Integrand size = 24, antiderivative size = 107 \[ \int x \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c^2}+\frac {B \left (b x^2+c x^4\right )^{3/2}}{6 c}+\frac {b^2 (b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{5/2}} \]
1/6*B*(c*x^4+b*x^2)^(3/2)/c+1/16*b^2*(-2*A*c+B*b)*arctanh(x^2*c^(1/2)/(c*x ^4+b*x^2)^(1/2))/c^(5/2)-1/16*(-2*A*c+B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2) /c^2
Time = 0.61 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.18 \[ \int x \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {c} x \left (-3 b^2 B+2 b c \left (3 A+B x^2\right )+4 c^2 x^2 \left (3 A+2 B x^2\right )\right )+\frac {6 b^2 (b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )}{\sqrt {b+c x^2}}\right )}{48 c^{5/2} x} \]
(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*x*(-3*b^2*B + 2*b*c*(3*A + B*x^2) + 4*c^2* x^2*(3*A + 2*B*x^2)) + (6*b^2*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])])/Sqrt[b + c*x^2]))/(48*c^(5/2)*x)
Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1940, 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.
| \(\displaystyle \int x \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx\) |
| \(\Big \downarrow \) 1940 |
| \(\displaystyle \frac {1}{2} \int \left (B x^2+A\right ) \sqrt {c x^4+b x^2}dx^2\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {(b B-2 A c) \int \sqrt {c x^4+b x^2}dx^2}{2 c}\right )\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {(b B-2 A c) \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c}\right )}{2 c}\right )\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {(b B-2 A c) \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{4 c}\right )}{2 c}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 c}-\frac {(b B-2 A c) \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{4 c^{3/2}}\right )}{2 c}\right )\) |
((B*(b*x^2 + c*x^4)^(3/2))/(3*c) - ((b*B - 2*A*c)*(((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
3.1.92.3.1 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_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) ^(n_))^(q_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1) *(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] && !IntegerQ[p] && NeQ[k, j] && I ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]
Time = 2.11 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.08
| method | result | size |
| risch | \(\frac {\left (8 B \,c^{2} x^{4}+12 A \,c^{2} x^{2}+2 B b c \,x^{2}+6 A b c -3 B \,b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{48 c^{2}}-\frac {b^{2} \left (2 A c -B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16 c^{\frac {5}{2}} x \sqrt {c \,x^{2}+b}}\) | \(116\) |
| pseudoelliptic | \(\frac {\left (-\frac {1}{2} b^{2} A c +\frac {1}{4} B \,b^{3}\right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )+\left (b \left (\frac {x^{2} B}{3}+A \right ) c^{\frac {3}{2}}+\left (\frac {4}{3} x^{4} B +2 A \,x^{2}\right ) c^{\frac {5}{2}}-\frac {B \sqrt {c}\, b^{2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}+\frac {\ln \left (2\right ) \left (A c -\frac {B b}{2}\right ) b^{2}}{2}}{8 c^{\frac {5}{2}}}\) | \(122\) |
| default | \(\frac {\sqrt {x^{4} c +b \,x^{2}}\, \left (8 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} x^{3}+12 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} x -6 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, b x -6 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b x +3 B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b^{2} x -6 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} c +3 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{3}\right )}{48 x \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}}}\) | \(164\) |
1/48*(8*B*c^2*x^4+12*A*c^2*x^2+2*B*b*c*x^2+6*A*b*c-3*B*b^2)/c^2*(x^2*(c*x^ 2+b))^(1/2)-1/16*b^2*(2*A*c-B*b)/c^(5/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*(x^ 2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
Time = 0.27 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.08 \[ \int x \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\left [-\frac {3 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{4} - 3 \, B b^{2} c + 6 \, A b c^{2} + 2 \, {\left (B b c^{2} + 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{3}}, -\frac {3 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (8 \, B c^{3} x^{4} - 3 \, B b^{2} c + 6 \, A b c^{2} + 2 \, {\left (B b c^{2} + 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{3}}\right ] \]
[-1/96*(3*(B*b^3 - 2*A*b^2*c)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b* x^2)*sqrt(c)) - 2*(8*B*c^3*x^4 - 3*B*b^2*c + 6*A*b*c^2 + 2*(B*b*c^2 + 6*A* c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^3, -1/48*(3*(B*b^3 - 2*A*b^2*c)*sqrt(-c)* arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (8*B*c^3*x^4 - 3*B*b^2* c + 6*A*b*c^2 + 2*(B*b*c^2 + 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^3]
Time = 0.63 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.38 \[ \int x \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {A \left (\begin {cases} - \frac {b^{2} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 c} + \left (\frac {b}{4 c} + \frac {x^{2}}{2}\right ) \sqrt {b x^{2} + c x^{4}} & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} + \frac {B \left (\begin {cases} \frac {b^{3} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 c^{2}} + \sqrt {b x^{2} + c x^{4}} \left (- \frac {b^{2}}{8 c^{2}} + \frac {b x^{2}}{12 c} + \frac {x^{4}}{3}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {5}{2}}}{5 b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} \]
A*Piecewise((-b**2*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) + 2* c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sqr t(c*(b/(2*c) + x**2)**2), True))/(8*c) + (b/(4*c) + x**2/2)*sqrt(b*x**2 + c*x**4), Ne(c, 0)), (2*(b*x**2)**(3/2)/(3*b), Ne(b, 0)), (0, True))/2 + B* Piecewise((b**3*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) + 2*c*x **2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sqrt(c *(b/(2*c) + x**2)**2), True))/(16*c**2) + sqrt(b*x**2 + c*x**4)*(-b**2/(8* c**2) + b*x**2/(12*c) + x**4/3), Ne(c, 0)), (2*(b*x**2)**(5/2)/(5*b**2), N e(b, 0)), (0, True))/2
Time = 0.21 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.65 \[ \int x \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {1}{16} \, {\left (4 \, \sqrt {c x^{4} + b x^{2}} x^{2} - \frac {b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}} + \frac {2 \, \sqrt {c x^{4} + b x^{2}} b}{c}\right )} A - \frac {1}{96} \, {\left (\frac {12 \, \sqrt {c x^{4} + b x^{2}} b x^{2}}{c} - \frac {3 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{c^{2}} - \frac {16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{c}\right )} B \]
1/16*(4*sqrt(c*x^4 + b*x^2)*x^2 - b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x ^2)*sqrt(c))/c^(3/2) + 2*sqrt(c*x^4 + b*x^2)*b/c)*A - 1/96*(12*sqrt(c*x^4 + b*x^2)*b*x^2/c - 3*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/ c^(5/2) + 6*sqrt(c*x^4 + b*x^2)*b^2/c^2 - 16*(c*x^4 + b*x^2)^(3/2)/c)*B
Time = 0.30 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.31 \[ \int x \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, B x^{2} \mathrm {sgn}\left (x\right ) + \frac {B b c^{3} \mathrm {sgn}\left (x\right ) + 6 \, A c^{4} \mathrm {sgn}\left (x\right )}{c^{4}}\right )} x^{2} - \frac {3 \, {\left (B b^{2} c^{2} \mathrm {sgn}\left (x\right ) - 2 \, A b c^{3} \mathrm {sgn}\left (x\right )\right )}}{c^{4}}\right )} \sqrt {c x^{2} + b} x - \frac {{\left (B b^{3} \mathrm {sgn}\left (x\right ) - 2 \, A b^{2} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{16 \, c^{\frac {5}{2}}} + \frac {{\left (B b^{3} \log \left ({\left | b \right |}\right ) - 2 \, A b^{2} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{32 \, c^{\frac {5}{2}}} \]
1/48*(2*(4*B*x^2*sgn(x) + (B*b*c^3*sgn(x) + 6*A*c^4*sgn(x))/c^4)*x^2 - 3*( B*b^2*c^2*sgn(x) - 2*A*b*c^3*sgn(x))/c^4)*sqrt(c*x^2 + b)*x - 1/16*(B*b^3* sgn(x) - 2*A*b^2*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(5/2) + 1/32*(B*b^3*log(abs(b)) - 2*A*b^2*c*log(abs(b)))*sgn(x)/c^(5/2)
Time = 9.67 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.31 \[ \int x \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {A\,\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2}}{2}+\frac {B\,b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\left |x\right |\,\sqrt {c\,x^2+b}\right )}{32\,c^{5/2}}-\frac {A\,b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{16\,c^{3/2}}+\frac {B\,\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{48\,c^2} \]