3.231 \(\int \frac {\tan ^{-1}(a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=118 \[ -\frac {a \sqrt {a^2 c x^2+c}}{6 c x^2}+\frac {2 a^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 c x}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 c x^3}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )}{6 \sqrt {c}} \]

[Out]

5/6*a^3*arctanh((a^2*c*x^2+c)^(1/2)/c^(1/2))/c^(1/2)-1/6*a*(a^2*c*x^2+c)^(1/2)/c/x^2-1/3*arctan(a*x)*(a^2*c*x^
2+c)^(1/2)/c/x^3+2/3*a^2*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/c/x

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Rubi [A]  time = 0.20, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4962, 266, 51, 63, 208, 4944} \[ -\frac {a \sqrt {a^2 c x^2+c}}{6 c x^2}+\frac {2 a^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 c x}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 c x^3}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )}{6 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]/(x^4*Sqrt[c + a^2*c*x^2]),x]

[Out]

-(a*Sqrt[c + a^2*c*x^2])/(6*c*x^2) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(3*c*x^3) + (2*a^2*Sqrt[c + a^2*c*x^2]*
ArcTan[a*x])/(3*c*x) + (5*a^3*ArcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/(6*Sqrt[c])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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 4962

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

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x^3}+\frac {1}{3} a \int \frac {1}{x^3 \sqrt {c+a^2 c x^2}} \, dx-\frac {1}{3} \left (2 a^2\right ) \int \frac {\tan ^{-1}(a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x^3}+\frac {2 a^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x}+\frac {1}{6} a \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {c+a^2 c x}} \, dx,x,x^2\right )-\frac {1}{3} \left (2 a^3\right ) \int \frac {1}{x \sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {a \sqrt {c+a^2 c x^2}}{6 c x^2}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x^3}+\frac {2 a^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x}-\frac {1}{12} a^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+a^2 c x}} \, dx,x,x^2\right )-\frac {1}{3} a^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+a^2 c x}} \, dx,x,x^2\right )\\ &=-\frac {a \sqrt {c+a^2 c x^2}}{6 c x^2}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x^3}+\frac {2 a^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x}-\frac {a \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 )}{6 c}-\frac {(2 a) \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 )}{3 c}\\ &=-\frac {a \sqrt {c+a^2 c x^2}}{6 c x^2}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x^3}+\frac {2 a^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 c x}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{6 \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 110, normalized size = 0.93 \[ \frac {-5 a^3 \sqrt {c} x^3 \log (x)-a x \sqrt {a^2 c x^2+c}+2 \left (2 a^2 x^2-1\right ) \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)+5 a^3 \sqrt {c} x^3 \log \left (\sqrt {c} \sqrt {a^2 c x^2+c}+c\right )}{6 c x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[a*x]/(x^4*Sqrt[c + a^2*c*x^2]),x]

[Out]

(-(a*x*Sqrt[c + a^2*c*x^2]) + 2*(-1 + 2*a^2*x^2)*Sqrt[c + a^2*c*x^2]*ArcTan[a*x] - 5*a^3*Sqrt[c]*x^3*Log[x] +
5*a^3*Sqrt[c]*x^3*Log[c + Sqrt[c]*Sqrt[c + a^2*c*x^2]])/(6*c*x^3)

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fricas [A]  time = 0.65, size = 89, normalized size = 0.75 \[ \frac {5 \, a^{3} \sqrt {c} x^{3} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {a^{2} c x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {a^{2} c x^{2} + c} {\left (a x - 2 \, {\left (2 \, a^{2} x^{2} - 1\right )} \arctan \left (a x\right )\right )}}{12 \, c x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

1/12*(5*a^3*sqrt(c)*x^3*log(-(a^2*c*x^2 + 2*sqrt(a^2*c*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*sqrt(a^2*c*x^2 + c)*(a
*x - 2*(2*a^2*x^2 - 1)*arctan(a*x)))/(c*x^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)/x^4/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 3.23, size = 163, normalized size = 1.38 \[ \frac {\left (4 \arctan \left (a x \right ) x^{2} a^{2}-a x -2 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 c \,x^{3}}+\frac {5 a^{3} \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 \sqrt {a^{2} x^{2}+1}\, c}-\frac {5 a^{3} \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 \sqrt {a^{2} x^{2}+1}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x)

[Out]

1/6*(4*arctan(a*x)*x^2*a^2-a*x-2*arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1/2)/c/x^3+5/6*a^3*ln(1+(1+I*a*x)/(a^2*x^2+
1)^(1/2))*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^(1/2)/c-5/6*a^3*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-1)*(c*(a*x-I)*(
I+a*x))^(1/2)/(a^2*x^2+1)^(1/2)/c

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maxima [A]  time = 0.50, size = 81, normalized size = 0.69 \[ \frac {{\left (5 \, a^{2} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {\sqrt {a^{2} x^{2} + 1}}{x^{2}}\right )} a + 2 \, {\left (\frac {2 \, \sqrt {a^{2} x^{2} + 1} a^{2}}{x} - \frac {\sqrt {a^{2} x^{2} + 1}}{x^{3}}\right )} \arctan \left (a x\right )}{6 \, \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

1/6*((5*a^2*arcsinh(1/(a*abs(x))) - sqrt(a^2*x^2 + 1)/x^2)*a + 2*(2*sqrt(a^2*x^2 + 1)*a^2/x - sqrt(a^2*x^2 + 1
)/x^3)*arctan(a*x))/sqrt(c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atan}\left (a\,x\right )}{x^4\,\sqrt {c\,a^2\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)/(x^4*(c + a^2*c*x^2)^(1/2)),x)

[Out]

int(atan(a*x)/(x^4*(c + a^2*c*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atan}{\left (a x \right )}}{x^{4} \sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)/x**4/(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(atan(a*x)/(x**4*sqrt(c*(a**2*x**2 + 1))), x)

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