3.14 \(\int \frac {(2+3 x^2) \sqrt {5+x^4}}{x^7} \, dx\)

Optimal. Leaf size=58 \[ -\frac {3 \sqrt {x^4+5}}{4 x^4}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )}{4 \sqrt {5}}-\frac {\left (x^4+5\right )^{3/2}}{15 x^6} \]

[Out]

-1/15*(x^4+5)^(3/2)/x^6-3/20*arctanh(1/5*(x^4+5)^(1/2)*5^(1/2))*5^(1/2)-3/4*(x^4+5)^(1/2)/x^4

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Rubi [A]  time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1252, 807, 266, 47, 63, 207} \[ -\frac {\left (x^4+5\right )^{3/2}}{15 x^6}-\frac {3 \sqrt {x^4+5}}{4 x^4}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )}{4 \sqrt {5}} \]

Antiderivative was successfully verified.

[In]

Int[((2 + 3*x^2)*Sqrt[5 + x^4])/x^7,x]

[Out]

(-3*Sqrt[5 + x^4])/(4*x^4) - (5 + x^4)^(3/2)/(15*x^6) - (3*ArcTanh[Sqrt[5 + x^4]/Sqrt[5]])/(4*Sqrt[5])

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rubi steps

\begin {align*} \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x^7} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(2+3 x) \sqrt {5+x^2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {\left (5+x^4\right )^{3/2}}{15 x^6}+\frac {3}{2} \operatorname {Subst}\left (\int \frac {\sqrt {5+x^2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\left (5+x^4\right )^{3/2}}{15 x^6}+\frac {3}{4} \operatorname {Subst}\left (\int \frac {\sqrt {5+x}}{x^2} \, dx,x,x^4\right )\\ &=-\frac {3 \sqrt {5+x^4}}{4 x^4}-\frac {\left (5+x^4\right )^{3/2}}{15 x^6}+\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {5+x}} \, dx,x,x^4\right )\\ &=-\frac {3 \sqrt {5+x^4}}{4 x^4}-\frac {\left (5+x^4\right )^{3/2}}{15 x^6}+\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{-5+x^2} \, dx,x,\sqrt {5+x^4}\right )\\ &=-\frac {3 \sqrt {5+x^4}}{4 x^4}-\frac {\left (5+x^4\right )^{3/2}}{15 x^6}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {5+x^4}}{\sqrt {5}}\right )}{4 \sqrt {5}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 72, normalized size = 1.24 \[ -\frac {3 \left (5 x^4+\sqrt {5} \sqrt {x^4+5} x^4 \tanh ^{-1}\left (\sqrt {\frac {x^4}{5}+1}\right )+25\right )}{20 x^4 \sqrt {x^4+5}}-\frac {\left (x^4+5\right )^{3/2}}{15 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((2 + 3*x^2)*Sqrt[5 + x^4])/x^7,x]

[Out]

-1/15*(5 + x^4)^(3/2)/x^6 - (3*(25 + 5*x^4 + Sqrt[5]*x^4*Sqrt[5 + x^4]*ArcTanh[Sqrt[1 + x^4/5]]))/(20*x^4*Sqrt
[5 + x^4])

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fricas [A]  time = 0.64, size = 59, normalized size = 1.02 \[ \frac {9 \, \sqrt {5} x^{6} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{x^{2}}\right ) - 4 \, x^{6} - {\left (4 \, x^{4} + 45 \, x^{2} + 20\right )} \sqrt {x^{4} + 5}}{60 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(1/2)/x^7,x, algorithm="fricas")

[Out]

1/60*(9*sqrt(5)*x^6*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2) - 4*x^6 - (4*x^4 + 45*x^2 + 20)*sqrt(x^4 + 5))/x^6

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giac [B]  time = 0.23, size = 116, normalized size = 2.00 \[ \frac {3}{20} \, \sqrt {5} \log \left (-\frac {x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}}{x^{2} - \sqrt {5} - \sqrt {x^{4} + 5}}\right ) + \frac {9 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{5} + 12 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{4} - 225 \, x^{2} + 225 \, \sqrt {x^{4} + 5} + 100}{6 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} - 5\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(1/2)/x^7,x, algorithm="giac")

[Out]

3/20*sqrt(5)*log(-(x^2 + sqrt(5) - sqrt(x^4 + 5))/(x^2 - sqrt(5) - sqrt(x^4 + 5))) + 1/6*(9*(x^2 - sqrt(x^4 +
5))^5 + 12*(x^2 - sqrt(x^4 + 5))^4 - 225*x^2 + 225*sqrt(x^4 + 5) + 100)/((x^2 - sqrt(x^4 + 5))^2 - 5)^3

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maple [A]  time = 0.01, size = 52, normalized size = 0.90 \[ -\frac {3 \sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{20}-\frac {3 \left (x^{4}+5\right )^{\frac {3}{2}}}{20 x^{4}}-\frac {\left (x^{4}+5\right )^{\frac {3}{2}}}{15 x^{6}}+\frac {3 \sqrt {x^{4}+5}}{20} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+2)*(x^4+5)^(1/2)/x^7,x)

[Out]

-1/15*(x^4+5)^(3/2)/x^6-3/20*(x^4+5)^(3/2)/x^4+3/20*(x^4+5)^(1/2)-3/20*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))

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maxima [A]  time = 1.61, size = 59, normalized size = 1.02 \[ \frac {3}{40} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{\sqrt {5} + \sqrt {x^{4} + 5}}\right ) - \frac {3 \, \sqrt {x^{4} + 5}}{4 \, x^{4}} - \frac {{\left (x^{4} + 5\right )}^{\frac {3}{2}}}{15 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(1/2)/x^7,x, algorithm="maxima")

[Out]

3/40*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) - 3/4*sqrt(x^4 + 5)/x^4 - 1/15*(x^4 + 5
)^(3/2)/x^6

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mupad [B]  time = 0.68, size = 43, normalized size = 0.74 \[ -\frac {3\,\sqrt {5}\,\mathrm {atanh}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}}{5}\right )}{20}-\frac {3\,\sqrt {x^4+5}}{4\,x^4}-\frac {{\left (x^4+5\right )}^{3/2}}{15\,x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 + 5)^(1/2)*(3*x^2 + 2))/x^7,x)

[Out]

- (3*5^(1/2)*atanh((5^(1/2)*(x^4 + 5)^(1/2))/5))/20 - (3*(x^4 + 5)^(1/2))/(4*x^4) - (x^4 + 5)^(3/2)/(15*x^6)

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sympy [A]  time = 5.96, size = 63, normalized size = 1.09 \[ - \frac {\sqrt {1 + \frac {5}{x^{4}}}}{15} - \frac {3 \sqrt {5} \operatorname {asinh}{\left (\frac {\sqrt {5}}{x^{2}} \right )}}{20} - \frac {3 \sqrt {1 + \frac {5}{x^{4}}}}{4 x^{2}} - \frac {\sqrt {1 + \frac {5}{x^{4}}}}{3 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2+2)*(x**4+5)**(1/2)/x**7,x)

[Out]

-sqrt(1 + 5/x**4)/15 - 3*sqrt(5)*asinh(sqrt(5)/x**2)/20 - 3*sqrt(1 + 5/x**4)/(4*x**2) - sqrt(1 + 5/x**4)/(3*x*
*4)

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