Integrand size = 27, antiderivative size = 136 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{7/2}} \, dx=-\frac {b B-a D}{5 b^2 \left (a+b x^2\right )^{5/2}}+\frac {(A b-a C) x}{5 a b \left (a+b x^2\right )^{5/2}}-\frac {D}{3 b^2 \left (a+b x^2\right )^{3/2}}+\frac {(4 A b+a C) x}{15 a^2 b \left (a+b x^2\right )^{3/2}}+\frac {2 (4 A b+a C) x}{15 a^3 b \sqrt {a+b x^2}} \] Output:
-1/5*(B*b-D*a)/b^2/(b*x^2+a)^(5/2)+1/5*(A*b-C*a)*x/a/b/(b*x^2+a)^(5/2)-1/3 *D/b^2/(b*x^2+a)^(3/2)+1/15*(4*A*b+C*a)*x/a^2/b/(b*x^2+a)^(3/2)+2/15*(4*A* b+C*a)*x/a^3/b/(b*x^2+a)^(1/2)
Time = 0.92 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {-2 a^4 D+8 A b^4 x^5+5 a^2 b^2 x \left (3 A+C x^2\right )+2 a b^3 x^3 \left (10 A+C x^2\right )-a^3 b \left (3 B+5 D x^2\right )}{15 a^3 b^2 \left (a+b x^2\right )^{5/2}} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(a + b*x^2)^(7/2),x]
Output:
(-2*a^4*D + 8*A*b^4*x^5 + 5*a^2*b^2*x*(3*A + C*x^2) + 2*a*b^3*x^3*(10*A + C*x^2) - a^3*b*(3*B + 5*D*x^2))/(15*a^3*b^2*(a + b*x^2)^(5/2))
Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2345, 25, 27, 454, 208}
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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {\int -\frac {b \left (4 A+\frac {a C}{b}\right )+5 a D x}{b \left (b x^2+a\right )^{5/2}}dx}{5 a}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{5 a b \left (a+b x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {4 A b+a C+5 a D x}{b \left (b x^2+a\right )^{5/2}}dx}{5 a}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{5 a b \left (a+b x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 A b+a C+5 a D x}{\left (b x^2+a\right )^{5/2}}dx}{5 a b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{5 a b \left (a+b x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 454 |
\(\displaystyle \frac {\frac {2 (a C+4 A b) \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{3 a}-\frac {5 a^2 D-b x (a C+4 A b)}{3 a b \left (a+b x^2\right )^{3/2}}}{5 a b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{5 a b \left (a+b x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {2 x (a C+4 A b)}{3 a^2 \sqrt {a+b x^2}}-\frac {5 a^2 D-b x (a C+4 A b)}{3 a b \left (a+b x^2\right )^{3/2}}}{5 a b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{5 a b \left (a+b x^2\right )^{5/2}}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(a + b*x^2)^(7/2),x]
Output:
-1/5*(a*(B - (a*D)/b) - (A*b - a*C)*x)/(a*b*(a + b*x^2)^(5/2)) + (-1/3*(5* a^2*D - b*(4*A*b + a*C)*x)/(a*b*(a + b*x^2)^(3/2)) + (2*(4*A*b + a*C)*x)/( 3*a^2*Sqrt[a + b*x^2]))/(5*a*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)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*d - b*c*x)/(2*a*b*(p + 1)))*(a + b*x^2)^(p + 1), x] + Simp[c*((2*p + 3)/(2*a *(p + 1))) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && L tQ[p, -1] && NeQ[p, -3/2]
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 = 0.94 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.68
method | result | size |
gosper | \(\frac {8 A \,b^{4} x^{5}+2 C a \,b^{3} x^{5}+20 A a \,b^{3} x^{3}+5 C \,a^{2} b^{2} x^{3}-5 D x^{2} a^{3} b +15 A \,a^{2} b^{2} x -3 B \,a^{3} b -2 D a^{4}}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3} b^{2}}\) | \(93\) |
trager | \(\frac {8 A \,b^{4} x^{5}+2 C a \,b^{3} x^{5}+20 A a \,b^{3} x^{3}+5 C \,a^{2} b^{2} x^{3}-5 D x^{2} a^{3} b +15 A \,a^{2} b^{2} x -3 B \,a^{3} b -2 D a^{4}}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3} b^{2}}\) | \(93\) |
orering | \(\frac {8 A \,b^{4} x^{5}+2 C a \,b^{3} x^{5}+20 A a \,b^{3} x^{3}+5 C \,a^{2} b^{2} x^{3}-5 D x^{2} a^{3} b +15 A \,a^{2} b^{2} x -3 B \,a^{3} b -2 D a^{4}}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3} b^{2}}\) | \(93\) |
default | \(A \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )-\frac {B}{5 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+C \left (-\frac {x}{4 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {a \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )}{4 b}\right )+D \left (-\frac {x^{2}}{3 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}-\frac {2 a}{15 b^{2} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}\right )\) | \(182\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/15*(8*A*b^4*x^5+2*C*a*b^3*x^5+20*A*a*b^3*x^3+5*C*a^2*b^2*x^3-5*D*a^3*b*x ^2+15*A*a^2*b^2*x-3*B*a^3*b-2*D*a^4)/(b*x^2+a)^(5/2)/a^3/b^2
Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{7/2}} \, dx=-\frac {{\left (5 \, D a^{3} b x^{2} - 15 \, A a^{2} b^{2} x - 2 \, {\left (C a b^{3} + 4 \, A b^{4}\right )} x^{5} + 2 \, D a^{4} + 3 \, B a^{3} b - 5 \, {\left (C a^{2} b^{2} + 4 \, A a b^{3}\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{15 \, {\left (a^{3} b^{5} x^{6} + 3 \, a^{4} b^{4} x^{4} + 3 \, a^{5} b^{3} x^{2} + a^{6} b^{2}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(7/2),x, algorithm="fricas")
Output:
-1/15*(5*D*a^3*b*x^2 - 15*A*a^2*b^2*x - 2*(C*a*b^3 + 4*A*b^4)*x^5 + 2*D*a^ 4 + 3*B*a^3*b - 5*(C*a^2*b^2 + 4*A*a*b^3)*x^3)*sqrt(b*x^2 + a)/(a^3*b^5*x^ 6 + 3*a^4*b^4*x^4 + 3*a^5*b^3*x^2 + a^6*b^2)
Leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (124) = 248\).
Time = 14.49 (sec) , antiderivative size = 777, normalized size of antiderivative = 5.71 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{7/2}} \, dx =\text {Too large to display} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x**2+a)**(7/2),x)
Output:
A*(15*a**5*x/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 15*a**(11/2)*b* *3*x**6*sqrt(1 + b*x**2/a)) + 35*a**4*b*x**3/(15*a**(17/2)*sqrt(1 + b*x**2 /a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt (1 + b*x**2/a) + 15*a**(11/2)*b**3*x**6*sqrt(1 + b*x**2/a)) + 28*a**3*b**2 *x**5/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x* *2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 15*a**(11/2)*b**3*x**6 *sqrt(1 + b*x**2/a)) + 8*a**2*b**3*x**7/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 15*a**(11/2)*b**3*x**6*sqrt(1 + b*x**2/a))) + B*Piecewise((-1/ (5*a**2*b*sqrt(a + b*x**2) + 10*a*b**2*x**2*sqrt(a + b*x**2) + 5*b**3*x**4 *sqrt(a + b*x**2)), Ne(b, 0)), (x**2/(2*a**(7/2)), True)) + C*(5*a*x**3/(1 5*a**(9/2)*sqrt(1 + b*x**2/a) + 30*a**(7/2)*b*x**2*sqrt(1 + b*x**2/a) + 15 *a**(5/2)*b**2*x**4*sqrt(1 + b*x**2/a)) + 2*b*x**5/(15*a**(9/2)*sqrt(1 + b *x**2/a) + 30*a**(7/2)*b*x**2*sqrt(1 + b*x**2/a) + 15*a**(5/2)*b**2*x**4*s qrt(1 + b*x**2/a))) + D*Piecewise((-2*a/(15*a**2*b**2*sqrt(a + b*x**2) + 3 0*a*b**3*x**2*sqrt(a + b*x**2) + 15*b**4*x**4*sqrt(a + b*x**2)) - 5*b*x**2 /(15*a**2*b**2*sqrt(a + b*x**2) + 30*a*b**3*x**2*sqrt(a + b*x**2) + 15*b** 4*x**4*sqrt(a + b*x**2)), Ne(b, 0)), (x**4/(4*a**(7/2)), True))
Time = 0.04 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{7/2}} \, dx=-\frac {D x^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b} + \frac {8 \, A x}{15 \, \sqrt {b x^{2} + a} a^{3}} + \frac {4 \, A x}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {A x}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a} - \frac {C x}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b} + \frac {2 \, C x}{15 \, \sqrt {b x^{2} + a} a^{2} b} + \frac {C x}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b} - \frac {2 \, D a}{15 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} - \frac {B}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(7/2),x, algorithm="maxima")
Output:
-1/3*D*x^2/((b*x^2 + a)^(5/2)*b) + 8/15*A*x/(sqrt(b*x^2 + a)*a^3) + 4/15*A *x/((b*x^2 + a)^(3/2)*a^2) + 1/5*A*x/((b*x^2 + a)^(5/2)*a) - 1/5*C*x/((b*x ^2 + a)^(5/2)*b) + 2/15*C*x/(sqrt(b*x^2 + a)*a^2*b) + 1/15*C*x/((b*x^2 + a )^(3/2)*a*b) - 2/15*D*a/((b*x^2 + a)^(5/2)*b^2) - 1/5*B/((b*x^2 + a)^(5/2) *b)
Time = 0.17 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {{\left ({\left (x {\left (\frac {2 \, {\left (C a b^{3} + 4 \, A b^{4}\right )} x^{2}}{a^{3} b^{2}} + \frac {5 \, {\left (C a^{2} b^{2} + 4 \, A a b^{3}\right )}}{a^{3} b^{2}}\right )} - \frac {5 \, D}{b}\right )} x + \frac {15 \, A}{a}\right )} x - \frac {2 \, D a^{4} + 3 \, B a^{3} b}{a^{3} b^{2}}}{15 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(7/2),x, algorithm="giac")
Output:
1/15*(((x*(2*(C*a*b^3 + 4*A*b^4)*x^2/(a^3*b^2) + 5*(C*a^2*b^2 + 4*A*a*b^3) /(a^3*b^2)) - 5*D/b)*x + 15*A/a)*x - (2*D*a^4 + 3*B*a^3*b)/(a^3*b^2))/(b*x ^2 + a)^(5/2)
Time = 17.66 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {8\,A\,x\,{\left (b\,x^2+a\right )}^2+3\,A\,a^2\,x+4\,A\,a\,x\,\left (b\,x^2+a\right )}{15\,a^3\,{\left (b\,x^2+a\right )}^{5/2}}-\frac {B}{5\,b\,{\left (b\,x^2+a\right )}^{5/2}}-\frac {\left (5\,b\,x^2+2\,a\right )\,D}{15\,b^2\,{\left (b\,x^2+a\right )}^{5/2}}+\frac {C\,x^3}{3\,a\,{\left (b\,x^2+a\right )}^{5/2}}+\frac {2\,C\,b\,x^5}{15\,a^2\,{\left (b\,x^2+a\right )}^{5/2}} \] Input:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x^2)^(7/2),x)
Output:
(8*A*x*(a + b*x^2)^2 + 3*A*a^2*x + 4*A*a*x*(a + b*x^2))/(15*a^3*(a + b*x^2 )^(5/2)) - B/(5*b*(a + b*x^2)^(5/2)) - ((2*a + 5*b*x^2)*D)/(15*b^2*(a + b* x^2)^(5/2)) + (C*x^3)/(3*a*(a + b*x^2)^(5/2)) + (2*C*b*x^5)/(15*a^2*(a + b *x^2)^(5/2))
Time = 0.17 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.90 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {-2 \sqrt {b \,x^{2}+a}\, a^{3} d +15 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x -3 \sqrt {b \,x^{2}+a}\, a^{2} b^{2}-5 \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{2}+20 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{3}+5 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{4} x^{5}+2 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{5}-8 \sqrt {b}\, a^{3} b -2 \sqrt {b}\, a^{3} c -24 \sqrt {b}\, a^{2} b^{2} x^{2}-6 \sqrt {b}\, a^{2} b c \,x^{2}-24 \sqrt {b}\, a \,b^{3} x^{4}-6 \sqrt {b}\, a \,b^{2} c \,x^{4}-8 \sqrt {b}\, b^{4} x^{6}-2 \sqrt {b}\, b^{3} c \,x^{6}}{15 a^{2} b^{2} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(7/2),x)
Output:
( - 2*sqrt(a + b*x**2)*a**3*d + 15*sqrt(a + b*x**2)*a**2*b**2*x - 3*sqrt(a + b*x**2)*a**2*b**2 - 5*sqrt(a + b*x**2)*a**2*b*d*x**2 + 20*sqrt(a + b*x* *2)*a*b**3*x**3 + 5*sqrt(a + b*x**2)*a*b**2*c*x**3 + 8*sqrt(a + b*x**2)*b* *4*x**5 + 2*sqrt(a + b*x**2)*b**3*c*x**5 - 8*sqrt(b)*a**3*b - 2*sqrt(b)*a* *3*c - 24*sqrt(b)*a**2*b**2*x**2 - 6*sqrt(b)*a**2*b*c*x**2 - 24*sqrt(b)*a* b**3*x**4 - 6*sqrt(b)*a*b**2*c*x**4 - 8*sqrt(b)*b**4*x**6 - 2*sqrt(b)*b**3 *c*x**6)/(15*a**2*b**2*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6))