Integrand size = 27, antiderivative size = 104 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {b B-a D}{3 b^2 \left (a+b x^2\right )^{3/2}}+\frac {(A b-a C) x}{3 a b \left (a+b x^2\right )^{3/2}}-\frac {D}{b^2 \sqrt {a+b x^2}}+\frac {(2 A b+a C) x}{3 a^2 b \sqrt {a+b x^2}} \] Output:
-1/3*(B*b-D*a)/b^2/(b*x^2+a)^(3/2)+1/3*(A*b-C*a)*x/a/b/(b*x^2+a)^(3/2)-D/b ^2/(b*x^2+a)^(1/2)+1/3*(2*A*b+C*a)*x/a^2/b/(b*x^2+a)^(1/2)
Time = 0.68 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.63 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-2 a^3 D+2 A b^3 x^3+a b^2 x \left (3 A+C x^2\right )-a^2 b \left (B+3 D x^2\right )}{3 a^2 b^2 \left (a+b x^2\right )^{3/2}} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(a + b*x^2)^(5/2),x]
Output:
(-2*a^3*D + 2*A*b^3*x^3 + a*b^2*x*(3*A + C*x^2) - a^2*b*(B + 3*D*x^2))/(3* a^2*b^2*(a + b*x^2)^(3/2))
Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.82, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2345, 25, 27, 453}
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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {\int -\frac {b \left (2 A+\frac {a C}{b}\right )+3 a D x}{b \left (b x^2+a\right )^{3/2}}dx}{3 a}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{3 a b \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {2 A b+a C+3 a D x}{b \left (b x^2+a\right )^{3/2}}dx}{3 a}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{3 a b \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 A b+a C+3 a D x}{\left (b x^2+a\right )^{3/2}}dx}{3 a b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{3 a b \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 453 |
\(\displaystyle -\frac {3 a^2 D-b x (a C+2 A b)}{3 a^2 b^2 \sqrt {a+b x^2}}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{3 a b \left (a+b x^2\right )^{3/2}}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(a + b*x^2)^(5/2),x]
Output:
-1/3*(a*(B - (a*D)/b) - (A*b - a*C)*x)/(a*b*(a + b*x^2)^(3/2)) - (3*a^2*D - b*(2*A*b + a*C)*x)/(3*a^2*b^2*Sqrt[a + b*x^2])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-(a* d - b*c*x)/(a*b*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b, c, d}, x]
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 = 0.95 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(\frac {2 A \,b^{3} x^{3}+C a \,b^{2} x^{3}-3 D x^{2} a^{2} b +3 A a \,b^{2} x -B \,a^{2} b -2 D a^{3}}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} b^{2}}\) | \(68\) |
trager | \(\frac {2 A \,b^{3} x^{3}+C a \,b^{2} x^{3}-3 D x^{2} a^{2} b +3 A a \,b^{2} x -B \,a^{2} b -2 D a^{3}}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} b^{2}}\) | \(68\) |
orering | \(\frac {2 A \,b^{3} x^{3}+C a \,b^{2} x^{3}-3 D x^{2} a^{2} b +3 A a \,b^{2} x -B \,a^{2} b -2 D a^{3}}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} b^{2}}\) | \(68\) |
default | \(A \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )-\frac {B}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+C \left (-\frac {x}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {a \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{2 b}\right )+D \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )\) | \(140\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3*(2*A*b^3*x^3+C*a*b^2*x^3-3*D*a^2*b*x^2+3*A*a*b^2*x-B*a^2*b-2*D*a^3)/(b *x^2+a)^(3/2)/a^2/b^2
Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {{\left (3 \, D a^{2} b x^{2} - 3 \, A a b^{2} x + 2 \, D a^{3} + B a^{2} b - {\left (C a b^{2} + 2 \, A b^{3}\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{2} b^{4} x^{4} + 2 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="fricas")
Output:
-1/3*(3*D*a^2*b*x^2 - 3*A*a*b^2*x + 2*D*a^3 + B*a^2*b - (C*a*b^2 + 2*A*b^3 )*x^3)*sqrt(b*x^2 + a)/(a^2*b^4*x^4 + 2*a^3*b^3*x^2 + a^4*b^2)
Time = 7.15 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.78 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/2}} \, dx=A \left (\frac {3 a x}{3 a^{\frac {7}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {5}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {2 b x^{3}}{3 a^{\frac {7}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {5}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B \left (\begin {cases} - \frac {1}{3 a b \sqrt {a + b x^{2}} + 3 b^{2} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) + \frac {C x^{3}}{3 a^{\frac {5}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {3}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + D \left (\begin {cases} - \frac {2 a}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} - \frac {3 b x^{2}}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x**2+a)**(5/2),x)
Output:
A*(3*a*x/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x** 2/a)) + 2*b*x**3/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x**2/a))) + B*Piecewise((-1/(3*a*b*sqrt(a + b*x**2) + 3*b**2*x**2*sqr t(a + b*x**2)), Ne(b, 0)), (x**2/(2*a**(5/2)), True)) + C*x**3/(3*a**(5/2) *sqrt(1 + b*x**2/a) + 3*a**(3/2)*b*x**2*sqrt(1 + b*x**2/a)) + D*Piecewise( (-2*a/(3*a*b**2*sqrt(a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2)) - 3*b*x** 2/(3*a*b**2*sqrt(a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2)), Ne(b, 0)), ( x**4/(4*a**(5/2)), True))
Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {D x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, A x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {A x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {C x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {C x}{3 \, \sqrt {b x^{2} + a} a b} - \frac {2 \, D a}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} - \frac {B}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")
Output:
-D*x^2/((b*x^2 + a)^(3/2)*b) + 2/3*A*x/(sqrt(b*x^2 + a)*a^2) + 1/3*A*x/((b *x^2 + a)^(3/2)*a) - 1/3*C*x/((b*x^2 + a)^(3/2)*b) + 1/3*C*x/(sqrt(b*x^2 + a)*a*b) - 2/3*D*a/((b*x^2 + a)^(3/2)*b^2) - 1/3*B/((b*x^2 + a)^(3/2)*b)
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.69 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {{\left (x {\left (\frac {3 \, D}{b} - \frac {{\left (C a b^{2} + 2 \, A b^{3}\right )} x}{a^{2} b^{2}}\right )} - \frac {3 \, A}{a}\right )} x + \frac {2 \, D a^{3} + B a^{2} b}{a^{2} b^{2}}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="giac")
Output:
-1/3*((x*(3*D/b - (C*a*b^2 + 2*A*b^3)*x/(a^2*b^2)) - 3*A/a)*x + (2*D*a^3 + B*a^2*b)/(a^2*b^2))/(b*x^2 + a)^(3/2)
Time = 17.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {2\,A\,x\,\left (b\,x^2+a\right )+A\,a\,x}{3\,a^2\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {B}{3\,b\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {\left (3\,b\,x^2+2\,a\right )\,D}{3\,b^2\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {C\,x^3}{3\,a\,{\left (b\,x^2+a\right )}^{3/2}} \] Input:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x^2)^(5/2),x)
Output:
(2*A*x*(a + b*x^2) + A*a*x)/(3*a^2*(a + b*x^2)^(3/2)) - B/(3*b*(a + b*x^2) ^(3/2)) - ((2*a + 3*b*x^2)*D)/(3*b^2*(a + b*x^2)^(3/2)) + (C*x^3)/(3*a*(a + b*x^2)^(3/2))
Time = 0.18 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.69 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-2 \sqrt {b \,x^{2}+a}\, a^{2} d +3 \sqrt {b \,x^{2}+a}\, a \,b^{2} x -\sqrt {b \,x^{2}+a}\, a \,b^{2}-3 \sqrt {b \,x^{2}+a}\, a b d \,x^{2}+2 \sqrt {b \,x^{2}+a}\, b^{3} x^{3}+\sqrt {b \,x^{2}+a}\, b^{2} c \,x^{3}-2 \sqrt {b}\, a^{2} b +\sqrt {b}\, a^{2} c -4 \sqrt {b}\, a \,b^{2} x^{2}+2 \sqrt {b}\, a b c \,x^{2}-2 \sqrt {b}\, b^{3} x^{4}+\sqrt {b}\, b^{2} c \,x^{4}}{3 a \,b^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x)
Output:
( - 2*sqrt(a + b*x**2)*a**2*d + 3*sqrt(a + b*x**2)*a*b**2*x - sqrt(a + b*x **2)*a*b**2 - 3*sqrt(a + b*x**2)*a*b*d*x**2 + 2*sqrt(a + b*x**2)*b**3*x**3 + sqrt(a + b*x**2)*b**2*c*x**3 - 2*sqrt(b)*a**2*b + sqrt(b)*a**2*c - 4*sq rt(b)*a*b**2*x**2 + 2*sqrt(b)*a*b*c*x**2 - 2*sqrt(b)*b**3*x**4 + sqrt(b)*b **2*c*x**4)/(3*a*b**2*(a**2 + 2*a*b*x**2 + b**2*x**4))