Integrand size = 13, antiderivative size = 74 \[ \int \frac {x^{3/2}}{1+x^2} \, dx=2 \sqrt {x}+\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {x}}{1+x}\right )}{\sqrt {2}} \] Output:
2*x^(1/2)-1/2*arctan(-1+2^(1/2)*x^(1/2))*2^(1/2)-1/2*arctan(1+2^(1/2)*x^(1 /2))*2^(1/2)-1/2*arctanh(2^(1/2)*x^(1/2)/(1+x))*2^(1/2)
Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.73 \[ \int \frac {x^{3/2}}{1+x^2} \, dx=2 \sqrt {x}-\frac {\arctan \left (\frac {-1+x}{\sqrt {2} \sqrt {x}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {x}}{1+x}\right )}{\sqrt {2}} \] Input:
Integrate[x^(3/2)/(1 + x^2),x]
Output:
2*Sqrt[x] - ArcTan[(-1 + x)/(Sqrt[2]*Sqrt[x])]/Sqrt[2] - ArcTanh[(Sqrt[2]* Sqrt[x])/(1 + x)]/Sqrt[2]
Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.51, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {262, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
\(\displaystyle \int \frac {x^{3/2}}{x^2+1} \, dx\) |
\(\Big \downarrow \) 262 |
\(\displaystyle 2 \sqrt {x}-\int \frac {1}{\sqrt {x} \left (x^2+1\right )}dx\) |
\(\Big \downarrow \) 266 |
\(\displaystyle 2 \sqrt {x}-2 \int \frac {1}{x^2+1}d\sqrt {x}\) |
\(\Big \downarrow \) 755 |
\(\displaystyle 2 \sqrt {x}-2 \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \int \frac {x+1}{x^2+1}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle 2 \sqrt {x}-2 \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}+\frac {1}{2} \int \frac {1}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}\right )\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle 2 \sqrt {x}-2 \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \left (\frac {\int \frac {1}{-x-1}d\left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-x-1}d\left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle 2 \sqrt {x}-2 \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle 2 \sqrt {x}-2 \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {x}+1\right )}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 2 \sqrt {x}-2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {x}+1\right )}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \sqrt {x}-2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {x}+1}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle 2 \sqrt {x}-2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x+\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (x-\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}\right )\right )\) |
Input:
Int[x^(3/2)/(1 + x^2),x]
Output:
2*Sqrt[x] - 2*((-(ArcTan[1 - Sqrt[2]*Sqrt[x]]/Sqrt[2]) + ArcTan[1 + Sqrt[2 ]*Sqrt[x]]/Sqrt[2])/2 + (-1/2*Log[1 - Sqrt[2]*Sqrt[x] + x]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[x] + x]/(2*Sqrt[2]))/2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Time = 0.83 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(2 \sqrt {x}-\frac {\sqrt {2}\, \left (\ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{4}\) | \(62\) |
default | \(2 \sqrt {x}-\frac {\sqrt {2}\, \left (\ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{4}\) | \(62\) |
risch | \(2 \sqrt {x}-\frac {\sqrt {2}\, \left (\ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{4}\) | \(62\) |
meijerg | \(2 \sqrt {x}-\frac {\sqrt {x}\, \left (-\frac {\sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{2 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{\left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{2 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{\left (x^{2}\right )^{\frac {1}{4}}}\right )}{2}\) | \(136\) |
trager | \(2 \sqrt {x}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+4 \sqrt {x}}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-x -1}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+4 \sqrt {x}}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}+x +1}\right )}{2}\) | \(187\) |
Input:
int(x^(3/2)/(x^2+1),x,method=_RETURNVERBOSE)
Output:
2*x^(1/2)-1/4*2^(1/2)*(ln((x+2^(1/2)*x^(1/2)+1)/(x-2^(1/2)*x^(1/2)+1))+2*a rctan(1+2^(1/2)*x^(1/2))+2*arctan(-1+2^(1/2)*x^(1/2)))
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93 \[ \int \frac {x^{3/2}}{1+x^2} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} \sqrt {x} + 1\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} \sqrt {x} - 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + 2 \, \sqrt {x} \] Input:
integrate(x^(3/2)/(x^2+1),x, algorithm="fricas")
Output:
-1/2*sqrt(2)*arctan(sqrt(2)*sqrt(x) + 1) - 1/2*sqrt(2)*arctan(sqrt(2)*sqrt (x) - 1) - 1/4*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) + 1/4*sqrt(2)*log(-sqr t(2)*sqrt(x) + x + 1) + 2*sqrt(x)
Time = 0.21 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.31 \[ \int \frac {x^{3/2}}{1+x^2} \, dx=2 \sqrt {x} + \frac {\sqrt {2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{4} - \frac {\sqrt {2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{4} - \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{2} - \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{2} \] Input:
integrate(x**(3/2)/(x**2+1),x)
Output:
2*sqrt(x) + sqrt(2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/4 - sqrt(2)*log(4*sq rt(2)*sqrt(x) + 4*x + 4)/4 - sqrt(2)*atan(sqrt(2)*sqrt(x) - 1)/2 - sqrt(2) *atan(sqrt(2)*sqrt(x) + 1)/2
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07 \[ \int \frac {x^{3/2}}{1+x^2} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + 2 \, \sqrt {x} \] Input:
integrate(x^(3/2)/(x^2+1),x, algorithm="maxima")
Output:
-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) - 1/2*sqrt(2)*arcta n(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x))) - 1/4*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) + 1/4*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) + 2*sqrt(x)
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07 \[ \int \frac {x^{3/2}}{1+x^2} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + 2 \, \sqrt {x} \] Input:
integrate(x^(3/2)/(x^2+1),x, algorithm="giac")
Output:
-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) - 1/2*sqrt(2)*arcta n(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x))) - 1/4*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) + 1/4*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) + 2*sqrt(x)
Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.57 \[ \int \frac {x^{3/2}}{1+x^2} \, dx=2\,\sqrt {x}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right ) \] Input:
int(x^(3/2)/(x^2 + 1),x)
Output:
2*x^(1/2) - 2^(1/2)*atan(2^(1/2)*x^(1/2)*(1/2 + 1i/2))*(1/2 - 1i/2) - 2^(1 /2)*atan(2^(1/2)*x^(1/2)*(1/2 - 1i/2))*(1/2 + 1i/2)
Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92 \[ \int \frac {x^{3/2}}{1+x^2} \, dx=-\frac {\sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x}-\sqrt {2}}{\sqrt {2}}\right )}{2}-\frac {\sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x}+\sqrt {2}}{\sqrt {2}}\right )}{2}+2 \sqrt {x}+\frac {\sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, \sqrt {2}+x +1\right )}{4}-\frac {\sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, \sqrt {2}+x +1\right )}{4} \] Input:
int(x^(3/2)/(x^2+1),x)
Output:
( - 2*sqrt(2)*atan((2*sqrt(x) - sqrt(2))/sqrt(2)) - 2*sqrt(2)*atan((2*sqrt (x) + sqrt(2))/sqrt(2)) + 8*sqrt(x) + sqrt(2)*log( - sqrt(x)*sqrt(2) + x + 1) - sqrt(2)*log(sqrt(x)*sqrt(2) + x + 1))/4