Integrand size = 22, antiderivative size = 74 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x^2}} \, dx=\frac {B \sqrt {a+b x^2}}{b}+\frac {C x \sqrt {a+b x^2}}{2 b}+\frac {(2 A b-a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \] Output:
B*(b*x^2+a)^(1/2)/b+1/2*C*x*(b*x^2+a)^(1/2)/b+1/2*(2*A*b-C*a)*arctanh(b^(1 /2)*x/(b*x^2+a)^(1/2))/b^(3/2)
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x^2}} \, dx=\frac {(2 B+C x) \sqrt {a+b x^2}}{2 b}+\frac {(2 A b-a C) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{b^{3/2}} \] Input:
Integrate[(A + B*x + C*x^2)/Sqrt[a + b*x^2],x]
Output:
((2*B + C*x)*Sqrt[a + b*x^2])/(2*b) + ((2*A*b - a*C)*ArcTanh[(Sqrt[b]*x)/( -Sqrt[a] + Sqrt[a + b*x^2])])/b^(3/2)
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2346, 455, 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.
\(\displaystyle \int \frac {A+B x+C x^2}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {\int \frac {2 A b+2 B x b-a C}{\sqrt {b x^2+a}}dx}{2 b}+\frac {C x \sqrt {a+b x^2}}{2 b}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {(2 A b-a C) \int \frac {1}{\sqrt {b x^2+a}}dx+2 B \sqrt {a+b x^2}}{2 b}+\frac {C x \sqrt {a+b x^2}}{2 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {(2 A b-a C) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+2 B \sqrt {a+b x^2}}{2 b}+\frac {C x \sqrt {a+b x^2}}{2 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {(2 A b-a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}+2 B \sqrt {a+b x^2}}{2 b}+\frac {C x \sqrt {a+b x^2}}{2 b}\) |
Input:
Int[(A + B*x + C*x^2)/Sqrt[a + b*x^2],x]
Output:
(C*x*Sqrt[a + b*x^2])/(2*b) + (2*B*Sqrt[a + b*x^2] + ((2*A*b - a*C)*ArcTan h[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b])/(2*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[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
Time = 0.49 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {\left (C x +2 B \right ) \sqrt {b \,x^{2}+a}}{2 b}+\frac {\left (2 A b -C a \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\) | \(53\) |
default | \(\frac {A \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+\frac {B \sqrt {b \,x^{2}+a}}{b}+C \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\) | \(77\) |
Input:
int((C*x^2+B*x+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2*(C*x+2*B)/b*(b*x^2+a)^(1/2)+1/2*(2*A*b-C*a)/b^(3/2)*ln(b^(1/2)*x+(b*x^ 2+a)^(1/2))
Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.64 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x^2}} \, dx=\left [-\frac {{\left (C a - 2 \, A b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (C b x + 2 \, B b\right )} \sqrt {b x^{2} + a}}{4 \, b^{2}}, \frac {{\left (C a - 2 \, A b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (C b x + 2 \, B b\right )} \sqrt {b x^{2} + a}}{2 \, b^{2}}\right ] \] Input:
integrate((C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
[-1/4*((C*a - 2*A*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(C*b*x + 2*B*b)*sqrt(b*x^2 + a))/b^2, 1/2*((C*a - 2*A*b)*sqrt(-b)*a rctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (C*b*x + 2*B*b)*sqrt(b*x^2 + a))/b^2]
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.24 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x^2}} \, dx=\begin {cases} \left (A - \frac {C a}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {a + b x^{2}} \left (\frac {B}{b} + \frac {C x}{2 b}\right ) & \text {for}\: b \neq 0 \\\frac {A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:
integrate((C*x**2+B*x+A)/(b*x**2+a)**(1/2),x)
Output:
Piecewise(((A - C*a/(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)) + sqrt(a + b*x**2)* (B/b + C*x/(2*b)), Ne(b, 0)), ((A*x + B*x**2/2 + C*x**3/3)/sqrt(a), True))
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} C x}{2 \, b} - \frac {C a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {A \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} + \frac {\sqrt {b x^{2} + a} B}{b} \] Input:
integrate((C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
1/2*sqrt(b*x^2 + a)*C*x/b - 1/2*C*a*arcsinh(b*x/sqrt(a*b))/b^(3/2) + A*arc sinh(b*x/sqrt(a*b))/sqrt(b) + sqrt(b*x^2 + a)*B/b
Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x^2}} \, dx=\frac {1}{2} \, \sqrt {b x^{2} + a} {\left (\frac {C x}{b} + \frac {2 \, B}{b}\right )} + \frac {{\left (C a - 2 \, A b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {3}{2}}} \] Input:
integrate((C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
1/2*sqrt(b*x^2 + a)*(C*x/b + 2*B/b) + 1/2*(C*a - 2*A*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)
Time = 0.65 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.45 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x^2}} \, dx=\left \{\begin {array}{cl} \frac {2\,C\,x^3+3\,B\,x^2+6\,A\,x}{6\,\sqrt {a}} & \text {\ if\ \ }b=0\\ \frac {B\,\sqrt {b\,x^2+a}}{b}+\frac {A\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{\sqrt {b}}-\frac {C\,a\,\ln \left (2\,\sqrt {b}\,x+2\,\sqrt {b\,x^2+a}\right )}{2\,b^{3/2}}+\frac {C\,x\,\sqrt {b\,x^2+a}}{2\,b} & \text {\ if\ \ }b\neq 0 \end {array}\right . \] Input:
int((A + B*x + C*x^2)/(a + b*x^2)^(1/2),x)
Output:
piecewise(b == 0, (6*A*x + 3*B*x^2 + 2*C*x^3)/(6*a^(1/2)), b ~= 0, (B*(a + b*x^2)^(1/2))/b + (A*log(b^(1/2)*x + (a + b*x^2)^(1/2)))/b^(1/2) - (C*a*l og(2*b^(1/2)*x + 2*(a + b*x^2)^(1/2)))/(2*b^(3/2)) + (C*x*(a + b*x^2)^(1/2 ))/(2*b))
Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x^2}} \, dx=\frac {2 \sqrt {b \,x^{2}+a}\, b^{2}+\sqrt {b \,x^{2}+a}\, b c x +2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b -\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a c}{2 b^{2}} \] Input:
int((C*x^2+B*x+A)/(b*x^2+a)^(1/2),x)
Output:
(2*sqrt(a + b*x**2)*b**2 + sqrt(a + b*x**2)*b*c*x + 2*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b - sqrt(b)*log((sqrt(a + b*x**2) + sqrt (b)*x)/sqrt(a))*a*c)/(2*b**2)