Integrand size = 27, antiderivative size = 100 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {a+b x^2}} \, dx=\frac {(b B-a D) \sqrt {a+b x^2}}{b^2}+\frac {C x \sqrt {a+b x^2}}{2 b}+\frac {D \left (a+b x^2\right )^{3/2}}{3 b^2}+\frac {(2 A b-a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \] Output:
(B*b-D*a)*(b*x^2+a)^(1/2)/b^2+1/2*C*x*(b*x^2+a)^(1/2)/b+1/3*D*(b*x^2+a)^(3 /2)/b^2+1/2*(2*A*b-C*a)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(3/2)
Time = 0.43 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (6 b B-4 a D+3 b C x+2 b D x^2\right )}{6 b^2}+\frac {(-2 A b+a C) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{3/2}} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/Sqrt[a + b*x^2],x]
Output:
(Sqrt[a + b*x^2]*(6*b*B - 4*a*D + 3*b*C*x + 2*b*D*x^2))/(6*b^2) + ((-2*A*b + a*C)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*b^(3/2))
Time = 0.32 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2346, 2346, 27, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {\int \frac {3 b C x^2+(3 b B-2 a D) x+3 A b}{\sqrt {b x^2+a}}dx}{3 b}+\frac {D x^2 \sqrt {a+b x^2}}{3 b}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {\frac {\int \frac {b (3 (2 A b-a C)+2 (3 b B-2 a D) x)}{\sqrt {b x^2+a}}dx}{2 b}+\frac {3}{2} C x \sqrt {a+b x^2}}{3 b}+\frac {D x^2 \sqrt {a+b x^2}}{3 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {3 (2 A b-a C)+2 (3 b B-2 a D) x}{\sqrt {b x^2+a}}dx+\frac {3}{2} C x \sqrt {a+b x^2}}{3 b}+\frac {D x^2 \sqrt {a+b x^2}}{3 b}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\frac {1}{2} \left (3 (2 A b-a C) \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {2 \sqrt {a+b x^2} (3 b B-2 a D)}{b}\right )+\frac {3}{2} C x \sqrt {a+b x^2}}{3 b}+\frac {D x^2 \sqrt {a+b x^2}}{3 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {1}{2} \left (3 (2 A b-a C) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {2 \sqrt {a+b x^2} (3 b B-2 a D)}{b}\right )+\frac {3}{2} C x \sqrt {a+b x^2}}{3 b}+\frac {D x^2 \sqrt {a+b x^2}}{3 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 (2 A b-a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}+\frac {2 \sqrt {a+b x^2} (3 b B-2 a D)}{b}\right )+\frac {3}{2} C x \sqrt {a+b x^2}}{3 b}+\frac {D x^2 \sqrt {a+b x^2}}{3 b}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/Sqrt[a + b*x^2],x]
Output:
(D*x^2*Sqrt[a + b*x^2])/(3*b) + ((3*C*x*Sqrt[a + b*x^2])/2 + ((2*(3*b*B - 2*a*D)*Sqrt[a + b*x^2])/b + (3*(2*A*b - a*C)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b])/2)/(3*b)
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[(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 = 0.62 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12
method | result | size |
default | \(\frac {A \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+\frac {B \sqrt {b \,x^{2}+a}}{b}+C \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+D \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )\) | \(112\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
A*ln(b^(1/2)*x+(b*x^2+a)^(1/2))/b^(1/2)+B*(b*x^2+a)^(1/2)/b+C*(1/2*x*(b*x^ 2+a)^(1/2)/b-1/2*a/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))+D*(1/3*x^2/b*(b* x^2+a)^(1/2)-2/3*a/b^2*(b*x^2+a)^(1/2))
Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.47 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (C a - 2 \, A b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (2 \, D b x^{2} + 3 \, C b x - 4 \, D a + 6 \, B b\right )} \sqrt {b x^{2} + a}}{12 \, b^{2}}, \frac {3 \, {\left (C a - 2 \, A b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, D b x^{2} + 3 \, C b x - 4 \, D a + 6 \, B b\right )} \sqrt {b x^{2} + a}}{6 \, b^{2}}\right ] \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
[-1/12*(3*(C*a - 2*A*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(2*D*b*x^2 + 3*C*b*x - 4*D*a + 6*B*b)*sqrt(b*x^2 + a))/b^2, 1/6* (3*(C*a - 2*A*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (2*D*b*x^2 + 3*C*b*x - 4*D*a + 6*B*b)*sqrt(b*x^2 + a))/b^2]
Time = 0.45 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {a+b x^2}} \, dx=\begin {cases} \left (A - \frac {C a}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {a + b x^{2}} \left (\frac {C x}{2 b} + \frac {D x^{2}}{3 b} + \frac {B - \frac {2 D a}{3 b}}{b}\right ) & \text {for}\: b \neq 0 \\\frac {A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:
integrate((D*x**3+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)* (C*x/(2*b) + D*x**2/(3*b) + (B - 2*D*a/(3*b))/b), Ne(b, 0)), ((A*x + B*x** 2/2 + C*x**3/3 + D*x**4/4)/sqrt(a), True))
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} D x^{2}}{3 \, b} + \frac {\sqrt {b x^{2} + a} C x}{2 \, b} - \frac {C a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {A \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {2 \, \sqrt {b x^{2} + a} D a}{3 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B}{b} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
1/3*sqrt(b*x^2 + a)*D*x^2/b + 1/2*sqrt(b*x^2 + a)*C*x/b - 1/2*C*a*arcsinh( b*x/sqrt(a*b))/b^(3/2) + A*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 2/3*sqrt(b*x^2 + a)*D*a/b^2 + sqrt(b*x^2 + a)*B/b
Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.77 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {a+b x^2}} \, dx=\frac {1}{6} \, \sqrt {b x^{2} + a} {\left ({\left (\frac {2 \, D x}{b} + \frac {3 \, C}{b}\right )} x - \frac {2 \, {\left (2 \, D a b - 3 \, B b^{2}\right )}}{b^{3}}\right )} + \frac {{\left (C a - 2 \, A b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {3}{2}}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
1/6*sqrt(b*x^2 + a)*((2*D*x/b + 3*C/b)*x - 2*(2*D*a*b - 3*B*b^2)/b^3) + 1/ 2*(C*a - 2*A*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)
Time = 3.13 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.49 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {a+b x^2}} \, dx=\left \{\begin {array}{cl} \frac {B\,x^2}{2\,\sqrt {a}}+\frac {C\,x^3}{3\,\sqrt {a}}+\frac {x^4\,D}{4\,\sqrt {a}}+\frac {A\,x}{\sqrt {a}} & \text {\ if\ \ }b=0\\ \frac {{\left (b\,x^2+a\right )}^{3/2}\,D-3\,a\,\sqrt {b\,x^2+a}\,D}{3\,b^2}+\frac {B\,\sqrt {b\,x^2+a}}{b}+\frac {A\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{\sqrt {b}}-\frac {C\,a\,\ln \left (2\,\sqrt {b}\,x+2\,\sqrt {b\,x^2+a}\right )}{2\,b^{3/2}}+\frac {C\,x\,\sqrt {b\,x^2+a}}{2\,b} & \text {\ if\ \ }b\neq 0 \end {array}\right . \] Input:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x^2)^(1/2),x)
Output:
piecewise(b == 0, (B*x^2)/(2*a^(1/2)) + (C*x^3)/(3*a^(1/2)) + (x^4*D)/(4*a ^(1/2)) + (A*x)/a^(1/2), b ~= 0, ((a + b*x^2)^(3/2)*D - 3*a*(a + b*x^2)^(1 /2)*D)/(3*b^2) + (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*log(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 = 0.15 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {a+b x^2}} \, dx=\frac {-4 \sqrt {b \,x^{2}+a}\, a d +6 \sqrt {b \,x^{2}+a}\, b^{2}+3 \sqrt {b \,x^{2}+a}\, b c x +2 \sqrt {b \,x^{2}+a}\, b d \,x^{2}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b -3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a c}{6 b^{2}} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x)
Output:
( - 4*sqrt(a + b*x**2)*a*d + 6*sqrt(a + b*x**2)*b**2 + 3*sqrt(a + b*x**2)* b*c*x + 2*sqrt(a + b*x**2)*b*d*x**2 + 6*sqrt(b)*log((sqrt(a + b*x**2) + sq rt(b)*x)/sqrt(a))*a*b - 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt( a))*a*c)/(6*b**2)