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]

-11/32*(2*x^2+5)*(x^4+5*x^2+3)^(3/2)+3/10*(x^4+5*x^2+3)^(5/2)-5577/512*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/
2))+429/256*(2*x^2+5)*(x^4+5*x^2+3)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 99, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.64, size = 61, normalized size = 0.62 \[ \frac {1}{1280} \, {\left (384 \, x^{8} + 2960 \, x^{6} + 5304 \, x^{4} + 2170 \, x^{2} + 7581\right )} \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {5577}{512} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) \]

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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giac [A]  time = 0.45, size = 151, normalized size = 1.53 \[ \frac {1}{1280} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x^{2} + 5\right )} x^{2} - 127\right )} x^{2} + 2635\right )} x^{2} - 33429\right )} + \frac {17}{384} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + 5\right )} x^{2} - 89\right )} x^{2} + 1095\right )} + \frac {19}{48} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, x^{2} + 5\right )} x^{2} - 51\right )} + \frac {3}{4} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, x^{2} + 5\right )} + \frac {5577}{512} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(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*(6*(8*x^2 + 5)*x^2 - 127)*x^2 + 2635)*x^2 - 33429) + 17/384*sqrt(x^4 + 5*x^
2 + 3)*(2*(4*(6*x^2 + 5)*x^2 - 89)*x^2 + 1095) + 19/48*sqrt(x^4 + 5*x^2 + 3)*(2*(4*x^2 + 5)*x^2 - 51) + 3/4*sq
rt(x^4 + 5*x^2 + 3)*(2*x^2 + 5) + 5577/512*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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maple [A]  time = 0.02, size = 104, normalized size = 1.05 \[ \frac {3 \sqrt {x^{4}+5 x^{2}+3}\, x^{8}}{10}+\frac {37 \sqrt {x^{4}+5 x^{2}+3}\, x^{6}}{16}+\frac {663 \sqrt {x^{4}+5 x^{2}+3}\, x^{4}}{160}+\frac {217 \sqrt {x^{4}+5 x^{2}+3}\, x^{2}}{128}-\frac {5577 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{512}+\frac {7581 \sqrt {x^{4}+5 x^{2}+3}}{1280} \]

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

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maxima [A]  time = 0.67, size = 101, normalized size = 1.02 \[ -\frac {11}{16} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} x^{2} + \frac {3}{10} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {5}{2}} + \frac {429}{128} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} - \frac {55}{32} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} + \frac {2145}{256} \, \sqrt {x^{4} + 5 \, x^{2} + 3} - \frac {5577}{512} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

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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mupad [B]  time = 0.53, size = 127, normalized size = 1.28 \[ \frac {\left (x^2+\frac {5}{2}\right )\,{\left (x^4+5\,x^2+3\right )}^{3/2}}{4}-\frac {15\,x^2\,{\left (x^4+5\,x^2+3\right )}^{3/2}}{16}-\frac {5577\,\ln \left (\sqrt {x^4+5\,x^2+3}+x^2+\frac {5}{2}\right )}{512}+\frac {585\,\left (2\,x^2+5\right )\,\sqrt {x^4+5\,x^2+3}}{256}-\frac {39\,\left (\frac {x^2}{2}+\frac {5}{4}\right )\,\sqrt {x^4+5\,x^2+3}}{16}-\frac {75\,{\left (x^4+5\,x^2+3\right )}^{3/2}}{32}+\frac {3\,{\left (x^4+5\,x^2+3\right )}^{5/2}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x^2 + 5/2)*(5*x^2 + x^4 + 3)^(3/2))/4 - (15*x^2*(5*x^2 + x^4 + 3)^(3/2))/16 - (5577*log((5*x^2 + x^4 + 3)^(1
/2) + x^2 + 5/2))/512 + (585*(2*x^2 + 5)*(5*x^2 + x^4 + 3)^(1/2))/256 - (39*(x^2/2 + 5/4)*(5*x^2 + x^4 + 3)^(1
/2))/16 - (75*(5*x^2 + x^4 + 3)^(3/2))/32 + (3*(5*x^2 + x^4 + 3)^(5/2))/10

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

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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