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

Optimal. Leaf size=43 \[ -\frac{2 x^{5/2}}{5}-2 \sqrt{x}-\log \left (1-\sqrt{x}\right )+\frac{1}{2} \log (x+1)+\tan ^{-1}\left (\sqrt{x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.068065, antiderivative size = 31, normalized size of antiderivative = 0.72, number of steps used = 10, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {1833, 275, 206, 302, 212, 203} \[ -\frac{2 x^{5/2}}{5}-2 \sqrt{x}+\tan ^{-1}\left (\sqrt{x}\right )+\tanh ^{-1}\left (\sqrt{x}\right )+\tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^(7/2))/(1 - x^2),x]

[Out]

-2*Sqrt[x] - (2*x^(5/2))/5 + ArcTan[Sqrt[x]] + ArcTanh[Sqrt[x]] + ArcTanh[x]

Rule 1833

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[((c*x)^(m + j)*Sum[Coeff[Pq, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p)/c^j, {
j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rule 275

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

Rule 206

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

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 212

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

Rule 203

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

Rubi steps

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

Mathematica [C]  time = 0.0291391, size = 67, normalized size = 1.56 \[ -\frac{2 x^{5/2}}{5}-2 \sqrt{x}+\left (\frac{1}{2}-\frac{i}{2}\right ) \log \left (-\sqrt{x}+i\right )-\log \left (1-\sqrt{x}\right )+\left (\frac{1}{2}+\frac{i}{2}\right ) \log \left (\sqrt{x}+i\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^(7/2))/(1 - x^2),x]

[Out]

-2*Sqrt[x] - (2*x^(5/2))/5 + (1/2 - I/2)*Log[I - Sqrt[x]] - Log[1 - Sqrt[x]] + (1/2 + I/2)*Log[I + Sqrt[x]]

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Maple [A]  time = 0.006, size = 34, normalized size = 0.8 \begin{align*} -{\frac{2}{5}{x}^{{\frac{5}{2}}}}-2\,\sqrt{x}-{\frac{1}{2}\ln \left ( -1+\sqrt{x} \right ) }+{\frac{1}{2}\ln \left ( 1+\sqrt{x} \right ) }+\arctan \left ( \sqrt{x} \right ) +{\it Artanh} \left ( x \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*ln(-1+x^(1/2))+1/2*ln(1+x^(1/2))+arctan(x^(1/2))+arctanh(x)

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Maxima [A]  time = 1.54976, size = 39, normalized size = 0.91 \begin{align*} -\frac{2}{5} \, x^{\frac{5}{2}} - 2 \, \sqrt{x} + \arctan \left (\sqrt{x}\right ) + \frac{1}{2} \, \log \left (x + 1\right ) - \log \left (\sqrt{x} - 1\right ) \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]

-2/5*x^(5/2) - 2*sqrt(x) + arctan(sqrt(x)) + 1/2*log(x + 1) - log(sqrt(x) - 1)

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Fricas [A]  time = 1.48442, size = 105, normalized size = 2.44 \begin{align*} -\frac{2}{5} \,{\left (x^{2} + 5\right )} \sqrt{x} + \arctan \left (\sqrt{x}\right ) + \frac{1}{2} \, \log \left (x + 1\right ) - \log \left (\sqrt{x} - 1\right ) \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]

-2/5*(x^2 + 5)*sqrt(x) + arctan(sqrt(x)) + 1/2*log(x + 1) - log(sqrt(x) - 1)

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Sympy [A]  time = 2.91003, size = 36, normalized size = 0.84 \begin{align*} - \frac{2 x^{\frac{5}{2}}}{5} - 2 \sqrt{x} - \log{\left (\sqrt{x} - 1 \right )} + \frac{\log{\left (x + 1 \right )}}{2} + \operatorname{atan}{\left (\sqrt{x} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x**(5/2)/5 - 2*sqrt(x) - log(sqrt(x) - 1) + log(x + 1)/2 + atan(sqrt(x))

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Giac [A]  time = 1.17964, size = 41, normalized size = 0.95 \begin{align*} -\frac{2}{5} \, x^{\frac{5}{2}} - 2 \, \sqrt{x} + \arctan \left (\sqrt{x}\right ) + \frac{1}{2} \, \log \left (x + 1\right ) - \log \left ({\left | \sqrt{x} - 1 \right |}\right ) \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]

-2/5*x^(5/2) - 2*sqrt(x) + arctan(sqrt(x)) + 1/2*log(x + 1) - log(abs(sqrt(x) - 1))