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

Optimal. Leaf size=157 \[ -\frac {40 x}{27 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 x \tan ^{-1}(a x)^2}{3 c^2 \sqrt {a^2 c x^2+c}}+\frac {4 \tan ^{-1}(a x)}{3 a c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x}{27 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {x \tan ^{-1}(a x)^2}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 \tan ^{-1}(a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

-2/27*x/c/(a^2*c*x^2+c)^(3/2)+2/9*arctan(a*x)/a/c/(a^2*c*x^2+c)^(3/2)+1/3*x*arctan(a*x)^2/c/(a^2*c*x^2+c)^(3/2
)-40/27*x/c^2/(a^2*c*x^2+c)^(1/2)+4/3*arctan(a*x)/a/c^2/(a^2*c*x^2+c)^(1/2)+2/3*x*arctan(a*x)^2/c^2/(a^2*c*x^2
+c)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4900, 4898, 191, 192} \[ -\frac {40 x}{27 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 x \tan ^{-1}(a x)^2}{3 c^2 \sqrt {a^2 c x^2+c}}+\frac {4 \tan ^{-1}(a x)}{3 a c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x}{27 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {x \tan ^{-1}(a x)^2}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 \tan ^{-1}(a x)}{9 a c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -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 4900

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 86, normalized size = 0.55 \[ \frac {\sqrt {a^2 c x^2+c} \left (-2 a x \left (20 a^2 x^2+21\right )+9 a x \left (2 a^2 x^2+3\right ) \tan ^{-1}(a x)^2+6 \left (6 a^2 x^2+7\right ) \tan ^{-1}(a x)\right )}{27 a c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 93, normalized size = 0.59 \[ -\frac {{\left (40 \, a^{3} x^{3} - 9 \, {\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \arctan \left (a x\right )^{2} + 42 \, a x - 6 \, {\left (6 \, a^{2} x^{2} + 7\right )} \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c}}{27 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*(40*a^3*x^3 - 9*(2*a^3*x^3 + 3*a*x)*arctan(a*x)^2 + 42*a*x - 6*(6*a^2*x^2 + 7)*arctan(a*x))*sqrt(a^2*c*x
^2 + c)/(a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*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(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 = 0.57, size = 272, normalized size = 1.73 \[ -\frac {\left (6 i \arctan \left (a x \right )+9 \arctan \left (a x \right )^{2}-2\right ) \left (a^{3} x^{3}-3 i x^{2} a^{2}-3 a x +i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{216 \left (a^{2} x^{2}+1\right )^{2} a \,c^{3}}+\frac {3 \left (\arctan \left (a x \right )^{2}-2+2 i \arctan \left (a x \right )\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 c^{3} a \left (a^{2} x^{2}+1\right )}+\frac {3 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )^{2}-2-2 i \arctan \left (a x \right )\right )}{8 c^{3} a \left (a^{2} x^{2}+1\right )}-\frac {\left (-6 i \arctan \left (a x \right )+9 \arctan \left (a x \right )^{2}-2\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a^{3} x^{3}+3 i x^{2} a^{2}-3 a x -i\right )}{216 \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) a \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(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)*(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/a/c^3+3/8*(arctan(a*x)^2-2+2*I*arctan(a*x))*(a*x-I)*(c*(a*x-I)*(I+a*x))^(1/2)/c^3/a/(a^2*x^2+1)+3/8*(c*(a*x
-I)*(I+a*x))^(1/2)*(I+a*x)*(arctan(a*x)^2-2-2*I*arctan(a*x))/c^3/a/(a^2*x^2+1)-1/216*(-6*I*arctan(a*x)+9*arcta
n(a*x)^2-2)*(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)/a/c^3

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maxima [A]  time = 0.43, size = 111, normalized size = 0.71 \[ \frac {1}{3} \, {\left (\frac {2 \, x}{\sqrt {a^{2} c x^{2} + c} c^{2}} + \frac {x}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c}\right )} \arctan \left (a x\right )^{2} - \frac {2 \, {\left (20 \, a^{3} x^{3} + 21 \, a x - 3 \, {\left (6 \, a^{2} x^{2} + 7\right )} \arctan \left (a x\right )\right )} a}{27 \, {\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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