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

Optimal. Leaf size=158 \[ -\frac {a \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )}{c^{5/2}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{c^3 x}-\frac {5 a}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {5 a^2 x \tan ^{-1}(a x)}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {a}{9 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {a^2 x \tan ^{-1}(a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

-1/9*a/c/(a^2*c*x^2+c)^(3/2)-1/3*a^2*x*arctan(a*x)/c/(a^2*c*x^2+c)^(3/2)-a*arctanh((a^2*c*x^2+c)^(1/2)/c^(1/2)
)/c^(5/2)-5/3*a/c^2/(a^2*c*x^2+c)^(1/2)-5/3*a^2*x*arctan(a*x)/c^2/(a^2*c*x^2+c)^(1/2)-arctan(a*x)*(a^2*c*x^2+c
)^(1/2)/c^3/x

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Rubi [A]  time = 0.34, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {4966, 4944, 266, 63, 208, 4894, 4896} \[ -\frac {5 a}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {5 a^2 x \tan ^{-1}(a x)}{3 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{c^3 x}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )}{c^{5/2}}-\frac {a}{9 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {a^2 x \tan ^{-1}(a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(9*c*(c + a^2*c*x^2)^(3/2)) - (5*a)/(3*c^2*Sqrt[c + a^2*c*x^2]) - (a^2*x*ArcTan[a*x])/(3*c*(c + a^2*c*x^2)^
(3/2)) - (5*a^2*x*ArcTan[a*x])/(3*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(c^3*x) - (a*Ar
cTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/c^(5/2)

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 4966

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/d, Int[
x^m*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/d, Int[x^(m + 2)*(d + e*x^2)^q*(a + b*ArcTan[c*
x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && ILtQ[m, 0] &
& NeQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 151, normalized size = 0.96 \[ \frac {a x \left (-\left (15 a^2 x^2+16\right ) \sqrt {a^2 c x^2+c}+9 \sqrt {c} \left (a^2 x^2+1\right )^2 \log (x)-9 \sqrt {c} \left (a^2 x^2+1\right )^2 \log \left (\sqrt {c} \sqrt {a^2 c x^2+c}+c\right )\right )-3 \left (8 a^4 x^4+12 a^2 x^2+3\right ) \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{9 c^3 x \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[c + a^2*c*x^2]*(3 + 12*a^2*x^2 + 8*a^4*x^4)*ArcTan[a*x] + a*x*(-((16 + 15*a^2*x^2)*Sqrt[c + a^2*c*x^2
]) + 9*Sqrt[c]*(1 + a^2*x^2)^2*Log[x] - 9*Sqrt[c]*(1 + a^2*x^2)^2*Log[c + Sqrt[c]*Sqrt[c + a^2*c*x^2]]))/(9*c^
3*x*(1 + a^2*x^2)^2)

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fricas [A]  time = 0.59, size = 142, normalized size = 0.90 \[ \frac {9 \, {\left (a^{5} x^{5} + 2 \, a^{3} x^{3} + a x\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} - 2 \, \sqrt {a^{2} c x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (15 \, a^{3} x^{3} + 16 \, a x + 3 \, {\left (8 \, a^{4} x^{4} + 12 \, a^{2} x^{2} + 3\right )} \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c}}{18 \, {\left (a^{4} c^{3} x^{5} + 2 \, a^{2} c^{3} x^{3} + c^{3} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(9*(a^5*x^5 + 2*a^3*x^3 + a*x)*sqrt(c)*log(-(a^2*c*x^2 - 2*sqrt(a^2*c*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*(1
5*a^3*x^3 + 16*a*x + 3*(8*a^4*x^4 + 12*a^2*x^2 + 3)*arctan(a*x))*sqrt(a^2*c*x^2 + c))/(a^4*c^3*x^5 + 2*a^2*c^3
*x^3 + c^3*x)

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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)/x^2/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 0.66, size = 369, normalized size = 2.34 \[ \frac {a \left (i+3 \arctan \left (a x \right )\right ) \left (a^{3} x^{3}-3 i x^{2} a^{2}-3 a x +i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{72 \left (a^{2} x^{2}+1\right )^{2} c^{3}}-\frac {7 a \left (i+\arctan \left (a x \right )\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 c^{3} \left (a^{2} x^{2}+1\right )}-\frac {7 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )-i\right ) a}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a^{3} x^{3}+3 i x^{2} a^{2}-3 a x -i\right ) \left (-i+3 \arctan \left (a x \right )\right ) a}{72 c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{x \,c^{3}}-\frac {a \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{3}}+\frac {a \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*a*(I+3*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/c^3-7/8*a*(I+ar
ctan(a*x))*(a*x-I)*(c*(a*x-I)*(I+a*x))^(1/2)/c^3/(a^2*x^2+1)-7/8*(c*(a*x-I)*(I+a*x))^(1/2)*(I+a*x)*(arctan(a*x
)-I)*a/c^3/(a^2*x^2+1)+1/72*(c*(a*x-I)*(I+a*x))^(1/2)*(a^3*x^3+3*I*x^2*a^2-3*a*x-I)*(-I+3*arctan(a*x))*a/c^3/(
a^4*x^4+2*a^2*x^2+1)-arctan(a*x)*(c*(a*x-I)*(I+a*x))^(1/2)/x/c^3-a*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))/(a^2*x^2+
1)^(1/2)*(c*(a*x-I)*(I+a*x))^(1/2)/c^3+a*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-1)/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(I+a*x
))^(1/2)/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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