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

Optimal. Leaf size=137 \[ \frac {4}{9 a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {4 x \tan ^{-1}(a x)}{9 a c^2 \sqrt {a^2 c x^2+c}}+\frac {2}{27 a^2 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\tan ^{-1}(a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 x \tan ^{-1}(a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 = {4930, 4896, 4894} \[ \frac {4}{9 a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {4 x \tan ^{-1}(a x)}{9 a c^2 \sqrt {a^2 c x^2+c}}+\frac {2}{27 a^2 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\tan ^{-1}(a x)^2}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 x \tan ^{-1}(a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4896

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(b*(d + e*x^2)^(q + 1))/(
4*c*d*(q + 1)^2), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x], x] - Si
mp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]))/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d
] && LtQ[q, -1] && NeQ[q, -3/2]

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]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 71, normalized size = 0.52 \[ \frac {\sqrt {a^2 c x^2+c} \left (2 \left (6 a^2 x^2+7\right )+6 a x \left (2 a^2 x^2+3\right ) \tan ^{-1}(a x)-9 \tan ^{-1}(a x)^2\right )}{27 c^3 \left (a^3 x^2+a\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 82, normalized size = 0.60 \[ \frac {\sqrt {a^{2} c x^{2} + c} {\left (12 \, a^{2} x^{2} + 6 \, {\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \arctan \left (a x\right ) - 9 \, \arctan \left (a x\right )^{2} + 14\right )}}{27 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*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)*(12*a^2*x^2 + 6*(2*a^3*x^3 + 3*a*x)*arctan(a*x) - 9*arctan(a*x)^2 + 14)/(a^6*c^3*x^4
+ 2*a^4*c^3*x^2 + a^2*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*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 = 1.06, size = 276, normalized size = 2.01 \[ \frac {\left (6 i \arctan \left (a x \right )+9 \arctan \left (a x \right )^{2}-2\right ) \left (i x^{3} a^{3}+3 a^{2} x^{2}-3 i a x -1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{216 \left (a^{2} x^{2}+1\right )^{2} c^{3} a^{2}}-\frac {\left (\arctan \left (a x \right )^{2}-2+2 i \arctan \left (a x \right )\right ) \left (i a x +1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 a^{2} c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a x -1\right ) \left (\arctan \left (a x \right )^{2}-2-2 i \arctan \left (a x \right )\right )}{8 a^{2} c^{3} \left (a^{2} x^{2}+1\right )}-\frac {\left (-6 i \arctan \left (a x \right )+9 \arctan \left (a x \right )^{2}-2\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i x^{3} a^{3}-3 a^{2} x^{2}-3 i a x +1\right )}{216 \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) c^{3} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*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)*(I+a*x))^(1/2)/(a^2*x^2+1
)^2/c^3/a^2-1/8*(arctan(a*x)^2-2+2*I*arctan(a*x))*(1+I*a*x)*(c*(a*x-I)*(I+a*x))^(1/2)/a^2/c^3/(a^2*x^2+1)+1/8*
(c*(a*x-I)*(I+a*x))^(1/2)*(-1+I*a*x)*(arctan(a*x)^2-2-2*I*arctan(a*x))/a^2/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)*(I*x^3*a^3-3*a^2*x^2-3*I*a*x+1)/(a^4*x^4+2*a^2*x^2+1)/c^3/a^
2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x*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 \[ \int \frac {x \operatorname {atan}^{2}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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