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

Optimal. Leaf size=122 \[ -\frac{\left (2-x^2\right ) \left (x^4+5 x^2+3\right )^{3/2}}{2 x^2}+\frac{3}{16} \left (18 x^2+109\right ) \sqrt{x^4+5 x^2+3}+\frac{609}{32} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )-12 \sqrt{3} \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right ) \]

[Out]

(3*(109 + 18*x^2)*Sqrt[3 + 5*x^2 + x^4])/16 - ((2 - x^2)*(3 + 5*x^2 + x^4)^(3/2))/(2*x^2) + (609*ArcTanh[(5 +
2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/32 - 12*Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])]

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Rubi [A]  time = 0.106506, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {1251, 812, 814, 843, 621, 206, 724} \[ -\frac{\left (2-x^2\right ) \left (x^4+5 x^2+3\right )^{3/2}}{2 x^2}+\frac{3}{16} \left (18 x^2+109\right ) \sqrt{x^4+5 x^2+3}+\frac{609}{32} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )-12 \sqrt{3} \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(109 + 18*x^2)*Sqrt[3 + 5*x^2 + x^4])/16 - ((2 - x^2)*(3 + 5*x^2 + x^4)^(3/2))/(2*x^2) + (609*ArcTanh[(5 +
2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/32 - 12*Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])]

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 812

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

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 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])

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]

Rubi steps

\begin{align*} \int \frac{\left (2+3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{x^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(2+3 x) \left (3+5 x+x^2\right )^{3/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{\left (2-x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{2 x^2}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{(-48-27 x) \sqrt{3+5 x+x^2}}{x} \, dx,x,x^2\right )\\ &=\frac{3}{16} \left (109+18 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{\left (2-x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{2 x^2}+\frac{1}{16} \operatorname{Subst}\left (\int \frac{576+\frac{609 x}{2}}{x \sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac{3}{16} \left (109+18 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{\left (2-x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{2 x^2}+\frac{609}{32} \operatorname{Subst}\left (\int \frac{1}{\sqrt{3+5 x+x^2}} \, dx,x,x^2\right )+36 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac{3}{16} \left (109+18 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{\left (2-x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{2 x^2}+\frac{609}{16} \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{5+2 x^2}{\sqrt{3+5 x^2+x^4}}\right )-72 \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{3}{16} \left (109+18 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{\left (2-x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{2 x^2}+\frac{609}{32} \tanh ^{-1}\left (\frac{5+2 x^2}{2 \sqrt{3+5 x^2+x^4}}\right )-12 \sqrt{3} \tanh ^{-1}\left (\frac{6+5 x^2}{2 \sqrt{3} \sqrt{3+5 x^2+x^4}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0542173, size = 107, normalized size = 0.88 \[ \frac{\sqrt{x^4+5 x^2+3} \left (8 x^6+78 x^4+271 x^2-48\right )}{16 x^2}+\frac{609}{32} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )-12 \sqrt{3} \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3 + 5*x^2 + x^4]*(-48 + 271*x^2 + 78*x^4 + 8*x^6))/(16*x^2) + (609*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2
+ x^4])])/32 - 12*Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])]

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Maple [A]  time = 0.015, size = 117, normalized size = 1. \begin{align*}{\frac{{x}^{4}}{2}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{39\,{x}^{2}}{8}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{271}{16}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{609}{32}\ln \left ({\frac{5}{2}}+{x}^{2}+\sqrt{{x}^{4}+5\,{x}^{2}+3} \right ) }-12\,{\it Artanh} \left ( 1/6\,{\frac{ \left ( 5\,{x}^{2}+6 \right ) \sqrt{3}}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}} \right ) \sqrt{3}-3\,{\frac{\sqrt{{x}^{4}+5\,{x}^{2}+3}}{{x}^{2}}} \end{align*}

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^3,x)

[Out]

1/2*x^4*(x^4+5*x^2+3)^(1/2)+39/8*x^2*(x^4+5*x^2+3)^(1/2)+271/16*(x^4+5*x^2+3)^(1/2)+609/32*ln(5/2+x^2+(x^4+5*x
^2+3)^(1/2))-12*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)-3*(x^4+5*x^2+3)^(1/2)/x^2

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Maxima [A]  time = 1.46772, size = 162, normalized size = 1.33 \begin{align*} \frac{27}{8} \, \sqrt{x^{4} + 5 \, x^{2} + 3} x^{2} + \frac{1}{2} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}} - 12 \, \sqrt{3} \log \left (\frac{2 \, \sqrt{3} \sqrt{x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac{6}{x^{2}} + 5\right ) + \frac{327}{16} \, \sqrt{x^{4} + 5 \, x^{2} + 3} - \frac{{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}}}{x^{2}} + \frac{609}{32} \, \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((3*x^2+2)*(x^4+5*x^2+3)^(3/2)/x^3,x, algorithm="maxima")

[Out]

27/8*sqrt(x^4 + 5*x^2 + 3)*x^2 + 1/2*(x^4 + 5*x^2 + 3)^(3/2) - 12*sqrt(3)*log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3)/
x^2 + 6/x^2 + 5) + 327/16*sqrt(x^4 + 5*x^2 + 3) - (x^4 + 5*x^2 + 3)^(3/2)/x^2 + 609/32*log(2*x^2 + 2*sqrt(x^4
+ 5*x^2 + 3) + 5)

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Fricas [A]  time = 1.34821, size = 325, normalized size = 2.66 \begin{align*} \frac{1536 \, \sqrt{3} x^{2} \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 ) - 2436 \, x^{2} \log \left (-2 \, x^{2} + 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3} - 5\right ) + 1541 \, x^{2} + 8 \,{\left (8 \, x^{6} + 78 \, x^{4} + 271 \, x^{2} - 48\right )} \sqrt{x^{4} + 5 \, x^{2} + 3}}{128 \, x^{2}} \end{align*}

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^3,x, algorithm="fricas")

[Out]

1/128*(1536*sqrt(3)*x^2*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) - 2436*x^2*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5) + 1541*x^2 + 8*(8*x^6 + 78*x^4 + 271*x^2 - 48)*sqrt(x^
4 + 5*x^2 + 3))/x^2

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

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**3,x)

[Out]

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

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

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^3,x, algorithm="giac")

[Out]

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