3.2.60 \(\int \frac {(2+3 x^2) (3+5 x^2+x^4)^{3/2}}{x^3} \, dx\) [160]

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

[Out]

-1/2*(-x^2+2)*(x^4+5*x^2+3)^(3/2)/x^2+609/32*arctanh(1/2*(2*x^2+5)/(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)+3/16*(18*x^2+109)*(x^4+5*x^2+3)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1265, 826, 828, 857, 635, 212, 738} \begin {gather*} -\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 ) \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 738

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 826

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 828

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 857

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 1265

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^3} \, dx &=\frac {1}{2} \text {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} \text {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} \text {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} \text {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )+36 \text {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} \text {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {5+2 x^2}{\sqrt {3+5 x^2+x^4}}\right )-72 \text {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*}

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Mathematica [A]
time = 0.23, size = 102, normalized size = 0.84 \begin {gather*} 24 \sqrt {3} \tanh ^{-1}\left (\frac {x^2-\sqrt {3+5 x^2+x^4}}{\sqrt {3}}\right )+\frac {1}{32} \left (\frac {2 \sqrt {3+5 x^2+x^4} \left (-48+271 x^2+78 x^4+8 x^6\right )}{x^2}-609 \log \left (-5-2 x^2+2 \sqrt {3+5 x^2+x^4}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.29, size = 117, normalized size = 0.96

method	result	size
trager	\(\frac {\left (8 x^{6}+78 x^{4}+271 x^{2}-48\right ) \sqrt {x^{4}+5 x^{2}+3}}{16 x^{2}}-\frac {609 \ln \left (2 x^{2}-2 \sqrt {x^{4}+5 x^{2}+3}+5\right )}{32}-12 \RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \RootOf \left (\textit {\_Z}^{2}-3\right )+6 \sqrt {x^{4}+5 x^{2}+3}}{x^{2}}\right )\)	\(108\)
default	\(\frac {39 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {271 \sqrt {x^{4}+5 x^{2}+3}}{16}+\frac {609 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{32}-\frac {3 \sqrt {x^{4}+5 x^{2}+3}}{x^{2}}-12 \arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}+\frac {x^{4} \sqrt {x^{4}+5 x^{2}+3}}{2}\)	\(117\)
risch	\(\frac {39 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {271 \sqrt {x^{4}+5 x^{2}+3}}{16}+\frac {609 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{32}-\frac {3 \sqrt {x^{4}+5 x^{2}+3}}{x^{2}}-12 \arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}+\frac {x^{4} \sqrt {x^{4}+5 x^{2}+3}}{2}\)	\(117\)
elliptic	\(\frac {39 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {271 \sqrt {x^{4}+5 x^{2}+3}}{16}+\frac {609 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{32}-\frac {3 \sqrt {x^{4}+5 x^{2}+3}}{x^{2}}-12 \arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}+\frac {x^{4} \sqrt {x^{4}+5 x^{2}+3}}{2}\)	\(117\)

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,method=_RETURNVERBOSE)

[Out]

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(x^2+5/2+(x^4+5*x^2+3)^(1/2))-3*(x^4+5*x^2+3)
^(1/2)/x^2-12*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)+1/2*x^4*(x^4+5*x^2+3)^(1/2)

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Maxima [A]
time = 0.51, size = 120, normalized size = 0.98 \begin {gather*} \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 {gather*}

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 = 0.40, size = 122, normalized size = 1.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (3 x^{2} + 2\right ) \left (x^{4} + 5 x^{2} + 3\right )^{\frac {3}{2}}}{x^{3}}\, dx \end {gather*}

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 [A]
time = 5.22, size = 153, normalized size = 1.25 \begin {gather*} \frac {1}{16} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, x^{2} + 39\right )} x^{2} + 271\right )} + 12 \, \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 {3 \, {\left (5 \, x^{2} - 5 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 6\right )}}{{\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{2} - 3} - \frac {609}{32} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \end {gather*}

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]

1/16*sqrt(x^4 + 5*x^2 + 3)*(2*(4*x^2 + 39)*x^2 + 271) + 12*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))) + 3*(5*x^2 - 5*sqrt(x^4 + 5*x^2 + 3) + 6)/((x^2 - sqrt(x^4 + 5*x^2 +
3))^2 - 3) - 609/32*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 \begin {gather*} \int \frac {\left (3\,x^2+2\right )\,{\left (x^4+5\,x^2+3\right )}^{3/2}}{x^3} \,d x \end {gather*}

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^3,x)

[Out]

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

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