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

Optimal. Leaf size=81 \[ -\frac{\left (2-x^2\right ) \left (x^4+5\right )^{3/2}}{2 x^2}+\frac{3}{2} \left (x^2+5\right ) \sqrt{x^4+5}+\frac{15}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{15}{2} \sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0748548, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {1252, 813, 815, 844, 215, 266, 63, 207} \[ -\frac{\left (2-x^2\right ) \left (x^4+5\right )^{3/2}}{2 x^2}+\frac{3}{2} \left (x^2+5\right ) \sqrt{x^4+5}+\frac{15}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{15}{2} \sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(5 + x^2)*Sqrt[5 + x^4])/2 - ((2 - x^2)*(5 + x^4)^(3/2))/(2*x^2) + (15*ArcSinh[x^2/Sqrt[5]])/2 - (15*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 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 ) \left (5+x^4\right )^{3/2}}{x^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(2+3 x) \left (5+x^2\right )^{3/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{\left (2-x^2\right ) \left (5+x^4\right )^{3/2}}{2 x^2}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{(-30-12 x) \sqrt{5+x^2}}{x} \, dx,x,x^2\right )\\ &=\frac{3}{2} \left (5+x^2\right ) \sqrt{5+x^4}-\frac{\left (2-x^2\right ) \left (5+x^4\right )^{3/2}}{2 x^2}-\frac{1}{8} \operatorname{Subst}\left (\int \frac{-300-60 x}{x \sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=\frac{3}{2} \left (5+x^2\right ) \sqrt{5+x^4}-\frac{\left (2-x^2\right ) \left (5+x^4\right )^{3/2}}{2 x^2}+\frac{15}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{5+x^2}} \, dx,x,x^2\right )+\frac{75}{2} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=\frac{3}{2} \left (5+x^2\right ) \sqrt{5+x^4}-\frac{\left (2-x^2\right ) \left (5+x^4\right )^{3/2}}{2 x^2}+\frac{15}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )+\frac{75}{4} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{5+x}} \, dx,x,x^4\right )\\ &=\frac{3}{2} \left (5+x^2\right ) \sqrt{5+x^4}-\frac{\left (2-x^2\right ) \left (5+x^4\right )^{3/2}}{2 x^2}+\frac{15}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )+\frac{75}{2} \operatorname{Subst}\left (\int \frac{1}{-5+x^2} \, dx,x,\sqrt{5+x^4}\right )\\ &=\frac{3}{2} \left (5+x^2\right ) \sqrt{5+x^4}-\frac{\left (2-x^2\right ) \left (5+x^4\right )^{3/2}}{2 x^2}+\frac{15}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{15}{2} \sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{5+x^4}}{\sqrt{5}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0568492, size = 71, normalized size = 0.88 \[ \frac{1}{2} \left (\sqrt{x^4+5} \left (x^4+20\right )-15 \sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )\right )-\frac{5 \sqrt{5} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};-\frac{x^4}{5}\right )}{x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[5 + x^4]*(20 + x^4) - 15*Sqrt[5]*ArcTanh[Sqrt[5 + x^4]/Sqrt[5]])/2 - (5*Sqrt[5]*Hypergeometric2F1[-3/2,
-1/2, 1/2, -x^4/5])/x^2

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Maple [A]  time = 0.017, size = 75, normalized size = 0.9 \begin{align*}{\frac{{x}^{4}}{2}\sqrt{{x}^{4}+5}}+10\,\sqrt{{x}^{4}+5}-{\frac{15\,\sqrt{5}}{2}{\it Artanh} \left ({\sqrt{5}{\frac{1}{\sqrt{{x}^{4}+5}}}} \right ) }+{\frac{{x}^{2}}{2}\sqrt{{x}^{4}+5}}+{\frac{15}{2}{\it Arcsinh} \left ({\frac{{x}^{2}\sqrt{5}}{5}} \right ) }-5\,{\frac{\sqrt{{x}^{4}+5}}{{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^4*(x^4+5)^(1/2)+10*(x^4+5)^(1/2)-15/2*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))+1/2*x^2*(x^4+5)^(1/2)+15/2*
arcsinh(1/5*x^2*5^(1/2))-5*(x^4+5)^(1/2)/x^2

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Maxima [B]  time = 1.42607, size = 165, normalized size = 2.04 \begin{align*} \frac{1}{2} \,{\left (x^{4} + 5\right )}^{\frac{3}{2}} + \frac{15}{4} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{\sqrt{5} + \sqrt{x^{4} + 5}}\right ) + \frac{15}{2} \, \sqrt{x^{4} + 5} - \frac{5 \, \sqrt{x^{4} + 5}}{x^{2}} + \frac{5 \, \sqrt{x^{4} + 5}}{2 \, x^{2}{\left (\frac{x^{4} + 5}{x^{4}} - 1\right )}} + \frac{15}{4} \, \log \left (\frac{\sqrt{x^{4} + 5}}{x^{2}} + 1\right ) - \frac{15}{4} \, \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)^(3/2)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.54216, size = 200, normalized size = 2.47 \begin{align*} \frac{15 \, \sqrt{5} x^{2} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{x^{2}}\right ) - 15 \, x^{2} \log \left (-x^{2} + \sqrt{x^{4} + 5}\right ) - 10 \, x^{2} +{\left (x^{6} + x^{4} + 20 \, x^{2} - 10\right )} \sqrt{x^{4} + 5}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(15*sqrt(5)*x^2*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2) - 15*x^2*log(-x^2 + sqrt(x^4 + 5)) - 10*x^2 + (x^6 + x
^4 + 20*x^2 - 10)*sqrt(x^4 + 5))/x^2

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Sympy [A]  time = 9.07861, size = 114, normalized size = 1.41 \begin{align*} \frac{x^{6}}{2 \sqrt{x^{4} + 5}} - \frac{5 x^{2}}{2 \sqrt{x^{4} + 5}} + \frac{\left (x^{4} + 5\right )^{\frac{3}{2}}}{2} + \frac{15 \sqrt{x^{4} + 5}}{2} + \frac{15 \sqrt{5} \log{\left (x^{4} \right )}}{4} - \frac{15 \sqrt{5} \log{\left (\sqrt{\frac{x^{4}}{5} + 1} + 1 \right )}}{2} + \frac{15 \operatorname{asinh}{\left (\frac{\sqrt{5} x^{2}}{5} \right )}}{2} - \frac{25}{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)**(3/2)/x**3,x)

[Out]

x**6/(2*sqrt(x**4 + 5)) - 5*x**2/(2*sqrt(x**4 + 5)) + (x**4 + 5)**(3/2)/2 + 15*sqrt(x**4 + 5)/2 + 15*sqrt(5)*l
og(x**4)/4 - 15*sqrt(5)*log(sqrt(x**4/5 + 1) + 1)/2 + 15*asinh(sqrt(5)*x**2/5)/2 - 25/(x**2*sqrt(x**4 + 5))

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

[Out]

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