3.315 \(\int \frac{x^{3/2}}{1+x^2} \, dx\)

Optimal. Leaf size=99 \[ 2 \sqrt{x}+\frac{\log \left (x-\sqrt{2} \sqrt{x}+1\right )}{2 \sqrt{2}}-\frac{\log \left (x+\sqrt{2} \sqrt{x}+1\right )}{2 \sqrt{2}}+\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{\sqrt{2}}-\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{\sqrt{2}} \]

[Out]

2*Sqrt[x] + ArcTan[1 - Sqrt[2]*Sqrt[x]]/Sqrt[2] - ArcTan[1 + Sqrt[2]*Sqrt[x]]/Sqrt[2] + Log[1 - Sqrt[2]*Sqrt[x
] + x]/(2*Sqrt[2]) - Log[1 + Sqrt[2]*Sqrt[x] + x]/(2*Sqrt[2])

________________________________________________________________________________________

Rubi [A]  time = 0.0550577, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {321, 329, 211, 1165, 628, 1162, 617, 204} \[ 2 \sqrt{x}+\frac{\log \left (x-\sqrt{2} \sqrt{x}+1\right )}{2 \sqrt{2}}-\frac{\log \left (x+\sqrt{2} \sqrt{x}+1\right )}{2 \sqrt{2}}+\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{\sqrt{2}}-\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(1 + x^2),x]

[Out]

2*Sqrt[x] + ArcTan[1 - Sqrt[2]*Sqrt[x]]/Sqrt[2] - ArcTan[1 + Sqrt[2]*Sqrt[x]]/Sqrt[2] + Log[1 - Sqrt[2]*Sqrt[x
] + x]/(2*Sqrt[2]) - Log[1 + Sqrt[2]*Sqrt[x] + x]/(2*Sqrt[2])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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
]]))

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^{3/2}}{1+x^2} \, dx &=2 \sqrt{x}-\int \frac{1}{\sqrt{x} \left (1+x^2\right )} \, dx\\ &=2 \sqrt{x}-2 \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\sqrt{x}\right )\\ &=2 \sqrt{x}-\operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{x}\right )-\operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{x}\right )\\ &=2 \sqrt{x}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{x}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{x}\right )+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{x}\right )}{2 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{x}\right )}{2 \sqrt{2}}\\ &=2 \sqrt{x}+\frac{\log \left (1-\sqrt{2} \sqrt{x}+x\right )}{2 \sqrt{2}}-\frac{\log \left (1+\sqrt{2} \sqrt{x}+x\right )}{2 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{x}\right )}{\sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{x}\right )}{\sqrt{2}}\\ &=2 \sqrt{x}+\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{\sqrt{2}}-\frac{\tan ^{-1}\left (1+\sqrt{2} \sqrt{x}\right )}{\sqrt{2}}+\frac{\log \left (1-\sqrt{2} \sqrt{x}+x\right )}{2 \sqrt{2}}-\frac{\log \left (1+\sqrt{2} \sqrt{x}+x\right )}{2 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0139088, size = 99, normalized size = 1. \[ 2 \sqrt{x}+\frac{\log \left (x-\sqrt{2} \sqrt{x}+1\right )}{2 \sqrt{2}}-\frac{\log \left (x+\sqrt{2} \sqrt{x}+1\right )}{2 \sqrt{2}}+\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{\sqrt{2}}-\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(1 + x^2),x]

[Out]

2*Sqrt[x] + ArcTan[1 - Sqrt[2]*Sqrt[x]]/Sqrt[2] - ArcTan[1 + Sqrt[2]*Sqrt[x]]/Sqrt[2] + Log[1 - Sqrt[2]*Sqrt[x
] + x]/(2*Sqrt[2]) - Log[1 + Sqrt[2]*Sqrt[x] + x]/(2*Sqrt[2])

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 67, normalized size = 0.7 \begin{align*} 2\,\sqrt{x}-{\frac{\sqrt{2}}{2}\arctan \left ( -1+\sqrt{2}\sqrt{x} \right ) }-{\frac{\sqrt{2}}{4}\ln \left ({ \left ( 1+x+\sqrt{2}\sqrt{x} \right ) \left ( 1+x-\sqrt{2}\sqrt{x} \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}}{2}\arctan \left ( 1+\sqrt{2}\sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(x^2+1),x)

[Out]

2*x^(1/2)-1/2*arctan(-1+2^(1/2)*x^(1/2))*2^(1/2)-1/4*2^(1/2)*ln((1+x+2^(1/2)*x^(1/2))/(1+x-2^(1/2)*x^(1/2)))-1
/2*arctan(1+2^(1/2)*x^(1/2))*2^(1/2)

________________________________________________________________________________________

Maxima [A]  time = 3.23855, size = 107, normalized size = 1.08 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(x^2+1),x, algorithm="maxima")

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) - 1/2*sqrt(2)*arctan(-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)

________________________________________________________________________________________

Fricas [A]  time = 1.37285, size = 366, normalized size = 3.7 \begin{align*} \sqrt{2} \arctan \left (\sqrt{2} \sqrt{\sqrt{2} \sqrt{x} + x + 1} - \sqrt{2} \sqrt{x} - 1\right ) + \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{-4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4} - \sqrt{2} \sqrt{x} + 1\right ) - \frac{1}{4} \, \sqrt{2} \log \left (4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4\right ) + \frac{1}{4} \, \sqrt{2} \log \left (-4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4\right ) + 2 \, \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(x^2+1),x, algorithm="fricas")

[Out]

sqrt(2)*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x) + x + 1) - sqrt(2)*sqrt(x) - 1) + sqrt(2)*arctan(1/2*sqrt(2)*sqrt(
-4*sqrt(2)*sqrt(x) + 4*x + 4) - sqrt(2)*sqrt(x) + 1) - 1/4*sqrt(2)*log(4*sqrt(2)*sqrt(x) + 4*x + 4) + 1/4*sqrt
(2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4) + 2*sqrt(x)

________________________________________________________________________________________

Sympy [A]  time = 0.922942, size = 97, normalized size = 0.98 \begin{align*} 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)/(x**2+1),x)

[Out]

2*sqrt(x) + sqrt(2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/4 - sqrt(2)*log(4*sqrt(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

________________________________________________________________________________________

Giac [A]  time = 2.4465, size = 107, normalized size = 1.08 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(x^2+1),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) - 1/2*sqrt(2)*arctan(-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)