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

Optimal. Leaf size=106 \[ -\frac {1}{40} \left (27-10 x^2\right ) \left (x^4+5 x^2+3\right )^{5/2}+\frac {123}{128} \left (2 x^2+5\right ) \left (x^4+5 x^2+3\right )^{3/2}-\frac {4797 \left (2 x^2+5\right ) \sqrt {x^4+5 x^2+3}}{1024}+\frac {62361 \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )}{2048} \]

[Out]

123/128*(2*x^2+5)*(x^4+5*x^2+3)^(3/2)-1/40*(-10*x^2+27)*(x^4+5*x^2+3)^(5/2)+62361/2048*arctanh(1/2*(2*x^2+5)/(
x^4+5*x^2+3)^(1/2))-4797/1024*(2*x^2+5)*(x^4+5*x^2+3)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1251, 779, 612, 621, 206} \[ -\frac {1}{40} \left (27-10 x^2\right ) \left (x^4+5 x^2+3\right )^{5/2}+\frac {123}{128} \left (2 x^2+5\right ) \left (x^4+5 x^2+3\right )^{3/2}-\frac {4797 \left (2 x^2+5\right ) \sqrt {x^4+5 x^2+3}}{1024}+\frac {62361 \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )}{2048} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4797*(5 + 2*x^2)*Sqrt[3 + 5*x^2 + x^4])/1024 + (123*(5 + 2*x^2)*(3 + 5*x^2 + x^4)^(3/2))/128 - ((27 - 10*x^2
)*(3 + 5*x^2 + x^4)^(5/2))/40 + (62361*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/2048

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int x^3 \left (2+3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x (2+3 x) \left (3+5 x+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=-\frac {1}{40} \left (27-10 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}+\frac {123}{16} \operatorname {Subst}\left (\int \left (3+5 x+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {123}{128} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac {1}{40} \left (27-10 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}-\frac {4797}{256} \operatorname {Subst}\left (\int \sqrt {3+5 x+x^2} \, dx,x,x^2\right )\\ &=-\frac {4797 \left (5+2 x^2\right ) \sqrt {3+5 x^2+x^4}}{1024}+\frac {123}{128} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac {1}{40} \left (27-10 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}+\frac {62361 \operatorname {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )}{2048}\\ &=-\frac {4797 \left (5+2 x^2\right ) \sqrt {3+5 x^2+x^4}}{1024}+\frac {123}{128} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac {1}{40} \left (27-10 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}+\frac {62361 \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {5+2 x^2}{\sqrt {3+5 x^2+x^4}}\right )}{1024}\\ &=-\frac {4797 \left (5+2 x^2\right ) \sqrt {3+5 x^2+x^4}}{1024}+\frac {123}{128} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac {1}{40} \left (27-10 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}+\frac {62361 \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )}{2048}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 76, normalized size = 0.72 \[ \frac {311805 \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )+2 \sqrt {x^4+5 x^2+3} \left (1280 x^{10}+9344 x^8+14960 x^6+5064 x^4+12390 x^2-77229\right )}{10240} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3 + 5*x^2 + x^4]*(-77229 + 12390*x^2 + 5064*x^4 + 14960*x^6 + 9344*x^8 + 1280*x^10) + 311805*ArcTanh[(
5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/10240

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 66, normalized size = 0.62 \[ \frac {1}{5120} \, {\left (1280 \, x^{10} + 9344 \, x^{8} + 14960 \, x^{6} + 5064 \, x^{4} + 12390 \, x^{2} - 77229\right )} \sqrt {x^{4} + 5 \, x^{2} + 3} - \frac {62361}{2048} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5120*(1280*x^10 + 9344*x^8 + 14960*x^6 + 5064*x^4 + 12390*x^2 - 77229)*sqrt(x^4 + 5*x^2 + 3) - 62361/2048*lo
g(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5)

________________________________________________________________________________________

giac [B]  time = 0.59, size = 179, normalized size = 1.69 \[ \frac {1}{1024} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, x^{2} + 1\right )} x^{2} - 33\right )} x^{2} + 321\right )} x^{2} - 6837\right )} x^{2} + 87147\right )} + \frac {17}{3840} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x^{2} + 5\right )} x^{2} - 127\right )} x^{2} + 2635\right )} x^{2} - 33429\right )} + \frac {19}{384} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + 5\right )} x^{2} - 89\right )} x^{2} + 1095\right )} + \frac {1}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, x^{2} + 5\right )} x^{2} - 51\right )} - \frac {62361}{2048} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1024*sqrt(x^4 + 5*x^2 + 3)*(2*(4*(2*(8*(2*x^2 + 1)*x^2 - 33)*x^2 + 321)*x^2 - 6837)*x^2 + 87147) + 17/3840*s
qrt(x^4 + 5*x^2 + 3)*(2*(4*(6*(8*x^2 + 5)*x^2 - 127)*x^2 + 2635)*x^2 - 33429) + 19/384*sqrt(x^4 + 5*x^2 + 3)*(
2*(4*(6*x^2 + 5)*x^2 - 89)*x^2 + 1095) + 1/8*sqrt(x^4 + 5*x^2 + 3)*(2*(4*x^2 + 5)*x^2 - 51) - 62361/2048*log(2
*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 121, normalized size = 1.14 \[ \frac {\sqrt {x^{4}+5 x^{2}+3}\, x^{10}}{4}+\frac {73 \sqrt {x^{4}+5 x^{2}+3}\, x^{8}}{40}+\frac {187 \sqrt {x^{4}+5 x^{2}+3}\, x^{6}}{64}+\frac {633 \sqrt {x^{4}+5 x^{2}+3}\, x^{4}}{640}+\frac {1239 \sqrt {x^{4}+5 x^{2}+3}\, x^{2}}{512}+\frac {62361 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{2048}-\frac {77229 \sqrt {x^{4}+5 x^{2}+3}}{5120} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

187/64*(x^4+5*x^2+3)^(1/2)*x^6+1239/512*(x^4+5*x^2+3)^(1/2)*x^2+62361/2048*ln(x^2+5/2+(x^4+5*x^2+3)^(1/2))+1/4
*(x^4+5*x^2+3)^(1/2)*x^10+73/40*(x^4+5*x^2+3)^(1/2)*x^8-77229/5120*(x^4+5*x^2+3)^(1/2)+633/640*(x^4+5*x^2+3)^(
1/2)*x^4

________________________________________________________________________________________

maxima [A]  time = 0.88, size = 118, normalized size = 1.11 \[ \frac {1}{4} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {5}{2}} x^{2} + \frac {123}{64} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} x^{2} - \frac {27}{40} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {5}{2}} - \frac {4797}{512} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} + \frac {615}{128} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} - \frac {23985}{1024} \, \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {62361}{2048} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x^4 + 5*x^2 + 3)^(5/2)*x^2 + 123/64*(x^4 + 5*x^2 + 3)^(3/2)*x^2 - 27/40*(x^4 + 5*x^2 + 3)^(5/2) - 4797/51
2*sqrt(x^4 + 5*x^2 + 3)*x^2 + 615/128*(x^4 + 5*x^2 + 3)^(3/2) - 23985/1024*sqrt(x^4 + 5*x^2 + 3) + 62361/2048*
log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (3\,x^2+2\right )\,{\left (x^4+5\,x^2+3\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (3 x^{2} + 2\right ) \left (x^{4} + 5 x^{2} + 3\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(3*x**2 + 2)*(x**4 + 5*x**2 + 3)**(3/2), x)

________________________________________________________________________________________