Integrand size = 19, antiderivative size = 86 \[ \int \frac {x^5}{\sqrt {b x^2+c x^4}} \, dx=-\frac {3 b \sqrt {b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {b x^2+c x^4}}{4 c}+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{5/2}} \]
3/8*b^2*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(5/2)-3/8*b*(c*x^4+b*x^ 2)^(1/2)/c^2+1/4*x^2*(c*x^4+b*x^2)^(1/2)/c
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int \frac {x^5}{\sqrt {b x^2+c x^4}} \, dx=\frac {x \left (\sqrt {c} x \left (-3 b^2-b c x^2+2 c^2 x^4\right )+6 b^2 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )\right )}{8 c^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \]
(x*(Sqrt[c]*x*(-3*b^2 - b*c*x^2 + 2*c^2*x^4) + 6*b^2*Sqrt[b + c*x^2]*ArcTa nh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])]))/(8*c^(5/2)*Sqrt[x^2*(b + c* x^2)])
Time = 0.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1424, 1134, 1160, 1091, 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.
\(\displaystyle \int \frac {x^5}{\sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle \frac {1}{2} \int \frac {x^4}{\sqrt {c x^4+b x^2}}dx^2\) |
\(\Big \downarrow \) 1134 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {b x^2+c x^4}}{2 c}-\frac {3 b \int \frac {x^2}{\sqrt {c x^4+b x^2}}dx^2}{4 c}\right )\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {b x^2+c x^4}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x^2+c x^4}}{c}-\frac {b \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{2 c}\right )}{4 c}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {b x^2+c x^4}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x^2+c x^4}}{c}-\frac {b \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{c}\right )}{4 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {b x^2+c x^4}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x^2+c x^4}}{c}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}\right )}{4 c}\right )\) |
((x^2*Sqrt[b*x^2 + c*x^4])/(2*c) - (3*b*(Sqrt[b*x^2 + c*x^4]/c - (b*ArcTan h[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/c^(3/2)))/(4*c))/2
3.3.61.3.1 Defintions of rubi rules used
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[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))) Int[(d + e*x)^ (m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[ c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 *p]
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[(x_)^(m_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{b, c, m, p}, x] && !IntegerQ[p] && IntegerQ[(m - 1)/2]
Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99
method | result | size |
default | \(\frac {x \sqrt {c \,x^{2}+b}\, \left (2 x^{3} \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}}-3 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b x +3 \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) b^{2} c \right )}{8 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {7}{2}}}\) | \(85\) |
risch | \(-\frac {x^{2} \left (-2 c \,x^{2}+3 b \right ) \left (c \,x^{2}+b \right )}{8 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {3 b^{2} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) x \sqrt {c \,x^{2}+b}}{8 c^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(87\) |
pseudoelliptic | \(\frac {4 c^{\frac {3}{2}} x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}+3 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2}-3 \ln \left (2\right ) b^{2}-6 b \sqrt {c}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16 c^{\frac {5}{2}}}\) | \(90\) |
1/8*x*(c*x^2+b)^(1/2)*(2*x^3*(c*x^2+b)^(1/2)*c^(5/2)-3*c^(3/2)*(c*x^2+b)^( 1/2)*b*x+3*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*b^2*c)/(c*x^4+b*x^2)^(1/2)/c^(7/2 )
Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.69 \[ \int \frac {x^5}{\sqrt {b x^2+c x^4}} \, dx=\left [\frac {3 \, b^{2} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, c^{2} x^{2} - 3 \, b c\right )}}{16 \, c^{3}}, -\frac {3 \, b^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (2 \, c^{2} x^{2} - 3 \, b c\right )}}{8 \, c^{3}}\right ] \]
[1/16*(3*b^2*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2 *sqrt(c*x^4 + b*x^2)*(2*c^2*x^2 - 3*b*c))/c^3, -1/8*(3*b^2*sqrt(-c)*arctan (sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - sqrt(c*x^4 + b*x^2)*(2*c^2*x^ 2 - 3*b*c))/c^3]
\[ \int \frac {x^5}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{5}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88 \[ \int \frac {x^5}{\sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {c x^{4} + b x^{2}} x^{2}}{4 \, c} + \frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{8 \, c^{2}} \]
1/4*sqrt(c*x^4 + b*x^2)*x^2/c + 3/16*b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(5/2) - 3/8*sqrt(c*x^4 + b*x^2)*b/c^2
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92 \[ \int \frac {x^5}{\sqrt {b x^2+c x^4}} \, dx=\frac {1}{8} \, \sqrt {c x^{2} + b} x {\left (\frac {2 \, x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {3 \, b}{c^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {5}{2}}} - \frac {3 \, b^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{8 \, c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \]
1/8*sqrt(c*x^2 + b)*x*(2*x^2/(c*sgn(x)) - 3*b/(c^2*sgn(x))) + 3/16*b^2*log (abs(b))*sgn(x)/c^(5/2) - 3/8*b^2*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/( c^(5/2)*sgn(x))
Timed out. \[ \int \frac {x^5}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^5}{\sqrt {c\,x^4+b\,x^2}} \,d x \]