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

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

[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]]

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Rubi [A]  time = 0.05466, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.0428864, size = 57, normalized size = 0.98 \[ \frac{1}{4} \left (\sqrt{x^4+5} \left (3 x^2+4\right )+15 \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-4 \sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )\right ) \]

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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Maple [A]  time = 0.014, size = 49, normalized size = 0.8 \begin{align*}{\frac{3\,{x}^{2}}{4}\sqrt{{x}^{4}+5}}+{\frac{15}{4}{\it Arcsinh} \left ({\frac{{x}^{2}\sqrt{5}}{5}} \right ) }+\sqrt{{x}^{4}+5}-\sqrt{5}{\it Artanh} \left ({\sqrt{5}{\frac{1}{\sqrt{{x}^{4}+5}}}} \right ) \end{align*}

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

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Maxima [B]  time = 1.44598, size = 134, normalized size = 2.31 \begin{align*} \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 ) \end{align*}

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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Fricas [A]  time = 1.58218, size = 149, normalized size = 2.57 \begin{align*} \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 ) \end{align*}

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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Sympy [A]  time = 12.3123, size = 83, normalized size = 1.43 \begin{align*} \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} \end{align*}

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

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]

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