Integrand size = 24, antiderivative size = 98 \[ \int \frac {A+B x^2+C x^4}{\sqrt {a+b x^2}} \, dx=\frac {(4 b B-3 a C) x \sqrt {a+b x^2}}{8 b^2}+\frac {C x^3 \sqrt {a+b x^2}}{4 b}+\frac {\left (8 A b^2-a (4 b B-3 a C)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] Output:
1/8*(4*B*b-3*C*a)*x*(b*x^2+a)^(1/2)/b^2+1/4*C*x^3*(b*x^2+a)^(1/2)/b+1/8*(8 *A*b^2-a*(4*B*b-3*C*a))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^2+C x^4}{\sqrt {a+b x^2}} \, dx=\frac {x \sqrt {a+b x^2} \left (4 b B-3 a C+2 b C x^2\right )}{8 b^2}+\frac {\left (8 A b^2-4 a b B+3 a^2 C\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{4 b^{5/2}} \] Input:
Integrate[(A + B*x^2 + C*x^4)/Sqrt[a + b*x^2],x]
Output:
(x*Sqrt[a + b*x^2]*(4*b*B - 3*a*C + 2*b*C*x^2))/(8*b^2) + ((8*A*b^2 - 4*a* b*B + 3*a^2*C)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(4*b^(5/ 2))
Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1473, 299, 224, 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.
\(\displaystyle \int \frac {A+B x^2+C x^4}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 1473 |
\(\displaystyle \frac {\int \frac {(4 b B-3 a C) x^2+4 A b}{\sqrt {b x^2+a}}dx}{4 b}+\frac {C x^3 \sqrt {a+b x^2}}{4 b}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {\left (8 A b^2-a (4 b B-3 a C)\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {x \sqrt {a+b x^2} (4 b B-3 a C)}{2 b}}{4 b}+\frac {C x^3 \sqrt {a+b x^2}}{4 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {\left (8 A b^2-a (4 b B-3 a C)\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {x \sqrt {a+b x^2} (4 b B-3 a C)}{2 b}}{4 b}+\frac {C x^3 \sqrt {a+b x^2}}{4 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (8 A b^2-a (4 b B-3 a C)\right )}{2 b^{3/2}}+\frac {x \sqrt {a+b x^2} (4 b B-3 a C)}{2 b}}{4 b}+\frac {C x^3 \sqrt {a+b x^2}}{4 b}\) |
Input:
Int[(A + B*x^2 + C*x^4)/Sqrt[a + b*x^2],x]
Output:
(C*x^3*Sqrt[a + b*x^2])/(4*b) + (((4*b*B - 3*a*C)*x*Sqrt[a + b*x^2])/(2*b) + ((8*A*b^2 - a*(4*b*B - 3*a*C))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2 *b^(3/2)))/(4*b)
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[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) , x] + Simp[1/(e*(4*p + 2*q + 1)) Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && !LtQ[q, -1]
Time = 0.54 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.73
method | result | size |
risch | \(\frac {x \left (2 C b \,x^{2}+4 B b -3 C a \right ) \sqrt {b \,x^{2}+a}}{8 b^{2}}+\frac {\left (8 b^{2} A -4 a b B +3 a^{2} C \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\) | \(72\) |
pseudoelliptic | \(\frac {\left (b^{2} A -\frac {1}{2} a b B +\frac {3}{8} a^{2} C \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\frac {\left (\left (\frac {C \,x^{2}}{2}+B \right ) b^{\frac {3}{2}}-\frac {3 C a \sqrt {b}}{4}\right ) \sqrt {b \,x^{2}+a}\, x}{2}}{b^{\frac {5}{2}}}\) | \(73\) |
default | \(\frac {A \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+C \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+B \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\) | \(127\) |
Input:
int((C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8*x*(2*C*b*x^2+4*B*b-3*C*a)/b^2*(b*x^2+a)^(1/2)+1/8*(8*A*b^2-4*B*a*b+3*C *a^2)/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
Time = 0.09 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.78 \[ \int \frac {A+B x^2+C x^4}{\sqrt {a+b x^2}} \, dx=\left [\frac {{\left (3 \, C a^{2} - 4 \, B a b + 8 \, A b^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, C b^{2} x^{3} - {\left (3 \, C a b - 4 \, B b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, b^{3}}, -\frac {{\left (3 \, C a^{2} - 4 \, B a b + 8 \, A b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, C b^{2} x^{3} - {\left (3 \, C a b - 4 \, B b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, b^{3}}\right ] \] Input:
integrate((C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
[1/16*((3*C*a^2 - 4*B*a*b + 8*A*b^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*C*b^2*x^3 - (3*C*a*b - 4*B*b^2)*x)*sqrt(b*x^2 + a))/b^3, -1/8*((3*C*a^2 - 4*B*a*b + 8*A*b^2)*sqrt(-b)*arctan(sqrt(-b)*x/sq rt(b*x^2 + a)) - (2*C*b^2*x^3 - (3*C*a*b - 4*B*b^2)*x)*sqrt(b*x^2 + a))/b^ 3]
Time = 0.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x^2+C x^4}{\sqrt {a+b x^2}} \, dx=\begin {cases} \left (A - \frac {a \left (B - \frac {3 C a}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {a + b x^{2}} \left (\frac {C x^{3}}{4 b} + \frac {x \left (B - \frac {3 C a}{4 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\\frac {A x + \frac {B x^{3}}{3} + \frac {C x^{5}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:
integrate((C*x**4+B*x**2+A)/(b*x**2+a)**(1/2),x)
Output:
Piecewise(((A - a*(B - 3*C*a/(4*b))/(2*b))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True)) + s qrt(a + b*x**2)*(C*x**3/(4*b) + x*(B - 3*C*a/(4*b))/(2*b)), Ne(b, 0)), ((A *x + B*x**3/3 + C*x**5/5)/sqrt(a), True))
Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x^2+C x^4}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} C x^{3}}{4 \, b} - \frac {3 \, \sqrt {b x^{2} + a} C a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B x}{2 \, b} + \frac {3 \, C a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {A \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} \] Input:
integrate((C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
1/4*sqrt(b*x^2 + a)*C*x^3/b - 3/8*sqrt(b*x^2 + a)*C*a*x/b^2 + 1/2*sqrt(b*x ^2 + a)*B*x/b + 3/8*C*a^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 1/2*B*a*arcsinh (b*x/sqrt(a*b))/b^(3/2) + A*arcsinh(b*x/sqrt(a*b))/sqrt(b)
Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x^2+C x^4}{\sqrt {a+b x^2}} \, dx=\frac {1}{8} \, \sqrt {b x^{2} + a} {\left (\frac {2 \, C x^{2}}{b} - \frac {3 \, C a b - 4 \, B b^{2}}{b^{3}}\right )} x - \frac {{\left (3 \, C a^{2} - 4 \, B a b + 8 \, A b^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \] Input:
integrate((C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
1/8*sqrt(b*x^2 + a)*(2*C*x^2/b - (3*C*a*b - 4*B*b^2)/b^3)*x - 1/8*(3*C*a^2 - 4*B*a*b + 8*A*b^2)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
Timed out. \[ \int \frac {A+B x^2+C x^4}{\sqrt {a+b x^2}} \, dx=\int \frac {C\,x^4+B\,x^2+A}{\sqrt {b\,x^2+a}} \,d x \] Input:
int((A + B*x^2 + C*x^4)/(a + b*x^2)^(1/2),x)
Output:
int((A + B*x^2 + C*x^4)/(a + b*x^2)^(1/2), x)
Time = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x^2+C x^4}{\sqrt {a+b x^2}} \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a b c x +4 \sqrt {b \,x^{2}+a}\, b^{3} x +2 \sqrt {b \,x^{2}+a}\, b^{2} c \,x^{3}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} c +4 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2}}{8 b^{3}} \] Input:
int((C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x)
Output:
( - 3*sqrt(a + b*x**2)*a*b*c*x + 4*sqrt(a + b*x**2)*b**3*x + 2*sqrt(a + b* x**2)*b**2*c*x**3 + 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))* a**2*c + 4*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2)/(8* b**3)