Integrand size = 27, antiderivative size = 94 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {b B-a D}{b^2 \sqrt {a+b x^2}}+\frac {(A b-a C) x}{a b \sqrt {a+b x^2}}+\frac {D \sqrt {a+b x^2}}{b^2}+\frac {C \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}} \] Output:
-(B*b-D*a)/b^2/(b*x^2+a)^(1/2)+(A*b-C*a)*x/a/b/(b*x^2+a)^(1/2)+D*(b*x^2+a) ^(1/2)/b^2+C*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(3/2)
Time = 0.51 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {2 a^2 D+A b^2 x-a b (B+x (C-D x))}{a b^2 \sqrt {a+b x^2}}-\frac {C \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{3/2}} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(a + b*x^2)^(3/2),x]
Output:
(2*a^2*D + A*b^2*x - a*b*(B + x*(C - D*x)))/(a*b^2*Sqrt[a + b*x^2]) - (C*L og[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/b^(3/2)
Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2345, 25, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {\int -\frac {a (C+D x)}{b \sqrt {b x^2+a}}dx}{a}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{a b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {a (C+D x)}{b \sqrt {b x^2+a}}dx}{a}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{a b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {C+D x}{\sqrt {b x^2+a}}dx}{b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{a b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {C \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {D \sqrt {a+b x^2}}{b}}{b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{a b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {C \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {D \sqrt {a+b x^2}}{b}}{b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{a b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {C \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}+\frac {D \sqrt {a+b x^2}}{b}}{b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{a b \sqrt {a+b x^2}}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(a + b*x^2)^(3/2),x]
Output:
-((a*(B - (a*D)/b) - (A*b - a*C)*x)/(a*b*Sqrt[a + b*x^2])) + ((D*Sqrt[a + b*x^2])/b + (C*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b])/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 = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 1.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11
method | result | size |
default | \(\frac {A x}{\sqrt {b \,x^{2}+a}\, a}-\frac {B}{b \sqrt {b \,x^{2}+a}}+C \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )+D \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )\) | \(104\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
A/(b*x^2+a)^(1/2)/a*x-B/b/(b*x^2+a)^(1/2)+C*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2 )*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))+D*(x^2/b/(b*x^2+a)^(1/2)+2*a/b^2/(b*x^2+a )^(1/2))
Time = 0.08 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.23 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{3/2}} \, dx=\left [\frac {{\left (C a b x^{2} + C a^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (D a b x^{2} + 2 \, D a^{2} - B a b - {\left (C a b - A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a b^{3} x^{2} + a^{2} b^{2}\right )}}, -\frac {{\left (C a b x^{2} + C a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (D a b x^{2} + 2 \, D a^{2} - B a b - {\left (C a b - A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{a b^{3} x^{2} + a^{2} b^{2}}\right ] \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x, algorithm="fricas")
Output:
[1/2*((C*a*b*x^2 + C*a^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b) *x - a) + 2*(D*a*b*x^2 + 2*D*a^2 - B*a*b - (C*a*b - A*b^2)*x)*sqrt(b*x^2 + a))/(a*b^3*x^2 + a^2*b^2), -((C*a*b*x^2 + C*a^2)*sqrt(-b)*arctan(sqrt(-b) *x/sqrt(b*x^2 + a)) - (D*a*b*x^2 + 2*D*a^2 - B*a*b - (C*a*b - A*b^2)*x)*sq rt(b*x^2 + a))/(a*b^3*x^2 + a^2*b^2)]
Time = 4.24 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.39 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {A x}{a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{2}}{a}}} + B \left (\begin {cases} - \frac {1}{b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + C \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {x}{\sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + D \left (\begin {cases} \frac {2 a}{b^{2} \sqrt {a + b x^{2}}} + \frac {x^{2}}{b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x**2+a)**(3/2),x)
Output:
A*x/(a**(3/2)*sqrt(1 + b*x**2/a)) + B*Piecewise((-1/(b*sqrt(a + b*x**2)), Ne(b, 0)), (x**2/(2*a**(3/2)), True)) + C*(asinh(sqrt(b)*x/sqrt(a))/b**(3/ 2) - x/(sqrt(a)*b*sqrt(1 + b*x**2/a))) + D*Piecewise((2*a/(b**2*sqrt(a + b *x**2)) + x**2/(b*sqrt(a + b*x**2)), Ne(b, 0)), (x**4/(4*a**(3/2)), True))
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {D x^{2}}{\sqrt {b x^{2} + a} b} + \frac {A x}{\sqrt {b x^{2} + a} a} - \frac {C x}{\sqrt {b x^{2} + a} b} + \frac {C \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, D a}{\sqrt {b x^{2} + a} b^{2}} - \frac {B}{\sqrt {b x^{2} + a} b} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x, algorithm="maxima")
Output:
D*x^2/(sqrt(b*x^2 + a)*b) + A*x/(sqrt(b*x^2 + a)*a) - C*x/(sqrt(b*x^2 + a) *b) + C*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 2*D*a/(sqrt(b*x^2 + a)*b^2) - B/( sqrt(b*x^2 + a)*b)
Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left (\frac {D x}{b} - \frac {C a b^{2} - A b^{3}}{a b^{3}}\right )} x + \frac {2 \, D a^{2} b - B a b^{2}}{a b^{3}}}{\sqrt {b x^{2} + a}} - \frac {C \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {3}{2}}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x, algorithm="giac")
Output:
((D*x/b - (C*a*b^2 - A*b^3)/(a*b^3))*x + (2*D*a^2*b - B*a*b^2)/(a*b^3))/sq rt(b*x^2 + a) - C*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)
Time = 16.93 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {C\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{b^{3/2}}-\frac {B}{b\,\sqrt {b\,x^2+a}}+\frac {\left (b\,x^2+2\,a\right )\,D}{b^2\,\sqrt {b\,x^2+a}}+\frac {A\,x}{a\,\sqrt {b\,x^2+a}}-\frac {C\,x}{b\,\sqrt {b\,x^2+a}} \] Input:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x^2)^(3/2),x)
Output:
(C*log(b^(1/2)*x + (a + b*x^2)^(1/2)))/b^(3/2) - B/(b*(a + b*x^2)^(1/2)) + ((2*a + b*x^2)*D)/(b^2*(a + b*x^2)^(1/2)) + (A*x)/(a*(a + b*x^2)^(1/2)) - (C*x)/(b*(a + b*x^2)^(1/2))
Time = 0.16 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.69 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {b \,x^{2}+a}\, a d +\sqrt {b \,x^{2}+a}\, b^{2} x -\sqrt {b \,x^{2}+a}\, b^{2}-\sqrt {b \,x^{2}+a}\, b c x +\sqrt {b \,x^{2}+a}\, b d \,x^{2}+\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a c +\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b c \,x^{2}+\sqrt {b}\, a b -\sqrt {b}\, a c +\sqrt {b}\, b^{2} x^{2}-\sqrt {b}\, b c \,x^{2}}{b^{2} \left (b \,x^{2}+a \right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x)
Output:
(2*sqrt(a + b*x**2)*a*d + sqrt(a + b*x**2)*b**2*x - sqrt(a + b*x**2)*b**2 - sqrt(a + b*x**2)*b*c*x + sqrt(a + b*x**2)*b*d*x**2 + sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*c + sqrt(b)*log((sqrt(a + b*x**2) + sqr t(b)*x)/sqrt(a))*b*c*x**2 + sqrt(b)*a*b - sqrt(b)*a*c + sqrt(b)*b**2*x**2 - sqrt(b)*b*c*x**2)/(b**2*(a + b*x**2))