3.329 \(\int \frac {x^{7/2}}{(1+x^2)^3} \, dx\)

Optimal. Leaf size=129 \[ -\frac {5 \sqrt {x}}{16 \left (x^2+1\right )}-\frac {x^{5/2}}{4 \left (x^2+1\right )^2}-\frac {5 \log \left (x-\sqrt {2} \sqrt {x}+1\right )}{64 \sqrt {2}}+\frac {5 \log \left (x+\sqrt {2} \sqrt {x}+1\right )}{64 \sqrt {2}}-\frac {5 \tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )}{32 \sqrt {2}}+\frac {5 \tan ^{-1}\left (\sqrt {2} \sqrt {x}+1\right )}{32 \sqrt {2}} \]

[Out]

-1/4*x^(5/2)/(x^2+1)^2+5/64*arctan(-1+2^(1/2)*x^(1/2))*2^(1/2)+5/64*arctan(1+2^(1/2)*x^(1/2))*2^(1/2)-5/128*ln
(1+x-2^(1/2)*x^(1/2))*2^(1/2)+5/128*ln(1+x+2^(1/2)*x^(1/2))*2^(1/2)-5/16*x^(1/2)/(x^2+1)

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Rubi [A]  time = 0.07, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {288, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {x^{5/2}}{4 \left (x^2+1\right )^2}-\frac {5 \sqrt {x}}{16 \left (x^2+1\right )}-\frac {5 \log \left (x-\sqrt {2} \sqrt {x}+1\right )}{64 \sqrt {2}}+\frac {5 \log \left (x+\sqrt {2} \sqrt {x}+1\right )}{64 \sqrt {2}}-\frac {5 \tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )}{32 \sqrt {2}}+\frac {5 \tan ^{-1}\left (\sqrt {2} \sqrt {x}+1\right )}{32 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(1 + x^2)^3,x]

[Out]

-x^(5/2)/(4*(1 + x^2)^2) - (5*Sqrt[x])/(16*(1 + x^2)) - (5*ArcTan[1 - Sqrt[2]*Sqrt[x]])/(32*Sqrt[2]) + (5*ArcT
an[1 + Sqrt[2]*Sqrt[x]])/(32*Sqrt[2]) - (5*Log[1 - Sqrt[2]*Sqrt[x] + x])/(64*Sqrt[2]) + (5*Log[1 + Sqrt[2]*Sqr
t[x] + x])/(64*Sqrt[2])

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])

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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {x^{7/2}}{\left (1+x^2\right )^3} \, dx &=-\frac {x^{5/2}}{4 \left (1+x^2\right )^2}+\frac {5}{8} \int \frac {x^{3/2}}{\left (1+x^2\right )^2} \, dx\\ &=-\frac {x^{5/2}}{4 \left (1+x^2\right )^2}-\frac {5 \sqrt {x}}{16 \left (1+x^2\right )}+\frac {5}{32} \int \frac {1}{\sqrt {x} \left (1+x^2\right )} \, dx\\ &=-\frac {x^{5/2}}{4 \left (1+x^2\right )^2}-\frac {5 \sqrt {x}}{16 \left (1+x^2\right )}+\frac {5}{16} \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {x}\right )\\ &=-\frac {x^{5/2}}{4 \left (1+x^2\right )^2}-\frac {5 \sqrt {x}}{16 \left (1+x^2\right )}+\frac {5}{32} \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {x}\right )+\frac {5}{32} \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {x}\right )\\ &=-\frac {x^{5/2}}{4 \left (1+x^2\right )^2}-\frac {5 \sqrt {x}}{16 \left (1+x^2\right )}+\frac {5}{64} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {x}\right )+\frac {5}{64} \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 )}{64 \sqrt {2}}-\frac {5 \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2}}\\ &=-\frac {x^{5/2}}{4 \left (1+x^2\right )^2}-\frac {5 \sqrt {x}}{16 \left (1+x^2\right )}-\frac {5 \log \left (1-\sqrt {2} \sqrt {x}+x\right )}{64 \sqrt {2}}+\frac {5 \log \left (1+\sqrt {2} \sqrt {x}+x\right )}{64 \sqrt {2}}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {x}\right )}{32 \sqrt {2}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {x}\right )}{32 \sqrt {2}}\\ &=-\frac {x^{5/2}}{4 \left (1+x^2\right )^2}-\frac {5 \sqrt {x}}{16 \left (1+x^2\right )}-\frac {5 \tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )}{32 \sqrt {2}}+\frac {5 \tan ^{-1}\left (1+\sqrt {2} \sqrt {x}\right )}{32 \sqrt {2}}-\frac {5 \log \left (1-\sqrt {2} \sqrt {x}+x\right )}{64 \sqrt {2}}+\frac {5 \log \left (1+\sqrt {2} \sqrt {x}+x\right )}{64 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 135, normalized size = 1.05 \[ \frac {1}{384} \left (\frac {40 \sqrt {x}}{x^2+1}-\frac {160 \sqrt {x}}{\left (x^2+1\right )^2}-\frac {256 x^{5/2}}{\left (x^2+1\right )^2}-15 \sqrt {2} \log \left (x-\sqrt {2} \sqrt {x}+1\right )+15 \sqrt {2} \log \left (x+\sqrt {2} \sqrt {x}+1\right )-30 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {x}\right )+30 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {x}+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(1 + x^2)^3,x]

[Out]

((-160*Sqrt[x])/(1 + x^2)^2 - (256*x^(5/2))/(1 + x^2)^2 + (40*Sqrt[x])/(1 + x^2) - 30*Sqrt[2]*ArcTan[1 - Sqrt[
2]*Sqrt[x]] + 30*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[x]] - 15*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[x] + x] + 15*Sqrt[2]*Lo
g[1 + Sqrt[2]*Sqrt[x] + x])/384

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fricas [A]  time = 0.93, size = 173, normalized size = 1.34 \[ -\frac {20 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (\sqrt {2} \sqrt {\sqrt {2} \sqrt {x} + x + 1} - \sqrt {2} \sqrt {x} - 1\right ) + 20 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} \sqrt {x} + 4 \, x + 4} - \sqrt {2} \sqrt {x} + 1\right ) - 5 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 1\right )} \log \left (4 \, \sqrt {2} \sqrt {x} + 4 \, x + 4\right ) + 5 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 1\right )} \log \left (-4 \, \sqrt {2} \sqrt {x} + 4 \, x + 4\right ) + 8 \, {\left (9 \, x^{2} + 5\right )} \sqrt {x}}{128 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(x^2+1)^3,x, algorithm="fricas")

[Out]

-1/128*(20*sqrt(2)*(x^4 + 2*x^2 + 1)*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x) + x + 1) - sqrt(2)*sqrt(x) - 1) + 20*
sqrt(2)*(x^4 + 2*x^2 + 1)*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*sqrt(x) + 4*x + 4) - sqrt(2)*sqrt(x) + 1) - 5*sqr
t(2)*(x^4 + 2*x^2 + 1)*log(4*sqrt(2)*sqrt(x) + 4*x + 4) + 5*sqrt(2)*(x^4 + 2*x^2 + 1)*log(-4*sqrt(2)*sqrt(x) +
4*x + 4) + 8*(9*x^2 + 5)*sqrt(x))/(x^4 + 2*x^2 + 1)

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giac [A]  time = 0.69, size = 94, normalized size = 0.73 \[ \frac {5}{64} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {5}{64} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {5}{128} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {5}{128} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {9 \, x^{\frac {5}{2}} + 5 \, \sqrt {x}}{16 \, {\left (x^{2} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(x^2+1)^3,x, algorithm="giac")

[Out]

5/64*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) + 5/64*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x)
)) + 5/128*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) - 5/128*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) - 1/16*(9*x^(5/2
) + 5*sqrt(x))/(x^2 + 1)^2

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maple [A]  time = 0.01, size = 82, normalized size = 0.64 \[ \frac {5 \sqrt {2}\, \arctan \left (\sqrt {2}\, \sqrt {x}-1\right )}{64}+\frac {5 \sqrt {2}\, \arctan \left (\sqrt {2}\, \sqrt {x}+1\right )}{64}+\frac {5 \sqrt {2}\, \ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )}{128}+\frac {-\frac {9 x^{\frac {5}{2}}}{16}-\frac {5 \sqrt {x}}{16}}{\left (x^{2}+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(x^2+1)^3,x)

[Out]

2*(-9/32*x^(5/2)-5/32*x^(1/2))/(x^2+1)^2+5/64*2^(1/2)*arctan(2^(1/2)*x^(1/2)-1)+5/128*2^(1/2)*ln((x+2^(1/2)*x^
(1/2)+1)/(x-2^(1/2)*x^(1/2)+1))+5/64*2^(1/2)*arctan(2^(1/2)*x^(1/2)+1)

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maxima [A]  time = 3.01, size = 99, normalized size = 0.77 \[ \frac {5}{64} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {5}{64} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {5}{128} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {5}{128} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {9 \, x^{\frac {5}{2}} + 5 \, \sqrt {x}}{16 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(x^2+1)^3,x, algorithm="maxima")

[Out]

5/64*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) + 5/64*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x)
)) + 5/128*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) - 5/128*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) - 1/16*(9*x^(5/2
) + 5*sqrt(x))/(x^4 + 2*x^2 + 1)

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mupad [B]  time = 4.73, size = 62, normalized size = 0.48 \[ -\frac {\frac {5\,\sqrt {x}}{16}+\frac {9\,x^{5/2}}{16}}{x^4+2\,x^2+1}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {5}{64}+\frac {5}{64}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {5}{64}-\frac {5}{64}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(x^2 + 1)^3,x)

[Out]

2^(1/2)*atan(2^(1/2)*x^(1/2)*(1/2 - 1i/2))*(5/64 + 5i/64) + 2^(1/2)*atan(2^(1/2)*x^(1/2)*(1/2 + 1i/2))*(5/64 -
5i/64) - ((5*x^(1/2))/16 + (9*x^(5/2))/16)/(2*x^2 + x^4 + 1)

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sympy [B]  time = 23.35, size = 481, normalized size = 3.73 \[ - \frac {72 x^{\frac {5}{2}}}{128 x^{4} + 256 x^{2} + 128} - \frac {40 \sqrt {x}}{128 x^{4} + 256 x^{2} + 128} - \frac {5 \sqrt {2} x^{4} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{128 x^{4} + 256 x^{2} + 128} + \frac {5 \sqrt {2} x^{4} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{128 x^{4} + 256 x^{2} + 128} + \frac {10 \sqrt {2} x^{4} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{128 x^{4} + 256 x^{2} + 128} + \frac {10 \sqrt {2} x^{4} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{128 x^{4} + 256 x^{2} + 128} - \frac {10 \sqrt {2} x^{2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{128 x^{4} + 256 x^{2} + 128} + \frac {10 \sqrt {2} x^{2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{128 x^{4} + 256 x^{2} + 128} + \frac {20 \sqrt {2} x^{2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{128 x^{4} + 256 x^{2} + 128} + \frac {20 \sqrt {2} x^{2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{128 x^{4} + 256 x^{2} + 128} - \frac {5 \sqrt {2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{128 x^{4} + 256 x^{2} + 128} + \frac {5 \sqrt {2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{128 x^{4} + 256 x^{2} + 128} + \frac {10 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{128 x^{4} + 256 x^{2} + 128} + \frac {10 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{128 x^{4} + 256 x^{2} + 128} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)/(x**2+1)**3,x)

[Out]

-72*x**(5/2)/(128*x**4 + 256*x**2 + 128) - 40*sqrt(x)/(128*x**4 + 256*x**2 + 128) - 5*sqrt(2)*x**4*log(-4*sqrt
(2)*sqrt(x) + 4*x + 4)/(128*x**4 + 256*x**2 + 128) + 5*sqrt(2)*x**4*log(4*sqrt(2)*sqrt(x) + 4*x + 4)/(128*x**4
+ 256*x**2 + 128) + 10*sqrt(2)*x**4*atan(sqrt(2)*sqrt(x) - 1)/(128*x**4 + 256*x**2 + 128) + 10*sqrt(2)*x**4*a
tan(sqrt(2)*sqrt(x) + 1)/(128*x**4 + 256*x**2 + 128) - 10*sqrt(2)*x**2*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(128*
x**4 + 256*x**2 + 128) + 10*sqrt(2)*x**2*log(4*sqrt(2)*sqrt(x) + 4*x + 4)/(128*x**4 + 256*x**2 + 128) + 20*sqr
t(2)*x**2*atan(sqrt(2)*sqrt(x) - 1)/(128*x**4 + 256*x**2 + 128) + 20*sqrt(2)*x**2*atan(sqrt(2)*sqrt(x) + 1)/(1
28*x**4 + 256*x**2 + 128) - 5*sqrt(2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(128*x**4 + 256*x**2 + 128) + 5*sqrt(2
)*log(4*sqrt(2)*sqrt(x) + 4*x + 4)/(128*x**4 + 256*x**2 + 128) + 10*sqrt(2)*atan(sqrt(2)*sqrt(x) - 1)/(128*x**
4 + 256*x**2 + 128) + 10*sqrt(2)*atan(sqrt(2)*sqrt(x) + 1)/(128*x**4 + 256*x**2 + 128)

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