3.2.56 \(\int x^5 (2+3 x^2) (3+5 x^2+x^4)^{3/2} \, dx\) [156]

Optimal. Leaf size=127 \[ \frac {28379 \left (5+2 x^2\right ) \sqrt {3+5 x^2+x^4}}{2048}-\frac {2183}{768} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac {\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}-\frac {368927 \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )}{4096} \]

[Out]

-2183/768*(2*x^2+5)*(x^4+5*x^2+3)^(3/2)+3/14*x^4*(x^4+5*x^2+3)^(5/2)+1/1680*(-1070*x^2+3313)*(x^4+5*x^2+3)^(5/
2)-368927/4096*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))+28379/2048*(2*x^2+5)*(x^4+5*x^2+3)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1265, 846, 793, 626, 635, 212} \begin {gather*} \frac {3}{14} \left (x^4+5 x^2+3\right )^{5/2} x^4+\frac {\left (3313-1070 x^2\right ) \left (x^4+5 x^2+3\right )^{5/2}}{1680}-\frac {2183}{768} \left (2 x^2+5\right ) \left (x^4+5 x^2+3\right )^{3/2}+\frac {28379 \left (2 x^2+5\right ) \sqrt {x^4+5 x^2+3}}{2048}-\frac {368927 \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )}{4096} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(28379*(5 + 2*x^2)*Sqrt[3 + 5*x^2 + x^4])/2048 - (2183*(5 + 2*x^2)*(3 + 5*x^2 + x^4)^(3/2))/768 + (3*x^4*(3 +
5*x^2 + x^4)^(5/2))/14 + ((3313 - 1070*x^2)*(3 + 5*x^2 + x^4)^(5/2))/1680 - (368927*ArcTanh[(5 + 2*x^2)/(2*Sqr
t[3 + 5*x^2 + x^4])])/4096

Rule 212

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

Rule 626

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 635

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 793

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 846

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

Rule 1265

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^5 \left (2+3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2} \, dx &=\frac {1}{2} \text {Subst}\left (\int x^2 (2+3 x) \left (3+5 x+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac {1}{14} \text {Subst}\left (\int \left (-18-\frac {107 x}{2}\right ) x \left (3+5 x+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac {\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}-\frac {2183}{96} \text {Subst}\left (\int \left (3+5 x+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=-\frac {2183}{768} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac {\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}+\frac {28379}{512} \text {Subst}\left (\int \sqrt {3+5 x+x^2} \, dx,x,x^2\right )\\ &=\frac {28379 \left (5+2 x^2\right ) \sqrt {3+5 x^2+x^4}}{2048}-\frac {2183}{768} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac {\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}-\frac {368927 \text {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )}{4096}\\ &=\frac {28379 \left (5+2 x^2\right ) \sqrt {3+5 x^2+x^4}}{2048}-\frac {2183}{768} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac {\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}-\frac {368927 \text {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {5+2 x^2}{\sqrt {3+5 x^2+x^4}}\right )}{2048}\\ &=\frac {28379 \left (5+2 x^2\right ) \sqrt {3+5 x^2+x^4}}{2048}-\frac {2183}{768} \left (5+2 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {3}{14} x^4 \left (3+5 x^2+x^4\right )^{5/2}+\frac {\left (3313-1070 x^2\right ) \left (3+5 x^2+x^4\right )^{5/2}}{1680}-\frac {368927 \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )}{4096}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 79, normalized size = 0.62 \begin {gather*} \frac {\sqrt {3+5 x^2+x^4} \left (9546951-1499570 x^2+283304 x^4+154800 x^6+482944 x^8+323840 x^{10}+46080 x^{12}\right )}{215040}+\frac {368927 \log \left (-5-2 x^2+2 \sqrt {3+5 x^2+x^4}\right )}{4096} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3 + 5*x^2 + x^4]*(9546951 - 1499570*x^2 + 283304*x^4 + 154800*x^6 + 482944*x^8 + 323840*x^10 + 46080*x^1
2))/215040 + (368927*Log[-5 - 2*x^2 + 2*Sqrt[3 + 5*x^2 + x^4]])/4096

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Maple [A]
time = 0.12, size = 138, normalized size = 1.09

method	result	size
risch	\(\frac {\left (46080 x^{12}+323840 x^{10}+482944 x^{8}+154800 x^{6}+283304 x^{4}-1499570 x^{2}+9546951\right ) \sqrt {x^{4}+5 x^{2}+3}}{215040}-\frac {368927 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{4096}\)	\(68\)
trager	\(\left (\frac {3}{14} x^{12}+\frac {253}{168} x^{10}+\frac {539}{240} x^{8}+\frac {645}{896} x^{6}+\frac {5059}{3840} x^{4}-\frac {149957}{21504} x^{2}+\frac {3182317}{71680}\right ) \sqrt {x^{4}+5 x^{2}+3}+\frac {368927 \ln \left (-2 x^{2}+2 \sqrt {x^{4}+5 x^{2}+3}-5\right )}{4096}\)	\(71\)
default	\(\frac {3 x^{12} \sqrt {x^{4}+5 x^{2}+3}}{14}+\frac {253 x^{10} \sqrt {x^{4}+5 x^{2}+3}}{168}+\frac {3182317 \sqrt {x^{4}+5 x^{2}+3}}{71680}-\frac {149957 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{21504}+\frac {539 x^{8} \sqrt {x^{4}+5 x^{2}+3}}{240}+\frac {645 x^{6} \sqrt {x^{4}+5 x^{2}+3}}{896}-\frac {368927 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{4096}+\frac {5059 x^{4} \sqrt {x^{4}+5 x^{2}+3}}{3840}\)	\(138\)
elliptic	\(\frac {3 x^{12} \sqrt {x^{4}+5 x^{2}+3}}{14}+\frac {253 x^{10} \sqrt {x^{4}+5 x^{2}+3}}{168}+\frac {3182317 \sqrt {x^{4}+5 x^{2}+3}}{71680}-\frac {149957 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{21504}+\frac {539 x^{8} \sqrt {x^{4}+5 x^{2}+3}}{240}+\frac {645 x^{6} \sqrt {x^{4}+5 x^{2}+3}}{896}-\frac {368927 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{4096}+\frac {5059 x^{4} \sqrt {x^{4}+5 x^{2}+3}}{3840}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*x^12*(x^4+5*x^2+3)^(1/2)+253/168*x^10*(x^4+5*x^2+3)^(1/2)+3182317/71680*(x^4+5*x^2+3)^(1/2)-149957/21504*
x^2*(x^4+5*x^2+3)^(1/2)+539/240*x^8*(x^4+5*x^2+3)^(1/2)+645/896*x^6*(x^4+5*x^2+3)^(1/2)-368927/4096*ln(x^2+5/2
+(x^4+5*x^2+3)^(1/2))+5059/3840*x^4*(x^4+5*x^2+3)^(1/2)

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Maxima [A]
time = 0.27, size = 135, normalized size = 1.06 \begin {gather*} \frac {3}{14} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {5}{2}} x^{4} - \frac {107}{168} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {5}{2}} x^{2} - \frac {2183}{384} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} x^{2} + \frac {3313}{1680} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {5}{2}} + \frac {28379}{1024} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} - \frac {10915}{768} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} + \frac {141895}{2048} \, \sqrt {x^{4} + 5 \, x^{2} + 3} - \frac {368927}{4096} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*(x^4 + 5*x^2 + 3)^(5/2)*x^4 - 107/168*(x^4 + 5*x^2 + 3)^(5/2)*x^2 - 2183/384*(x^4 + 5*x^2 + 3)^(3/2)*x^2
+ 3313/1680*(x^4 + 5*x^2 + 3)^(5/2) + 28379/1024*sqrt(x^4 + 5*x^2 + 3)*x^2 - 10915/768*(x^4 + 5*x^2 + 3)^(3/2)
+ 141895/2048*sqrt(x^4 + 5*x^2 + 3) - 368927/4096*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Fricas [A]
time = 0.33, size = 71, normalized size = 0.56 \begin {gather*} \frac {1}{215040} \, {\left (46080 \, x^{12} + 323840 \, x^{10} + 482944 \, x^{8} + 154800 \, x^{6} + 283304 \, x^{4} - 1499570 \, x^{2} + 9546951\right )} \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {368927}{4096} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/215040*(46080*x^12 + 323840*x^10 + 482944*x^8 + 154800*x^6 + 283304*x^4 - 1499570*x^2 + 9546951)*sqrt(x^4 +
5*x^2 + 3) + 368927/4096*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{5} \cdot \left (3 x^{2} + 2\right ) \left (x^{4} + 5 x^{2} + 3\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.22, size = 207, normalized size = 1.63 \begin {gather*} \frac {1}{71680} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, {\left (12 \, x^{2} + 5\right )} x^{2} - 203\right )} x^{2} + 7635\right )} x^{2} - 76083\right )} x^{2} + 1627215\right )} x^{2} - 20756241\right )} + \frac {17}{3072} \, \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 {19}{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 {1}{64} \, \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 {368927}{4096} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/71680*sqrt(x^4 + 5*x^2 + 3)*(2*(4*(2*(8*(10*(12*x^2 + 5)*x^2 - 203)*x^2 + 7635)*x^2 - 76083)*x^2 + 1627215)*
x^2 - 20756241) + 17/3072*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) + 19/3840*sqrt(x^4 + 5*x^2 + 3)*(2*(4*(6*(8*x^2 + 5)*x^2 - 127)*x^2 + 2635)*x^2 - 33429) + 1/64*sqrt(x
^4 + 5*x^2 + 3)*(2*(4*(6*x^2 + 5)*x^2 - 89)*x^2 + 1095) + 368927/4096*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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