3.14.59 \(\int \frac {(2+x^2) (-4+x+2 x^2) \sqrt {8-7 x^2+2 x^4}}{x^4} \, dx\)

Optimal. Leaf size=98 \[ \frac {\log \left (\sqrt {2} x^2+\sqrt {2 x^4-7 x^2+8}-2 \sqrt {2}\right )}{2 \sqrt {2}}+\frac {\sqrt {2 x^4-7 x^2+8} \left (4 x^4+3 x^3-14 x^2-6 x+16\right )}{6 x^3}-\frac {\log (x)}{2 \sqrt {2}} \]

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Rubi [A]  time = 0.14, antiderivative size = 93, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1662, 1590, 1251, 812, 838, 206} \begin {gather*} -\frac {\left (2-x^2\right ) \sqrt {2 x^4-7 x^2+8}}{2 x^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \left (2-x^2\right )}{\sqrt {2 x^4-7 x^2+8}}\right )}{2 \sqrt {2}}+\frac {\left (2 x^4-7 x^2+8\right )^{3/2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2 + x^2)*(-4 + x + 2*x^2)*Sqrt[8 - 7*x^2 + 2*x^4])/x^4,x]

[Out]

-1/2*((2 - x^2)*Sqrt[8 - 7*x^2 + 2*x^4])/x^2 + (8 - 7*x^2 + 2*x^4)^(3/2)/(3*x^3) - ArcTanh[(Sqrt[2]*(2 - x^2))
/Sqrt[8 - 7*x^2 + 2*x^4]]/(2*Sqrt[2])

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 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 838

Int[((f_) + (g_.)*(x_))/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[(4*f
*(a - d))/(b*d - a*e), Subst[Int[1/(4*(a - d) - x^2), x], x, (2*(a - d) + (b - e)*x)/Sqrt[a + b*x + c*x^2]], x
] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[4*c*(a - d) - (b - e)^2, 0] && EqQ[e*f*(b - e) - 2*g*(b*d - a*e),
0] && NeQ[b*d - a*e, 0]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 1590

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[(Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*Rr^(n + 1))/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x
, r]), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rule 1662

Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x],
k}, Int[(d*x)^m*Sum[Coeff[Pq, x, 2*k]*x^(2*k), {k, 0, q/2 + 1}]*(a + b*x^2 + c*x^4)^p, x] + Dist[1/d, Int[(d*
x)^(m + 1)*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0, (q - 1)/2 + 1}]*(a + b*x^2 + c*x^4)^p, x], x]] /; FreeQ[{
a, b, c, d, m, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 105, normalized size = 1.07 \begin {gather*} \frac {\sinh ^{-1}\left (\frac {4 x^2-7}{\sqrt {15}}\right )-\tanh ^{-1}\left (\frac {16-7 x^2}{4 \sqrt {2} \sqrt {2 x^4-7 x^2+8}}\right )}{4 \sqrt {2}}+\sqrt {2 x^4-7 x^2+8} \left (\frac {8}{3 x^3}-\frac {1}{x^2}+\frac {2 x}{3}-\frac {7}{3 x}+\frac {1}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2 + x^2)*(-4 + x + 2*x^2)*Sqrt[8 - 7*x^2 + 2*x^4])/x^4,x]

[Out]

(1/2 + 8/(3*x^3) - x^(-2) - 7/(3*x) + (2*x)/3)*Sqrt[8 - 7*x^2 + 2*x^4] + (ArcSinh[(-7 + 4*x^2)/Sqrt[15]] - Arc
Tanh[(16 - 7*x^2)/(4*Sqrt[2]*Sqrt[8 - 7*x^2 + 2*x^4])])/(4*Sqrt[2])

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IntegrateAlgebraic [A]  time = 0.70, size = 98, normalized size = 1.00 \begin {gather*} \frac {\sqrt {8-7 x^2+2 x^4} \left (16-6 x-14 x^2+3 x^3+4 x^4\right )}{6 x^3}-\frac {\log (x)}{2 \sqrt {2}}+\frac {\log \left (-2 \sqrt {2}+\sqrt {2} x^2+\sqrt {8-7 x^2+2 x^4}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((2 + x^2)*(-4 + x + 2*x^2)*Sqrt[8 - 7*x^2 + 2*x^4])/x^4,x]

[Out]

(Sqrt[8 - 7*x^2 + 2*x^4]*(16 - 6*x - 14*x^2 + 3*x^3 + 4*x^4))/(6*x^3) - Log[x]/(2*Sqrt[2]) + Log[-2*Sqrt[2] +
Sqrt[2]*x^2 + Sqrt[8 - 7*x^2 + 2*x^4]]/(2*Sqrt[2])

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fricas [A]  time = 0.69, size = 91, normalized size = 0.93 \begin {gather*} \frac {3 \, \sqrt {2} x^{3} \log \left (\frac {4 \, x^{4} + 2 \, \sqrt {2} \sqrt {2 \, x^{4} - 7 \, x^{2} + 8} {\left (x^{2} - 2\right )} - 15 \, x^{2} + 16}{x^{2}}\right ) + 4 \, {\left (4 \, x^{4} + 3 \, x^{3} - 14 \, x^{2} - 6 \, x + 16\right )} \sqrt {2 \, x^{4} - 7 \, x^{2} + 8}}{24 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)*(2*x^2+x-4)*(2*x^4-7*x^2+8)^(1/2)/x^4,x, algorithm="fricas")

[Out]

1/24*(3*sqrt(2)*x^3*log((4*x^4 + 2*sqrt(2)*sqrt(2*x^4 - 7*x^2 + 8)*(x^2 - 2) - 15*x^2 + 16)/x^2) + 4*(4*x^4 +
3*x^3 - 14*x^2 - 6*x + 16)*sqrt(2*x^4 - 7*x^2 + 8))/x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2 \, x^{4} - 7 \, x^{2} + 8} {\left (2 \, x^{2} + x - 4\right )} {\left (x^{2} + 2\right )}}{x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)*(2*x^2+x-4)*(2*x^4-7*x^2+8)^(1/2)/x^4,x, algorithm="giac")

[Out]

integrate(sqrt(2*x^4 - 7*x^2 + 8)*(2*x^2 + x - 4)*(x^2 + 2)/x^4, x)

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maple [C]  time = 0.64, size = 88, normalized size = 0.90

method	result	size
trager	\(\frac {\sqrt {2 x^{4}-7 x^{2}+8}\, \left (4 x^{4}+3 x^{3}-14 x^{2}-6 x +16\right )}{6 x^{3}}-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}-2\right ) x^{2}+\sqrt {2 x^{4}-7 x^{2}+8}+2 \RootOf \left (\textit {\_Z}^{2}-2\right )}{x}\right )}{4}\)	\(88\)
elliptic	\(\frac {\sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {15}\, \left (x^{2}-\frac {7}{4}\right )}{15}\right )}{8}+\frac {\sqrt {2 x^{4}-7 x^{2}+8}}{2}-\frac {\sqrt {2 x^{4}-7 x^{2}+8}}{x^{2}}-\frac {\sqrt {2}\, \arctanh \left (\frac {\left (-7 x^{2}+16\right ) \sqrt {2}}{8 \sqrt {2 x^{4}-7 x^{2}+8}}\right )}{8}+\frac {\left (2 x^{4}-7 x^{2}+8\right )^{\frac {3}{2}}}{3 x^{3}}\)	\(104\)
risch	\(-\frac {14 x^{6}+6 x^{5}-65 x^{4}-21 x^{3}+112 x^{2}+24 x -64}{3 x^{3} \sqrt {2 x^{4}-7 x^{2}+8}}+\frac {2 x \sqrt {2 x^{4}-7 x^{2}+8}}{3}+\frac {\sqrt {2 x^{4}-7 x^{2}+8}}{2}+\frac {\sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {15}\, \left (x^{2}-\frac {7}{4}\right )}{15}\right )}{8}-\frac {\sqrt {2}\, \arctanh \left (\frac {\left (-7 x^{2}+16\right ) \sqrt {2}}{8 \sqrt {2 x^{4}-7 x^{2}+8}}\right )}{8}\)	\(132\)
default	\(\frac {2 x \sqrt {2 x^{4}-7 x^{2}+8}}{3}+\frac {8 \sqrt {2 x^{4}-7 x^{2}+8}}{3 x^{3}}-\frac {7 \sqrt {2 x^{4}-7 x^{2}+8}}{3 x}+\frac {\sqrt {2 x^{4}-7 x^{2}+8}}{16}+\frac {\sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {15}\, \left (x^{2}-\frac {7}{4}\right )}{15}\right )}{8}-\frac {\sqrt {2}\, \arctanh \left (\frac {\left (-7 x^{2}+16\right ) \sqrt {2}}{8 \sqrt {2 x^{4}-7 x^{2}+8}}\right )}{8}-\frac {\left (2 x^{4}-7 x^{2}+8\right )^{\frac {3}{2}}}{8 x^{2}}+\frac {\left (4 x^{2}-7\right ) \sqrt {2 x^{4}-7 x^{2}+8}}{16}\)	\(163\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+2)*(2*x^2+x-4)*(2*x^4-7*x^2+8)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/6*(2*x^4-7*x^2+8)^(1/2)*(4*x^4+3*x^3-14*x^2-6*x+16)/x^3-1/4*RootOf(_Z^2-2)*ln((-RootOf(_Z^2-2)*x^2+(2*x^4-7*
x^2+8)^(1/2)+2*RootOf(_Z^2-2))/x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2 \, x^{4} - 7 \, x^{2} + 8} {\left (2 \, x^{2} + x - 4\right )} {\left (x^{2} + 2\right )}}{x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)*(2*x^2+x-4)*(2*x^4-7*x^2+8)^(1/2)/x^4,x, algorithm="maxima")

[Out]

integrate(sqrt(2*x^4 - 7*x^2 + 8)*(2*x^2 + x - 4)*(x^2 + 2)/x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (x^2+2\right )\,\left (2\,x^2+x-4\right )\,\sqrt {2\,x^4-7\,x^2+8}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2 + 2)*(x + 2*x^2 - 4)*(2*x^4 - 7*x^2 + 8)^(1/2))/x^4,x)

[Out]

int(((x^2 + 2)*(x + 2*x^2 - 4)*(2*x^4 - 7*x^2 + 8)^(1/2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} + 2\right ) \left (2 x^{2} + x - 4\right ) \sqrt {2 x^{4} - 7 x^{2} + 8}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+2)*(2*x**2+x-4)*(2*x**4-7*x**2+8)**(1/2)/x**4,x)

[Out]

Integral((x**2 + 2)*(2*x**2 + x - 4)*sqrt(2*x**4 - 7*x**2 + 8)/x**4, x)

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