3.181 \(\int \frac {x^7 (2+3 x^2)}{\sqrt {3+5 x^2+x^4}} \, dx\)

Optimal. Leaf size=98 \[ -\frac {89}{48} \sqrt {x^4+5 x^2+3} x^4-\frac {1}{384} \left (24243-3802 x^2\right ) \sqrt {x^4+5 x^2+3}+\frac {32801}{256} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )+\frac {3}{8} \sqrt {x^4+5 x^2+3} x^6 \]

[Out]

32801/256*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))-89/48*x^4*(x^4+5*x^2+3)^(1/2)+3/8*x^6*(x^4+5*x^2+3)^(1/2)
-1/384*(-3802*x^2+24243)*(x^4+5*x^2+3)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1251, 832, 779, 621, 206} \[ \frac {3}{8} \sqrt {x^4+5 x^2+3} x^6-\frac {89}{48} \sqrt {x^4+5 x^2+3} x^4-\frac {1}{384} \left (24243-3802 x^2\right ) \sqrt {x^4+5 x^2+3}+\frac {32801}{256} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x^7*(2 + 3*x^2))/Sqrt[3 + 5*x^2 + x^4],x]

[Out]

(-89*x^4*Sqrt[3 + 5*x^2 + x^4])/48 + (3*x^6*Sqrt[3 + 5*x^2 + x^4])/8 - ((24243 - 3802*x^2)*Sqrt[3 + 5*x^2 + x^
4])/384 + (32801*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/256

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 621

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

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 832

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

Rubi steps

\begin {align*} \int \frac {x^7 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3 (2+3 x)}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {3}{8} x^6 \sqrt {3+5 x^2+x^4}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {\left (-27-\frac {89 x}{2}\right ) x^2}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {89}{48} x^4 \sqrt {3+5 x^2+x^4}+\frac {3}{8} x^6 \sqrt {3+5 x^2+x^4}+\frac {1}{24} \operatorname {Subst}\left (\int \frac {x \left (267+\frac {1901 x}{4}\right )}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {89}{48} x^4 \sqrt {3+5 x^2+x^4}+\frac {3}{8} x^6 \sqrt {3+5 x^2+x^4}-\frac {1}{384} \left (24243-3802 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {32801}{256} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {89}{48} x^4 \sqrt {3+5 x^2+x^4}+\frac {3}{8} x^6 \sqrt {3+5 x^2+x^4}-\frac {1}{384} \left (24243-3802 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {32801}{128} \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {5+2 x^2}{\sqrt {3+5 x^2+x^4}}\right )\\ &=-\frac {89}{48} x^4 \sqrt {3+5 x^2+x^4}+\frac {3}{8} x^6 \sqrt {3+5 x^2+x^4}-\frac {1}{384} \left (24243-3802 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {32801}{256} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.67 \[ \frac {1}{768} \left (98403 \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )+2 \sqrt {x^4+5 x^2+3} \left (144 x^6-712 x^4+3802 x^2-24243\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(2 + 3*x^2))/Sqrt[3 + 5*x^2 + x^4],x]

[Out]

(2*Sqrt[3 + 5*x^2 + x^4]*(-24243 + 3802*x^2 - 712*x^4 + 144*x^6) + 98403*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2
+ x^4])])/768

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fricas [A]  time = 0.55, size = 56, normalized size = 0.57 \[ \frac {1}{384} \, {\left (144 \, x^{6} - 712 \, x^{4} + 3802 \, x^{2} - 24243\right )} \sqrt {x^{4} + 5 \, x^{2} + 3} - \frac {32801}{256} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x, algorithm="fricas")

[Out]

1/384*(144*x^6 - 712*x^4 + 3802*x^2 - 24243)*sqrt(x^4 + 5*x^2 + 3) - 32801/256*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2
+ 3) - 5)

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giac [A]  time = 0.37, size = 60, normalized size = 0.61 \[ \frac {1}{384} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, {\left (18 \, x^{2} - 89\right )} x^{2} + 1901\right )} x^{2} - 24243\right )} - \frac {32801}{256} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x, algorithm="giac")

[Out]

1/384*sqrt(x^4 + 5*x^2 + 3)*(2*(4*(18*x^2 - 89)*x^2 + 1901)*x^2 - 24243) - 32801/256*log(2*x^2 - 2*sqrt(x^4 +
5*x^2 + 3) + 5)

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maple [A]  time = 0.02, size = 87, normalized size = 0.89 \[ \frac {3 \sqrt {x^{4}+5 x^{2}+3}\, x^{6}}{8}-\frac {89 \sqrt {x^{4}+5 x^{2}+3}\, x^{4}}{48}+\frac {1901 \sqrt {x^{4}+5 x^{2}+3}\, x^{2}}{192}+\frac {32801 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{256}-\frac {8081 \sqrt {x^{4}+5 x^{2}+3}}{128} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x)

[Out]

3/8*(x^4+5*x^2+3)^(1/2)*x^6-89/48*(x^4+5*x^2+3)^(1/2)*x^4+1901/192*(x^4+5*x^2+3)^(1/2)*x^2-8081/128*(x^4+5*x^2
+3)^(1/2)+32801/256*ln(x^2+5/2+(x^4+5*x^2+3)^(1/2))

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maxima [A]  time = 0.93, size = 90, normalized size = 0.92 \[ \frac {3}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{6} - \frac {89}{48} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{4} + \frac {1901}{192} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} - \frac {8081}{128} \, \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {32801}{256} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

3/8*sqrt(x^4 + 5*x^2 + 3)*x^6 - 89/48*sqrt(x^4 + 5*x^2 + 3)*x^4 + 1901/192*sqrt(x^4 + 5*x^2 + 3)*x^2 - 8081/12
8*sqrt(x^4 + 5*x^2 + 3) + 32801/256*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^7\,\left (3\,x^2+2\right )}{\sqrt {x^4+5\,x^2+3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^7*(3*x^2 + 2))/(5*x^2 + x^4 + 3)^(1/2),x)

[Out]

int((x^7*(3*x^2 + 2))/(5*x^2 + x^4 + 3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{7} \left (3 x^{2} + 2\right )}{\sqrt {x^{4} + 5 x^{2} + 3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(3*x**2+2)/(x**4+5*x**2+3)**(1/2),x)

[Out]

Integral(x**7*(3*x**2 + 2)/sqrt(x**4 + 5*x**2 + 3), x)

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