Integrand size = 21, antiderivative size = 59 \[ \int -\frac {\arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{x^4} \, dx=-\frac {\pi }{12 x^3}+\frac {1}{30 x^{5/2}}-\frac {1}{18 x^{3/2}}+\frac {1}{6 \sqrt {x}}+\frac {\arctan \left (\sqrt {x}\right )}{6}+\frac {\arctan \left (\sqrt {x}\right )}{6 x^3} \] Output:
-1/12*Pi/x^3+1/30/x^(5/2)-1/18/x^(3/2)+1/6/x^(1/2)+1/6*arctan(x^(1/2))+1/6 *arctan(x^(1/2))/x^3
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int -\frac {\arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{x^4} \, dx=\frac {1}{90} \left (-\frac {-3+5 x-15 x^2}{x^{5/2}}+15 \arctan \left (\sqrt {x}\right )+\frac {30 \arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{x^3}\right ) \] Input:
Integrate[-(ArcTan[Sqrt[x] - Sqrt[1 + x]]/x^4),x]
Output:
(-((-3 + 5*x - 15*x^2)/x^(5/2)) + 15*ArcTan[Sqrt[x]] + (30*ArcTan[Sqrt[x] - Sqrt[1 + x]])/x^3)/90
Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {25, 5682, 15, 5361, 61, 61, 61, 73, 216}
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.
\(\displaystyle \int -\frac {\arctan \left (\sqrt {x}-\sqrt {x+1}\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\arctan \left (\sqrt {x}-\sqrt {x+1}\right )}{x^4}dx\) |
\(\Big \downarrow \) 5682 |
\(\displaystyle \frac {1}{4} \pi \int \frac {1}{x^4}dx-\frac {1}{2} \int \frac {\arctan \left (\sqrt {x}\right )}{x^4}dx\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {1}{2} \int \frac {\arctan \left (\sqrt {x}\right )}{x^4}dx-\frac {\pi }{12 x^3}\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\sqrt {x}\right )}{3 x^3}-\frac {1}{6} \int \frac {1}{x^{7/2} (x+1)}dx\right )-\frac {\pi }{12 x^3}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{6} \left (\int \frac {1}{x^{5/2} (x+1)}dx+\frac {2}{5 x^{5/2}}\right )+\frac {\arctan \left (\sqrt {x}\right )}{3 x^3}\right )-\frac {\pi }{12 x^3}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{6} \left (-\int \frac {1}{x^{3/2} (x+1)}dx-\frac {2}{3 x^{3/2}}+\frac {2}{5 x^{5/2}}\right )+\frac {\arctan \left (\sqrt {x}\right )}{3 x^3}\right )-\frac {\pi }{12 x^3}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{6} \left (\int \frac {1}{\sqrt {x} (x+1)}dx-\frac {2}{3 x^{3/2}}+\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\right )+\frac {\arctan \left (\sqrt {x}\right )}{3 x^3}\right )-\frac {\pi }{12 x^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{6} \left (2 \int \frac {1}{x+1}d\sqrt {x}-\frac {2}{3 x^{3/2}}+\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\right )+\frac {\arctan \left (\sqrt {x}\right )}{3 x^3}\right )-\frac {\pi }{12 x^3}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{6} \left (2 \arctan \left (\sqrt {x}\right )-\frac {2}{3 x^{3/2}}+\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\right )+\frac {\arctan \left (\sqrt {x}\right )}{3 x^3}\right )-\frac {\pi }{12 x^3}\) |
Input:
Int[-(ArcTan[Sqrt[x] - Sqrt[1 + x]]/x^4),x]
Output:
-1/12*Pi/x^3 + (ArcTan[Sqrt[x]]/(3*x^3) + (2/(5*x^(5/2)) - 2/(3*x^(3/2)) + 2/Sqrt[x] + 2*ArcTan[Sqrt[x]])/6)/2
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & & IntegerQ[m])) && NeQ[m, -1]
Int[ArcTan[(v_) + (s_.)*Sqrt[w_]]*(u_.), x_Symbol] :> Simp[Pi*(s/4) Int[u , x], x] + Simp[1/2 Int[u*ArcTan[v], x], x] /; EqQ[s^2, 1] && EqQ[w, v^2 + 1]
Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.68
method | result | size |
default | \(\frac {\arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{3 x^{3}}+\frac {1}{30 x^{\frac {5}{2}}}-\frac {1}{18 x^{\frac {3}{2}}}+\frac {1}{6 \sqrt {x}}+\frac {\arctan \left (\sqrt {x}\right )}{6}\) | \(40\) |
parts | \(\frac {\arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{3 x^{3}}+\frac {1}{30 x^{\frac {5}{2}}}-\frac {1}{18 x^{\frac {3}{2}}}+\frac {1}{6 \sqrt {x}}+\frac {\arctan \left (\sqrt {x}\right )}{6}\) | \(40\) |
Input:
int(-arctan(x^(1/2)-(1+x)^(1/2))/x^4,x,method=_RETURNVERBOSE)
Output:
1/3*arctan(x^(1/2)-(1+x)^(1/2))/x^3+1/30/x^(5/2)-1/18/x^(3/2)+1/6/x^(1/2)+ 1/6*arctan(x^(1/2))
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.68 \[ \int -\frac {\arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{x^4} \, dx=-\frac {30 \, {\left (x^{3} + 1\right )} \arctan \left (\sqrt {x + 1} - \sqrt {x}\right ) - {\left (15 \, x^{2} - 5 \, x + 3\right )} \sqrt {x}}{90 \, x^{3}} \] Input:
integrate(-arctan(x^(1/2)-(1+x)^(1/2))/x^4,x, algorithm="fricas")
Output:
-1/90*(30*(x^3 + 1)*arctan(sqrt(x + 1) - sqrt(x)) - (15*x^2 - 5*x + 3)*sqr t(x))/x^3
Timed out. \[ \int -\frac {\arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{x^4} \, dx=\text {Timed out} \] Input:
integrate(-atan(x**(1/2)-(1+x)**(1/2))/x**4,x)
Output:
Timed out
Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.66 \[ \int -\frac {\arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{x^4} \, dx=\frac {1}{6 \, \sqrt {x}} - \frac {1}{18 \, x^{\frac {3}{2}}} - \frac {\arctan \left (\sqrt {x + 1} - \sqrt {x}\right )}{3 \, x^{3}} + \frac {1}{30 \, x^{\frac {5}{2}}} + \frac {1}{6} \, \arctan \left (\sqrt {x}\right ) \] Input:
integrate(-arctan(x^(1/2)-(1+x)^(1/2))/x^4,x, algorithm="maxima")
Output:
1/6/sqrt(x) - 1/18/x^(3/2) - 1/3*arctan(sqrt(x + 1) - sqrt(x))/x^3 + 1/30/ x^(5/2) + 1/6*arctan(sqrt(x))
Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.66 \[ \int -\frac {\arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{x^4} \, dx=\frac {15 \, x^{2} - 5 \, x + 3}{90 \, x^{\frac {5}{2}}} + \frac {\arctan \left (-\sqrt {x + 1} + \sqrt {x}\right )}{3 \, x^{3}} + \frac {1}{6} \, \arctan \left (\sqrt {x}\right ) \] Input:
integrate(-arctan(x^(1/2)-(1+x)^(1/2))/x^4,x, algorithm="giac")
Output:
1/90*(15*x^2 - 5*x + 3)/x^(5/2) + 1/3*arctan(-sqrt(x + 1) + sqrt(x))/x^3 + 1/6*arctan(sqrt(x))
Time = 0.87 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.95 \[ \int -\frac {\arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{x^4} \, dx=-\frac {\frac {\mathrm {atan}\left (\sqrt {x+1}-\sqrt {x}\right )}{3}-\frac {\sqrt {x}}{30}+\frac {x^{3/2}}{18}-\frac {x^{5/2}}{6}}{x^3}+\frac {\ln \left (\frac {{\left (-1+\sqrt {x}\,1{}\mathrm {i}\right )}^2}{x+1}\right )\,1{}\mathrm {i}}{12} \] Input:
int(atan((x + 1)^(1/2) - x^(1/2))/x^4,x)
Output:
(log((x^(1/2)*1i - 1)^2/(x + 1))*1i)/12 - (atan((x + 1)^(1/2) - x^(1/2))/3 - x^(1/2)/30 + x^(3/2)/18 - x^(5/2)/6)/x^3
Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int -\frac {\arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{x^4} \, dx=\frac {-30 \mathit {atan} \left (\frac {1}{\sqrt {x +1}+\sqrt {x}}\right ) x^{3}-30 \mathit {atan} \left (\frac {1}{\sqrt {x +1}+\sqrt {x}}\right )+15 \sqrt {x}\, x^{2}-5 \sqrt {x}\, x +3 \sqrt {x}}{90 x^{3}} \] Input:
int(-atan(x^(1/2)-(1+x)^(1/2))/x^4,x)
Output:
( - 30*atan(1/(sqrt(x + 1) + sqrt(x)))*x**3 - 30*atan(1/(sqrt(x + 1) + sqr t(x))) + 15*sqrt(x)*x**2 - 5*sqrt(x)*x + 3*sqrt(x))/(90*x**3)