Integrand size = 22, antiderivative size = 106 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {(4 A b-a C) x \sqrt {a+b x^2}}{8 b}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b}+\frac {C x \left (a+b x^2\right )^{3/2}}{4 b}+\frac {a (4 A b-a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \] Output:
1/8*(4*A*b-C*a)*x*(b*x^2+a)^(1/2)/b+1/3*B*(b*x^2+a)^(3/2)/b+1/4*C*x*(b*x^2 +a)^(3/2)/b+1/8*a*(4*A*b-C*a)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(3/2)
Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.82 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {\sqrt {a+b x^2} \left (8 a B+12 A b x+3 a C x+8 b B x^2+6 b C x^3\right )}{24 b}+\frac {a (-4 A b+a C) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{3/2}} \] Input:
Integrate[Sqrt[a + b*x^2]*(A + B*x + C*x^2),x]
Output:
(Sqrt[a + b*x^2]*(8*a*B + 12*A*b*x + 3*a*C*x + 8*b*B*x^2 + 6*b*C*x^3))/(24 *b) + (a*(-4*A*b + a*C)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*b^(3/2))
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2346, 455, 211, 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.
\(\displaystyle \int \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {\int (4 A b+4 B x b-a C) \sqrt {b x^2+a}dx}{4 b}+\frac {C x \left (a+b x^2\right )^{3/2}}{4 b}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {(4 A b-a C) \int \sqrt {b x^2+a}dx+\frac {4}{3} B \left (a+b x^2\right )^{3/2}}{4 b}+\frac {C x \left (a+b x^2\right )^{3/2}}{4 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {(4 A b-a C) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {4}{3} B \left (a+b x^2\right )^{3/2}}{4 b}+\frac {C x \left (a+b x^2\right )^{3/2}}{4 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {(4 A b-a C) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {4}{3} B \left (a+b x^2\right )^{3/2}}{4 b}+\frac {C x \left (a+b x^2\right )^{3/2}}{4 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(4 A b-a C) \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {4}{3} B \left (a+b x^2\right )^{3/2}}{4 b}+\frac {C x \left (a+b x^2\right )^{3/2}}{4 b}\) |
Input:
Int[Sqrt[a + b*x^2]*(A + B*x + C*x^2),x]
Output:
(C*x*(a + b*x^2)^(3/2))/(4*b) + ((4*B*(a + b*x^2)^(3/2))/3 + (4*A*b - a*C) *((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt [b])))/(4*b)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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 = 0.50 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {\left (6 C b \,x^{3}+8 b B \,x^{2}+12 A b x +3 C a x +8 B a \right ) \sqrt {b \,x^{2}+a}}{24 b}+\frac {a \left (4 A b -C a \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {3}{2}}}\) | \(76\) |
default | \(A \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+\frac {B \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}+C \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )\) | \(113\) |
Input:
int((b*x^2+a)^(1/2)*(C*x^2+B*x+A),x,method=_RETURNVERBOSE)
Output:
1/24*(6*C*b*x^3+8*B*b*x^2+12*A*b*x+3*C*a*x+8*B*a)/b*(b*x^2+a)^(1/2)+1/8*a* (4*A*b-C*a)/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
Time = 0.09 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.76 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\left [-\frac {3 \, {\left (C a^{2} - 4 \, A a b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (6 \, C b^{2} x^{3} + 8 \, B b^{2} x^{2} + 8 \, B a b + 3 \, {\left (C a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{2}}, \frac {3 \, {\left (C a^{2} - 4 \, A a b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, C b^{2} x^{3} + 8 \, B b^{2} x^{2} + 8 \, B a b + 3 \, {\left (C a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{24 \, b^{2}}\right ] \] Input:
integrate((b*x^2+a)^(1/2)*(C*x^2+B*x+A),x, algorithm="fricas")
Output:
[-1/48*(3*(C*a^2 - 4*A*a*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt( b)*x - a) - 2*(6*C*b^2*x^3 + 8*B*b^2*x^2 + 8*B*a*b + 3*(C*a*b + 4*A*b^2)*x )*sqrt(b*x^2 + a))/b^2, 1/24*(3*(C*a^2 - 4*A*a*b)*sqrt(-b)*arctan(sqrt(-b) *x/sqrt(b*x^2 + a)) + (6*C*b^2*x^3 + 8*B*b^2*x^2 + 8*B*a*b + 3*(C*a*b + 4* A*b^2)*x)*sqrt(b*x^2 + a))/b^2]
Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.17 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {B a}{3 b} + \frac {B x^{2}}{3} + \frac {C x^{3}}{4} + \frac {x \left (A b + \frac {C a}{4}\right )}{2 b}\right ) + \left (A a - \frac {a \left (A b + \frac {C a}{4}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x**2+a)**(1/2)*(C*x**2+B*x+A),x)
Output:
Piecewise((sqrt(a + b*x**2)*(B*a/(3*b) + B*x**2/3 + C*x**3/4 + x*(A*b + C* a/4)/(2*b)) + (A*a - a*(A*b + C*a/4)/(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)), N e(b, 0)), (sqrt(a)*(A*x + B*x**2/2 + C*x**3/3), True))
Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {1}{2} \, \sqrt {b x^{2} + a} A x + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C x}{4 \, b} - \frac {\sqrt {b x^{2} + a} C a x}{8 \, b} - \frac {C a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{3 \, b} \] Input:
integrate((b*x^2+a)^(1/2)*(C*x^2+B*x+A),x, algorithm="maxima")
Output:
1/2*sqrt(b*x^2 + a)*A*x + 1/4*(b*x^2 + a)^(3/2)*C*x/b - 1/8*sqrt(b*x^2 + a )*C*a*x/b - 1/8*C*a^2*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 1/2*A*a*arcsinh(b*x /sqrt(a*b))/sqrt(b) + 1/3*(b*x^2 + a)^(3/2)*B/b
Time = 0.13 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.79 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {1}{24} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, C x + 4 \, B\right )} x + \frac {3 \, {\left (C a b + 4 \, A b^{2}\right )}}{b^{2}}\right )} x + \frac {8 \, B a}{b}\right )} + \frac {{\left (C a^{2} - 4 \, A a b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} \] Input:
integrate((b*x^2+a)^(1/2)*(C*x^2+B*x+A),x, algorithm="giac")
Output:
1/24*sqrt(b*x^2 + a)*((2*(3*C*x + 4*B)*x + 3*(C*a*b + 4*A*b^2)/b^2)*x + 8* B*a/b) + 1/8*(C*a^2 - 4*A*a*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3 /2)
Timed out. \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\int \sqrt {b\,x^2+a}\,\left (C\,x^2+B\,x+A\right ) \,d x \] Input:
int((a + b*x^2)^(1/2)*(A + B*x + C*x^2),x)
Output:
int((a + b*x^2)^(1/2)*(A + B*x + C*x^2), x)
Time = 0.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.28 \[ \int \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {12 \sqrt {b \,x^{2}+a}\, a \,b^{2} x +8 \sqrt {b \,x^{2}+a}\, a \,b^{2}+3 \sqrt {b \,x^{2}+a}\, a b c x +8 \sqrt {b \,x^{2}+a}\, b^{3} x^{2}+6 \sqrt {b \,x^{2}+a}\, b^{2} c \,x^{3}+12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b -3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} c}{24 b^{2}} \] Input:
int((b*x^2+a)^(1/2)*(C*x^2+B*x+A),x)
Output:
(12*sqrt(a + b*x**2)*a*b**2*x + 8*sqrt(a + b*x**2)*a*b**2 + 3*sqrt(a + b*x **2)*a*b*c*x + 8*sqrt(a + b*x**2)*b**3*x**2 + 6*sqrt(a + b*x**2)*b**2*c*x* *3 + 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b - 3*sqr t(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*c)/(24*b**2)