Integrand size = 19, antiderivative size = 91 \[ \int x^3 \sqrt {b x^2+c x^4} \, dx=-\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c^2}+\frac {\left (b x^2+c x^4\right )^{3/2}}{6 c}+\frac {b^3 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{5/2}} \]
1/6*(c*x^4+b*x^2)^(3/2)/c+1/16*b^3*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2) )/c^(5/2)-1/16*b*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^2
Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.21 \[ \int x^3 \sqrt {b x^2+c x^4} \, dx=\frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (-3 b^2+2 b c x^2+8 c^2 x^4\right )+6 b^3 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )\right )}{48 c^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \]
(x*Sqrt[b + c*x^2]*(Sqrt[c]*x*Sqrt[b + c*x^2]*(-3*b^2 + 2*b*c*x^2 + 8*c^2* x^4) + 6*b^3*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])]))/(48*c^(5/ 2)*Sqrt[x^2*(b + c*x^2)])
Time = 0.24 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1424, 1160, 1087, 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.
\(\displaystyle \int x^3 \sqrt {b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle \frac {1}{2} \int x^2 \sqrt {c x^4+b x^2}dx^2\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {b \int \sqrt {c x^4+b x^2}dx^2}{2 c}\right )\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {b \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c}\right )}{2 c}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {b \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{4 c}\right )}{2 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {b \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{4 c^{3/2}}\right )}{2 c}\right )\) |
((b*x^2 + c*x^4)^(3/2)/(3*c) - (b*(((b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(4* c) - (b^2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(4*c^(3/2))))/(2*c)) /2
3.3.22.3.1 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[((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] - Simp[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] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
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_))*((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 = 90, normalized size of antiderivative = 0.99
method | result | size |
risch | \(-\frac {\left (-8 c^{2} x^{4}-2 b c \,x^{2}+3 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{48 c^{2}}+\frac {b^{3} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16 c^{\frac {5}{2}} x \sqrt {c \,x^{2}+b}}\) | \(90\) |
default | \(\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (8 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}}-6 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b x +3 \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{2} x +3 \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) b^{3}\right )}{48 x \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}}}\) | \(104\) |
pseudoelliptic | \(\frac {16 c^{\frac {5}{2}} x^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}+4 c^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, b \,x^{2}+3 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{3}-3 \ln \left (2\right ) b^{3}-6 \sqrt {c}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, b^{2}}{96 c^{\frac {5}{2}}}\) | \(114\) |
-1/48*(-8*c^2*x^4-2*b*c*x^2+3*b^2)/c^2*(x^2*(c*x^2+b))^(1/2)+1/16*b^3/c^(5 /2)*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
Time = 0.26 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.84 \[ \int x^3 \sqrt {b x^2+c x^4} \, dx=\left [\frac {3 \, b^{3} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} x^{4} + 2 \, b c^{2} x^{2} - 3 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{3}}, -\frac {3 \, b^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (8 \, c^{3} x^{4} + 2 \, b c^{2} x^{2} - 3 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{3}}\right ] \]
[1/96*(3*b^3*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2 *(8*c^3*x^4 + 2*b*c^2*x^2 - 3*b^2*c)*sqrt(c*x^4 + b*x^2))/c^3, -1/48*(3*b^ 3*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (8*c^3*x^4 + 2*b*c^2*x^2 - 3*b^2*c)*sqrt(c*x^4 + b*x^2))/c^3]
\[ \int x^3 \sqrt {b x^2+c x^4} \, dx=\int x^{3} \sqrt {x^{2} \left (b + c x^{2}\right )}\, dx \]
Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07 \[ \int x^3 \sqrt {b x^2+c x^4} \, dx=-\frac {\sqrt {c x^{4} + b x^{2}} b x^{2}}{8 \, c} + \frac {b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{32 \, c^{\frac {5}{2}}} - \frac {\sqrt {c x^{4} + b x^{2}} b^{2}}{16 \, c^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{6 \, c} \]
-1/8*sqrt(c*x^4 + b*x^2)*b*x^2/c + 1/32*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(5/2) - 1/16*sqrt(c*x^4 + b*x^2)*b^2/c^2 + 1/6*(c*x^4 + b*x^2)^(3/2)/c
Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.93 \[ \int x^3 \sqrt {b x^2+c x^4} \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, x^{2} \mathrm {sgn}\left (x\right ) + \frac {b \mathrm {sgn}\left (x\right )}{c}\right )} x^{2} - \frac {3 \, b^{2} \mathrm {sgn}\left (x\right )}{c^{2}}\right )} \sqrt {c x^{2} + b} x - \frac {b^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {5}{2}}} + \frac {b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{32 \, c^{\frac {5}{2}}} \]
1/48*(2*(4*x^2*sgn(x) + b*sgn(x)/c)*x^2 - 3*b^2*sgn(x)/c^2)*sqrt(c*x^2 + b )*x - 1/16*b^3*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))*sgn(x)/c^(5/2) + 1/3 2*b^3*log(abs(b))*sgn(x)/c^(5/2)
Time = 13.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.85 \[ \int x^3 \sqrt {b x^2+c x^4} \, dx=\frac {b^3\,\ln \left (\frac {2\,c\,x^2+b}{\sqrt {c}}+2\,\sqrt {c\,x^4+b\,x^2}\right )}{32\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{48\,c^2} \]