Integrand size = 21, antiderivative size = 68 \[ \int -x^3 \arctan \left (\sqrt {x}-\sqrt {1+x}\right ) \, dx=-\frac {\sqrt {x}}{8}+\frac {x^{3/2}}{24}-\frac {x^{5/2}}{40}+\frac {x^{7/2}}{56}+\frac {\pi x^4}{16}+\frac {\arctan \left (\sqrt {x}\right )}{8}-\frac {1}{8} x^4 \arctan \left (\sqrt {x}\right ) \]
1/24*x^(3/2)-1/40*x^(5/2)+1/56*x^(7/2)+1/16*Pi*x^4+1/8*arctan(x^(1/2))-1/8 *x^4*arctan(x^(1/2))-1/8*x^(1/2)
Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.85 \[ \int -x^3 \arctan \left (\sqrt {x}-\sqrt {1+x}\right ) \, dx=\frac {\arctan \left (\sqrt {x}\right )}{8}-\frac {1}{840} \sqrt {x} \left (105-35 x+21 x^2-15 x^3+210 x^{7/2} \arctan \left (\sqrt {x}-\sqrt {1+x}\right )\right ) \]
ArcTan[Sqrt[x]]/8 - (Sqrt[x]*(105 - 35*x + 21*x^2 - 15*x^3 + 210*x^(7/2)*A rcTan[Sqrt[x] - Sqrt[1 + x]]))/840
Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {25, 5682, 15, 5361, 60, 60, 60, 60, 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.
\(\displaystyle \int -x^3 \arctan \left (\sqrt {x}-\sqrt {x+1}\right ) \, dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int x^3 \arctan \left (\sqrt {x}-\sqrt {x+1}\right )dx\) |
\(\Big \downarrow \) 5682 |
\(\displaystyle \frac {\pi \int x^3dx}{4}-\frac {1}{2} \int x^3 \arctan \left (\sqrt {x}\right )dx\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\pi x^4}{16}-\frac {1}{2} \int x^3 \arctan \left (\sqrt {x}\right )dx\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \int \frac {x^{7/2}}{x+1}dx-\frac {1}{4} x^4 \arctan \left (\sqrt {x}\right )\right )+\frac {\pi x^4}{16}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \left (\frac {2 x^{7/2}}{7}-\int \frac {x^{5/2}}{x+1}dx\right )-\frac {1}{4} x^4 \arctan \left (\sqrt {x}\right )\right )+\frac {\pi x^4}{16}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \left (\int \frac {x^{3/2}}{x+1}dx+\frac {2 x^{7/2}}{7}-\frac {2 x^{5/2}}{5}\right )-\frac {1}{4} x^4 \arctan \left (\sqrt {x}\right )\right )+\frac {\pi x^4}{16}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \left (-\int \frac {\sqrt {x}}{x+1}dx+\frac {2 x^{7/2}}{7}-\frac {2 x^{5/2}}{5}+\frac {2 x^{3/2}}{3}\right )-\frac {1}{4} x^4 \arctan \left (\sqrt {x}\right )\right )+\frac {\pi x^4}{16}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \left (\int \frac {1}{\sqrt {x} (x+1)}dx+\frac {2 x^{7/2}}{7}-\frac {2 x^{5/2}}{5}+\frac {2 x^{3/2}}{3}-2 \sqrt {x}\right )-\frac {1}{4} x^4 \arctan \left (\sqrt {x}\right )\right )+\frac {\pi x^4}{16}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \left (2 \int \frac {1}{x+1}d\sqrt {x}+\frac {2 x^{7/2}}{7}-\frac {2 x^{5/2}}{5}+\frac {2 x^{3/2}}{3}-2 \sqrt {x}\right )-\frac {1}{4} x^4 \arctan \left (\sqrt {x}\right )\right )+\frac {\pi x^4}{16}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{8} \left (2 \arctan \left (\sqrt {x}\right )+\frac {2 x^{7/2}}{7}-\frac {2 x^{5/2}}{5}+\frac {2 x^{3/2}}{3}-2 \sqrt {x}\right )-\frac {1}{4} x^4 \arctan \left (\sqrt {x}\right )\right )+\frac {\pi x^4}{16}\) |
(Pi*x^4)/16 + (-1/4*(x^4*ArcTan[Sqrt[x]]) + (-2*Sqrt[x] + (2*x^(3/2))/3 - (2*x^(5/2))/5 + (2*x^(7/2))/7 + 2*ArcTan[Sqrt[x]])/8)/2
3.2.26.3.1 Defintions of rubi rules used
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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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 = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.66
method | result | size |
default | \(-\frac {x^{4} \arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{4}+\frac {x^{\frac {7}{2}}}{56}-\frac {x^{\frac {5}{2}}}{40}+\frac {x^{\frac {3}{2}}}{24}-\frac {\sqrt {x}}{8}+\frac {\arctan \left (\sqrt {x}\right )}{8}\) | \(45\) |
parts | \(-\frac {x^{4} \arctan \left (\sqrt {x}-\sqrt {1+x}\right )}{4}+\frac {x^{\frac {7}{2}}}{56}-\frac {x^{\frac {5}{2}}}{40}+\frac {x^{\frac {3}{2}}}{24}-\frac {\sqrt {x}}{8}+\frac {\arctan \left (\sqrt {x}\right )}{8}\) | \(45\) |
-1/4*x^4*arctan(x^(1/2)-(1+x)^(1/2))+1/56*x^(7/2)-1/40*x^(5/2)+1/24*x^(3/2 )-1/8*x^(1/2)+1/8*arctan(x^(1/2))
Time = 0.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.59 \[ \int -x^3 \arctan \left (\sqrt {x}-\sqrt {1+x}\right ) \, dx=\frac {1}{4} \, {\left (x^{4} - 1\right )} \arctan \left (\sqrt {x + 1} - \sqrt {x}\right ) + \frac {1}{840} \, {\left (15 \, x^{3} - 21 \, x^{2} + 35 \, x - 105\right )} \sqrt {x} \]
Timed out. \[ \int -x^3 \arctan \left (\sqrt {x}-\sqrt {1+x}\right ) \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.65 \[ \int -x^3 \arctan \left (\sqrt {x}-\sqrt {1+x}\right ) \, dx=\frac {1}{4} \, x^{4} \arctan \left (\sqrt {x + 1} - \sqrt {x}\right ) + \frac {1}{56} \, x^{\frac {7}{2}} - \frac {1}{40} \, x^{\frac {5}{2}} + \frac {1}{24} \, x^{\frac {3}{2}} - \frac {1}{8} \, \sqrt {x} + \frac {1}{8} \, \arctan \left (\sqrt {x}\right ) \]
1/4*x^4*arctan(sqrt(x + 1) - sqrt(x)) + 1/56*x^(7/2) - 1/40*x^(5/2) + 1/24 *x^(3/2) - 1/8*sqrt(x) + 1/8*arctan(sqrt(x))
Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.65 \[ \int -x^3 \arctan \left (\sqrt {x}-\sqrt {1+x}\right ) \, dx=-\frac {1}{4} \, x^{4} \arctan \left (-\sqrt {x + 1} + \sqrt {x}\right ) + \frac {1}{56} \, x^{\frac {7}{2}} - \frac {1}{40} \, x^{\frac {5}{2}} + \frac {1}{24} \, x^{\frac {3}{2}} - \frac {1}{8} \, \sqrt {x} + \frac {1}{8} \, \arctan \left (\sqrt {x}\right ) \]
-1/4*x^4*arctan(-sqrt(x + 1) + sqrt(x)) + 1/56*x^(7/2) - 1/40*x^(5/2) + 1/ 24*x^(3/2) - 1/8*sqrt(x) + 1/8*arctan(sqrt(x))
Time = 1.92 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.06 \[ \int -x^3 \arctan \left (\sqrt {x}-\sqrt {1+x}\right ) \, dx=\frac {x^{3/2}}{24}-\frac {\sqrt {x}}{8}-\frac {x^{5/2}}{40}+\frac {x^{7/2}}{56}+\frac {\mathrm {atan}\left (\sqrt {x+1}-\sqrt {x}\right )\,\left (\frac {x^5}{2}+\frac {x^4}{2}\right )}{2\,x+2}+\frac {\ln \left (\frac {{\left (-1+\sqrt {x}\,1{}\mathrm {i}\right )}^2}{x+1}\right )\,1{}\mathrm {i}}{16} \]