Integrand size = 26, antiderivative size = 83 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {\left (3 b B-4 A c-2 B c x^2\right ) \sqrt {b x^2+c x^4}}{8 c^2}+\frac {b (3 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{5/2}} \] Output:
-1/8*(-2*B*c*x^2-4*A*c+3*B*b)*(c*x^4+b*x^2)^(1/2)/c^2+1/8*b*(-4*A*c+3*B*b) *arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(5/2)
Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.30 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {x \left (\sqrt {c} x \left (b+c x^2\right ) \left (-3 b B+4 A c+2 B c x^2\right )+2 b (3 b B-4 A c) \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )\right )}{8 c^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(x^3*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]
Output:
(x*(Sqrt[c]*x*(b + c*x^2)*(-3*b*B + 4*A*c + 2*B*c*x^2) + 2*b*(3*b*B - 4*A* c)*Sqrt[b + c*x^2]*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])]))/(8* c^(5/2)*Sqrt[x^2*(b + c*x^2)])
Time = 0.41 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1940, 1225, 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.
\(\displaystyle \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {x^2 \left (B x^2+A\right )}{\sqrt {c x^4+b x^2}}dx^2\) |
\(\Big \downarrow \) 1225 |
\(\displaystyle \frac {1}{2} \left (\frac {b (3 b B-4 A c) \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c^2}-\frac {\sqrt {b x^2+c x^4} \left (-4 A c+3 b B-2 B c x^2\right )}{4 c^2}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {b (3 b B-4 A c) \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{4 c^2}-\frac {\sqrt {b x^2+c x^4} \left (-4 A c+3 b B-2 B c x^2\right )}{4 c^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {b (3 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{4 c^{5/2}}-\frac {\sqrt {b x^2+c x^4} \left (-4 A c+3 b B-2 B c x^2\right )}{4 c^2}\right )\) |
Input:
Int[(x^3*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]
Output:
(-1/4*((3*b*B - 4*A*c - 2*B*c*x^2)*Sqrt[b*x^2 + c*x^4])/c^2 + (b*(3*b*B - 4*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(4*c^(5/2)))/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_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[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 = 0.45 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.20
method | result | size |
risch | \(\frac {x^{2} \left (2 B c \,x^{2}+4 A c -3 B b \right ) \left (c \,x^{2}+b \right )}{8 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {b \left (4 A c -3 B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) x \sqrt {c \,x^{2}+b}}{8 c^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(100\) |
default | \(\frac {x \sqrt {c \,x^{2}+b}\, \left (2 B \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} x^{3}+4 A \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} x -3 B \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b x -4 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b \,c^{2}+3 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} c \right )}{8 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {7}{2}}}\) | \(127\) |
pseudoelliptic | \(\frac {4 B \,c^{\frac {3}{2}} x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}-4 A \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b c +4 A \ln \left (2\right ) b c +8 A \,c^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}+3 B \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2}-3 B \ln \left (2\right ) b^{2}-6 B b \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}}{16 c^{\frac {5}{2}}}\) | \(156\) |
Input:
int(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8*x^2*(2*B*c*x^2+4*A*c-3*B*b)*(c*x^2+b)/c^2/(x^2*(c*x^2+b))^(1/2)-1/8*b* (4*A*c-3*B*b)/c^(5/2)*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 = 177, normalized size of antiderivative = 2.13 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\left [-\frac {{\left (3 \, B b^{2} - 4 \, A b c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (2 \, B c^{2} x^{2} - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, c^{3}}, -\frac {{\left (3 \, B b^{2} - 4 \, A b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (2 \, B c^{2} x^{2} - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, c^{3}}\right ] \] Input:
integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
[-1/16*((3*B*b^2 - 4*A*b*c)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^ 2)*sqrt(c)) - 2*(2*B*c^2*x^2 - 3*B*b*c + 4*A*c^2)*sqrt(c*x^4 + b*x^2))/c^3 , -1/8*((3*B*b^2 - 4*A*b*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/( c*x^2 + b)) - (2*B*c^2*x^2 - 3*B*b*c + 4*A*c^2)*sqrt(c*x^4 + b*x^2))/c^3]
Time = 0.79 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.05 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {\begin {cases} - \frac {b \left (A - \frac {3 B b}{4 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{2 c} + \sqrt {b x^{2} + c x^{4}} \left (\frac {B x^{2}}{2 c} + \frac {A - \frac {3 B b}{4 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {\frac {2 A \left (b x^{2}\right )^{\frac {3}{2}}}{3 b} + \frac {2 B \left (b x^{2}\right )^{\frac {5}{2}}}{5 b^{2}}}{b} & \text {for}\: b \neq 0 \\\tilde {\infty } \left (\frac {A x^{4}}{2} + \frac {B x^{6}}{3}\right ) & \text {otherwise} \end {cases}}{2} \] Input:
integrate(x**3*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)
Output:
Piecewise((-b*(A - 3*B*b/(4*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))/(2*c) + sqrt(b*x**2 + c*x**4 )*(B*x**2/(2*c) + (A - 3*B*b/(4*c))/c), Ne(c, 0)), ((2*A*(b*x**2)**(3/2)/( 3*b) + 2*B*(b*x**2)**(5/2)/(5*b**2))/b, Ne(b, 0)), (zoo*(A*x**4/2 + B*x**6 /3), True))/2
Time = 0.04 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.61 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {1}{16} \, {\left (\frac {4 \, \sqrt {c x^{4} + b x^{2}} x^{2}}{c} + \frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b}{c^{2}}\right )} B - \frac {1}{4} \, A {\left (\frac {b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}} - \frac {2 \, \sqrt {c x^{4} + b x^{2}}}{c}\right )} \] Input:
integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
1/16*(4*sqrt(c*x^4 + b*x^2)*x^2/c + 3*b^2*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/c^2)*B - 1/4*A*(b*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 )/c)
Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.35 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {1}{8} \, \sqrt {c x^{2} + b} x {\left (\frac {2 \, B x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {3 \, B b c \mathrm {sgn}\left (x\right ) - 4 \, A c^{2} \mathrm {sgn}\left (x\right )}{c^{3}}\right )} + \frac {{\left (3 \, B b^{2} \log \left ({\left | b \right |}\right ) - 4 \, A b c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {5}{2}}} - \frac {{\left (3 \, B b^{2} - 4 \, A b c\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{8 \, c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
1/8*sqrt(c*x^2 + b)*x*(2*B*x^2/(c*sgn(x)) - (3*B*b*c*sgn(x) - 4*A*c^2*sgn( x))/c^3) + 1/16*(3*B*b^2*log(abs(b)) - 4*A*b*c*log(abs(b)))*sgn(x)/c^(5/2) - 1/8*(3*B*b^2 - 4*A*b*c)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(c^(5/2) *sgn(x))
Timed out. \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^3\,\left (B\,x^2+A\right )}{\sqrt {c\,x^4+b\,x^2}} \,d x \] Input:
int((x^3*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2),x)
Output:
int((x^3*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2), x)
Time = 0.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {4 \sqrt {c \,x^{2}+b}\, a \,c^{2} x -3 \sqrt {c \,x^{2}+b}\, b^{2} c x +2 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{3}-4 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b c +3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3}}{8 c^{3}} \] Input:
int(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)
Output:
(4*sqrt(b + c*x**2)*a*c**2*x - 3*sqrt(b + c*x**2)*b**2*c*x + 2*sqrt(b + c* x**2)*b*c**2*x**3 - 4*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))* a*b*c + 3*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**3)/(8*c** 3)