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

Optimal. Leaf size=51 \[ -\frac{15}{16} \sqrt{x^4+5} x^2+\frac{1}{24} \left (9 x^2+8\right ) \left (x^4+5\right )^{3/2}-\frac{75}{16} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right ) \]

[Out]

(-15*x^2*Sqrt[5 + x^4])/16 + ((8 + 9*x^2)*(5 + x^4)^(3/2))/24 - (75*ArcSinh[x^2/Sqrt[5]])/16

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Rubi [A]  time = 0.0307754, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1252, 780, 195, 215} \[ -\frac{15}{16} \sqrt{x^4+5} x^2+\frac{1}{24} \left (9 x^2+8\right ) \left (x^4+5\right )^{3/2}-\frac{75}{16} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*(2 + 3*x^2)*Sqrt[5 + x^4],x]

[Out]

(-15*x^2*Sqrt[5 + x^4])/16 + ((8 + 9*x^2)*(5 + x^4)^(3/2))/24 - (75*ArcSinh[x^2/Sqrt[5]])/16

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 780

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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]

Rubi steps

\begin{align*} \int x^3 \left (2+3 x^2\right ) \sqrt{5+x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (2+3 x) \sqrt{5+x^2} \, dx,x,x^2\right )\\ &=\frac{1}{24} \left (8+9 x^2\right ) \left (5+x^4\right )^{3/2}-\frac{15}{8} \operatorname{Subst}\left (\int \sqrt{5+x^2} \, dx,x,x^2\right )\\ &=-\frac{15}{16} x^2 \sqrt{5+x^4}+\frac{1}{24} \left (8+9 x^2\right ) \left (5+x^4\right )^{3/2}-\frac{75}{16} \operatorname{Subst}\left (\int \frac{1}{\sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=-\frac{15}{16} x^2 \sqrt{5+x^4}+\frac{1}{24} \left (8+9 x^2\right ) \left (5+x^4\right )^{3/2}-\frac{75}{16} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0288524, size = 44, normalized size = 0.86 \[ \frac{1}{48} \left (\sqrt{x^4+5} \left (18 x^6+16 x^4+45 x^2+80\right )-225 \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[5 + x^4]*(80 + 45*x^2 + 16*x^4 + 18*x^6) - 225*ArcSinh[x^2/Sqrt[5]])/48

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Maple [A]  time = 0.004, size = 46, normalized size = 0.9 \begin{align*}{\frac{3\,{x}^{2}}{8} \left ({x}^{4}+5 \right ) ^{{\frac{3}{2}}}}-{\frac{15\,{x}^{2}}{16}\sqrt{{x}^{4}+5}}-{\frac{75}{16}{\it Arcsinh} \left ({\frac{{x}^{2}\sqrt{5}}{5}} \right ) }+{\frac{1}{3} \left ({x}^{4}+5 \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*x^2*(x^4+5)^(3/2)-15/16*x^2*(x^4+5)^(1/2)-75/16*arcsinh(1/5*x^2*5^(1/2))+1/3*(x^4+5)^(3/2)

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Maxima [B]  time = 1.44704, size = 126, normalized size = 2.47 \begin{align*} \frac{1}{3} \,{\left (x^{4} + 5\right )}^{\frac{3}{2}} - \frac{75 \,{\left (\frac{\sqrt{x^{4} + 5}}{x^{2}} + \frac{{\left (x^{4} + 5\right )}^{\frac{3}{2}}}{x^{6}}\right )}}{16 \,{\left (\frac{2 \,{\left (x^{4} + 5\right )}}{x^{4}} - \frac{{\left (x^{4} + 5\right )}^{2}}{x^{8}} - 1\right )}} - \frac{75}{32} \, \log \left (\frac{\sqrt{x^{4} + 5}}{x^{2}} + 1\right ) + \frac{75}{32} \, \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(x^3*(3*x^2+2)*(x^4+5)^(1/2),x, algorithm="maxima")

[Out]

1/3*(x^4 + 5)^(3/2) - 75/16*(sqrt(x^4 + 5)/x^2 + (x^4 + 5)^(3/2)/x^6)/(2*(x^4 + 5)/x^4 - (x^4 + 5)^2/x^8 - 1)
- 75/32*log(sqrt(x^4 + 5)/x^2 + 1) + 75/32*log(sqrt(x^4 + 5)/x^2 - 1)

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Fricas [A]  time = 1.55788, size = 116, normalized size = 2.27 \begin{align*} \frac{1}{48} \,{\left (18 \, x^{6} + 16 \, x^{4} + 45 \, x^{2} + 80\right )} \sqrt{x^{4} + 5} + \frac{75}{16} \, \log \left (-x^{2} + \sqrt{x^{4} + 5}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(18*x^6 + 16*x^4 + 45*x^2 + 80)*sqrt(x^4 + 5) + 75/16*log(-x^2 + sqrt(x^4 + 5))

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Sympy [A]  time = 4.08094, size = 70, normalized size = 1.37 \begin{align*} \frac{3 x^{10}}{8 \sqrt{x^{4} + 5}} + \frac{45 x^{6}}{16 \sqrt{x^{4} + 5}} + \frac{75 x^{2}}{16 \sqrt{x^{4} + 5}} + \frac{\left (x^{4} + 5\right )^{\frac{3}{2}}}{3} - \frac{75 \operatorname{asinh}{\left (\frac{\sqrt{5} x^{2}}{5} \right )}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x**10/(8*sqrt(x**4 + 5)) + 45*x**6/(16*sqrt(x**4 + 5)) + 75*x**2/(16*sqrt(x**4 + 5)) + (x**4 + 5)**(3/2)/3 -
75*asinh(sqrt(5)*x**2/5)/16

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Giac [A]  time = 1.14915, size = 62, normalized size = 1.22 \begin{align*} \frac{1}{48} \, \sqrt{x^{4} + 5}{\left ({\left (2 \,{\left (9 \, x^{2} + 8\right )} x^{2} + 45\right )} x^{2} + 80\right )} + \frac{75}{16} \, \log \left (-x^{2} + \sqrt{x^{4} + 5}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*sqrt(x^4 + 5)*((2*(9*x^2 + 8)*x^2 + 45)*x^2 + 80) + 75/16*log(-x^2 + sqrt(x^4 + 5))