3.69 \(\int \frac {\cot ^{-1}(x)}{(a+a x^2)^{7/2}} \, dx\)

Optimal. Leaf size=118 \[ -\frac {8}{15 a^3 \sqrt {a x^2+a}}+\frac {8 x \cot ^{-1}(x)}{15 a^3 \sqrt {a x^2+a}}-\frac {4}{45 a^2 \left (a x^2+a\right )^{3/2}}+\frac {4 x \cot ^{-1}(x)}{15 a^2 \left (a x^2+a\right )^{3/2}}-\frac {1}{25 a \left (a x^2+a\right )^{5/2}}+\frac {x \cot ^{-1}(x)}{5 a \left (a x^2+a\right )^{5/2}} \]

[Out]

-1/25/a/(a*x^2+a)^(5/2)-4/45/a^2/(a*x^2+a)^(3/2)+1/5*x*arccot(x)/a/(a*x^2+a)^(5/2)+4/15*x*arccot(x)/a^2/(a*x^2
+a)^(3/2)-8/15/a^3/(a*x^2+a)^(1/2)+8/15*x*arccot(x)/a^3/(a*x^2+a)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4897, 4895} \[ -\frac {8}{15 a^3 \sqrt {a x^2+a}}-\frac {4}{45 a^2 \left (a x^2+a\right )^{3/2}}+\frac {8 x \cot ^{-1}(x)}{15 a^3 \sqrt {a x^2+a}}+\frac {4 x \cot ^{-1}(x)}{15 a^2 \left (a x^2+a\right )^{3/2}}-\frac {1}{25 a \left (a x^2+a\right )^{5/2}}+\frac {x \cot ^{-1}(x)}{5 a \left (a x^2+a\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[x]/(a + a*x^2)^(7/2),x]

[Out]

-1/(25*a*(a + a*x^2)^(5/2)) - 4/(45*a^2*(a + a*x^2)^(3/2)) - 8/(15*a^3*Sqrt[a + a*x^2]) + (x*ArcCot[x])/(5*a*(
a + a*x^2)^(5/2)) + (4*x*ArcCot[x])/(15*a^2*(a + a*x^2)^(3/2)) + (8*x*ArcCot[x])/(15*a^3*Sqrt[a + a*x^2])

Rule 4895

Int[((a_.) + ArcCot[(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*ArcCot[c*x]))/(d*Sqrt[d + e*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]

Rule 4897

Int[((a_.) + ArcCot[(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*ArcCot[c*x]), x], x] - S
imp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcCot[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]

Rubi steps

\begin {align*} \int \frac {\cot ^{-1}(x)}{\left (a+a x^2\right )^{7/2}} \, dx &=-\frac {1}{25 a \left (a+a x^2\right )^{5/2}}+\frac {x \cot ^{-1}(x)}{5 a \left (a+a x^2\right )^{5/2}}+\frac {4 \int \frac {\cot ^{-1}(x)}{\left (a+a x^2\right )^{5/2}} \, dx}{5 a}\\ &=-\frac {1}{25 a \left (a+a x^2\right )^{5/2}}-\frac {4}{45 a^2 \left (a+a x^2\right )^{3/2}}+\frac {x \cot ^{-1}(x)}{5 a \left (a+a x^2\right )^{5/2}}+\frac {4 x \cot ^{-1}(x)}{15 a^2 \left (a+a x^2\right )^{3/2}}+\frac {8 \int \frac {\cot ^{-1}(x)}{\left (a+a x^2\right )^{3/2}} \, dx}{15 a^2}\\ &=-\frac {1}{25 a \left (a+a x^2\right )^{5/2}}-\frac {4}{45 a^2 \left (a+a x^2\right )^{3/2}}-\frac {8}{15 a^3 \sqrt {a+a x^2}}+\frac {x \cot ^{-1}(x)}{5 a \left (a+a x^2\right )^{5/2}}+\frac {4 x \cot ^{-1}(x)}{15 a^2 \left (a+a x^2\right )^{3/2}}+\frac {8 x \cot ^{-1}(x)}{15 a^3 \sqrt {a+a x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 47, normalized size = 0.40 \[ \frac {-120 x^4-260 x^2+15 \left (8 x^4+20 x^2+15\right ) x \cot ^{-1}(x)-149}{225 a \left (a \left (x^2+1\right )\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[x]/(a + a*x^2)^(7/2),x]

[Out]

(-149 - 260*x^2 - 120*x^4 + 15*x*(15 + 20*x^2 + 8*x^4)*ArcCot[x])/(225*a*(a*(1 + x^2))^(5/2))

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fricas [A]  time = 0.62, size = 70, normalized size = 0.59 \[ -\frac {{\left (120 \, x^{4} + 260 \, x^{2} - 15 \, {\left (8 \, x^{5} + 20 \, x^{3} + 15 \, x\right )} \operatorname {arccot}\relax (x) + 149\right )} \sqrt {a x^{2} + a}}{225 \, {\left (a^{4} x^{6} + 3 \, a^{4} x^{4} + 3 \, a^{4} x^{2} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/(a*x^2+a)^(7/2),x, algorithm="fricas")

[Out]

-1/225*(120*x^4 + 260*x^2 - 15*(8*x^5 + 20*x^3 + 15*x)*arccot(x) + 149)*sqrt(a*x^2 + a)/(a^4*x^6 + 3*a^4*x^4 +
3*a^4*x^2 + a^4)

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giac [A]  time = 0.17, size = 83, normalized size = 0.70 \[ \frac {{\left (4 \, x^{2} {\left (\frac {2 \, x^{2}}{a} + \frac {5}{a}\right )} + \frac {15}{a}\right )} x \arctan \left (\frac {1}{x}\right )}{15 \, {\left (a x^{2} + a\right )}^{\frac {5}{2}}} - \frac {120 \, {\left (a x^{2} + a\right )}^{2} + 20 \, {\left (a x^{2} + a\right )} a + 9 \, a^{2}}{225 \, {\left (a x^{2} + a\right )}^{\frac {5}{2}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/(a*x^2+a)^(7/2),x, algorithm="giac")

[Out]

1/15*(4*x^2*(2*x^2/a + 5/a) + 15/a)*x*arctan(1/x)/(a*x^2 + a)^(5/2) - 1/225*(120*(a*x^2 + a)^2 + 20*(a*x^2 + a
)*a + 9*a^2)/((a*x^2 + a)^(5/2)*a^3)

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maple [C]  time = 0.85, size = 258, normalized size = 2.19 \[ \frac {\left (i+5 \,\mathrm {arccot}\relax (x )\right ) \left (x^{5}+5 i x^{4}-10 x^{3}-10 i x^{2}+5 x +i\right ) \sqrt {a \left (x +i\right ) \left (x -i\right )}}{800 \left (x^{2}+1\right )^{3} a^{4}}+\frac {5 \left (i+\mathrm {arccot}\relax (x )\right ) \left (x +i\right ) \sqrt {a \left (x +i\right ) \left (x -i\right )}}{16 \left (x^{2}+1\right ) a^{4}}+\frac {5 \sqrt {a \left (x +i\right ) \left (x -i\right )}\, \left (x -i\right ) \left (\mathrm {arccot}\relax (x )-i\right )}{16 \left (x^{2}+1\right ) a^{4}}-\frac {5 \left (-i+3 \,\mathrm {arccot}\relax (x )\right ) \sqrt {a \left (x +i\right ) \left (x -i\right )}\, \left (x^{3}-3 i x^{2}-3 x +i\right )}{288 \left (x^{4}+2 x^{2}+1\right ) a^{4}}-\frac {\sqrt {a \left (x +i\right ) \left (x -i\right )}\, \left (x -i\right ) \left (67 i+165 \,\mathrm {arccot}\relax (x )\right ) \cos \left (4 \,\mathrm {arccot}\relax (x )\right )}{3600 \left (x^{2}+1\right ) a^{4}}-\frac {\sqrt {a \left (x +i\right ) \left (x -i\right )}\, \left (i x +1\right ) \left (29 i+105 \,\mathrm {arccot}\relax (x )\right ) \sin \left (4 \,\mathrm {arccot}\relax (x )\right )}{1800 \left (x^{2}+1\right ) a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(x)/(a*x^2+a)^(7/2),x)

[Out]

1/800*(I+5*arccot(x))*(5*I*x^4+x^5-10*I*x^2-10*x^3+I+5*x)*(a*(x+I)*(x-I))^(1/2)/(x^2+1)^3/a^4+5/16*(I+arccot(x
))*(x+I)*(a*(x+I)*(x-I))^(1/2)/(x^2+1)/a^4+5/16*(a*(x+I)*(x-I))^(1/2)*(x-I)*(arccot(x)-I)/(x^2+1)/a^4-5/288*(-
I+3*arccot(x))*(a*(x+I)*(x-I))^(1/2)*(-3*x-3*I*x^2+x^3+I)/(x^4+2*x^2+1)/a^4-1/3600*(a*(x+I)*(x-I))^(1/2)*(x-I)
*(67*I+165*arccot(x))*cos(4*arccot(x))/(x^2+1)/a^4-1/1800*(a*(x+I)*(x-I))^(1/2)*(1+I*x)*(29*I+105*arccot(x))*s
in(4*arccot(x))/(x^2+1)/a^4

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maxima [A]  time = 0.45, size = 93, normalized size = 0.79 \[ \frac {1}{15} \, {\left (\frac {8 \, x}{\sqrt {a x^{2} + a} a^{3}} + \frac {4 \, x}{{\left (a x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {3 \, x}{{\left (a x^{2} + a\right )}^{\frac {5}{2}} a}\right )} \operatorname {arccot}\relax (x) - \frac {8}{15 \, \sqrt {a x^{2} + a} a^{3}} - \frac {4}{45 \, {\left (a x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {1}{25 \, {\left (a x^{2} + a\right )}^{\frac {5}{2}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/(a*x^2+a)^(7/2),x, algorithm="maxima")

[Out]

1/15*(8*x/(sqrt(a*x^2 + a)*a^3) + 4*x/((a*x^2 + a)^(3/2)*a^2) + 3*x/((a*x^2 + a)^(5/2)*a))*arccot(x) - 8/15/(s
qrt(a*x^2 + a)*a^3) - 4/45/((a*x^2 + a)^(3/2)*a^2) - 1/25/((a*x^2 + a)^(5/2)*a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acot}\relax (x)}{{\left (a\,x^2+a\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(x)/(a + a*x^2)^(7/2),x)

[Out]

int(acot(x)/(a + a*x^2)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acot}{\relax (x )}}{\left (a \left (x^{2} + 1\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(x)/(a*x**2+a)**(7/2),x)

[Out]

Integral(acot(x)/(a*(x**2 + 1))**(7/2), x)

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