Integrand size = 24, antiderivative size = 66 \[ \int \frac {x \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {B \sqrt {b x^2+c x^4}}{2 c}-\frac {(b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 c^{3/2}} \] Output:
1/2*B*(c*x^4+b*x^2)^(1/2)/c-1/2*(-2*A*c+B*b)*arctanh(c^(1/2)*x^2/(c*x^4+b* x^2)^(1/2))/c^(3/2)
Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.38 \[ \int \frac {x \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {x \left (B \sqrt {c} x \left (b+c x^2\right )+2 (b B-2 A c) \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{2 c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(x*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]
Output:
(x*(B*Sqrt[c]*x*(b + c*x^2) + 2*(b*B - 2*A*c)*Sqrt[b + c*x^2]*ArcTanh[(Sqr t[c]*x)/(Sqrt[b] - Sqrt[b + c*x^2])]))/(2*c^(3/2)*Sqrt[x^2*(b + c*x^2)])
Time = 0.36 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1940, 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.
\(\displaystyle \int \frac {x \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{\sqrt {c x^4+b x^2}}dx^2\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{2} \left (\frac {B \sqrt {b x^2+c x^4}}{c}-\frac {(b B-2 A c) \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{2 c}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {B \sqrt {b x^2+c x^4}}{c}-\frac {(b B-2 A c) \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {B \sqrt {b x^2+c x^4}}{c}-\frac {(b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}\right )\) |
Input:
Int[(x*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]
Output:
((B*Sqrt[b*x^2 + c*x^4])/c - ((b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x ^2 + c*x^4]])/c^(3/2))/2
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_))*((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 = 0.44 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.27
method | result | size |
risch | \(\frac {B \,x^{2} \left (c \,x^{2}+b \right )}{2 c \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (2 A c -B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) x \sqrt {c \,x^{2}+b}}{2 c^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(84\) |
default | \(\frac {x \sqrt {c \,x^{2}+b}\, \left (B \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} x +2 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) c^{2}-B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b c \right )}{2 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {5}{2}}}\) | \(88\) |
pseudoelliptic | \(\frac {2 A \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) c -2 A \ln \left (2\right ) c -B \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b +B \ln \left (2\right ) b +2 B \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}}{4 c^{\frac {3}{2}}}\) | \(107\) |
Input:
int(x*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2*B*x^2/c*(c*x^2+b)/(x^2*(c*x^2+b))^(1/2)+1/2*(2*A*c-B*b)/c^(3/2)*ln(c^( 1/2)*x+(c*x^2+b)^(1/2))*x/(x^2*(c*x^2+b))^(1/2)*(c*x^2+b)^(1/2)
Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.98 \[ \int \frac {x \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\left [\frac {2 \, \sqrt {c x^{4} + b x^{2}} B c - {\left (B b - 2 \, A c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{4 \, c^{2}}, \frac {\sqrt {c x^{4} + b x^{2}} B c + {\left (B b - 2 \, A c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right )}{2 \, c^{2}}\right ] \] Input:
integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
[1/4*(2*sqrt(c*x^4 + b*x^2)*B*c - (B*b - 2*A*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)))/c^2, 1/2*(sqrt(c*x^4 + b*x^2)*B*c + (B*b - 2*A*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)))/c^2]
Time = 0.80 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.11 \[ \int \frac {x \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {\begin {cases} \frac {B \sqrt {b x^{2} + c x^{4}}}{c} + \left (A - \frac {B b}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 A \sqrt {b x^{2}} + \frac {2 B \left (b x^{2}\right )^{\frac {3}{2}}}{3 b}}{b} & \text {for}\: b \neq 0 \\\tilde {\infty } \left (A x^{2} + \frac {B x^{4}}{2}\right ) & \text {otherwise} \end {cases}}{2} \] Input:
integrate(x*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)
Output:
Piecewise((B*sqrt(b*x**2 + c*x**4)/c + (A - B*b/(2*c))*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)), Ne(c , 0)), ((2*A*sqrt(b*x**2) + 2*B*(b*x**2)**(3/2)/(3*b))/b, Ne(b, 0)), (zoo* (A*x**2 + B*x**4/2), True))/2
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.33 \[ \int \frac {x \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {1}{4} \, B {\left (\frac {b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}} - \frac {2 \, \sqrt {c x^{4} + b x^{2}}}{c}\right )} + \frac {A \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{2 \, \sqrt {c}} \] Input:
integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
-1/4*B*(b*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(3/2) - 2*sqr t(c*x^4 + b*x^2)/c) + 1/2*A*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c ))/sqrt(c)
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.17 \[ \int \frac {x \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {c x^{2} + b} B x}{2 \, c \mathrm {sgn}\left (x\right )} - \frac {{\left (B b \log \left ({\left | b \right |}\right ) - 2 \, A c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{4 \, c^{\frac {3}{2}}} + \frac {{\left (B b - 2 \, A c\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{2 \, c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
1/2*sqrt(c*x^2 + b)*B*x/(c*sgn(x)) - 1/4*(B*b*log(abs(b)) - 2*A*c*log(abs( b)))*sgn(x)/c^(3/2) + 1/2*(B*b - 2*A*c)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(c^(3/2)*sgn(x))
Time = 9.54 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.35 \[ \int \frac {x \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {B\,\sqrt {c\,x^4+b\,x^2}}{2\,c}+\frac {A\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{2\,\sqrt {c}}-\frac {B\,b\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{4\,c^{3/2}} \] Input:
int((x*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2),x)
Output:
(B*(b*x^2 + c*x^4)^(1/2))/(2*c) + (A*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(2*c^(1/2)) - (B*b*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x ^4)^(1/2)))/(4*c^(3/2))
Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.05 \[ \int \frac {x \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {c \,x^{2}+b}\, b c x +2 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c -\sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2}}{2 c^{2}} \] Input:
int(x*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)
Output:
(sqrt(b + c*x**2)*b*c*x + 2*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqr t(b))*a*c - sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**2)/(2*c **2)