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]

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

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Rubi [A]  time = 0.07, 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 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])

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 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 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 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 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)]

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*}

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Mathematica [C]  time = 0.03, 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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fricas [A]  time = 0.46, size = 29, normalized size = 0.67 \[ -\frac {2}{5} \, {\left (x^{2} + 5\right )} \sqrt {x} + \arctan \left (\sqrt {x}\right ) + \frac {1}{2} \, \log \left (x + 1\right ) - \log \left (\sqrt {x} - 1\right ) \]

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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giac [A]  time = 0.34, size = 30, normalized size = 0.70 \[ -\frac {2}{5} \, x^{\frac {5}{2}} - 2 \, \sqrt {x} + \arctan \left (\sqrt {x}\right ) + \frac {1}{2} \, \log \left (x + 1\right ) - \log \left ({\left | \sqrt {x} - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.01, size = 34, normalized size = 0.79 \[ -\frac {2 x^{\frac {5}{2}}}{5}+\arctanh \relax (x )+\arctan \left (\sqrt {x}\right )-\frac {\ln \left (\sqrt {x}-1\right )}{2}+\frac {\ln \left (\sqrt {x}+1\right )}{2}-2 \sqrt {x} \]

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

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maxima [A]  time = 1.80, size = 29, normalized size = 0.67 \[ -\frac {2}{5} \, x^{\frac {5}{2}} - 2 \, \sqrt {x} + \arctan \left (\sqrt {x}\right ) + \frac {1}{2} \, \log \left (x + 1\right ) - \log \left (\sqrt {x} - 1\right ) \]

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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mupad [B]  time = 3.11, size = 53, normalized size = 1.23 \[ -\ln \left (10\,\sqrt {x}-10\right )-2\,\sqrt {x}-\frac {2\,x^{5/2}}{5}+\ln \left (1+\sqrt {x}\,\left (-3-\mathrm {i}\right )-3{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\ln \left (1+\sqrt {x}\,\left (-3+1{}\mathrm {i}\right )+3{}\mathrm {i}\right )\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^(7/2) + 1)/(x^2 - 1),x)

[Out]

log((1 - 3i) - x^(1/2)*(3 + 1i))*(1/2 + 1i/2) - log(10*x^(1/2) - 10) + log((1 + 3i) - x^(1/2)*(3 - 1i))*(1/2 -
1i/2) - 2*x^(1/2) - (2*x^(5/2))/5

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sympy [A]  time = 2.43, size = 36, normalized size = 0.84 \[ - \frac {2 x^{\frac {5}{2}}}{5} - 2 \sqrt {x} - \log {\left (\sqrt {x} - 1 \right )} + \frac {\log {\left (x + 1 \right )}}{2} + \operatorname {atan}{\left (\sqrt {x} \right )} \]

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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