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{2 \tan ^{-1}(a x)^3}{3 a^4 c^2 \sqrt{c+a^2 c x^2}}+\frac{2 \int \frac{\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^3 c}-\frac{4 \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{9 a^2 c}\\ &=-\frac{2 x^3}{27 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{2 x^2 \tan ^{-1}(a x)}{9 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{40 \tan ^{-1}(a x)}{9 a^4 c^2 \sqrt{c+a^2 c x^2}}+\frac{x^3 \tan ^{-1}(a x)^2}{3 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{2 x \tan ^{-1}(a x)^2}{a^3 c^2 \sqrt{c+a^2 c x^2}}-\frac{x^2 \tan ^{-1}(a x)^3}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{2 \tan ^{-1}(a x)^3}{3 a^4 c^2 \sqrt{c+a^2 c x^2}}-\frac{4 \int \frac{1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{9 a^3 c}-\frac{4 \int \frac{1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{a^3 c}\\ &=-\frac{2 x^3}{27 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac{40 x}{9 a^3 c^2 \sqrt{c+a^2 c x^2}}+\frac{2 x^2 \tan ^{-1}(a x)}{9 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{40 \tan ^{-1}(a x)}{9 a^4 c^2 \sqrt{c+a^2 c x^2}}+\frac{x^3 \tan ^{-1}(a x)^2}{3 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{2 x \tan ^{-1}(a x)^2}{a^3 c^2 \sqrt{c+a^2 c x^2}}-\frac{x^2 \tan ^{-1}(a x)^3}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac{2 \tan ^{-1}(a x)^3}{3 a^4 c^2 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.13304, size = 104, normalized size = 0.44 \[ \frac{\sqrt{a^2 c x^2+c} \left (-2 a x \left (61 a^2 x^2+60\right )-9 \left (3 a^2 x^2+2\right ) \tan ^{-1}(a x)^3+9 a x \left (7 a^2 x^2+6\right ) \tan ^{-1}(a x)^2+6 \left (21 a^2 x^2+20\right ) \tan ^{-1}(a x)\right )}{27 a^4 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + a^2*c*x^2]*(-2*a*x*(60 + 61*a^2*x^2) + 6*(20 + 21*a^2*x^2)*ArcTan[a*x] + 9*a*x*(6 + 7*a^2*x^2)*ArcTa
n[a*x]^2 - 9*(2 + 3*a^2*x^2)*ArcTan[a*x]^3))/(27*a^4*c^3*(1 + a^2*x^2)^2)

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Maple [C]  time = 0.969, size = 312, normalized size = 1.3 \begin{align*} -{\frac{ \left ( 9\,i \left ( \arctan \left ( ax \right ) \right ) ^{2}+9\, \left ( \arctan \left ( ax \right ) \right ) ^{3}-2\,i-6\,\arctan \left ( ax \right ) \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 ) ^{3}-18\,\arctan \left ( ax \right ) +9\,i \left ( \arctan \left ( ax \right ) \right ) ^{2}-18\,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 ( \left ( \arctan \left ( ax \right ) \right ) ^{3}-6\,\arctan \left ( ax \right ) -3\,i \left ( \arctan \left ( ax \right ) \right ) ^{2}+6\,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 ( -9\,i \left ( \arctan \left ( ax \right ) \right ) ^{2}+9\, \left ( \arctan \left ( ax \right ) \right ) ^{3}+2\,i-6\,\arctan \left ( ax \right ) \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)^3/(a^2*c*x^2+c)^(5/2),x)

[Out]

-1/216*(9*I*arctan(a*x)^2+9*arctan(a*x)^3-2*I-6*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*(arctan(a*x)^3-6*arctan(a*x)+3*I*arctan(a*x)^2-6*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)^3-6*arctan(a*x)-3*I*arc
tan(a*x)^2+6*I)/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)*(-9*I*arct
an(a*x)^2+9*arctan(a*x)^3+2*I-6*arctan(a*x))/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 )^{3}}{{\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)^3/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.00609, size = 266, normalized size = 1.12 \begin{align*} -\frac{{\left (122 \, a^{3} x^{3} + 9 \,{\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )^{3} - 9 \,{\left (7 \, a^{3} x^{3} + 6 \, a x\right )} \arctan \left (a x\right )^{2} + 120 \, a x - 6 \,{\left (21 \, a^{2} x^{2} + 20\right )} \arctan \left (a x\right )\right )} \sqrt{a^{2} c x^{2} + c}}{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)^3/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

-1/27*(122*a^3*x^3 + 9*(3*a^2*x^2 + 2)*arctan(a*x)^3 - 9*(7*a^3*x^3 + 6*a*x)*arctan(a*x)^2 + 120*a*x - 6*(21*a
^2*x^2 + 20)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \operatorname{atan}^{3}{\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)**3/(a**2*c*x**2+c)**(5/2),x)

[Out]

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

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Giac [A]  time = 1.3581, size = 207, normalized size = 0.87 \begin{align*} \frac{x{\left (\frac{7 \, x^{2}}{a c} + \frac{6}{a^{3} c}\right )} \arctan \left (a x\right )^{2}}{3 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}} - \frac{2 \, x{\left (\frac{61 \, x^{2}}{a c} + \frac{60}{a^{3} c}\right )}}{27 \,{\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}}{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 )} \arctan \left (a x\right )}{9 \,{\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)^3/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/3*x*(7*x^2/(a*c) + 6/(a^3*c))*arctan(a*x)^2/(a^2*c*x^2 + c)^(3/2) - 2/27*x*(61*x^2/(a*c) + 60/(a^3*c))/(a^2*
c*x^2 + c)^(3/2) - 1/3*(3*a^2*c*x^2 + 2*c)*arctan(a*x)^3/((a^2*c*x^2 + c)^(3/2)*a^4*c^2) + 2/9*(21*a^2*c*x^2 +
20*c)*arctan(a*x)/((a^2*c*x^2 + c)^(3/2)*a^4*c^2)