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

Optimal. Leaf size=215 \[ -\frac {40}{9 a c^2 \sqrt {a^2 c x^2+c}}+\frac {2 x \tan ^{-1}(a x)^3}{3 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 \tan ^{-1}(a x)^2}{a c^2 \sqrt {a^2 c x^2+c}}-\frac {40 x \tan ^{-1}(a x)}{9 c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{27 a c \left (a^2 c x^2+c\right )^{3/2}}+\frac {x \tan ^{-1}(a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {\tan ^{-1}(a x)^2}{3 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 x \tan ^{-1}(a x)}{9 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 215, 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, 4894, 4896} \[ -\frac {40}{9 a c^2 \sqrt {a^2 c x^2+c}}+\frac {2 x \tan ^{-1}(a x)^3}{3 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 \tan ^{-1}(a x)^2}{a c^2 \sqrt {a^2 c x^2+c}}-\frac {40 x \tan ^{-1}(a x)}{9 c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{27 a c \left (a^2 c x^2+c\right )^{3/2}}+\frac {x \tan ^{-1}(a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {\tan ^{-1}(a x)^2}{3 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 x \tan ^{-1}(a x)}{9 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.10, size = 104, normalized size = 0.48 \[ \frac {\sqrt {a^2 c x^2+c} \left (-2 \left (60 a^2 x^2+61\right )+9 a x \left (2 a^2 x^2+3\right ) \tan ^{-1}(a x)^3+9 \left (6 a^2 x^2+7\right ) \tan ^{-1}(a x)^2-6 a x \left (20 a^2 x^2+21\right ) \tan ^{-1}(a x)\right )}{27 a c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 111, normalized size = 0.52 \[ -\frac {\sqrt {a^{2} c x^{2} + c} {\left (120 \, a^{2} x^{2} - 9 \, {\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \arctan \left (a x\right )^{3} - 9 \, {\left (6 \, a^{2} x^{2} + 7\right )} \arctan \left (a x\right )^{2} + 6 \, {\left (20 \, a^{3} x^{3} + 21 \, a x\right )} \arctan \left (a x\right ) + 122\right )}}{27 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*sqrt(a^2*c*x^2 + c)*(120*a^2*x^2 - 9*(2*a^3*x^3 + 3*a*x)*arctan(a*x)^3 - 9*(6*a^2*x^2 + 7)*arctan(a*x)^2
+ 6*(20*a^3*x^3 + 21*a*x)*arctan(a*x) + 122)/(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)^3/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 0.53, size = 308, normalized size = 1.43 \[ -\frac {\left (9 i \arctan \left (a x \right )^{2}+9 \arctan \left (a x \right )^{3}-2 i-6 \arctan \left (a x \right )\right ) \left (a^{3} x^{3}-3 i x^{2} a^{2}-3 a x +i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{216 \left (a^{2} x^{2}+1\right )^{2} a \,c^{3}}+\frac {3 \left (\arctan \left (a x \right )^{3}-6 \arctan \left (a x \right )+3 i \arctan \left (a x \right )^{2}-6 i\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 c^{3} a \left (a^{2} x^{2}+1\right )}+\frac {3 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )^{3}-6 \arctan \left (a x \right )-3 i \arctan \left (a x \right )^{2}+6 i\right )}{8 c^{3} a \left (a^{2} x^{2}+1\right )}-\frac {\left (-9 i \arctan \left (a x \right )^{2}+9 \arctan \left (a x \right )^{3}+2 i-6 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a^{3} x^{3}+3 i x^{2} a^{2}-3 a x -i\right )}{216 \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) a \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)^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))*(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)^3-6*arctan(a*x)+3*I*arctan(a*x)^2-6*I)*(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)^3-6*arctan(a*x)-3*I*arctan(a*x)^2+
6*I)/c^3/a/(a^2*x^2+1)-1/216*(-9*I*arctan(a*x)^2+9*arctan(a*x)^3+2*I-6*arctan(a*x))*(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 [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^3}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atan}^{3}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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