3.73 \(\int \frac{a+b \tan ^{-1}(c x^2)}{x^6} \, dx\)

Optimal. Leaf size=159 \[ -\frac{a+b \tan ^{-1}\left (c x^2\right )}{5 x^5}+\frac{b c^{5/2} \log \left (c x^2-\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2}}-\frac{b c^{5/2} \log \left (c x^2+\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2}}+\frac{b c^{5/2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2}}-\frac{b c^{5/2} \tan ^{-1}\left (\sqrt{2} \sqrt{c} x+1\right )}{5 \sqrt{2}}-\frac{2 b c}{15 x^3} \]

[Out]

(-2*b*c)/(15*x^3) - (a + b*ArcTan[c*x^2])/(5*x^5) + (b*c^(5/2)*ArcTan[1 - Sqrt[2]*Sqrt[c]*x])/(5*Sqrt[2]) - (b
*c^(5/2)*ArcTan[1 + Sqrt[2]*Sqrt[c]*x])/(5*Sqrt[2]) + (b*c^(5/2)*Log[1 - Sqrt[2]*Sqrt[c]*x + c*x^2])/(10*Sqrt[
2]) - (b*c^(5/2)*Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2])/(10*Sqrt[2])

________________________________________________________________________________________

Rubi [A]  time = 0.102719, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {5033, 325, 211, 1165, 628, 1162, 617, 204} \[ -\frac{a+b \tan ^{-1}\left (c x^2\right )}{5 x^5}+\frac{b c^{5/2} \log \left (c x^2-\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2}}-\frac{b c^{5/2} \log \left (c x^2+\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2}}+\frac{b c^{5/2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2}}-\frac{b c^{5/2} \tan ^{-1}\left (\sqrt{2} \sqrt{c} x+1\right )}{5 \sqrt{2}}-\frac{2 b c}{15 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^2])/x^6,x]

[Out]

(-2*b*c)/(15*x^3) - (a + b*ArcTan[c*x^2])/(5*x^5) + (b*c^(5/2)*ArcTan[1 - Sqrt[2]*Sqrt[c]*x])/(5*Sqrt[2]) - (b
*c^(5/2)*ArcTan[1 + Sqrt[2]*Sqrt[c]*x])/(5*Sqrt[2]) + (b*c^(5/2)*Log[1 - Sqrt[2]*Sqrt[c]*x + c*x^2])/(10*Sqrt[
2]) - (b*c^(5/2)*Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2])/(10*Sqrt[2])

Rule 5033

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{a+b \tan ^{-1}\left (c x^2\right )}{x^6} \, dx &=-\frac{a+b \tan ^{-1}\left (c x^2\right )}{5 x^5}+\frac{1}{5} (2 b c) \int \frac{1}{x^4 \left (1+c^2 x^4\right )} \, dx\\ &=-\frac{2 b c}{15 x^3}-\frac{a+b \tan ^{-1}\left (c x^2\right )}{5 x^5}-\frac{1}{5} \left (2 b c^3\right ) \int \frac{1}{1+c^2 x^4} \, dx\\ &=-\frac{2 b c}{15 x^3}-\frac{a+b \tan ^{-1}\left (c x^2\right )}{5 x^5}-\frac{1}{5} \left (b c^3\right ) \int \frac{1-c x^2}{1+c^2 x^4} \, dx-\frac{1}{5} \left (b c^3\right ) \int \frac{1+c x^2}{1+c^2 x^4} \, dx\\ &=-\frac{2 b c}{15 x^3}-\frac{a+b \tan ^{-1}\left (c x^2\right )}{5 x^5}-\frac{1}{10} \left (b c^2\right ) \int \frac{1}{\frac{1}{c}-\frac{\sqrt{2} x}{\sqrt{c}}+x^2} \, dx-\frac{1}{10} \left (b c^2\right ) \int \frac{1}{\frac{1}{c}+\frac{\sqrt{2} x}{\sqrt{c}}+x^2} \, dx+\frac{\left (b c^{5/2}\right ) \int \frac{\frac{\sqrt{2}}{\sqrt{c}}+2 x}{-\frac{1}{c}-\frac{\sqrt{2} x}{\sqrt{c}}-x^2} \, dx}{10 \sqrt{2}}+\frac{\left (b c^{5/2}\right ) \int \frac{\frac{\sqrt{2}}{\sqrt{c}}-2 x}{-\frac{1}{c}+\frac{\sqrt{2} x}{\sqrt{c}}-x^2} \, dx}{10 \sqrt{2}}\\ &=-\frac{2 b c}{15 x^3}-\frac{a+b \tan ^{-1}\left (c x^2\right )}{5 x^5}+\frac{b c^{5/2} \log \left (1-\sqrt{2} \sqrt{c} x+c x^2\right )}{10 \sqrt{2}}-\frac{b c^{5/2} \log \left (1+\sqrt{2} \sqrt{c} x+c x^2\right )}{10 \sqrt{2}}-\frac{\left (b c^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2}}+\frac{\left (b c^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2}}\\ &=-\frac{2 b c}{15 x^3}-\frac{a+b \tan ^{-1}\left (c x^2\right )}{5 x^5}+\frac{b c^{5/2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2}}-\frac{b c^{5/2} \tan ^{-1}\left (1+\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2}}+\frac{b c^{5/2} \log \left (1-\sqrt{2} \sqrt{c} x+c x^2\right )}{10 \sqrt{2}}-\frac{b c^{5/2} \log \left (1+\sqrt{2} \sqrt{c} x+c x^2\right )}{10 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0501376, size = 177, normalized size = 1.11 \[ -\frac{a}{5 x^5}+\frac{b c^{5/2} \log \left (c x^2-\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2}}-\frac{b c^{5/2} \log \left (c x^2+\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2}}-\frac{b c^{5/2} \tan ^{-1}\left (\frac{2 \sqrt{c} x-\sqrt{2}}{\sqrt{2}}\right )}{5 \sqrt{2}}-\frac{b c^{5/2} \tan ^{-1}\left (\frac{2 \sqrt{c} x+\sqrt{2}}{\sqrt{2}}\right )}{5 \sqrt{2}}-\frac{2 b c}{15 x^3}-\frac{b \tan ^{-1}\left (c x^2\right )}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^2])/x^6,x]

[Out]

-a/(5*x^5) - (2*b*c)/(15*x^3) - (b*ArcTan[c*x^2])/(5*x^5) - (b*c^(5/2)*ArcTan[(-Sqrt[2] + 2*Sqrt[c]*x)/Sqrt[2]
])/(5*Sqrt[2]) - (b*c^(5/2)*ArcTan[(Sqrt[2] + 2*Sqrt[c]*x)/Sqrt[2]])/(5*Sqrt[2]) + (b*c^(5/2)*Log[1 - Sqrt[2]*
Sqrt[c]*x + c*x^2])/(10*Sqrt[2]) - (b*c^(5/2)*Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2])/(10*Sqrt[2])

________________________________________________________________________________________

Maple [A]  time = 0.03, size = 138, normalized size = 0.9 \begin{align*} -{\frac{a}{5\,{x}^{5}}}-{\frac{b\arctan \left ( c{x}^{2} \right ) }{5\,{x}^{5}}}-{\frac{b{c}^{3}\sqrt{2}}{10}\sqrt [4]{{c}^{-2}}\arctan \left ({\sqrt{2}x{\frac{1}{\sqrt [4]{{c}^{-2}}}}}-1 \right ) }-{\frac{b{c}^{3}\sqrt{2}}{20}\sqrt [4]{{c}^{-2}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{c}^{-2}}x\sqrt{2}+\sqrt{{c}^{-2}} \right ) \left ({x}^{2}-\sqrt [4]{{c}^{-2}}x\sqrt{2}+\sqrt{{c}^{-2}} \right ) ^{-1}} \right ) }-{\frac{b{c}^{3}\sqrt{2}}{10}\sqrt [4]{{c}^{-2}}\arctan \left ({\sqrt{2}x{\frac{1}{\sqrt [4]{{c}^{-2}}}}}+1 \right ) }-{\frac{2\,bc}{15\,{x}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^2))/x^6,x)

[Out]

-1/5*a/x^5-1/5*b/x^5*arctan(c*x^2)-1/10*b*c^3*(1/c^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c^2)^(1/4)*x-1)-1/20*b*c
^3*(1/c^2)^(1/4)*2^(1/2)*ln((x^2+(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(1/2))/(x^2-(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(
1/2)))-1/10*b*c^3*(1/c^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c^2)^(1/4)*x+1)-2/15*b*c/x^3

________________________________________________________________________________________

Maxima [B]  time = 1.51966, size = 366, normalized size = 2.3 \begin{align*} -\frac{1}{60} \,{\left ({\left (\frac{3 \, \sqrt{2} c^{2} \log \left (\sqrt{c^{2}} x^{2} + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} x + 1\right )}{{\left (c^{2}\right )}^{\frac{1}{4}}} - \frac{3 \, \sqrt{2} c^{2} \log \left (\sqrt{c^{2}} x^{2} - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} x + 1\right )}{{\left (c^{2}\right )}^{\frac{1}{4}}} + \frac{3 \, \sqrt{2} c^{2} \log \left (\frac{2 \, \sqrt{c^{2}} x - \sqrt{2} \sqrt{-\sqrt{c^{2}}} + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}{2 \, \sqrt{c^{2}} x + \sqrt{2} \sqrt{-\sqrt{c^{2}}} + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}\right )}{\sqrt{-\sqrt{c^{2}}}} + \frac{3 \, \sqrt{2} c^{2} \log \left (\frac{2 \, \sqrt{c^{2}} x - \sqrt{2} \sqrt{-\sqrt{c^{2}}} - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}{2 \, \sqrt{c^{2}} x + \sqrt{2} \sqrt{-\sqrt{c^{2}}} - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}\right )}{\sqrt{-\sqrt{c^{2}}}} + \frac{8}{x^{3}}\right )} c + \frac{12 \, \arctan \left (c x^{2}\right )}{x^{5}}\right )} b - \frac{a}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))/x^6,x, algorithm="maxima")

[Out]

-1/60*((3*sqrt(2)*c^2*log(sqrt(c^2)*x^2 + sqrt(2)*(c^2)^(1/4)*x + 1)/(c^2)^(1/4) - 3*sqrt(2)*c^2*log(sqrt(c^2)
*x^2 - sqrt(2)*(c^2)^(1/4)*x + 1)/(c^2)^(1/4) + 3*sqrt(2)*c^2*log((2*sqrt(c^2)*x - sqrt(2)*sqrt(-sqrt(c^2)) +
sqrt(2)*(c^2)^(1/4))/(2*sqrt(c^2)*x + sqrt(2)*sqrt(-sqrt(c^2)) + sqrt(2)*(c^2)^(1/4)))/sqrt(-sqrt(c^2)) + 3*sq
rt(2)*c^2*log((2*sqrt(c^2)*x - sqrt(2)*sqrt(-sqrt(c^2)) - sqrt(2)*(c^2)^(1/4))/(2*sqrt(c^2)*x + sqrt(2)*sqrt(-
sqrt(c^2)) - sqrt(2)*(c^2)^(1/4)))/sqrt(-sqrt(c^2)) + 8/x^3)*c + 12*arctan(c*x^2)/x^5)*b - 1/5*a/x^5

________________________________________________________________________________________

Fricas [B]  time = 2.83075, size = 873, normalized size = 5.49 \begin{align*} \frac{12 \, \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{1}{4}} x^{5} \arctan \left (-\frac{b^{4} c^{10} + \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{3}{4}} b c^{3} x - \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{3}{4}} \sqrt{b^{2} c^{6} x^{2} + \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{1}{4}} b c^{3} x + \sqrt{b^{4} c^{10}}}}{b^{4} c^{10}}\right ) + 12 \, \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{1}{4}} x^{5} \arctan \left (\frac{b^{4} c^{10} - \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{3}{4}} b c^{3} x + \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{3}{4}} \sqrt{b^{2} c^{6} x^{2} - \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{1}{4}} b c^{3} x + \sqrt{b^{4} c^{10}}}}{b^{4} c^{10}}\right ) - 3 \, \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{1}{4}} x^{5} \log \left (b^{2} c^{6} x^{2} + \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{1}{4}} b c^{3} x + \sqrt{b^{4} c^{10}}\right ) + 3 \, \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{1}{4}} x^{5} \log \left (b^{2} c^{6} x^{2} - \sqrt{2} \left (b^{4} c^{10}\right )^{\frac{1}{4}} b c^{3} x + \sqrt{b^{4} c^{10}}\right ) - 8 \, b c x^{2} - 12 \, b \arctan \left (c x^{2}\right ) - 12 \, a}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))/x^6,x, algorithm="fricas")

[Out]

1/60*(12*sqrt(2)*(b^4*c^10)^(1/4)*x^5*arctan(-(b^4*c^10 + sqrt(2)*(b^4*c^10)^(3/4)*b*c^3*x - sqrt(2)*(b^4*c^10
)^(3/4)*sqrt(b^2*c^6*x^2 + sqrt(2)*(b^4*c^10)^(1/4)*b*c^3*x + sqrt(b^4*c^10)))/(b^4*c^10)) + 12*sqrt(2)*(b^4*c
^10)^(1/4)*x^5*arctan((b^4*c^10 - sqrt(2)*(b^4*c^10)^(3/4)*b*c^3*x + sqrt(2)*(b^4*c^10)^(3/4)*sqrt(b^2*c^6*x^2
- sqrt(2)*(b^4*c^10)^(1/4)*b*c^3*x + sqrt(b^4*c^10)))/(b^4*c^10)) - 3*sqrt(2)*(b^4*c^10)^(1/4)*x^5*log(b^2*c^
6*x^2 + sqrt(2)*(b^4*c^10)^(1/4)*b*c^3*x + sqrt(b^4*c^10)) + 3*sqrt(2)*(b^4*c^10)^(1/4)*x^5*log(b^2*c^6*x^2 -
sqrt(2)*(b^4*c^10)^(1/4)*b*c^3*x + sqrt(b^4*c^10)) - 8*b*c*x^2 - 12*b*arctan(c*x^2) - 12*a)/x^5

________________________________________________________________________________________

Sympy [A]  time = 107.518, size = 1265, normalized size = 7.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**2))/x**6,x)

[Out]

Piecewise((-(a - b*atan((-sqrt(2)/2 - sqrt(2)*I/2)**(-2)))/(5*x**5), Eq(c, -1/(x**2*(-sqrt(2)/2 - sqrt(2)*I/2)
**2))), (-(a - b*atan((-sqrt(2)/2 + sqrt(2)*I/2)**(-2)))/(5*x**5), Eq(c, -1/(x**2*(-sqrt(2)/2 + sqrt(2)*I/2)**
2))), (-(a - b*atan((sqrt(2)/2 - sqrt(2)*I/2)**(-2)))/(5*x**5), Eq(c, -1/(x**2*(sqrt(2)/2 - sqrt(2)*I/2)**2)))
, (-(a - b*atan((sqrt(2)/2 + sqrt(2)*I/2)**(-2)))/(5*x**5), Eq(c, -1/(x**2*(sqrt(2)/2 + sqrt(2)*I/2)**2))), (-
a/(5*x**5), Eq(c, 0)), (6*(-1)**(3/4)*a*c**15*x**4*(c**(-2))**(39/4)/(-30*(-1)**(3/4)*c**15*x**9*(c**(-2))**(3
9/4) - 30*(-1)**(3/4)*c**13*x**5*(c**(-2))**(39/4)) + 6*(-1)**(3/4)*a*c**13*(c**(-2))**(39/4)/(-30*(-1)**(3/4)
*c**15*x**9*(c**(-2))**(39/4) - 30*(-1)**(3/4)*c**13*x**5*(c**(-2))**(39/4)) + 4*(-1)**(3/4)*b*c**16*x**6*(c**
(-2))**(39/4)/(-30*(-1)**(3/4)*c**15*x**9*(c**(-2))**(39/4) - 30*(-1)**(3/4)*c**13*x**5*(c**(-2))**(39/4)) + 6
*(-1)**(3/4)*b*c**15*x**4*(c**(-2))**(39/4)*atan(c*x**2)/(-30*(-1)**(3/4)*c**15*x**9*(c**(-2))**(39/4) - 30*(-
1)**(3/4)*c**13*x**5*(c**(-2))**(39/4)) + 4*(-1)**(3/4)*b*c**14*x**2*(c**(-2))**(39/4)/(-30*(-1)**(3/4)*c**15*
x**9*(c**(-2))**(39/4) - 30*(-1)**(3/4)*c**13*x**5*(c**(-2))**(39/4)) + 6*(-1)**(3/4)*b*c**13*(c**(-2))**(39/4
)*atan(c*x**2)/(-30*(-1)**(3/4)*c**15*x**9*(c**(-2))**(39/4) - 30*(-1)**(3/4)*c**13*x**5*(c**(-2))**(39/4)) -
6*I*b*c**11*x**9*(c**(-2))**(13/2)*atan(c*x**2)/(-30*(-1)**(3/4)*c**15*x**9*(c**(-2))**(39/4) - 30*(-1)**(3/4)
*c**13*x**5*(c**(-2))**(39/4)) - 6*I*b*c**9*x**5*(c**(-2))**(13/2)*atan(c*x**2)/(-30*(-1)**(3/4)*c**15*x**9*(c
**(-2))**(39/4) - 30*(-1)**(3/4)*c**13*x**5*(c**(-2))**(39/4)) + 6*b*x**9*log(x - (-1)**(1/4)*(c**(-2))**(1/4)
)/(-30*(-1)**(3/4)*c**17*x**9*(c**(-2))**(39/4) - 30*(-1)**(3/4)*c**15*x**5*(c**(-2))**(39/4)) - 3*b*x**9*log(
x**2 + I*sqrt(c**(-2)))/(-30*(-1)**(3/4)*c**17*x**9*(c**(-2))**(39/4) - 30*(-1)**(3/4)*c**15*x**5*(c**(-2))**(
39/4)) + 6*b*x**9*atan((-1)**(3/4)*x/(c**(-2))**(1/4))/(-30*(-1)**(3/4)*c**17*x**9*(c**(-2))**(39/4) - 30*(-1)
**(3/4)*c**15*x**5*(c**(-2))**(39/4)) + 6*b*x**5*log(x - (-1)**(1/4)*(c**(-2))**(1/4))/(-30*(-1)**(3/4)*c**19*
x**9*(c**(-2))**(39/4) - 30*(-1)**(3/4)*c**17*x**5*(c**(-2))**(39/4)) - 3*b*x**5*log(x**2 + I*sqrt(c**(-2)))/(
-30*(-1)**(3/4)*c**19*x**9*(c**(-2))**(39/4) - 30*(-1)**(3/4)*c**17*x**5*(c**(-2))**(39/4)) + 6*b*x**5*atan((-
1)**(3/4)*x/(c**(-2))**(1/4))/(-30*(-1)**(3/4)*c**19*x**9*(c**(-2))**(39/4) - 30*(-1)**(3/4)*c**17*x**5*(c**(-
2))**(39/4)), True))

________________________________________________________________________________________

Giac [A]  time = 1.36396, size = 203, normalized size = 1.28 \begin{align*} -\frac{1}{20} \, b c^{3}{\left (\frac{2 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x + \frac{\sqrt{2}}{\sqrt{{\left | c \right |}}}\right )} \sqrt{{\left | c \right |}}\right )}{\sqrt{{\left | c \right |}}} + \frac{2 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x - \frac{\sqrt{2}}{\sqrt{{\left | c \right |}}}\right )} \sqrt{{\left | c \right |}}\right )}{\sqrt{{\left | c \right |}}} + \frac{\sqrt{2} \log \left (x^{2} + \frac{\sqrt{2} x}{\sqrt{{\left | c \right |}}} + \frac{1}{{\left | c \right |}}\right )}{\sqrt{{\left | c \right |}}} - \frac{\sqrt{2} \log \left (x^{2} - \frac{\sqrt{2} x}{\sqrt{{\left | c \right |}}} + \frac{1}{{\left | c \right |}}\right )}{\sqrt{{\left | c \right |}}}\right )} - \frac{2 \, b c x^{2} + 3 \, b \arctan \left (c x^{2}\right ) + 3 \, a}{15 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))/x^6,x, algorithm="giac")

[Out]

-1/20*b*c^3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/sqrt(abs(c)) + 2*sqrt(2)*
arctan(1/2*sqrt(2)*(2*x - sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/sqrt(abs(c)) + sqrt(2)*log(x^2 + sqrt(2)*x/sqrt(
abs(c)) + 1/abs(c))/sqrt(abs(c)) - sqrt(2)*log(x^2 - sqrt(2)*x/sqrt(abs(c)) + 1/abs(c))/sqrt(abs(c))) - 1/15*(
2*b*c*x^2 + 3*b*arctan(c*x^2) + 3*a)/x^5