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

Optimal. Leaf size=99 \[ \frac{3}{10} \left (x^4+5 x^2+3\right )^{5/2}-\frac{11}{32} \left (2 x^2+5\right ) \left (x^4+5 x^2+3\right )^{3/2}+\frac{429}{256} \left (2 x^2+5\right ) \sqrt{x^4+5 x^2+3}-\frac{5577}{512} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right ) \]

[Out]

(429*(5 + 2*x^2)*Sqrt[3 + 5*x^2 + x^4])/256 - (11*(5 + 2*x^2)*(3 + 5*x^2 + x^4)^(3/2))/32 + (3*(3 + 5*x^2 + x^
4)^(5/2))/10 - (5577*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/512

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Rubi [A]  time = 0.0583216, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {1247, 640, 612, 621, 206} \[ \frac{3}{10} \left (x^4+5 x^2+3\right )^{5/2}-\frac{11}{32} \left (2 x^2+5\right ) \left (x^4+5 x^2+3\right )^{3/2}+\frac{429}{256} \left (2 x^2+5\right ) \sqrt{x^4+5 x^2+3}-\frac{5577}{512} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(429*(5 + 2*x^2)*Sqrt[3 + 5*x^2 + x^4])/256 - (11*(5 + 2*x^2)*(3 + 5*x^2 + x^4)^(3/2))/32 + (3*(3 + 5*x^2 + x^
4)^(5/2))/10 - (5577*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/512

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(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]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[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 \left (2+3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int (2+3 x) \left (3+5 x+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{3}{10} \left (3+5 x^2+x^4\right )^{5/2}-\frac{11}{4} \operatorname{Subst}\left (\int \left (3+5 x+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=-\frac{11}{32} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{3}{10} \left (3+5 x^2+x^4\right )^{5/2}+\frac{429}{64} \operatorname{Subst}\left (\int \sqrt{3+5 x+x^2} \, dx,x,x^2\right )\\ &=\frac{429}{256} \left (5+2 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{11}{32} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{3}{10} \left (3+5 x^2+x^4\right )^{5/2}-\frac{5577}{512} \operatorname{Subst}\left (\int \frac{1}{\sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac{429}{256} \left (5+2 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{11}{32} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{3}{10} \left (3+5 x^2+x^4\right )^{5/2}-\frac{5577}{256} \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{429}{256} \left (5+2 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{11}{32} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{3}{10} \left (3+5 x^2+x^4\right )^{5/2}-\frac{5577}{512} \tanh ^{-1}\left (\frac{5+2 x^2}{2 \sqrt{3+5 x^2+x^4}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0259414, size = 71, normalized size = 0.72 \[ \frac{2 \sqrt{x^4+5 x^2+3} \left (384 x^8+2960 x^6+5304 x^4+2170 x^2+7581\right )-27885 \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )}{2560} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3 + 5*x^2 + x^4]*(7581 + 2170*x^2 + 5304*x^4 + 2960*x^6 + 384*x^8) - 27885*ArcTanh[(5 + 2*x^2)/(2*Sqrt
[3 + 5*x^2 + x^4])])/2560

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Maple [A]  time = 0.014, size = 104, normalized size = 1.1 \begin{align*}{\frac{3\,{x}^{8}}{10}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{37\,{x}^{6}}{16}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{663\,{x}^{4}}{160}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{217\,{x}^{2}}{128}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{7581}{1280}\sqrt{{x}^{4}+5\,{x}^{2}+3}}-{\frac{5577}{512}\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*(3*x^2+2)*(x^4+5*x^2+3)^(3/2),x)

[Out]

3/10*x^8*(x^4+5*x^2+3)^(1/2)+37/16*x^6*(x^4+5*x^2+3)^(1/2)+663/160*x^4*(x^4+5*x^2+3)^(1/2)+217/128*x^2*(x^4+5*
x^2+3)^(1/2)+7581/1280*(x^4+5*x^2+3)^(1/2)-5577/512*ln(5/2+x^2+(x^4+5*x^2+3)^(1/2))

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Maxima [A]  time = 0.945292, size = 136, normalized size = 1.37 \begin{align*} -\frac{11}{16} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}} x^{2} + \frac{3}{10} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{5}{2}} + \frac{429}{128} \, \sqrt{x^{4} + 5 \, x^{2} + 3} x^{2} - \frac{55}{32} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}} + \frac{2145}{256} \, \sqrt{x^{4} + 5 \, x^{2} + 3} - \frac{5577}{512} \, \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*(3*x^2+2)*(x^4+5*x^2+3)^(3/2),x, algorithm="maxima")

[Out]

-11/16*(x^4 + 5*x^2 + 3)^(3/2)*x^2 + 3/10*(x^4 + 5*x^2 + 3)^(5/2) + 429/128*sqrt(x^4 + 5*x^2 + 3)*x^2 - 55/32*
(x^4 + 5*x^2 + 3)^(3/2) + 2145/256*sqrt(x^4 + 5*x^2 + 3) - 5577/512*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Fricas [A]  time = 1.24524, size = 180, normalized size = 1.82 \begin{align*} \frac{1}{1280} \,{\left (384 \, x^{8} + 2960 \, x^{6} + 5304 \, x^{4} + 2170 \, x^{2} + 7581\right )} \sqrt{x^{4} + 5 \, x^{2} + 3} + \frac{5577}{512} \, \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*(3*x^2+2)*(x^4+5*x^2+3)^(3/2),x, algorithm="fricas")

[Out]

1/1280*(384*x^8 + 2960*x^6 + 5304*x^4 + 2170*x^2 + 7581)*sqrt(x^4 + 5*x^2 + 3) + 5577/512*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 \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*(3*x**2+2)*(x**4+5*x**2+3)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.10413, size = 90, normalized size = 0.91 \begin{align*} \frac{1}{1280} \, \sqrt{x^{4} + 5 \, x^{2} + 3}{\left (2 \,{\left (4 \,{\left (2 \,{\left (24 \, x^{2} + 185\right )} x^{2} + 663\right )} x^{2} + 1085\right )} x^{2} + 7581\right )} + \frac{5577}{512} \, \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*(3*x^2+2)*(x^4+5*x^2+3)^(3/2),x, algorithm="giac")

[Out]

1/1280*sqrt(x^4 + 5*x^2 + 3)*(2*(4*(2*(24*x^2 + 185)*x^2 + 663)*x^2 + 1085)*x^2 + 7581) + 5577/512*log(2*x^2 -
2*sqrt(x^4 + 5*x^2 + 3) + 5)