3.1.20 \(\int x^5 (2+3 x^2) (5+x^4)^{3/2} \, dx\) [20]

Optimal. Leaf size=83 \[ -\frac {25}{16} x^2 \sqrt {5+x^4}-\frac {5}{24} x^2 \left (5+x^4\right )^{3/2}+\frac {3}{14} x^4 \left (5+x^4\right )^{5/2}-\frac {1}{42} \left (18-7 x^2\right ) \left (5+x^4\right )^{5/2}-\frac {125}{16} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

[Out]

-5/24*x^2*(x^4+5)^(3/2)+3/14*x^4*(x^4+5)^(5/2)-1/42*(-7*x^2+18)*(x^4+5)^(5/2)-125/16*arcsinh(1/5*x^2*5^(1/2))-
25/16*x^2*(x^4+5)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1266, 847, 794, 201, 221} \begin {gather*} \frac {3}{14} \left (x^4+5\right )^{5/2} x^4-\frac {125}{16} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {5}{24} \left (x^4+5\right )^{3/2} x^2-\frac {25}{16} \sqrt {x^4+5} x^2-\frac {1}{42} \left (18-7 x^2\right ) \left (x^4+5\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*x^2*Sqrt[5 + x^4])/16 - (5*x^2*(5 + x^4)^(3/2))/24 + (3*x^4*(5 + x^4)^(5/2))/14 - ((18 - 7*x^2)*(5 + x^4)
^(5/2))/42 - (125*ArcSinh[x^2/Sqrt[5]])/16

Rule 201

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 221

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 794

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 847

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^
m*((a + 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 + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + 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 1266

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]

Rubi steps

\begin {align*} \int x^5 \left (2+3 x^2\right ) \left (5+x^4\right )^{3/2} \, dx &=\frac {1}{2} \text {Subst}\left (\int x^2 (2+3 x) \left (5+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {3}{14} x^4 \left (5+x^4\right )^{5/2}+\frac {1}{14} \text {Subst}\left (\int x (-30+14 x) \left (5+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {3}{14} x^4 \left (5+x^4\right )^{5/2}-\frac {1}{42} \left (18-7 x^2\right ) \left (5+x^4\right )^{5/2}-\frac {5}{6} \text {Subst}\left (\int \left (5+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=-\frac {5}{24} x^2 \left (5+x^4\right )^{3/2}+\frac {3}{14} x^4 \left (5+x^4\right )^{5/2}-\frac {1}{42} \left (18-7 x^2\right ) \left (5+x^4\right )^{5/2}-\frac {25}{8} \text {Subst}\left (\int \sqrt {5+x^2} \, dx,x,x^2\right )\\ &=-\frac {25}{16} x^2 \sqrt {5+x^4}-\frac {5}{24} x^2 \left (5+x^4\right )^{3/2}+\frac {3}{14} x^4 \left (5+x^4\right )^{5/2}-\frac {1}{42} \left (18-7 x^2\right ) \left (5+x^4\right )^{5/2}-\frac {125}{16} \text {Subst}\left (\int \frac {1}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=-\frac {25}{16} x^2 \sqrt {5+x^4}-\frac {5}{24} x^2 \left (5+x^4\right )^{3/2}+\frac {3}{14} x^4 \left (5+x^4\right )^{5/2}-\frac {1}{42} \left (18-7 x^2\right ) \left (5+x^4\right )^{5/2}-\frac {125}{16} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 64, normalized size = 0.77 \begin {gather*} \frac {1}{336} \sqrt {5+x^4} \left (-3600+525 x^2+360 x^4+490 x^6+576 x^8+56 x^{10}+72 x^{12}\right )-\frac {125}{16} \tanh ^{-1}\left (\frac {x^2}{\sqrt {5+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[5 + x^4]*(-3600 + 525*x^2 + 360*x^4 + 490*x^6 + 576*x^8 + 56*x^10 + 72*x^12))/336 - (125*ArcTanh[x^2/Sqr
t[5 + x^4]])/16

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Maple [A]
time = 0.14, size = 73, normalized size = 0.88

method	result	size
risch	\(\frac {\left (72 x^{12}+56 x^{10}+576 x^{8}+490 x^{6}+360 x^{4}+525 x^{2}-3600\right ) \sqrt {x^{4}+5}}{336}-\frac {125 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{16}\)	\(54\)
trager	\(\left (\frac {3}{14} x^{12}+\frac {1}{6} x^{10}+\frac {12}{7} x^{8}+\frac {35}{24} x^{6}+\frac {15}{14} x^{4}+\frac {25}{16} x^{2}-\frac {75}{7}\right ) \sqrt {x^{4}+5}-\frac {125 \ln \left (x^{2}+\sqrt {x^{4}+5}\right )}{16}\)	\(56\)
default	\(\frac {3 \sqrt {x^{4}+5}\, \left (x^{4}-2\right ) \left (x^{8}+10 x^{4}+25\right )}{14}+\frac {x^{10} \sqrt {x^{4}+5}}{6}+\frac {35 x^{6} \sqrt {x^{4}+5}}{24}+\frac {25 x^{2} \sqrt {x^{4}+5}}{16}-\frac {125 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{16}\)	\(73\)
elliptic	\(\frac {3 x^{12} \sqrt {x^{4}+5}}{14}+\frac {12 x^{8} \sqrt {x^{4}+5}}{7}+\frac {15 x^{4} \sqrt {x^{4}+5}}{14}-\frac {75 \sqrt {x^{4}+5}}{7}+\frac {x^{10} \sqrt {x^{4}+5}}{6}+\frac {35 x^{6} \sqrt {x^{4}+5}}{24}+\frac {25 x^{2} \sqrt {x^{4}+5}}{16}-\frac {125 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{16}\)	\(94\)
meijerg	\(\frac {1125 \sqrt {5}\, \left (\frac {16 \sqrt {\pi }}{105}-\frac {2 \sqrt {\pi }\, \left (-\frac {4}{25} x^{12}-\frac {32}{25} x^{8}-\frac {4}{5} x^{4}+8\right ) \sqrt {1+\frac {x^{4}}{5}}}{105}\right )}{16 \sqrt {\pi }}+\frac {\frac {25 \sqrt {\pi }\, x^{2} \sqrt {5}\, \left (\frac {8}{25} x^{8}+\frac {14}{5} x^{4}+3\right ) \sqrt {1+\frac {x^{4}}{5}}}{48}-\frac {125 \sqrt {\pi }\, \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{16}}{\sqrt {\pi }}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*(x^4+5)^(1/2)*(x^4-2)*(x^8+10*x^4+25)+1/6*x^10*(x^4+5)^(1/2)+35/24*x^6*(x^4+5)^(1/2)+25/16*x^2*(x^4+5)^(1
/2)-125/16*arcsinh(1/5*x^2*5^(1/2))

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Maxima [A]
time = 0.50, size = 127, normalized size = 1.53 \begin {gather*} \frac {3}{14} \, {\left (x^{4} + 5\right )}^{\frac {7}{2}} - \frac {3}{2} \, {\left (x^{4} + 5\right )}^{\frac {5}{2}} - \frac {125 \, {\left (\frac {3 \, \sqrt {x^{4} + 5}}{x^{2}} - \frac {8 \, {\left (x^{4} + 5\right )}^{\frac {3}{2}}}{x^{6}} - \frac {3 \, {\left (x^{4} + 5\right )}^{\frac {5}{2}}}{x^{10}}\right )}}{48 \, {\left (\frac {3 \, {\left (x^{4} + 5\right )}}{x^{4}} - \frac {3 \, {\left (x^{4} + 5\right )}^{2}}{x^{8}} + \frac {{\left (x^{4} + 5\right )}^{3}}{x^{12}} - 1\right )}} - \frac {125}{32} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) + \frac {125}{32} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*(x^4 + 5)^(7/2) - 3/2*(x^4 + 5)^(5/2) - 125/48*(3*sqrt(x^4 + 5)/x^2 - 8*(x^4 + 5)^(3/2)/x^6 - 3*(x^4 + 5)
^(5/2)/x^10)/(3*(x^4 + 5)/x^4 - 3*(x^4 + 5)^2/x^8 + (x^4 + 5)^3/x^12 - 1) - 125/32*log(sqrt(x^4 + 5)/x^2 + 1)
+ 125/32*log(sqrt(x^4 + 5)/x^2 - 1)

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Fricas [A]
time = 0.34, size = 58, normalized size = 0.70 \begin {gather*} \frac {1}{336} \, {\left (72 \, x^{12} + 56 \, x^{10} + 576 \, x^{8} + 490 \, x^{6} + 360 \, x^{4} + 525 \, x^{2} - 3600\right )} \sqrt {x^{4} + 5} + \frac {125}{16} \, \log \left (-x^{2} + \sqrt {x^{4} + 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)^(3/2),x, algorithm="fricas")

[Out]

1/336*(72*x^12 + 56*x^10 + 576*x^8 + 490*x^6 + 360*x^4 + 525*x^2 - 3600)*sqrt(x^4 + 5) + 125/16*log(-x^2 + sqr
t(x^4 + 5))

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Sympy [A]
time = 8.97, size = 131, normalized size = 1.58 \begin {gather*} \frac {x^{14}}{6 \sqrt {x^{4} + 5}} + \frac {3 x^{12} \sqrt {x^{4} + 5}}{14} + \frac {55 x^{10}}{24 \sqrt {x^{4} + 5}} + \frac {12 x^{8} \sqrt {x^{4} + 5}}{7} + \frac {425 x^{6}}{48 \sqrt {x^{4} + 5}} + \frac {15 x^{4} \sqrt {x^{4} + 5}}{14} + \frac {125 x^{2}}{16 \sqrt {x^{4} + 5}} - \frac {75 \sqrt {x^{4} + 5}}{7} - \frac {125 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**14/(6*sqrt(x**4 + 5)) + 3*x**12*sqrt(x**4 + 5)/14 + 55*x**10/(24*sqrt(x**4 + 5)) + 12*x**8*sqrt(x**4 + 5)/7
+ 425*x**6/(48*sqrt(x**4 + 5)) + 15*x**4*sqrt(x**4 + 5)/14 + 125*x**2/(16*sqrt(x**4 + 5)) - 75*sqrt(x**4 + 5)
/7 - 125*asinh(sqrt(5)*x**2/5)/16

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Giac [A]
time = 3.62, size = 80, normalized size = 0.96 \begin {gather*} \frac {3}{14} \, {\left (x^{4} + 5\right )}^{\frac {7}{2}} + \frac {1}{48} \, {\left (2 \, {\left (4 \, x^{4} + 5\right )} x^{4} - 75\right )} \sqrt {x^{4} + 5} x^{2} + \frac {5}{8} \, {\left (2 \, x^{4} + 5\right )} \sqrt {x^{4} + 5} x^{2} - \frac {3}{2} \, {\left (x^{4} + 5\right )}^{\frac {5}{2}} + \frac {125}{16} \, \log \left (-x^{2} + \sqrt {x^{4} + 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)^(3/2),x, algorithm="giac")

[Out]

3/14*(x^4 + 5)^(7/2) + 1/48*(2*(4*x^4 + 5)*x^4 - 75)*sqrt(x^4 + 5)*x^2 + 5/8*(2*x^4 + 5)*sqrt(x^4 + 5)*x^2 - 3
/2*(x^4 + 5)^(5/2) + 125/16*log(-x^2 + sqrt(x^4 + 5))

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Mupad [B]
time = 0.32, size = 52, normalized size = 0.63 \begin {gather*} \sqrt {x^4+5}\,\left (\frac {3\,x^{12}}{14}+\frac {x^{10}}{6}+\frac {12\,x^8}{7}+\frac {35\,x^6}{24}+\frac {15\,x^4}{14}+\frac {25\,x^2}{16}-\frac {75}{7}\right )-\frac {125\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^4 + 5)^(1/2)*((25*x^2)/16 + (15*x^4)/14 + (35*x^6)/24 + (12*x^8)/7 + x^10/6 + (3*x^12)/14 - 75/7) - (125*as
inh((5^(1/2)*x^2)/5))/16

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