Integrand size = 27, antiderivative size = 153 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {B x^4 \sqrt {a+b x^2+c x^4}}{6 c}+\frac {\left (15 b^2 B-18 A b c-16 a B c-2 c (5 b B-6 A c) x^2\right ) \sqrt {a+b x^2+c x^4}}{48 c^3}-\frac {\left (5 b^3 B-6 A b^2 c-12 a b B c+8 a A c^2\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{7/2}} \]
-1/32*(8*A*a*c^2-6*A*b^2*c-12*B*a*b*c+5*B*b^3)*arctanh(1/2*(2*c*x^2+b)/c^( 1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(7/2)+1/6*B*x^4*(c*x^4+b*x^2+a)^(1/2)/c+1/48 *(15*B*b^2-18*A*b*c-16*B*a*c-2*c*(-6*A*c+5*B*b)*x^2)*(c*x^4+b*x^2+a)^(1/2) /c^3
Time = 0.52 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.88 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {\sqrt {a+b x^2+c x^4} \left (15 b^2 B-18 A b c-16 a B c-10 b B c x^2+12 A c^2 x^2+8 B c^2 x^4\right )}{48 c^3}+\frac {\left (5 b^3 B-6 A b^2 c-12 a b B c+8 a A c^2\right ) \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{32 c^{7/2}} \]
(Sqrt[a + b*x^2 + c*x^4]*(15*b^2*B - 18*A*b*c - 16*a*B*c - 10*b*B*c*x^2 + 12*A*c^2*x^2 + 8*B*c^2*x^4))/(48*c^3) + ((5*b^3*B - 6*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/(32*c^ (7/2))
Time = 0.35 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1578, 1236, 27, 1225, 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.
| \(\displaystyle \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle \frac {1}{2} \int \frac {x^4 \left (B x^2+A\right )}{\sqrt {c x^4+b x^2+a}}dx^2\) |
| \(\Big \downarrow \) 1236 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {x^2 \left ((5 b B-6 A c) x^2+4 a B\right )}{2 \sqrt {c x^4+b x^2+a}}dx^2}{3 c}+\frac {B x^4 \sqrt {a+b x^2+c x^4}}{3 c}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {B x^4 \sqrt {a+b x^2+c x^4}}{3 c}-\frac {\int \frac {x^2 \left ((5 b B-6 A c) x^2+4 a B\right )}{\sqrt {c x^4+b x^2+a}}dx^2}{6 c}\right )\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {1}{2} \left (\frac {B x^4 \sqrt {a+b x^2+c x^4}}{3 c}-\frac {\frac {3 \left (8 a A c^2-12 a b B c-6 A b^2 c+5 b^3 B\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2}{8 c^2}-\frac {\sqrt {a+b x^2+c x^4} \left (-16 a B c-2 c x^2 (5 b B-6 A c)-18 A b c+15 b^2 B\right )}{4 c^2}}{6 c}\right )\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {1}{2} \left (\frac {B x^4 \sqrt {a+b x^2+c x^4}}{3 c}-\frac {\frac {3 \left (8 a A c^2-12 a b B c-6 A b^2 c+5 b^3 B\right ) \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}}{4 c^2}-\frac {\sqrt {a+b x^2+c x^4} \left (-16 a B c-2 c x^2 (5 b B-6 A c)-18 A b c+15 b^2 B\right )}{4 c^2}}{6 c}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (\frac {B x^4 \sqrt {a+b x^2+c x^4}}{3 c}-\frac {\frac {3 \left (8 a A c^2-12 a b B c-6 A b^2 c+5 b^3 B\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{8 c^{5/2}}-\frac {\sqrt {a+b x^2+c x^4} \left (-16 a B c-2 c x^2 (5 b B-6 A c)-18 A b c+15 b^2 B\right )}{4 c^2}}{6 c}\right )\) |
((B*x^4*Sqrt[a + b*x^2 + c*x^4])/(3*c) - (-1/4*((15*b^2*B - 18*A*b*c - 16* a*B*c - 2*c*(5*b*B - 6*A*c)*x^2)*Sqrt[a + b*x^2 + c*x^4])/c^2 + (3*(5*b^3* B - 6*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*S qrt[a + b*x^2 + c*x^4])])/(8*c^(5/2)))/(6*c))/2
3.2.69.3.1 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_))*((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] + Simp[(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] && !LeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[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] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && EqQ[f, 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 = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.82
| method | result | size |
| risch | \(-\frac {\left (-8 B \,c^{2} x^{4}-12 A \,c^{2} x^{2}+10 B b c \,x^{2}+18 A b c +16 B a c -15 B \,b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{48 c^{3}}-\frac {\left (8 A a \,c^{2}-6 A \,b^{2} c -12 B a b c +5 B \,b^{3}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {7}{2}}}\) | \(125\) |
| pseudoelliptic | \(-\frac {3 \left (\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (\left (\frac {5 B \,x^{2}}{9}+A \right ) b +\frac {8 B a}{9}\right ) c^{\frac {3}{2}}-\frac {2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (\frac {2 B \,x^{2}}{3}+A \right ) x^{2} c^{\frac {5}{2}}}{3}-\frac {5 B \,b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}}{6}+\frac {2 \left (A a \,c^{2}-\frac {3}{4} A \,b^{2} c -\frac {3}{2} B a b c +\frac {5}{8} B \,b^{3}\right ) \left (-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )\right )}{3}\right )}{8 c^{\frac {7}{2}}}\) | \(158\) |
| default | \(B \left (\frac {x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{6 c}-\frac {5 b \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{24 c^{2}}+\frac {5 b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 c^{3}}-\frac {5 b^{3} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {7}{2}}}+\frac {3 b a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 c^{\frac {5}{2}}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{3 c^{2}}\right )+A \left (\frac {x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}-\frac {3 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}-\frac {a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}\right )\) | \(282\) |
| elliptic | \(\frac {B \,x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{6 c}-\frac {5 B b \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{24 c^{2}}+\frac {5 B \,b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 c^{3}}-\frac {5 B \,b^{3} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {7}{2}}}+\frac {3 B b a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 c^{\frac {5}{2}}}-\frac {B a \sqrt {c \,x^{4}+b \,x^{2}+a}}{3 c^{2}}+\frac {A \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}-\frac {3 A b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {3 A \,b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}-\frac {A a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {3}{2}}}\) | \(286\) |
-1/48*(-8*B*c^2*x^4-12*A*c^2*x^2+10*B*b*c*x^2+18*A*b*c+16*B*a*c-15*B*b^2)* (c*x^4+b*x^2+a)^(1/2)/c^3-1/32*(8*A*a*c^2-6*A*b^2*c-12*B*a*b*c+5*B*b^3)/c^ (7/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))
Time = 0.27 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.06 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\left [\frac {3 \, {\left (5 \, B b^{3} + 8 \, A a c^{2} - 6 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, B c^{3} x^{4} + 15 \, B b^{2} c - 2 \, {\left (8 \, B a + 9 \, A b\right )} c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{3} + 8 \, A a c^{2} - 6 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, {\left (8 \, B c^{3} x^{4} + 15 \, B b^{2} c - 2 \, {\left (8 \, B a + 9 \, A b\right )} c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, c^{4}}\right ] \]
[1/192*(3*(5*B*b^3 + 8*A*a*c^2 - 6*(2*B*a*b + A*b^2)*c)*sqrt(c)*log(-8*c^2 *x^4 - 8*b*c*x^2 - b^2 + 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) + 4*(8*B*c^3*x^4 + 15*B*b^2*c - 2*(8*B*a + 9*A*b)*c^2 - 2*(5*B*b*c ^2 - 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4, 1/96*(3*(5*B*b^3 + 8*A*a* c^2 - 6*(2*B*a*b + A*b^2)*c)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*( 2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*(8*B*c^3*x^4 + 15*B*b ^2*c - 2*(8*B*a + 9*A*b)*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x^2)*sqrt(c*x^4 + b *x^2 + a))/c^4]
Time = 1.05 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.12 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {\begin {cases} \left (- \frac {a \left (A - \frac {5 B b}{6 c}\right )}{2 c} - \frac {b \left (- \frac {2 B a}{3 c} - \frac {3 b \left (A - \frac {5 B b}{6 c}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {a + b x^{2} + c x^{4}} \left (\frac {B x^{4}}{3 c} + \frac {x^{2} \left (A - \frac {5 B b}{6 c}\right )}{2 c} + \frac {- \frac {2 B a}{3 c} - \frac {3 b \left (A - \frac {5 B b}{6 c}\right )}{4 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {\frac {2 A \left (a^{2} \sqrt {a + b x^{2}} - \frac {2 a \left (a + b x^{2}\right )^{\frac {3}{2}}}{3} + \frac {\left (a + b x^{2}\right )^{\frac {5}{2}}}{5}\right )}{b^{2}} + \frac {2 B \left (- a^{3} \sqrt {a + b x^{2}} + a^{2} \left (a + b x^{2}\right )^{\frac {3}{2}} - \frac {3 a \left (a + b x^{2}\right )^{\frac {5}{2}}}{5} + \frac {\left (a + b x^{2}\right )^{\frac {7}{2}}}{7}\right )}{b^{3}}}{b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{6}}{3} + \frac {B x^{8}}{4}}{\sqrt {a}} & \text {otherwise} \end {cases}}{2} \]
Piecewise(((-a*(A - 5*B*b/(6*c))/(2*c) - b*(-2*B*a/(3*c) - 3*b*(A - 5*B*b/ (6*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x**2 + c*x**4 ) + 2*c*x**2)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x**2)*log(b/(2* c) + x**2)/sqrt(c*(b/(2*c) + x**2)**2), True)) + sqrt(a + b*x**2 + c*x**4) *(B*x**4/(3*c) + x**2*(A - 5*B*b/(6*c))/(2*c) + (-2*B*a/(3*c) - 3*b*(A - 5 *B*b/(6*c))/(4*c))/c), Ne(c, 0)), ((2*A*(a**2*sqrt(a + b*x**2) - 2*a*(a + b*x**2)**(3/2)/3 + (a + b*x**2)**(5/2)/5)/b**2 + 2*B*(-a**3*sqrt(a + b*x** 2) + a**2*(a + b*x**2)**(3/2) - 3*a*(a + b*x**2)**(5/2)/5 + (a + b*x**2)** (7/2)/7)/b**3)/b, Ne(b, 0)), ((A*x**6/3 + B*x**8/4)/sqrt(a), True))/2
Exception generated. \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.33 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.89 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {1}{48} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (\frac {4 \, B x^{2}}{c} - \frac {5 \, B b c - 6 \, A c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, B b^{2} - 16 \, B a c - 18 \, A b c}{c^{3}}\right )} + \frac {{\left (5 \, B b^{3} - 12 \, B a b c - 6 \, A b^{2} c + 8 \, A a c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{32 \, c^{\frac {7}{2}}} \]
1/48*sqrt(c*x^4 + b*x^2 + a)*(2*(4*B*x^2/c - (5*B*b*c - 6*A*c^2)/c^3)*x^2 + (15*B*b^2 - 16*B*a*c - 18*A*b*c)/c^3) + 1/32*(5*B*b^3 - 12*B*a*b*c - 6*A *b^2*c + 8*A*a*c^2)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt (c) + b))/c^(7/2)
Timed out. \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {x^5\,\left (B\,x^2+A\right )}{\sqrt {c\,x^4+b\,x^2+a}} \,d x \]