Integrand size = 26, antiderivative size = 64 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {(b B-A c) x}{b c \sqrt {b x^2+c x^4}}-\frac {A \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{b^{3/2}} \] Output:
-(-A*c+B*b)*x/b/c/(c*x^4+b*x^2)^(1/2)-A*arctanh(b^(1/2)*x/(c*x^4+b*x^2)^(1 /2))/b^(3/2)
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x \left (\sqrt {b} (b B-A c)+A c \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{b^{3/2} c \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
Output:
-((x*(Sqrt[b]*(b*B - A*c) + A*c*Sqrt[b + c*x^2]*ArcTanh[Sqrt[b + c*x^2]/Sq rt[b]]))/(b^(3/2)*c*Sqrt[x^2*(b + c*x^2)]))
Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1943, 1400, 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.
\(\displaystyle \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1943 |
\(\displaystyle \frac {A \int \frac {1}{\sqrt {c x^4+b x^2}}dx}{b}-\frac {x (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle -\frac {A \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}}{b}-\frac {x (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {A \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{b^{3/2}}-\frac {x (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
Input:
Int[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
Output:
-(((b*B - A*c)*x)/(b*c*Sqrt[b*x^2 + c*x^4])) - (A*ArcTanh[(Sqrt[b]*x)/Sqrt [b*x^2 + c*x^4]])/b^(3/2)
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_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x ^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1))) Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 ] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
Time = 0.75 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.23
method | result | size |
default | \(-\frac {x^{3} \left (c \,x^{2}+b \right ) \left (A \sqrt {c \,x^{2}+b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b c -A \,b^{\frac {3}{2}} c +B \,b^{\frac {5}{2}}\right )}{\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {5}{2}} c}\) | \(79\) |
Input:
int(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-x^3*(c*x^2+b)*(A*(c*x^2+b)^(1/2)*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*b*c- A*b^(3/2)*c+B*b^(5/2))/(c*x^4+b*x^2)^(3/2)/b^(5/2)/c
Time = 0.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 3.03 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\left [\frac {{\left (A c^{2} x^{3} + A b c x\right )} \sqrt {b} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} {\left (B b^{2} - A b c\right )}}{2 \, {\left (b^{2} c^{2} x^{3} + b^{3} c x\right )}}, \frac {{\left (A c^{2} x^{3} + A b c x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (B b^{2} - A b c\right )}}{b^{2} c^{2} x^{3} + b^{3} c x}\right ] \] Input:
integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
[1/2*((A*c^2*x^3 + A*b*c*x)*sqrt(b)*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b *x^2)*sqrt(b))/x^3) - 2*sqrt(c*x^4 + b*x^2)*(B*b^2 - A*b*c))/(b^2*c^2*x^3 + b^3*c*x), ((A*c^2*x^3 + A*b*c*x)*sqrt(-b)*arctan(sqrt(c*x^4 + b*x^2)*sqr t(-b)/(b*x)) - sqrt(c*x^4 + b*x^2)*(B*b^2 - A*b*c))/(b^2*c^2*x^3 + b^3*c*x )]
\[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{2} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**2*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)
Output:
Integral(x**2*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)
\[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{2}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate((B*x^2 + A)*x^2/(c*x^4 + b*x^2)^(3/2), x)
Time = 0.22 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.70 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {A \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b \mathrm {sgn}\left (x\right )} - \frac {{\left (A \sqrt {b} c \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) - B \sqrt {-b} b + A \sqrt {-b} c\right )} \mathrm {sgn}\left (x\right )}{\sqrt {-b} b^{\frac {3}{2}} c} - \frac {B b - A c}{\sqrt {c x^{2} + b} b c \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
A*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b*sgn(x)) - (A*sqrt(b)*c*arct an(sqrt(b)/sqrt(-b)) - B*sqrt(-b)*b + A*sqrt(-b)*c)*sgn(x)/(sqrt(-b)*b^(3/ 2)*c) - (B*b - A*c)/(sqrt(c*x^2 + b)*b*c*sgn(x))
Timed out. \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^2\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:
int((x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)
Output:
int((x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)
Time = 0.23 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.56 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {c \,x^{2}+b}\, a b c -\sqrt {c \,x^{2}+b}\, b^{3}+\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b c +\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{2} x^{2}-\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b c -\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{2} x^{2}}{b^{2} c \left (c \,x^{2}+b \right )} \] Input:
int(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)
Output:
(sqrt(b + c*x**2)*a*b*c - sqrt(b + c*x**2)*b**3 + sqrt(b)*log((sqrt(b + c* x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*a*b*c + sqrt(b)*log((sqrt(b + c*x**2 ) - sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c**2*x**2 - sqrt(b)*log((sqrt(b + c*x* *2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*a*b*c - sqrt(b)*log((sqrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c**2*x**2)/(b**2*c*(b + c*x**2))