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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1252, 815, 844, 215, 266, 63, 207} \[ \frac {1}{4} \sqrt {x^4+5} \left (3 x^2+4\right )+\frac {15}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((2 + 3*x^2)*Sqrt[5 + x^4])/x,x]

[Out]

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

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 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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 815

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 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} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(2+3 x) \sqrt {5+x^2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{4} \left (4+3 x^2\right ) \sqrt {5+x^4}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {20+15 x}{x \sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{4} \left (4+3 x^2\right ) \sqrt {5+x^4}+\frac {15}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {5+x^2}} \, dx,x,x^2\right )+5 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{4} \left (4+3 x^2\right ) \sqrt {5+x^4}+\frac {15}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )+\frac {5}{2} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {5+x}} \, dx,x,x^4\right )\\ &=\frac {1}{4} \left (4+3 x^2\right ) \sqrt {5+x^4}+\frac {15}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )+5 \operatorname {Subst}\left (\int \frac {1}{-5+x^2} \, dx,x,\sqrt {5+x^4}\right )\\ &=\frac {1}{4} \left (4+3 x^2\right ) \sqrt {5+x^4}+\frac {15}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {5+x^4}}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 57, normalized size = 0.98 \[ \frac {1}{4} \left (-4 \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )+15 \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )+\sqrt {x^4+5} \left (3 x^2+4\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 56, normalized size = 0.97 \[ \frac {1}{4} \, \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 4\right )} + \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{x^{2}}\right ) - \frac {15}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 76, normalized size = 1.31 \[ \frac {1}{4} \, \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 4\right )} + \sqrt {5} \log \left (-\frac {x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}}{x^{2} - \sqrt {5} - \sqrt {x^{4} + 5}}\right ) - \frac {15}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 49, normalized size = 0.84 \[ \frac {3 \sqrt {x^{4}+5}\, x^{2}}{4}+\frac {15 \arcsinh \left (\frac {\sqrt {5}\, x^{2}}{5}\right )}{4}-\sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )+\sqrt {x^{4}+5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.56, size = 99, normalized size = 1.71 \[ \frac {1}{2} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{\sqrt {5} + \sqrt {x^{4} + 5}}\right ) + \sqrt {x^{4} + 5} + \frac {15 \, \sqrt {x^{4} + 5}}{4 \, x^{2} {\left (\frac {x^{4} + 5}{x^{4}} - 1\right )}} + \frac {15}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) - \frac {15}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) + sqrt(x^4 + 5) + 15/4*sqrt(x^4 + 5)/(x^
2*((x^4 + 5)/x^4 - 1)) + 15/8*log(sqrt(x^4 + 5)/x^2 + 1) - 15/8*log(sqrt(x^4 + 5)/x^2 - 1)

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mupad [B]  time = 0.15, size = 45, normalized size = 0.78 \[ \frac {15\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{4}-\sqrt {5}\,\mathrm {atanh}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}}{5}\right )+\sqrt {x^4+5}\,\left (\frac {3\,x^2}{4}+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 15.59, size = 83, normalized size = 1.43 \[ \frac {3 x^{6}}{4 \sqrt {x^{4} + 5}} + \frac {15 x^{2}}{4 \sqrt {x^{4} + 5}} + \sqrt {x^{4} + 5} + \frac {\sqrt {5} \log {\left (x^{4} \right )}}{2} - \sqrt {5} \log {\left (\sqrt {\frac {x^{4}}{5} + 1} + 1 \right )} + \frac {15 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x**6/(4*sqrt(x**4 + 5)) + 15*x**2/(4*sqrt(x**4 + 5)) + sqrt(x**4 + 5) + sqrt(5)*log(x**4)/2 - sqrt(5)*log(sq
rt(x**4/5 + 1) + 1) + 15*asinh(sqrt(5)*x**2/5)/4

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