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

Optimal. Leaf size=119 \[ \frac {1}{48} \left (18 x^2+61\right ) \left (x^4+5 x^2+3\right )^{3/2}+\frac {1}{128} \left (199-74 x^2\right ) \sqrt {x^4+5 x^2+3}+\frac {2401}{256} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-3 \sqrt {3} \tanh ^{-1}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right ) \]

[Out]

1/48*(18*x^2+61)*(x^4+5*x^2+3)^(3/2)+2401/256*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))-3*arctanh(1/6*(5*x^2+
6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)+1/128*(-74*x^2+199)*(x^4+5*x^2+3)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1251, 814, 843, 621, 206, 724} \[ \frac {1}{48} \left (18 x^2+61\right ) \left (x^4+5 x^2+3\right )^{3/2}+\frac {1}{128} \left (199-74 x^2\right ) \sqrt {x^4+5 x^2+3}+\frac {2401}{256} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-3 \sqrt {3} \tanh ^{-1}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((199 - 74*x^2)*Sqrt[3 + 5*x^2 + x^4])/128 + ((61 + 18*x^2)*(3 + 5*x^2 + x^4)^(3/2))/48 + (2401*ArcTanh[(5 + 2
*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/256 - 3*Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])]

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 724

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

Rule 814

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

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 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 {\left (2+3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(2+3 x) \left (3+5 x+x^2\right )^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{48} \left (61+18 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac {1}{16} \operatorname {Subst}\left (\int \frac {\left (-48+\frac {37 x}{2}\right ) \sqrt {3+5 x+x^2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{128} \left (199-74 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{48} \left (61+18 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {1}{64} \operatorname {Subst}\left (\int \frac {576+\frac {2401 x}{4}}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{128} \left (199-74 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{48} \left (61+18 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+9 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )+\frac {2401}{256} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{128} \left (199-74 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{48} \left (61+18 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-18 \operatorname {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {6+5 x^2}{\sqrt {3+5 x^2+x^4}}\right )+\frac {2401}{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 {1}{128} \left (199-74 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{48} \left (61+18 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {2401}{256} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )-3 \sqrt {3} \tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 104, normalized size = 0.87 \[ \frac {2401}{256} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-3 \sqrt {3} \tanh ^{-1}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right )+\frac {1}{384} \sqrt {x^4+5 x^2+3} \left (144 x^6+1208 x^4+2650 x^2+2061\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3 + 5*x^2 + x^4]*(2061 + 2650*x^2 + 1208*x^4 + 144*x^6))/384 + (2401*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x
^2 + x^4])])/256 - 3*Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])]

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fricas [A]  time = 0.69, size = 106, normalized size = 0.89 \[ \frac {1}{384} \, {\left (144 \, x^{6} + 1208 \, x^{4} + 2650 \, x^{2} + 2061\right )} \sqrt {x^{4} + 5 \, x^{2} + 3} + 3 \, \sqrt {3} \log \left (\frac {25 \, x^{2} - 2 \, \sqrt {3} {\left (5 \, x^{2} + 6\right )} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (5 \, \sqrt {3} - 6\right )} + 30}{x^{2}}\right ) - \frac {2401}{256} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(144*x^6 + 1208*x^4 + 2650*x^2 + 2061)*sqrt(x^4 + 5*x^2 + 3) + 3*sqrt(3)*log((25*x^2 - 2*sqrt(3)*(5*x^2
+ 6) - 2*sqrt(x^4 + 5*x^2 + 3)*(5*sqrt(3) - 6) + 30)/x^2) - 2401/256*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5)

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giac [A]  time = 0.50, size = 113, normalized size = 0.95 \[ \frac {1}{384} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, {\left (18 \, x^{2} + 151\right )} x^{2} + 1325\right )} x^{2} + 2061\right )} + 3 \, \sqrt {3} \log \left (\frac {x^{2} + \sqrt {3} - \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2} - \sqrt {3} - \sqrt {x^{4} + 5 \, x^{2} + 3}}\right ) - \frac {2401}{256} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*sqrt(x^4 + 5*x^2 + 3)*(2*(4*(18*x^2 + 151)*x^2 + 1325)*x^2 + 2061) + 3*sqrt(3)*log((x^2 + sqrt(3) - sqrt
(x^4 + 5*x^2 + 3))/(x^2 - sqrt(3) - sqrt(x^4 + 5*x^2 + 3))) - 2401/256*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5
)

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maple [A]  time = 0.02, size = 117, normalized size = 0.98 \[ \frac {3 \sqrt {x^{4}+5 x^{2}+3}\, x^{6}}{8}+\frac {151 \sqrt {x^{4}+5 x^{2}+3}\, x^{4}}{48}+\frac {1325 \sqrt {x^{4}+5 x^{2}+3}\, x^{2}}{192}-3 \sqrt {3}\, \arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right )+\frac {2401 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{256}+\frac {687 \sqrt {x^{4}+5 x^{2}+3}}{128} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(x^4+5*x^2+3)^(1/2)*x^6+151/48*(x^4+5*x^2+3)^(1/2)*x^4+1325/192*(x^4+5*x^2+3)^(1/2)*x^2+687/128*(x^4+5*x^2
+3)^(1/2)+2401/256*ln(x^2+5/2+(x^4+5*x^2+3)^(1/2))-3*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2
)

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maxima [A]  time = 1.40, size = 120, normalized size = 1.01 \[ \frac {3}{8} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} x^{2} - \frac {37}{64} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} + \frac {61}{48} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} - 3 \, \sqrt {3} \log \left (\frac {2 \, \sqrt {3} \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac {6}{x^{2}} + 5\right ) + \frac {199}{128} \, \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {2401}{256} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(x^4 + 5*x^2 + 3)^(3/2)*x^2 - 37/64*sqrt(x^4 + 5*x^2 + 3)*x^2 + 61/48*(x^4 + 5*x^2 + 3)^(3/2) - 3*sqrt(3)*
log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3)/x^2 + 6/x^2 + 5) + 199/128*sqrt(x^4 + 5*x^2 + 3) + 2401/256*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 \[ \int \frac {\left (3\,x^2+2\right )\,{\left (x^4+5\,x^2+3\right )}^{3/2}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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