Integrand size = 25, antiderivative size = 77 \[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=-\frac {x^2 \left (33+47 x^2\right )}{13 \sqrt {3+5 x^2+x^4}}+\frac {133}{26} \sqrt {3+5 x^2+x^4}-\frac {41}{4} \text {arctanh}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right ) \]
-41/4*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))-1/13*x^2*(47*x^2+33)/(x^4 +5*x^2+3)^(1/2)+133/26*(x^4+5*x^2+3)^(1/2)
Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.77 \[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\frac {399+599 x^2+39 x^4}{26 \sqrt {3+5 x^2+x^4}}+\frac {41}{4} \log \left (-5-2 x^2+2 \sqrt {3+5 x^2+x^4}\right ) \]
(399 + 599*x^2 + 39*x^4)/(26*Sqrt[3 + 5*x^2 + x^4]) + (41*Log[-5 - 2*x^2 + 2*Sqrt[3 + 5*x^2 + x^4]])/4
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1578, 1233, 27, 1160, 1092, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^5 \left (3 x^2+2\right )}{\left (x^4+5 x^2+3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {1}{2} \int \frac {x^4 \left (3 x^2+2\right )}{\left (x^4+5 x^2+3\right )^{3/2}}dx^2\) |
\(\Big \downarrow \) 1233 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{13} \int \frac {133 x^2+66}{2 \sqrt {x^4+5 x^2+3}}dx^2-\frac {2 x^2 \left (47 x^2+33\right )}{13 \sqrt {x^4+5 x^2+3}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{13} \int \frac {133 x^2+66}{\sqrt {x^4+5 x^2+3}}dx^2-\frac {2 x^2 \left (47 x^2+33\right )}{13 \sqrt {x^4+5 x^2+3}}\right )\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{13} \left (133 \sqrt {x^4+5 x^2+3}-\frac {533}{2} \int \frac {1}{\sqrt {x^4+5 x^2+3}}dx^2\right )-\frac {2 x^2 \left (47 x^2+33\right )}{13 \sqrt {x^4+5 x^2+3}}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{13} \left (133 \sqrt {x^4+5 x^2+3}-533 \int \frac {1}{4-x^4}d\frac {2 x^2+5}{\sqrt {x^4+5 x^2+3}}\right )-\frac {2 x^2 \left (47 x^2+33\right )}{13 \sqrt {x^4+5 x^2+3}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{13} \left (133 \sqrt {x^4+5 x^2+3}-\frac {533}{2} \text {arctanh}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )\right )-\frac {2 x^2 \left (47 x^2+33\right )}{13 \sqrt {x^4+5 x^2+3}}\right )\) |
((-2*x^2*(33 + 47*x^2))/(13*Sqrt[3 + 5*x^2 + x^4]) + (133*Sqrt[3 + 5*x^2 + x^4] - (533*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/2)/13)/2
3.2.94.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) ^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c *(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f *(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | | !ILtQ[m + 2*p + 3, 0])
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[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] && Int egerQ[(m - 1)/2]
Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.62
method | result | size |
risch | \(\frac {39 x^{4}+599 x^{2}+399}{26 \sqrt {x^{4}+5 x^{2}+3}}-\frac {41 \ln \left (\frac {5}{2}+x^{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{4}\) | \(48\) |
trager | \(\frac {39 x^{4}+599 x^{2}+399}{26 \sqrt {x^{4}+5 x^{2}+3}}+\frac {41 \ln \left (-2 x^{2}+2 \sqrt {x^{4}+5 x^{2}+3}-5\right )}{4}\) | \(52\) |
pseudoelliptic | \(\frac {78 x^{4}-533 \ln \left (2 x^{2}+5+2 \sqrt {x^{4}+5 x^{2}+3}\right ) \sqrt {x^{4}+5 x^{2}+3}+1198 x^{2}+798}{52 \sqrt {x^{4}+5 x^{2}+3}}\) | \(63\) |
default | \(\frac {3 x^{4}}{2 \sqrt {x^{4}+5 x^{2}+3}}+\frac {41 x^{2}}{4 \sqrt {x^{4}+5 x^{2}+3}}-\frac {133}{8 \sqrt {x^{4}+5 x^{2}+3}}+\frac {\frac {665 x^{2}}{52}+\frac {3325}{104}}{\sqrt {x^{4}+5 x^{2}+3}}-\frac {41 \ln \left (\frac {5}{2}+x^{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{4}\) | \(91\) |
elliptic | \(\frac {3 x^{4}}{2 \sqrt {x^{4}+5 x^{2}+3}}+\frac {41 x^{2}}{4 \sqrt {x^{4}+5 x^{2}+3}}-\frac {133}{8 \sqrt {x^{4}+5 x^{2}+3}}+\frac {\frac {665 x^{2}}{52}+\frac {3325}{104}}{\sqrt {x^{4}+5 x^{2}+3}}-\frac {41 \ln \left (\frac {5}{2}+x^{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{4}\) | \(91\) |
Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.12 \[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\frac {1811 \, x^{4} + 9055 \, x^{2} + 1066 \, {\left (x^{4} + 5 \, x^{2} + 3\right )} \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) + 4 \, {\left (39 \, x^{4} + 599 \, x^{2} + 399\right )} \sqrt {x^{4} + 5 \, x^{2} + 3} + 5433}{104 \, {\left (x^{4} + 5 \, x^{2} + 3\right )}} \]
1/104*(1811*x^4 + 9055*x^2 + 1066*(x^4 + 5*x^2 + 3)*log(-2*x^2 + 2*sqrt(x^ 4 + 5*x^2 + 3) - 5) + 4*(39*x^4 + 599*x^2 + 399)*sqrt(x^4 + 5*x^2 + 3) + 5 433)/(x^4 + 5*x^2 + 3)
\[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\int \frac {x^{5} \cdot \left (3 x^{2} + 2\right )}{\left (x^{4} + 5 x^{2} + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\frac {3 \, x^{4}}{2 \, \sqrt {x^{4} + 5 \, x^{2} + 3}} + \frac {599 \, x^{2}}{26 \, \sqrt {x^{4} + 5 \, x^{2} + 3}} + \frac {399}{26 \, \sqrt {x^{4} + 5 \, x^{2} + 3}} - \frac {41}{4} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]
3/2*x^4/sqrt(x^4 + 5*x^2 + 3) + 599/26*x^2/sqrt(x^4 + 5*x^2 + 3) + 399/26/ sqrt(x^4 + 5*x^2 + 3) - 41/4*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.68 \[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\frac {{\left (39 \, x^{2} + 599\right )} x^{2} + 399}{26 \, \sqrt {x^{4} + 5 \, x^{2} + 3}} + \frac {41}{4} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]
1/26*((39*x^2 + 599)*x^2 + 399)/sqrt(x^4 + 5*x^2 + 3) + 41/4*log(2*x^2 - 2 *sqrt(x^4 + 5*x^2 + 3) + 5)
Timed out. \[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx=\int \frac {x^5\,\left (3\,x^2+2\right )}{{\left (x^4+5\,x^2+3\right )}^{3/2}} \,d x \]