Integrand size = 14, antiderivative size = 118 \[ \int \frac {\cot ^{-1}(x)}{\left (a+a x^2\right )^{7/2}} \, dx=-\frac {1}{25 a \left (a+a x^2\right )^{5/2}}-\frac {4}{45 a^2 \left (a+a x^2\right )^{3/2}}-\frac {8}{15 a^3 \sqrt {a+a x^2}}+\frac {x \cot ^{-1}(x)}{5 a \left (a+a x^2\right )^{5/2}}+\frac {4 x \cot ^{-1}(x)}{15 a^2 \left (a+a x^2\right )^{3/2}}+\frac {8 x \cot ^{-1}(x)}{15 a^3 \sqrt {a+a x^2}} \]
-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)+8/1 5*x*arccot(x)/a^3/(a*x^2+a)^(1/2)
Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.40 \[ \int \frac {\cot ^{-1}(x)}{\left (a+a x^2\right )^{7/2}} \, dx=\frac {-149-260 x^2-120 x^4+15 x \left (15+20 x^2+8 x^4\right ) \cot ^{-1}(x)}{225 a \left (a \left (1+x^2\right )\right )^{5/2}} \]
(-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))
Time = 0.39 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5432, 5432, 5430}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cot ^{-1}(x)}{\left (a x^2+a\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 5432 |
\(\displaystyle \frac {4 \int \frac {\cot ^{-1}(x)}{\left (a x^2+a\right )^{5/2}}dx}{5 a}-\frac {1}{25 a \left (a x^2+a\right )^{5/2}}+\frac {x \cot ^{-1}(x)}{5 a \left (a x^2+a\right )^{5/2}}\) |
\(\Big \downarrow \) 5432 |
\(\displaystyle \frac {4 \left (\frac {2 \int \frac {\cot ^{-1}(x)}{\left (a x^2+a\right )^{3/2}}dx}{3 a}-\frac {1}{9 a \left (a x^2+a\right )^{3/2}}+\frac {x \cot ^{-1}(x)}{3 a \left (a x^2+a\right )^{3/2}}\right )}{5 a}-\frac {1}{25 a \left (a x^2+a\right )^{5/2}}+\frac {x \cot ^{-1}(x)}{5 a \left (a x^2+a\right )^{5/2}}\) |
\(\Big \downarrow \) 5430 |
\(\displaystyle -\frac {1}{25 a \left (a x^2+a\right )^{5/2}}+\frac {x \cot ^{-1}(x)}{5 a \left (a x^2+a\right )^{5/2}}+\frac {4 \left (-\frac {1}{9 a \left (a x^2+a\right )^{3/2}}+\frac {x \cot ^{-1}(x)}{3 a \left (a x^2+a\right )^{3/2}}+\frac {2 \left (\frac {x \cot ^{-1}(x)}{a \sqrt {a x^2+a}}-\frac {1}{a \sqrt {a x^2+a}}\right )}{3 a}\right )}{5 a}\) |
-1/25*1/(a*(a + a*x^2)^(5/2)) + (x*ArcCot[x])/(5*a*(a + a*x^2)^(5/2)) + (4 *(-1/9*1/(a*(a + a*x^2)^(3/2)) + (x*ArcCot[x])/(3*a*(a + a*x^2)^(3/2)) + ( 2*(-(1/(a*Sqrt[a + a*x^2])) + (x*ArcCot[x])/(a*Sqrt[a + a*x^2])))/(3*a)))/ (5*a)
3.1.69.3.1 Defintions of rubi rules used
Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbo l] :> Simp[-b/(c*d*Sqrt[d + e*x^2]), x] + Simp[x*((a + b*ArcCot[c*x])/(d*Sq rt[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]
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] + (-Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcCot[c*x])/(2*d*(q + 1))), x] + Simp[(2*q + 3)/(2* d*(q + 1)) Int[(d + e*x^2)^(q + 1)*(a + b*ArcCot[c*x]), x], x]) /; FreeQ[ {a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -3/2]
Result contains complex when optimal does not.
Time = 1.41 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.10
method | result | size |
risch | \(\frac {i x \left (8 x^{4}+20 x^{2}+15\right ) \ln \left (i x +1\right )}{30 a^{3} \left (x^{2}+1\right )^{2} \sqrt {a \left (x^{2}+1\right )}}+\frac {-120 i x^{5} \ln \left (-i x +1\right )+120 \pi \,x^{5}-300 i x^{3} \ln \left (-i x +1\right )+300 \pi \,x^{3}-240 x^{4}-225 i x \ln \left (-i x +1\right )+225 \pi x -520 x^{2}-298}{450 a^{3} \left (x^{2}+1\right )^{2} \sqrt {a \left (x^{2}+1\right )}}\) | \(130\) |
default | \(\frac {\left (i+5 \,\operatorname {arccot}\left (x \right )\right ) \left (x^{5}+5 i x^{4}-10 x^{3}-10 i x^{2}+5 x +i\right ) \sqrt {a \left (i+x \right ) \left (x -i\right )}}{800 \left (x^{2}+1\right )^{3} a^{4}}+\frac {5 \left (\operatorname {arccot}\left (x \right )+i\right ) \left (i+x \right ) \sqrt {a \left (i+x \right ) \left (x -i\right )}}{16 \left (x^{2}+1\right ) a^{4}}+\frac {5 \sqrt {a \left (i+x \right ) \left (x -i\right )}\, \left (x -i\right ) \left (\operatorname {arccot}\left (x \right )-i\right )}{16 \left (x^{2}+1\right ) a^{4}}-\frac {5 \left (-i+3 \,\operatorname {arccot}\left (x \right )\right ) \sqrt {a \left (i+x \right ) \left (x -i\right )}\, \left (x^{3}-3 i x^{2}-3 x +i\right )}{288 \left (x^{4}+2 x^{2}+1\right ) a^{4}}-\frac {\sqrt {a \left (i+x \right ) \left (x -i\right )}\, \left (x -i\right ) \left (67 i+165 \,\operatorname {arccot}\left (x \right )\right ) \cos \left (4 \,\operatorname {arccot}\left (x \right )\right )}{3600 \left (x^{2}+1\right ) a^{4}}-\frac {\sqrt {a \left (i+x \right ) \left (x -i\right )}\, \left (i x +1\right ) \left (29 i+105 \,\operatorname {arccot}\left (x \right )\right ) \sin \left (4 \,\operatorname {arccot}\left (x \right )\right )}{1800 \left (x^{2}+1\right ) a^{4}}\) | \(258\) |
1/30*I/a^3*x*(8*x^4+20*x^2+15)/(x^2+1)^2/(a*(x^2+1))^(1/2)*ln(1+I*x)+1/450 /a^3*(-120*I*x^5*ln(1-I*x)+120*Pi*x^5-300*I*x^3*ln(1-I*x)+300*Pi*x^3-240*x ^4-225*I*x*ln(1-I*x)+225*Pi*x-520*x^2-298)/(x^2+1)^2/(a*(x^2+1))^(1/2)
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.59 \[ \int \frac {\cot ^{-1}(x)}{\left (a+a x^2\right )^{7/2}} \, dx=-\frac {{\left (120 \, x^{4} + 260 \, x^{2} - 15 \, {\left (8 \, x^{5} + 20 \, x^{3} + 15 \, x\right )} \operatorname {arccot}\left (x\right ) + 149\right )} \sqrt {a x^{2} + a}}{225 \, {\left (a^{4} x^{6} + 3 \, a^{4} x^{4} + 3 \, a^{4} x^{2} + a^{4}\right )}} \]
-1/225*(120*x^4 + 260*x^2 - 15*(8*x^5 + 20*x^3 + 15*x)*arccot(x) + 149)*sq rt(a*x^2 + a)/(a^4*x^6 + 3*a^4*x^4 + 3*a^4*x^2 + a^4)
\[ \int \frac {\cot ^{-1}(x)}{\left (a+a x^2\right )^{7/2}} \, dx=\int \frac {\operatorname {acot}{\left (x \right )}}{\left (a \left (x^{2} + 1\right )\right )^{\frac {7}{2}}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79 \[ \int \frac {\cot ^{-1}(x)}{\left (a+a x^2\right )^{7/2}} \, dx=\frac {1}{15} \, {\left (\frac {8 \, x}{\sqrt {a x^{2} + a} a^{3}} + \frac {4 \, x}{{\left (a x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {3 \, x}{{\left (a x^{2} + a\right )}^{\frac {5}{2}} a}\right )} \operatorname {arccot}\left (x\right ) - \frac {8}{15 \, \sqrt {a x^{2} + a} a^{3}} - \frac {4}{45 \, {\left (a x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {1}{25 \, {\left (a x^{2} + a\right )}^{\frac {5}{2}} a} \]
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/(sqrt(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)
Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.70 \[ \int \frac {\cot ^{-1}(x)}{\left (a+a x^2\right )^{7/2}} \, dx=\frac {{\left (4 \, x^{2} {\left (\frac {2 \, x^{2}}{a} + \frac {5}{a}\right )} + \frac {15}{a}\right )} x \arctan \left (\frac {1}{x}\right )}{15 \, {\left (a x^{2} + a\right )}^{\frac {5}{2}}} - \frac {120 \, {\left (a x^{2} + a\right )}^{2} + 20 \, {\left (a x^{2} + a\right )} a + 9 \, a^{2}}{225 \, {\left (a x^{2} + a\right )}^{\frac {5}{2}} a^{3}} \]
1/15*(4*x^2*(2*x^2/a + 5/a) + 15/a)*x*arctan(1/x)/(a*x^2 + a)^(5/2) - 1/22 5*(120*(a*x^2 + a)^2 + 20*(a*x^2 + a)*a + 9*a^2)/((a*x^2 + a)^(5/2)*a^3)
Timed out. \[ \int \frac {\cot ^{-1}(x)}{\left (a+a x^2\right )^{7/2}} \, dx=\int \frac {\mathrm {acot}\left (x\right )}{{\left (a\,x^2+a\right )}^{7/2}} \,d x \]