3.80 \(\int x^4 \cot ^{-1}(a x^2) \, dx\)

Optimal. Leaf size=152 \[ -\frac {\log \left (a x^2-\sqrt {2} \sqrt {a} x+1\right )}{10 \sqrt {2} a^{5/2}}+\frac {\log \left (a x^2+\sqrt {2} \sqrt {a} x+1\right )}{10 \sqrt {2} a^{5/2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {a} x\right )}{5 \sqrt {2} a^{5/2}}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {a} x+1\right )}{5 \sqrt {2} a^{5/2}}+\frac {2 x^3}{15 a}+\frac {1}{5} x^5 \cot ^{-1}\left (a x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5034, 321, 297, 1162, 617, 204, 1165, 628} \[ -\frac {\log \left (a x^2-\sqrt {2} \sqrt {a} x+1\right )}{10 \sqrt {2} a^{5/2}}+\frac {\log \left (a x^2+\sqrt {2} \sqrt {a} x+1\right )}{10 \sqrt {2} a^{5/2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {a} x\right )}{5 \sqrt {2} a^{5/2}}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {a} x+1\right )}{5 \sqrt {2} a^{5/2}}+\frac {2 x^3}{15 a}+\frac {1}{5} x^5 \cot ^{-1}\left (a x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcCot[a*x^2],x]

[Out]

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

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])

Rule 297

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

Rule 321

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

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 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 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 5034

Int[((a_.) + ArcCot[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCot
[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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 136, normalized size = 0.89 \[ \frac {8 a^{3/2} x^3+12 a^{5/2} x^5 \cot ^{-1}\left (a x^2\right )-3 \sqrt {2} \log \left (a x^2-\sqrt {2} \sqrt {a} x+1\right )+3 \sqrt {2} \log \left (a x^2+\sqrt {2} \sqrt {a} x+1\right )+6 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {a} x\right )-6 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {a} x+1\right )}{60 a^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcCot[a*x^2],x]

[Out]

(8*a^(3/2)*x^3 + 12*a^(5/2)*x^5*ArcCot[a*x^2] + 6*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[a]*x] - 6*Sqrt[2]*ArcTan[1 +
Sqrt[2]*Sqrt[a]*x] - 3*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[a]*x + a*x^2] + 3*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[a]*x + a*x
^2])/(60*a^(5/2))

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fricas [B]  time = 0.56, size = 239, normalized size = 1.57 \[ \frac {12 \, a x^{5} \arctan \left (\frac {1}{a x^{2}}\right ) + 8 \, x^{3} + 12 \, \sqrt {2} a \frac {1}{a^{10}}^{\frac {1}{4}} \arctan \left (-\sqrt {2} a^{3} \frac {1}{a^{10}}^{\frac {1}{4}} x + \sqrt {2} \sqrt {\sqrt {2} a^{7} \frac {1}{a^{10}}^{\frac {3}{4}} x + a^{4} \sqrt {\frac {1}{a^{10}}} + x^{2}} a^{3} \frac {1}{a^{10}}^{\frac {1}{4}} - 1\right ) + 12 \, \sqrt {2} a \frac {1}{a^{10}}^{\frac {1}{4}} \arctan \left (-\sqrt {2} a^{3} \frac {1}{a^{10}}^{\frac {1}{4}} x + \sqrt {2} \sqrt {-\sqrt {2} a^{7} \frac {1}{a^{10}}^{\frac {3}{4}} x + a^{4} \sqrt {\frac {1}{a^{10}}} + x^{2}} a^{3} \frac {1}{a^{10}}^{\frac {1}{4}} + 1\right ) + 3 \, \sqrt {2} a \frac {1}{a^{10}}^{\frac {1}{4}} \log \left (\sqrt {2} a^{7} \frac {1}{a^{10}}^{\frac {3}{4}} x + a^{4} \sqrt {\frac {1}{a^{10}}} + x^{2}\right ) - 3 \, \sqrt {2} a \frac {1}{a^{10}}^{\frac {1}{4}} \log \left (-\sqrt {2} a^{7} \frac {1}{a^{10}}^{\frac {3}{4}} x + a^{4} \sqrt {\frac {1}{a^{10}}} + x^{2}\right )}{60 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arccot(a*x^2),x, algorithm="fricas")

[Out]

1/60*(12*a*x^5*arctan(1/(a*x^2)) + 8*x^3 + 12*sqrt(2)*a*(a^(-10))^(1/4)*arctan(-sqrt(2)*a^3*(a^(-10))^(1/4)*x
+ sqrt(2)*sqrt(sqrt(2)*a^7*(a^(-10))^(3/4)*x + a^4*sqrt(a^(-10)) + x^2)*a^3*(a^(-10))^(1/4) - 1) + 12*sqrt(2)*
a*(a^(-10))^(1/4)*arctan(-sqrt(2)*a^3*(a^(-10))^(1/4)*x + sqrt(2)*sqrt(-sqrt(2)*a^7*(a^(-10))^(3/4)*x + a^4*sq
rt(a^(-10)) + x^2)*a^3*(a^(-10))^(1/4) + 1) + 3*sqrt(2)*a*(a^(-10))^(1/4)*log(sqrt(2)*a^7*(a^(-10))^(3/4)*x +
a^4*sqrt(a^(-10)) + x^2) - 3*sqrt(2)*a*(a^(-10))^(1/4)*log(-sqrt(2)*a^7*(a^(-10))^(3/4)*x + a^4*sqrt(a^(-10))
+ x^2))/a

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giac [A]  time = 0.15, size = 156, normalized size = 1.03 \[ \frac {1}{5} \, x^{5} \arctan \left (\frac {1}{a x^{2}}\right ) + \frac {1}{60} \, a {\left (\frac {8 \, x^{3}}{a^{2}} - \frac {6 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | a \right |}}}\right )} \sqrt {{\left | a \right |}}\right )}{a^{2} {\left | a \right |}^{\frac {3}{2}}} - \frac {6 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | a \right |}}}\right )} \sqrt {{\left | a \right |}}\right )}{a^{2} {\left | a \right |}^{\frac {3}{2}}} + \frac {3 \, \sqrt {2} \sqrt {{\left | a \right |}} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | a \right |}}} + \frac {1}{{\left | a \right |}}\right )}{a^{4}} - \frac {3 \, \sqrt {2} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | a \right |}}} + \frac {1}{{\left | a \right |}}\right )}{a^{2} {\left | a \right |}^{\frac {3}{2}}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arccot(a*x^2),x, algorithm="giac")

[Out]

1/5*x^5*arctan(1/(a*x^2)) + 1/60*a*(8*x^3/a^2 - 6*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)/sqrt(abs(a)))*sqrt
(abs(a)))/(a^2*abs(a)^(3/2)) - 6*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)/sqrt(abs(a)))*sqrt(abs(a)))/(a^2*ab
s(a)^(3/2)) + 3*sqrt(2)*sqrt(abs(a))*log(x^2 + sqrt(2)*x/sqrt(abs(a)) + 1/abs(a))/a^4 - 3*sqrt(2)*log(x^2 - sq
rt(2)*x/sqrt(abs(a)) + 1/abs(a))/(a^2*abs(a)^(3/2)))

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maple [A]  time = 0.06, size = 129, normalized size = 0.85 \[ \frac {x^{5} \mathrm {arccot}\left (a \,x^{2}\right )}{5}+\frac {2 x^{3}}{15 a}-\frac {\sqrt {2}\, \ln \left (\frac {x^{2}-\left (\frac {1}{a^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{a^{2}}}}{x^{2}+\left (\frac {1}{a^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{a^{2}}}}\right )}{20 a^{3} \left (\frac {1}{a^{2}}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{a^{2}}\right )^{\frac {1}{4}}}+1\right )}{10 a^{3} \left (\frac {1}{a^{2}}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{a^{2}}\right )^{\frac {1}{4}}}-1\right )}{10 a^{3} \left (\frac {1}{a^{2}}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arccot(a*x^2),x)

[Out]

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

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maxima [A]  time = 0.41, size = 137, normalized size = 0.90 \[ \frac {1}{5} \, x^{5} \operatorname {arccot}\left (a x^{2}\right ) + \frac {1}{60} \, a {\left (\frac {8 \, x^{3}}{a^{2}} - \frac {3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, a x + \sqrt {2} \sqrt {a}\right )}}{2 \, \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, a x - \sqrt {2} \sqrt {a}\right )}}{2 \, \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (a x^{2} + \sqrt {2} \sqrt {a} x + 1\right )}{a^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (a x^{2} - \sqrt {2} \sqrt {a} x + 1\right )}{a^{\frac {3}{2}}}\right )}}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arccot(a*x^2),x, algorithm="maxima")

[Out]

1/5*x^5*arccot(a*x^2) + 1/60*a*(8*x^3/a^2 - 3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*a*x + sqrt(2)*sqrt(a))/sqrt(a))
/a^(3/2) + 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*a*x - sqrt(2)*sqrt(a))/sqrt(a))/a^(3/2) - sqrt(2)*log(a*x^2 + sqrt(
2)*sqrt(a)*x + 1)/a^(3/2) + sqrt(2)*log(a*x^2 - sqrt(2)*sqrt(a)*x + 1)/a^(3/2))/a^2)

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mupad [B]  time = 0.44, size = 54, normalized size = 0.36 \[ \frac {x^5\,\mathrm {acot}\left (a\,x^2\right )}{5}+\frac {2\,x^3}{15\,a}-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {a}\,x\right )}{5\,a^{5/2}}-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {a}\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{5\,a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*acot(a*x^2),x)

[Out]

(x^5*acot(a*x^2))/5 + (2*x^3)/(15*a) - ((-1)^(1/4)*atan((-1)^(1/4)*a^(1/2)*x))/(5*a^(5/2)) - ((-1)^(1/4)*atan(
(-1)^(1/4)*a^(1/2)*x*1i)*1i)/(5*a^(5/2))

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sympy [A]  time = 30.12, size = 167, normalized size = 1.10 \[ \begin {cases} \frac {x^{5} \operatorname {acot}{\left (a x^{2} \right )}}{5} + \frac {2 x^{3}}{15 a} - \frac {\sqrt [4]{-1} \sqrt [4]{\frac {1}{a^{2}}} \operatorname {acot}{\left (a x^{2} \right )}}{5 a^{2}} + \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (x - \sqrt [4]{-1} \sqrt [4]{\frac {1}{a^{2}}} \right )}}{5 a^{3} \sqrt [4]{\frac {1}{a^{2}}}} - \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (x^{2} + i \sqrt {\frac {1}{a^{2}}} \right )}}{10 a^{3} \sqrt [4]{\frac {1}{a^{2}}}} - \frac {\left (-1\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} x}{\sqrt [4]{\frac {1}{a^{2}}}} \right )}}{5 a^{3} \sqrt [4]{\frac {1}{a^{2}}}} & \text {for}\: a \neq 0 \\\frac {\pi x^{5}}{10} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*acot(a*x**2),x)

[Out]

Piecewise((x**5*acot(a*x**2)/5 + 2*x**3/(15*a) - (-1)**(1/4)*(a**(-2))**(1/4)*acot(a*x**2)/(5*a**2) + (-1)**(3
/4)*log(x - (-1)**(1/4)*(a**(-2))**(1/4))/(5*a**3*(a**(-2))**(1/4)) - (-1)**(3/4)*log(x**2 + I*sqrt(a**(-2)))/
(10*a**3*(a**(-2))**(1/4)) - (-1)**(3/4)*atan((-1)**(3/4)*x/(a**(-2))**(1/4))/(5*a**3*(a**(-2))**(1/4)), Ne(a,
0)), (pi*x**5/10, True))

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