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

Optimal. Leaf size=84 \[ -\frac {a \sqrt {a^2 c x^2+c}}{6 x^2}-\frac {\left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{3 c x^3}-\frac {1}{6} a^3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4944, 266, 47, 63, 208} \[ -\frac {a \sqrt {a^2 c x^2+c}}{6 x^2}-\frac {\left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{3 c x^3}-\frac {1}{6} a^3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 105, normalized size = 1.25 \[ \frac {a^3 \sqrt {c} x^3 \log (x)-a x \left (\sqrt {a^2 c x^2+c}+a^2 \sqrt {c} x^2 \log \left (\sqrt {c} \sqrt {a^2 c x^2+c}+c\right )\right )-2 \left (a^2 x^2+1\right ) \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.79, size = 84, normalized size = 1.00 \[ \frac {a^{3} \sqrt {c} x^{3} \log \left (-\frac {a^{2} c x^{2} - 2 \, \sqrt {a^{2} c x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {a^{2} c x^{2} + c} {\left (a x + 2 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )}}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(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*(a^2*x^2 + 1)*arctan(a*x)))/x^3

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 1.22, size = 153, normalized size = 1.82 \[ -\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (2 \arctan \left (a x \right ) x^{2} a^{2}+a x +2 \arctan \left (a x \right )\right )}{6 x^{3}}+\frac {a^{3} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right )}{6 \sqrt {a^{2} x^{2}+1}}-\frac {a^{3} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{6 \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 73, normalized size = 0.87 \[ -\frac {1}{6} \, {\left ({\left (a^{2} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \sqrt {a^{2} x^{2} + 1} a^{2} + \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{x^{2}}\right )} a + \frac {2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \arctan \left (a x\right )}{x^{3}}\right )} \sqrt {c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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