3.2.61 \(\int \frac {(2+3 x^2) (3+5 x^2+x^4)^{3/2}}{x^5} \, dx\) [161]

Optimal. Leaf size=127 \[ -\frac {3 \left (28-19 x^2\right ) \sqrt {3+5 x^2+x^4}}{8 x^2}-\frac {\left (2-3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{4 x^4}+\frac {453}{16} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )-\frac {127}{8} \sqrt {3} \tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right ) \]

[Out]

-1/4*(-3*x^2+2)*(x^4+5*x^2+3)^(3/2)/x^4+453/16*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))-127/8*arctanh(1/6*(5
*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)-3/8*(-19*x^2+28)*(x^4+5*x^2+3)^(1/2)/x^2

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Rubi [A]
time = 0.07, antiderivative size = 127, 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 = {1265, 826, 857, 635, 212, 738} \begin {gather*} -\frac {\left (2-3 x^2\right ) \left (x^4+5 x^2+3\right )^{3/2}}{4 x^4}-\frac {3 \left (28-19 x^2\right ) \sqrt {x^4+5 x^2+3}}{8 x^2}+\frac {453}{16} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-\frac {127}{8} \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^5,x]

[Out]

(-3*(28 - 19*x^2)*Sqrt[3 + 5*x^2 + x^4])/(8*x^2) - ((2 - 3*x^2)*(3 + 5*x^2 + x^4)^(3/2))/(4*x^4) + (453*ArcTan
h[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/16 - (127*Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^
4])])/8

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 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^5} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(2+3 x) \left (3+5 x+x^2\right )^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\left (2-3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{4 x^4}-\frac {3}{16} \text {Subst}\left (\int \frac {(-56-38 x) \sqrt {3+5 x+x^2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {3 \left (28-19 x^2\right ) \sqrt {3+5 x^2+x^4}}{8 x^2}-\frac {\left (2-3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{4 x^4}+\frac {3}{32} \text {Subst}\left (\int \frac {508+302 x}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {3 \left (28-19 x^2\right ) \sqrt {3+5 x^2+x^4}}{8 x^2}-\frac {\left (2-3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{4 x^4}+\frac {453}{16} \text {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )+\frac {381}{8} \text {Subst}\left (\int \frac {1}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {3 \left (28-19 x^2\right ) \sqrt {3+5 x^2+x^4}}{8 x^2}-\frac {\left (2-3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{4 x^4}+\frac {453}{8} \text {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {5+2 x^2}{\sqrt {3+5 x^2+x^4}}\right )-\frac {381}{4} \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 \left (28-19 x^2\right ) \sqrt {3+5 x^2+x^4}}{8 x^2}-\frac {\left (2-3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{4 x^4}+\frac {453}{16} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )-\frac {127}{8} \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 = 0.24, size = 101, normalized size = 0.80 \begin {gather*} \frac {1}{16} \left (\frac {2 \sqrt {3+5 x^2+x^4} \left (-12-86 x^2+83 x^4+6 x^6\right )}{x^4}+508 \sqrt {3} \tanh ^{-1}\left (\frac {x^2-\sqrt {3+5 x^2+x^4}}{\sqrt {3}}\right )-453 \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^5,x]

[Out]

((2*Sqrt[3 + 5*x^2 + x^4]*(-12 - 86*x^2 + 83*x^4 + 6*x^6))/x^4 + 508*Sqrt[3]*ArcTanh[(x^2 - Sqrt[3 + 5*x^2 + x
^4])/Sqrt[3]] - 453*Log[-5 - 2*x^2 + 2*Sqrt[3 + 5*x^2 + x^4]])/16

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Maple [A]
time = 0.27, size = 117, normalized size = 0.92

method	result	size
trager	\(\frac {\left (6 x^{6}+83 x^{4}-86 x^{2}-12\right ) \sqrt {x^{4}+5 x^{2}+3}}{8 x^{4}}+\frac {453 \ln \left (-2 x^{2}-2 \sqrt {x^{4}+5 x^{2}+3}-5\right )}{16}-\frac {127 \RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \RootOf \left (\textit {\_Z}^{2}-3\right )+6 \sqrt {x^{4}+5 x^{2}+3}}{x^{2}}\right )}{8}\)	\(108\)
default	\(\frac {83 \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {453 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{16}-\frac {3 \sqrt {x^{4}+5 x^{2}+3}}{2 x^{4}}-\frac {43 \sqrt {x^{4}+5 x^{2}+3}}{4 x^{2}}-\frac {127 \arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{8}+\frac {3 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{4}\)	\(117\)
risch	\(-\frac {43 x^{6}+221 x^{4}+159 x^{2}+18}{4 x^{4} \sqrt {x^{4}+5 x^{2}+3}}+\frac {3 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{4}+\frac {83 \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {453 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{16}-\frac {127 \arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{8}\)	\(117\)
elliptic	\(\frac {83 \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {453 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{16}-\frac {3 \sqrt {x^{4}+5 x^{2}+3}}{2 x^{4}}-\frac {43 \sqrt {x^{4}+5 x^{2}+3}}{4 x^{2}}-\frac {127 \arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{8}+\frac {3 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{4}\)	\(117\)

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^5,x,method=_RETURNVERBOSE)

[Out]

83/8*(x^4+5*x^2+3)^(1/2)+453/16*ln(x^2+5/2+(x^4+5*x^2+3)^(1/2))-3/2*(x^4+5*x^2+3)^(1/2)/x^4-43/4*(x^4+5*x^2+3)
^(1/2)/x^2-127/8*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)+3/4*x^2*(x^4+5*x^2+3)^(1/2)

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Maxima [A]
time = 0.49, size = 137, normalized size = 1.08 \begin {gather*} \frac {7}{2} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} + \frac {1}{6} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} - \frac {127}{8} \, \sqrt {3} \log \left (\frac {2 \, \sqrt {3} \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac {6}{x^{2}} + 5\right ) + \frac {197}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} - \frac {23 \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}}}{12 \, x^{2}} - \frac {{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {5}{2}}}{6 \, x^{4}} + \frac {453}{16} \, \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^5,x, algorithm="maxima")

[Out]

7/2*sqrt(x^4 + 5*x^2 + 3)*x^2 + 1/6*(x^4 + 5*x^2 + 3)^(3/2) - 127/8*sqrt(3)*log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3
)/x^2 + 6/x^2 + 5) + 197/8*sqrt(x^4 + 5*x^2 + 3) - 23/12*(x^4 + 5*x^2 + 3)^(3/2)/x^2 - 1/6*(x^4 + 5*x^2 + 3)^(
5/2)/x^4 + 453/16*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Fricas [A]
time = 0.37, size = 122, normalized size = 0.96 \begin {gather*} \frac {1016 \, \sqrt {3} x^{4} \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 ) - 1812 \, x^{4} \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) + 67 \, x^{4} + 8 \, {\left (6 \, x^{6} + 83 \, x^{4} - 86 \, x^{2} - 12\right )} \sqrt {x^{4} + 5 \, x^{2} + 3}}{64 \, x^{4}} \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^5,x, algorithm="fricas")

[Out]

1/64*(1016*sqrt(3)*x^4*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
) - 1812*x^4*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5) + 67*x^4 + 8*(6*x^6 + 83*x^4 - 86*x^2 - 12)*sqrt(x^4 +
5*x^2 + 3))/x^4

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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^{5}}\, 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**5,x)

[Out]

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

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Giac [A]
time = 5.66, size = 190, normalized size = 1.50 \begin {gather*} \frac {1}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (6 \, x^{2} + 83\right )} + \frac {127}{8} \, \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 {227 \, {\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{3} + 348 \, {\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{2} - 459 \, x^{2} + 459 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 684}{4 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{2} - 3\right )}^{2}} - \frac {453}{16} \, \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^5,x, algorithm="giac")

[Out]

1/8*sqrt(x^4 + 5*x^2 + 3)*(6*x^2 + 83) + 127/8*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))) + 1/4*(227*(x^2 - sqrt(x^4 + 5*x^2 + 3))^3 + 348*(x^2 - sqrt(x^4 + 5*x^2 + 3))^2
- 459*x^2 + 459*sqrt(x^4 + 5*x^2 + 3) - 684)/((x^2 - sqrt(x^4 + 5*x^2 + 3))^2 - 3)^2 - 453/16*log(2*x^2 - 2*s
qrt(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^5} \,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^5,x)

[Out]

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

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