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

Optimal. Leaf size=108 \[ -\frac{2}{5 x^{5/2}}+\frac{2}{\sqrt{x}}+\frac{\log \left (x-\sqrt{2} \sqrt{x}+1\right )}{2 \sqrt{2}}-\frac{\log \left (x+\sqrt{2} \sqrt{x}+1\right )}{2 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{\sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{\sqrt{2}} \]

[Out]

-2/(5*x^(5/2)) + 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])

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Rubi [A]  time = 0.0573483, antiderivative size = 108, normalized size of antiderivative = 1., 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 = {325, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{2}{5 x^{5/2}}+\frac{2}{\sqrt{x}}+\frac{\log \left (x-\sqrt{2} \sqrt{x}+1\right )}{2 \sqrt{2}}-\frac{\log \left (x+\sqrt{2} \sqrt{x}+1\right )}{2 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{\sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*(1 + x^2)),x]

[Out]

-2/(5*x^(5/2)) + 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

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])

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]

Rubi steps

\begin{align*} \int \frac{1}{x^{7/2} \left (1+x^2\right )} \, dx &=-\frac{2}{5 x^{5/2}}-\int \frac{1}{x^{3/2} \left (1+x^2\right )} \, dx\\ &=-\frac{2}{5 x^{5/2}}+\frac{2}{\sqrt{x}}+\int \frac{\sqrt{x}}{1+x^2} \, dx\\ &=-\frac{2}{5 x^{5/2}}+\frac{2}{\sqrt{x}}+2 \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2}{5 x^{5/2}}+\frac{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 )\\ &=-\frac{2}{5 x^{5/2}}+\frac{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}}\\ &=-\frac{2}{5 x^{5/2}}+\frac{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}}\\ &=-\frac{2}{5 x^{5/2}}+\frac{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 [C]  time = 0.0049442, size = 22, normalized size = 0.2 \[ -\frac{2 \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-x^2\right )}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*(1 + x^2)),x]

[Out]

(-2*Hypergeometric2F1[-5/4, 1, -1/4, -x^2])/(5*x^(5/2))

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Maple [A]  time = 0.007, size = 72, normalized size = 0.7 \begin{align*} -{\frac{2}{5}{x}^{-{\frac{5}{2}}}}+2\,{\frac{1}{\sqrt{x}}}+{\frac{\sqrt{2}}{2}\arctan \left ( 1+\sqrt{2}\sqrt{x} \right ) }+{\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/x^(5/2)+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)+1/4*2^(1/2
)*ln((1+x-2^(1/2)*x^(1/2))/(1+x+2^(1/2)*x^(1/2)))

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Maxima [A]  time = 4.31251, size = 116, normalized size = 1.07 \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 ) + \frac{2 \,{\left (5 \, x^{2} - 1\right )}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(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) + 2/5*(5*x^2 - 1)/x^(5
/2)

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Fricas [A]  time = 1.31501, size = 423, normalized size = 3.92 \begin{align*} -\frac{20 \, \sqrt{2} x^{3} \arctan \left (\sqrt{2} \sqrt{\sqrt{2} \sqrt{x} + x + 1} - \sqrt{2} \sqrt{x} - 1\right ) + 20 \, \sqrt{2} x^{3} \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} x^{3} \log \left (4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4\right ) - 5 \, \sqrt{2} x^{3} \log \left (-4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4\right ) - 8 \,{\left (5 \, x^{2} - 1\right )} \sqrt{x}}{20 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(20*sqrt(2)*x^3*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x) + x + 1) - sqrt(2)*sqrt(x) - 1) + 20*sqrt(2)*x^3*arc
tan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*sqrt(x) + 4*x + 4) - sqrt(2)*sqrt(x) + 1) + 5*sqrt(2)*x^3*log(4*sqrt(2)*sqrt(x
) + 4*x + 4) - 5*sqrt(2)*x^3*log(-4*sqrt(2)*sqrt(x) + 4*x + 4) - 8*(5*x^2 - 1)*sqrt(x))/x^3

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Sympy [A]  time = 7.1108, size = 105, normalized size = 0.97 \begin{align*} \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} + \frac{2}{\sqrt{x}} - \frac{2}{5 x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 + 2/sqrt(x) - 2/(5*x**(5/2))

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Giac [A]  time = 1.69134, size = 116, normalized size = 1.07 \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 ) + \frac{2 \,{\left (5 \, x^{2} - 1\right )}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(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) + 2/5*(5*x^2 - 1)/x^(5
/2)