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

Optimal. Leaf size=59 \[ -\frac{\sqrt{x^4+5} \left (2-3 x^2\right )}{2 x^2}+\sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{3}{2} \sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0557464, antiderivative size = 59, 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, 813, 844, 215, 266, 63, 207} \[ -\frac{\sqrt{x^4+5} \left (2-3 x^2\right )}{2 x^2}+\sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{3}{2} \sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + 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 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^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(2+3 x) \sqrt{5+x^2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{\left (2-3 x^2\right ) \sqrt{5+x^4}}{2 x^2}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{-30-4 x}{x \sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=-\frac{\left (2-3 x^2\right ) \sqrt{5+x^4}}{2 x^2}+\frac{15}{2} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{5+x^2}} \, dx,x,x^2\right )+\operatorname{Subst}\left (\int \frac{1}{\sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=-\frac{\left (2-3 x^2\right ) \sqrt{5+x^4}}{2 x^2}+\sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )+\frac{15}{4} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{5+x}} \, dx,x,x^4\right )\\ &=-\frac{\left (2-3 x^2\right ) \sqrt{5+x^4}}{2 x^2}+\sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )+\frac{15}{2} \operatorname{Subst}\left (\int \frac{1}{-5+x^2} \, dx,x,\sqrt{5+x^4}\right )\\ &=-\frac{\left (2-3 x^2\right ) \sqrt{5+x^4}}{2 x^2}+\sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{3}{2} \sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{5+x^4}}{\sqrt{5}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0476781, size = 59, normalized size = 1. \[ \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )+\frac{1}{2} \left (\frac{\left (3 x^2-2\right ) \sqrt{x^4+5}}{x^2}-3 \sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 61, normalized size = 1. \begin{align*}{\frac{3}{2}\sqrt{{x}^{4}+5}}-{\frac{3\,\sqrt{5}}{2}{\it Artanh} \left ({\sqrt{5}{\frac{1}{\sqrt{{x}^{4}+5}}}} \right ) }-{\frac{1}{5\,{x}^{2}} \left ({x}^{4}+5 \right ) ^{{\frac{3}{2}}}}+{\frac{{x}^{2}}{5}\sqrt{{x}^{4}+5}}+{\it Arcsinh} \left ({\frac{{x}^{2}\sqrt{5}}{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^3,x)

[Out]

3/2*(x^4+5)^(1/2)-3/2*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))-1/5/x^2*(x^4+5)^(3/2)+1/5*x^2*(x^4+5)^(1/2)+arcsi
nh(1/5*x^2*5^(1/2))

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Maxima [A]  time = 1.45379, size = 119, normalized size = 2.02 \begin{align*} \frac{3}{4} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{\sqrt{5} + \sqrt{x^{4} + 5}}\right ) + \frac{3}{2} \, \sqrt{x^{4} + 5} - \frac{\sqrt{x^{4} + 5}}{x^{2}} + \frac{1}{2} \, \log \left (\frac{\sqrt{x^{4} + 5}}{x^{2}} + 1\right ) - \frac{1}{2} \, \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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.57317, size = 177, normalized size = 3. \begin{align*} \frac{3 \, \sqrt{5} x^{2} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{x^{2}}\right ) - 2 \, x^{2} \log \left (-x^{2} + \sqrt{x^{4} + 5}\right ) - 2 \, x^{2} + \sqrt{x^{4} + 5}{\left (3 \, x^{2} - 2\right )}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 6.00339, size = 83, normalized size = 1.41 \begin{align*} - \frac{x^{2}}{\sqrt{x^{4} + 5}} + \frac{3 \sqrt{x^{4} + 5}}{2} + \frac{3 \sqrt{5} \log{\left (x^{4} \right )}}{4} - \frac{3 \sqrt{5} \log{\left (\sqrt{\frac{x^{4}}{5} + 1} + 1 \right )}}{2} + \operatorname{asinh}{\left (\frac{\sqrt{5} x^{2}}{5} \right )} - \frac{5}{x^{2} \sqrt{x^{4} + 5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2/sqrt(x**4 + 5) + 3*sqrt(x**4 + 5)/2 + 3*sqrt(5)*log(x**4)/4 - 3*sqrt(5)*log(sqrt(x**4/5 + 1) + 1)/2 + as
inh(sqrt(5)*x**2/5) - 5/(x**2*sqrt(x**4 + 5))

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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^{3}}\,{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^3,x, algorithm="giac")

[Out]

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