3.321 \(\int \frac{x^{7/2}}{(1+x^2)^2} \, dx\)

Optimal. Leaf size=122 \[ -\frac{x^{5/2}}{2 \left (x^2+1\right )}+\frac{5 \sqrt{x}}{2}+\frac{5 \log \left (x-\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}-\frac{5 \log \left (x+\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}+\frac{5 \tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}-\frac{5 \tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{4 \sqrt{2}} \]

[Out]

(5*Sqrt[x])/2 - x^(5/2)/(2*(1 + x^2)) + (5*ArcTan[1 - Sqrt[2]*Sqrt[x]])/(4*Sqrt[2]) - (5*ArcTan[1 + Sqrt[2]*Sq
rt[x]])/(4*Sqrt[2]) + (5*Log[1 - Sqrt[2]*Sqrt[x] + x])/(8*Sqrt[2]) - (5*Log[1 + Sqrt[2]*Sqrt[x] + x])/(8*Sqrt[
2])

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Rubi [A]  time = 0.0614054, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {288, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{x^{5/2}}{2 \left (x^2+1\right )}+\frac{5 \sqrt{x}}{2}+\frac{5 \log \left (x-\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}-\frac{5 \log \left (x+\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}+\frac{5 \tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}-\frac{5 \tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(1 + x^2)^2,x]

[Out]

(5*Sqrt[x])/2 - x^(5/2)/(2*(1 + x^2)) + (5*ArcTan[1 - Sqrt[2]*Sqrt[x]])/(4*Sqrt[2]) - (5*ArcTan[1 + Sqrt[2]*Sq
rt[x]])/(4*Sqrt[2]) + (5*Log[1 - Sqrt[2]*Sqrt[x] + x])/(8*Sqrt[2]) - (5*Log[1 + Sqrt[2]*Sqrt[x] + x])/(8*Sqrt[
2])

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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^{7/2}}{\left (1+x^2\right )^2} \, dx &=-\frac{x^{5/2}}{2 \left (1+x^2\right )}+\frac{5}{4} \int \frac{x^{3/2}}{1+x^2} \, dx\\ &=\frac{5 \sqrt{x}}{2}-\frac{x^{5/2}}{2 \left (1+x^2\right )}-\frac{5}{4} \int \frac{1}{\sqrt{x} \left (1+x^2\right )} \, dx\\ &=\frac{5 \sqrt{x}}{2}-\frac{x^{5/2}}{2 \left (1+x^2\right )}-\frac{5}{2} \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\sqrt{x}\right )\\ &=\frac{5 \sqrt{x}}{2}-\frac{x^{5/2}}{2 \left (1+x^2\right )}-\frac{5}{4} \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{x}\right )-\frac{5}{4} \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{x}\right )\\ &=\frac{5 \sqrt{x}}{2}-\frac{x^{5/2}}{2 \left (1+x^2\right )}-\frac{5}{8} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{x}\right )-\frac{5}{8} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{x}\right )+\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{x}\right )}{8 \sqrt{2}}+\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{x}\right )}{8 \sqrt{2}}\\ &=\frac{5 \sqrt{x}}{2}-\frac{x^{5/2}}{2 \left (1+x^2\right )}+\frac{5 \log \left (1-\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}-\frac{5 \log \left (1+\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}\\ &=\frac{5 \sqrt{x}}{2}-\frac{x^{5/2}}{2 \left (1+x^2\right )}+\frac{5 \tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}-\frac{5 \tan ^{-1}\left (1+\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}+\frac{5 \log \left (1-\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}-\frac{5 \log \left (1+\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0462298, size = 121, normalized size = 0.99 \[ \frac{1}{16} \left (\frac{32 x^{5/2}}{x^2+1}+\frac{40 \sqrt{x}}{x^2+1}+5 \sqrt{2} \log \left (x-\sqrt{2} \sqrt{x}+1\right )-5 \sqrt{2} \log \left (x+\sqrt{2} \sqrt{x}+1\right )+10 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )-10 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(1 + x^2)^2,x]

[Out]

((40*Sqrt[x])/(1 + x^2) + (32*x^(5/2))/(1 + x^2) + 10*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[x]] - 10*Sqrt[2]*ArcTan[
1 + Sqrt[2]*Sqrt[x]] + 5*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[x] + x] - 5*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[x] + x])/16

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Maple [A]  time = 0.01, size = 79, normalized size = 0.7 \begin{align*} 2\,\sqrt{x}+{\frac{1}{2\,{x}^{2}+2}\sqrt{x}}-{\frac{5\,\sqrt{2}}{8}\arctan \left ( -1+\sqrt{2}\sqrt{x} \right ) }-{\frac{5\,\sqrt{2}}{16}\ln \left ({ \left ( 1+x+\sqrt{2}\sqrt{x} \right ) \left ( 1+x-\sqrt{2}\sqrt{x} \right ) ^{-1}} \right ) }-{\frac{5\,\sqrt{2}}{8}\arctan \left ( 1+\sqrt{2}\sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(x^2+1)^2,x)

[Out]

2*x^(1/2)+1/2*x^(1/2)/(x^2+1)-5/8*arctan(-1+2^(1/2)*x^(1/2))*2^(1/2)-5/16*2^(1/2)*ln((1+x+2^(1/2)*x^(1/2))/(1+
x-2^(1/2)*x^(1/2)))-5/8*arctan(1+2^(1/2)*x^(1/2))*2^(1/2)

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Maxima [A]  time = 2.35135, size = 123, normalized size = 1.01 \begin{align*} -\frac{5}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{x}\right )}\right ) - \frac{5}{8} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{x}\right )}\right ) - \frac{5}{16} \, \sqrt{2} \log \left (\sqrt{2} \sqrt{x} + x + 1\right ) + \frac{5}{16} \, \sqrt{2} \log \left (-\sqrt{2} \sqrt{x} + x + 1\right ) + 2 \, \sqrt{x} + \frac{\sqrt{x}}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(x^2+1)^2,x, algorithm="maxima")

[Out]

-5/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) - 5/8*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x))
) - 5/16*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) + 5/16*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) + 2*sqrt(x) + 1/2*s
qrt(x)/(x^2 + 1)

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Fricas [A]  time = 1.3944, size = 462, normalized size = 3.79 \begin{align*} \frac{20 \, \sqrt{2}{\left (x^{2} + 1\right )} \arctan \left (\sqrt{2} \sqrt{\sqrt{2} \sqrt{x} + x + 1} - \sqrt{2} \sqrt{x} - 1\right ) + 20 \, \sqrt{2}{\left (x^{2} + 1\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{-4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4} - \sqrt{2} \sqrt{x} + 1\right ) - 5 \, \sqrt{2}{\left (x^{2} + 1\right )} \log \left (4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4\right ) + 5 \, \sqrt{2}{\left (x^{2} + 1\right )} \log \left (-4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4\right ) + 8 \,{\left (4 \, x^{2} + 5\right )} \sqrt{x}}{16 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(x^2+1)^2,x, algorithm="fricas")

[Out]

1/16*(20*sqrt(2)*(x^2 + 1)*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x) + x + 1) - sqrt(2)*sqrt(x) - 1) + 20*sqrt(2)*(x
^2 + 1)*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*sqrt(x) + 4*x + 4) - sqrt(2)*sqrt(x) + 1) - 5*sqrt(2)*(x^2 + 1)*log
(4*sqrt(2)*sqrt(x) + 4*x + 4) + 5*sqrt(2)*(x^2 + 1)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4) + 8*(4*x^2 + 5)*sqrt(x))
/(x^2 + 1)

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Sympy [B]  time = 13.795, size = 277, normalized size = 2.27 \begin{align*} \frac{32 x^{\frac{5}{2}}}{16 x^{2} + 16} + \frac{40 \sqrt{x}}{16 x^{2} + 16} + \frac{5 \sqrt{2} x^{2} \log{\left (- 4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{2} + 16} - \frac{5 \sqrt{2} x^{2} \log{\left (4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{2} + 16} - \frac{10 \sqrt{2} x^{2} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} - 1 \right )}}{16 x^{2} + 16} - \frac{10 \sqrt{2} x^{2} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} + 1 \right )}}{16 x^{2} + 16} + \frac{5 \sqrt{2} \log{\left (- 4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{2} + 16} - \frac{5 \sqrt{2} \log{\left (4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{2} + 16} - \frac{10 \sqrt{2} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} - 1 \right )}}{16 x^{2} + 16} - \frac{10 \sqrt{2} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} + 1 \right )}}{16 x^{2} + 16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)/(x**2+1)**2,x)

[Out]

32*x**(5/2)/(16*x**2 + 16) + 40*sqrt(x)/(16*x**2 + 16) + 5*sqrt(2)*x**2*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*
x**2 + 16) - 5*sqrt(2)*x**2*log(4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) - 10*sqrt(2)*x**2*atan(sqrt(2)*sqr
t(x) - 1)/(16*x**2 + 16) - 10*sqrt(2)*x**2*atan(sqrt(2)*sqrt(x) + 1)/(16*x**2 + 16) + 5*sqrt(2)*log(-4*sqrt(2)
*sqrt(x) + 4*x + 4)/(16*x**2 + 16) - 5*sqrt(2)*log(4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) - 10*sqrt(2)*at
an(sqrt(2)*sqrt(x) - 1)/(16*x**2 + 16) - 10*sqrt(2)*atan(sqrt(2)*sqrt(x) + 1)/(16*x**2 + 16)

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Giac [A]  time = 2.35365, size = 123, normalized size = 1.01 \begin{align*} -\frac{5}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{x}\right )}\right ) - \frac{5}{8} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{x}\right )}\right ) - \frac{5}{16} \, \sqrt{2} \log \left (\sqrt{2} \sqrt{x} + x + 1\right ) + \frac{5}{16} \, \sqrt{2} \log \left (-\sqrt{2} \sqrt{x} + x + 1\right ) + 2 \, \sqrt{x} + \frac{\sqrt{x}}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(x^2+1)^2,x, algorithm="giac")

[Out]

-5/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) - 5/8*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x))
) - 5/16*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) + 5/16*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) + 2*sqrt(x) + 1/2*s
qrt(x)/(x^2 + 1)