Integrand size = 26, antiderivative size = 67 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {(b B-A c) x^2}{b c \sqrt {b x^2+c x^4}}+\frac {B \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}} \] Output:
-(-A*c+B*b)*x^2/b/c/(c*x^4+b*x^2)^(1/2)+B*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2 )^(1/2))/c^(3/2)
Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x \left (\sqrt {c} (b B-A c) x+b B \sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{b c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(x^3*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
Output:
-((x*(Sqrt[c]*(b*B - A*c)*x + b*B*Sqrt[b + c*x^2]*Log[-(Sqrt[c]*x) + Sqrt[ b + c*x^2]]))/(b*c^(3/2)*Sqrt[x^2*(b + c*x^2)]))
Time = 0.39 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1940, 1211, 27, 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.
\(\displaystyle \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {x^2 \left (B x^2+A\right )}{\left (c x^4+b x^2\right )^{3/2}}dx^2\) |
\(\Big \downarrow \) 1211 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {B}{\sqrt {c x^4+b x^2}}dx^2}{c}-\frac {2 x^2 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {B \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{c}-\frac {2 x^2 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {2 B \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{c}-\frac {2 x^2 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {2 B \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}-\frac {2 x^2 (b B-A c)}{b c \sqrt {b x^2+c x^4}}\right )\) |
Input:
Int[(x^3*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
Output:
((-2*(b*B - A*c)*x^2)/(b*c*Sqrt[b*x^2 + c*x^4]) + (2*B*ArcTanh[(Sqrt[c]*x^ 2)/Sqrt[b*x^2 + c*x^4]])/c^(3/2))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-2*(2*c*d - b*e)^(m - 2)*(c*( e*f + d*g) - b*e*g)^n*((d + e*x)/(c^(m + n - 1)*e^(n - 1)*Sqrt[a + b*x + c* x^2])), x] + Simp[1/(c^(m + n - 1)*e^(n - 2)) Int[ExpandToSum[((2*c*d - b *e)^(m - 1)*(c*(e*f + d*g) - b*e*g)^n - c^(m + n - 1)*e^n*(d + e*x)^(m - 1) *(f + g*x)^n)/(c*d - b*e - c*e*x), x]/Sqrt[a + b*x + c*x^2], x], x] /; Free Q[{a, b, c, d, e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]
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 = 75, normalized size of antiderivative = 1.12
method | result | size |
default | \(\frac {x^{3} \left (c \,x^{2}+b \right ) \left (A \,c^{\frac {5}{2}} x -B x b \,c^{\frac {3}{2}}+B \sqrt {c \,x^{2}+b}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b c \right )}{\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b \,c^{\frac {5}{2}}}\) | \(75\) |
pseudoelliptic | \(\frac {2 A \,c^{\frac {3}{2}} x^{2}-2 B b \,x^{2} \sqrt {c}+B \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b \sqrt {x^{2} \left (c \,x^{2}+b \right )}-B \ln \left (2\right ) b \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{2 c^{\frac {3}{2}} b \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(108\) |
Input:
int(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
x^3*(c*x^2+b)*(A*c^(5/2)*x-B*x*b*c^(3/2)+B*(c*x^2+b)^(1/2)*ln(c^(1/2)*x+(c *x^2+b)^(1/2))*b*c)/(c*x^4+b*x^2)^(3/2)/b/c^(5/2)
Time = 0.10 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.81 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\left [\frac {{\left (B b c x^{2} + B b^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} {\left (B b c - A c^{2}\right )}}{2 \, {\left (b c^{3} x^{2} + b^{2} c^{2}\right )}}, -\frac {{\left (B b c x^{2} + B b^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (B b c - A c^{2}\right )}}{b c^{3} x^{2} + b^{2} c^{2}}\right ] \] Input:
integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
[1/2*((B*b*c*x^2 + B*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)*(B*b*c - A*c^2))/(b*c^3*x^2 + b^2*c^2), -((B*b*c*x^2 + B*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)*(B*b*c - A*c^2))/(b*c^3*x^2 + b^2*c^2)]
\[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{3} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**3*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)
Output:
Integral(x**3*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.18 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {1}{2} \, B {\left (\frac {2 \, x^{2}}{\sqrt {c x^{4} + b x^{2}} c} - \frac {\log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}}\right )} + \frac {A x^{2}}{\sqrt {c x^{4} + b x^{2}} b} \] Input:
integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
-1/2*B*(2*x^2/(sqrt(c*x^4 + b*x^2)*c) - log(2*c*x^2 + b + 2*sqrt(c*x^4 + b *x^2)*sqrt(c))/c^(3/2)) + A*x^2/(sqrt(c*x^4 + b*x^2)*b)
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.04 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {B \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{2 \, c^{\frac {3}{2}}} - \frac {{\left (B b \mathrm {sgn}\left (x\right ) - A c \mathrm {sgn}\left (x\right )\right )} x}{\sqrt {c x^{2} + b} b c} - \frac {B \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
1/2*B*log(abs(b))*sgn(x)/c^(3/2) - (B*b*sgn(x) - A*c*sgn(x))*x/(sqrt(c*x^2 + b)*b*c) - B*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(c^(3/2)*sgn(x))
Time = 9.72 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {B\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{2\,c^{3/2}}+\frac {A\,x^2}{b\,\sqrt {c\,x^4+b\,x^2}}-\frac {B\,x^2}{c\,\sqrt {c\,x^4+b\,x^2}} \] Input:
int((x^3*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)
Output:
(B*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(2*c^(3/2)) + (A*x^ 2)/(b*(b*x^2 + c*x^4)^(1/2)) - (B*x^2)/(c*(b*x^2 + c*x^4)^(1/2))
Time = 0.20 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.00 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {c \,x^{2}+b}\, a \,c^{2} x -\sqrt {c \,x^{2}+b}\, b^{2} c x +\sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3}+\sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c \,x^{2}+\sqrt {c}\, a b c +\sqrt {c}\, a \,c^{2} x^{2}-\sqrt {c}\, b^{3}-\sqrt {c}\, b^{2} c \,x^{2}}{b \,c^{2} \left (c \,x^{2}+b \right )} \] Input:
int(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)
Output:
(sqrt(b + c*x**2)*a*c**2*x - sqrt(b + c*x**2)*b**2*c*x + sqrt(c)*log((sqrt (b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**3 + sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**2*c*x**2 + sqrt(c)*a*b*c + sqrt(c)*a*c**2*x**2 - sq rt(c)*b**3 - sqrt(c)*b**2*c*x**2)/(b*c**2*(b + c*x**2))