3.156 \(\int x^5 (2+3 x^2) (3+5 x^2+x^4)^{3/2} \, dx\)

Optimal. Leaf size=127 \[ \frac{3}{14} \left (x^4+5 x^2+3\right )^{5/2} x^4+\frac{\left (3313-1070 x^2\right ) \left (x^4+5 x^2+3\right )^{5/2}}{1680}-\frac{2183}{768} \left (2 x^2+5\right ) \left (x^4+5 x^2+3\right )^{3/2}+\frac{28379 \left (2 x^2+5\right ) \sqrt{x^4+5 x^2+3}}{2048}-\frac{368927 \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )}{4096} \]

[Out]

(28379*(5 + 2*x^2)*Sqrt[3 + 5*x^2 + x^4])/2048 - (2183*(5 + 2*x^2)*(3 + 5*x^2 + x^4)^(3/2))/768 + (3*x^4*(3 +
5*x^2 + x^4)^(5/2))/14 + ((3313 - 1070*x^2)*(3 + 5*x^2 + x^4)^(5/2))/1680 - (368927*ArcTanh[(5 + 2*x^2)/(2*Sqr
t[3 + 5*x^2 + x^4])])/4096

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Rubi [A]  time = 0.0958111, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1251, 832, 779, 612, 621, 206} \[ \frac{3}{14} \left (x^4+5 x^2+3\right )^{5/2} x^4+\frac{\left (3313-1070 x^2\right ) \left (x^4+5 x^2+3\right )^{5/2}}{1680}-\frac{2183}{768} \left (2 x^2+5\right ) \left (x^4+5 x^2+3\right )^{3/2}+\frac{28379 \left (2 x^2+5\right ) \sqrt{x^4+5 x^2+3}}{2048}-\frac{368927 \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )}{4096} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(28379*(5 + 2*x^2)*Sqrt[3 + 5*x^2 + x^4])/2048 - (2183*(5 + 2*x^2)*(3 + 5*x^2 + x^4)^(3/2))/768 + (3*x^4*(3 +
5*x^2 + x^4)^(5/2))/14 + ((3313 - 1070*x^2)*(3 + 5*x^2 + x^4)^(5/2))/1680 - (368927*ArcTanh[(5 + 2*x^2)/(2*Sqr
t[3 + 5*x^2 + x^4])])/4096

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 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 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 612

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

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 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])

Rubi steps

\begin{align*} \int x^5 \left (2+3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 (2+3 x) \left (3+5 x+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac{1}{14} \operatorname{Subst}\left (\int \left (-18-\frac{107 x}{2}\right ) x \left (3+5 x+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac{\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}-\frac{2183}{96} \operatorname{Subst}\left (\int \left (3+5 x+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=-\frac{2183}{768} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac{\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}+\frac{28379}{512} \operatorname{Subst}\left (\int \sqrt{3+5 x+x^2} \, dx,x,x^2\right )\\ &=\frac{28379 \left (5+2 x^2\right ) \sqrt{3+5 x^2+x^4}}{2048}-\frac{2183}{768} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac{\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}-\frac{368927 \operatorname{Subst}\left (\int \frac{1}{\sqrt{3+5 x+x^2}} \, dx,x,x^2\right )}{4096}\\ &=\frac{28379 \left (5+2 x^2\right ) \sqrt{3+5 x^2+x^4}}{2048}-\frac{2183}{768} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac{\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}-\frac{368927 \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{5+2 x^2}{\sqrt{3+5 x^2+x^4}}\right )}{2048}\\ &=\frac{28379 \left (5+2 x^2\right ) \sqrt{3+5 x^2+x^4}}{2048}-\frac{2183}{768} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac{\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}-\frac{368927 \tanh ^{-1}\left (\frac{5+2 x^2}{2 \sqrt{3+5 x^2+x^4}}\right )}{4096}\\ \end{align*}

Mathematica [A]  time = 0.0384189, size = 81, normalized size = 0.64 \[ \frac{2 \sqrt{x^4+5 x^2+3} \left (46080 x^{12}+323840 x^{10}+482944 x^8+154800 x^6+283304 x^4-1499570 x^2+9546951\right )-38737335 \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )}{430080} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3 + 5*x^2 + x^4]*(9546951 - 1499570*x^2 + 283304*x^4 + 154800*x^6 + 482944*x^8 + 323840*x^10 + 46080*x
^12) - 38737335*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/430080

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Maple [A]  time = 0.034, size = 138, normalized size = 1.1 \begin{align*}{\frac{3\,{x}^{12}}{14}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{253\,{x}^{10}}{168}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{539\,{x}^{8}}{240}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{3182317}{71680}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{5059\,{x}^{4}}{3840}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{645\,{x}^{6}}{896}\sqrt{{x}^{4}+5\,{x}^{2}+3}}-{\frac{149957\,{x}^{2}}{21504}\sqrt{{x}^{4}+5\,{x}^{2}+3}}-{\frac{368927}{4096}\ln \left ({\frac{5}{2}}+{x}^{2}+\sqrt{{x}^{4}+5\,{x}^{2}+3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*x^12*(x^4+5*x^2+3)^(1/2)+253/168*x^10*(x^4+5*x^2+3)^(1/2)+539/240*x^8*(x^4+5*x^2+3)^(1/2)+3182317/71680*(
x^4+5*x^2+3)^(1/2)+5059/3840*x^4*(x^4+5*x^2+3)^(1/2)+645/896*x^6*(x^4+5*x^2+3)^(1/2)-149957/21504*x^2*(x^4+5*x
^2+3)^(1/2)-368927/4096*ln(5/2+x^2+(x^4+5*x^2+3)^(1/2))

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Maxima [A]  time = 1.05815, size = 182, normalized size = 1.43 \begin{align*} \frac{3}{14} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{5}{2}} x^{4} - \frac{107}{168} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{5}{2}} x^{2} - \frac{2183}{384} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}} x^{2} + \frac{3313}{1680} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{5}{2}} + \frac{28379}{1024} \, \sqrt{x^{4} + 5 \, x^{2} + 3} x^{2} - \frac{10915}{768} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}} + \frac{141895}{2048} \, \sqrt{x^{4} + 5 \, x^{2} + 3} - \frac{368927}{4096} \, \log \left (2 \, x^{2} + 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3} + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*(x^4 + 5*x^2 + 3)^(5/2)*x^4 - 107/168*(x^4 + 5*x^2 + 3)^(5/2)*x^2 - 2183/384*(x^4 + 5*x^2 + 3)^(3/2)*x^2
+ 3313/1680*(x^4 + 5*x^2 + 3)^(5/2) + 28379/1024*sqrt(x^4 + 5*x^2 + 3)*x^2 - 10915/768*(x^4 + 5*x^2 + 3)^(3/2)
+ 141895/2048*sqrt(x^4 + 5*x^2 + 3) - 368927/4096*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Fricas [A]  time = 1.2053, size = 240, normalized size = 1.89 \begin{align*} \frac{1}{215040} \,{\left (46080 \, x^{12} + 323840 \, x^{10} + 482944 \, x^{8} + 154800 \, x^{6} + 283304 \, x^{4} - 1499570 \, x^{2} + 9546951\right )} \sqrt{x^{4} + 5 \, x^{2} + 3} + \frac{368927}{4096} \, \log \left (-2 \, x^{2} + 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3} - 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/215040*(46080*x^12 + 323840*x^10 + 482944*x^8 + 154800*x^6 + 283304*x^4 - 1499570*x^2 + 9546951)*sqrt(x^4 +
5*x^2 + 3) + 368927/4096*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \left (3 x^{2} + 2\right ) \left (x^{4} + 5 x^{2} + 3\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12456, size = 109, normalized size = 0.86 \begin{align*} \frac{1}{215040} \, \sqrt{x^{4} + 5 \, x^{2} + 3}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (10 \,{\left (36 \, x^{2} + 253\right )} x^{2} + 3773\right )} x^{2} + 9675\right )} x^{2} + 35413\right )} x^{2} - 749785\right )} x^{2} + 9546951\right )} + \frac{368927}{4096} \, \log \left (2 \, x^{2} - 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3} + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/215040*sqrt(x^4 + 5*x^2 + 3)*(2*(4*(2*(8*(10*(36*x^2 + 253)*x^2 + 3773)*x^2 + 9675)*x^2 + 35413)*x^2 - 74978
5)*x^2 + 9546951) + 368927/4096*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)