Integrand size = 30, antiderivative size = 119 \[ \int \frac {x \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {A b^3-a \left (b^2 B-a b C+a^2 D\right )}{b^4 \sqrt {a+b x^2}}+\frac {\left (b^2 B-2 a b C+3 a^2 D\right ) \sqrt {a+b x^2}}{b^4}+\frac {(b C-3 a D) \left (a+b x^2\right )^{3/2}}{3 b^4}+\frac {D \left (a+b x^2\right )^{5/2}}{5 b^4} \] Output:
-(A*b^3-a*(B*b^2-C*a*b+D*a^2))/b^4/(b*x^2+a)^(1/2)+(B*b^2-2*C*a*b+3*D*a^2) *(b*x^2+a)^(1/2)/b^4+1/3*(C*b-3*D*a)*(b*x^2+a)^(3/2)/b^4+1/5*D*(b*x^2+a)^( 5/2)/b^4
Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.77 \[ \int \frac {x \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-15 A b^3+48 a^3 D-8 a^2 b \left (5 C-3 D x^2\right )+2 a b^2 \left (15 B-10 C x^2-3 D x^4\right )+b^3 x^2 \left (15 B+5 C x^2+3 D x^4\right )}{15 b^4 \sqrt {a+b x^2}} \] Input:
Integrate[(x*(A + B*x^2 + C*x^4 + D*x^6))/(a + b*x^2)^(3/2),x]
Output:
(-15*A*b^3 + 48*a^3*D - 8*a^2*b*(5*C - 3*D*x^2) + 2*a*b^2*(15*B - 10*C*x^2 - 3*D*x^4) + b^3*x^2*(15*B + 5*C*x^2 + 3*D*x^4))/(15*b^4*Sqrt[a + b*x^2])
Time = 0.39 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2331, 2389, 2009}
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.
\(\displaystyle \int \frac {x \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2331 |
\(\displaystyle \frac {1}{2} \int \frac {D x^6+C x^4+B x^2+A}{\left (b x^2+a\right )^{3/2}}dx^2\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \frac {1}{2} \int \left (\frac {D \left (b x^2+a\right )^{3/2}}{b^3}+\frac {(b C-3 a D) \sqrt {b x^2+a}}{b^3}+\frac {3 D a^2-2 b C a+b^2 B}{b^3 \sqrt {b x^2+a}}+\frac {A b^3-a \left (D a^2-b C a+b^2 B\right )}{b^3 \left (b x^2+a\right )^{3/2}}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{b^4 \sqrt {a+b x^2}}+\frac {2 \sqrt {a+b x^2} \left (3 a^2 D-2 a b C+b^2 B\right )}{b^4}+\frac {2 \left (a+b x^2\right )^{3/2} (b C-3 a D)}{3 b^4}+\frac {2 D \left (a+b x^2\right )^{5/2}}{5 b^4}\right )\) |
Input:
Int[(x*(A + B*x^2 + C*x^4 + D*x^6))/(a + b*x^2)^(3/2),x]
Output:
((-2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D)))/(b^4*Sqrt[a + b*x^2]) + (2*(b^2* B - 2*a*b*C + 3*a^2*D)*Sqrt[a + b*x^2])/b^4 + (2*(b*C - 3*a*D)*(a + b*x^2) ^(3/2))/(3*b^4) + (2*D*(a + b*x^2)^(5/2))/(5*b^4))/2
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/2 S ubst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Time = 0.50 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(\frac {\left (3 D x^{6}+5 C \,x^{4}+15 x^{2} B -15 A \right ) b^{3}+30 \left (-\frac {1}{5} D x^{4}-\frac {2}{3} C \,x^{2}+B \right ) a \,b^{2}-40 \left (-\frac {3 D x^{2}}{5}+C \right ) a^{2} b +48 a^{3} D}{15 \sqrt {b \,x^{2}+a}\, b^{4}}\) | \(82\) |
gosper | \(-\frac {-3 b^{3} D x^{6}-5 b^{3} C \,x^{4}+6 D a \,b^{2} x^{4}-15 b^{3} B \,x^{2}+20 C a \,b^{2} x^{2}-24 D a^{2} b \,x^{2}+15 b^{3} A -30 a \,b^{2} B +40 a^{2} b C -48 a^{3} D}{15 \sqrt {b \,x^{2}+a}\, b^{4}}\) | \(99\) |
trager | \(-\frac {-3 b^{3} D x^{6}-5 b^{3} C \,x^{4}+6 D a \,b^{2} x^{4}-15 b^{3} B \,x^{2}+20 C a \,b^{2} x^{2}-24 D a^{2} b \,x^{2}+15 b^{3} A -30 a \,b^{2} B +40 a^{2} b C -48 a^{3} D}{15 \sqrt {b \,x^{2}+a}\, b^{4}}\) | \(99\) |
orering | \(-\frac {-3 b^{3} D x^{6}-5 b^{3} C \,x^{4}+6 D a \,b^{2} x^{4}-15 b^{3} B \,x^{2}+20 C a \,b^{2} x^{2}-24 D a^{2} b \,x^{2}+15 b^{3} A -30 a \,b^{2} B +40 a^{2} b C -48 a^{3} D}{15 \sqrt {b \,x^{2}+a}\, b^{4}}\) | \(99\) |
default | \(B \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )+C \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )+D \left (\frac {x^{6}}{5 b \sqrt {b \,x^{2}+a}}-\frac {6 a \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )}{5 b}\right )-\frac {A}{b \sqrt {b \,x^{2}+a}}\) | \(191\) |
Input:
int(x*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/15*((3*D*x^6+5*C*x^4+15*B*x^2-15*A)*b^3+30*(-1/5*D*x^4-2/3*C*x^2+B)*a*b^ 2-40*(-3/5*D*x^2+C)*a^2*b+48*a^3*D)/(b*x^2+a)^(1/2)/b^4
Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.89 \[ \int \frac {x \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left (3 \, D b^{3} x^{6} - {\left (6 \, D a b^{2} - 5 \, C b^{3}\right )} x^{4} + 48 \, D a^{3} - 40 \, C a^{2} b + 30 \, B a b^{2} - 15 \, A b^{3} + {\left (24 \, D a^{2} b - 20 \, C a b^{2} + 15 \, B b^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, {\left (b^{5} x^{2} + a b^{4}\right )}} \] Input:
integrate(x*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="fricas")
Output:
1/15*(3*D*b^3*x^6 - (6*D*a*b^2 - 5*C*b^3)*x^4 + 48*D*a^3 - 40*C*a^2*b + 30 *B*a*b^2 - 15*A*b^3 + (24*D*a^2*b - 20*C*a*b^2 + 15*B*b^3)*x^2)*sqrt(b*x^2 + a)/(b^5*x^2 + a*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (105) = 210\).
Time = 0.31 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.97 \[ \int \frac {x \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\begin {cases} - \frac {A}{b \sqrt {a + b x^{2}}} + \frac {2 B a}{b^{2} \sqrt {a + b x^{2}}} + \frac {B x^{2}}{b \sqrt {a + b x^{2}}} - \frac {8 C a^{2}}{3 b^{3} \sqrt {a + b x^{2}}} - \frac {4 C a x^{2}}{3 b^{2} \sqrt {a + b x^{2}}} + \frac {C x^{4}}{3 b \sqrt {a + b x^{2}}} + \frac {16 D a^{3}}{5 b^{4} \sqrt {a + b x^{2}}} + \frac {8 D a^{2} x^{2}}{5 b^{3} \sqrt {a + b x^{2}}} - \frac {2 D a x^{4}}{5 b^{2} \sqrt {a + b x^{2}}} + \frac {D x^{6}}{5 b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{2}}{2} + \frac {B x^{4}}{4} + \frac {C x^{6}}{6} + \frac {D x^{8}}{8}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(x*(D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(3/2),x)
Output:
Piecewise((-A/(b*sqrt(a + b*x**2)) + 2*B*a/(b**2*sqrt(a + b*x**2)) + B*x** 2/(b*sqrt(a + b*x**2)) - 8*C*a**2/(3*b**3*sqrt(a + b*x**2)) - 4*C*a*x**2/( 3*b**2*sqrt(a + b*x**2)) + C*x**4/(3*b*sqrt(a + b*x**2)) + 16*D*a**3/(5*b* *4*sqrt(a + b*x**2)) + 8*D*a**2*x**2/(5*b**3*sqrt(a + b*x**2)) - 2*D*a*x** 4/(5*b**2*sqrt(a + b*x**2)) + D*x**6/(5*b*sqrt(a + b*x**2)), Ne(b, 0)), (( A*x**2/2 + B*x**4/4 + C*x**6/6 + D*x**8/8)/a**(3/2), True))
Time = 0.04 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.51 \[ \int \frac {x \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {D x^{6}}{5 \, \sqrt {b x^{2} + a} b} - \frac {2 \, D a x^{4}}{5 \, \sqrt {b x^{2} + a} b^{2}} + \frac {C x^{4}}{3 \, \sqrt {b x^{2} + a} b} + \frac {8 \, D a^{2} x^{2}}{5 \, \sqrt {b x^{2} + a} b^{3}} - \frac {4 \, C a x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {B x^{2}}{\sqrt {b x^{2} + a} b} + \frac {16 \, D a^{3}}{5 \, \sqrt {b x^{2} + a} b^{4}} - \frac {8 \, C a^{2}}{3 \, \sqrt {b x^{2} + a} b^{3}} + \frac {2 \, B a}{\sqrt {b x^{2} + a} b^{2}} - \frac {A}{\sqrt {b x^{2} + a} b} \] Input:
integrate(x*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="maxima")
Output:
1/5*D*x^6/(sqrt(b*x^2 + a)*b) - 2/5*D*a*x^4/(sqrt(b*x^2 + a)*b^2) + 1/3*C* x^4/(sqrt(b*x^2 + a)*b) + 8/5*D*a^2*x^2/(sqrt(b*x^2 + a)*b^3) - 4/3*C*a*x^ 2/(sqrt(b*x^2 + a)*b^2) + B*x^2/(sqrt(b*x^2 + a)*b) + 16/5*D*a^3/(sqrt(b*x ^2 + a)*b^4) - 8/3*C*a^2/(sqrt(b*x^2 + a)*b^3) + 2*B*a/(sqrt(b*x^2 + a)*b^ 2) - A/(sqrt(b*x^2 + a)*b)
Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.18 \[ \int \frac {x \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {D a^{3} - C a^{2} b + B a b^{2} - A b^{3}}{\sqrt {b x^{2} + a} b^{4}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} D b^{16} - 15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} D a b^{16} + 45 \, \sqrt {b x^{2} + a} D a^{2} b^{16} + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} C b^{17} - 30 \, \sqrt {b x^{2} + a} C a b^{17} + 15 \, \sqrt {b x^{2} + a} B b^{18}}{15 \, b^{20}} \] Input:
integrate(x*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="giac")
Output:
(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)/(sqrt(b*x^2 + a)*b^4) + 1/15*(3*(b*x^2 + a)^(5/2)*D*b^16 - 15*(b*x^2 + a)^(3/2)*D*a*b^16 + 45*sqrt(b*x^2 + a)*D* a^2*b^16 + 5*(b*x^2 + a)^(3/2)*C*b^17 - 30*sqrt(b*x^2 + a)*C*a*b^17 + 15*s qrt(b*x^2 + a)*B*b^18)/b^20
Timed out. \[ \int \frac {x \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x\,\left (A+B\,x^2+C\,x^4+x^6\,D\right )}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:
int((x*(A + B*x^2 + C*x^4 + x^6*D))/(a + b*x^2)^(3/2),x)
Output:
int((x*(A + B*x^2 + C*x^4 + x^6*D))/(a + b*x^2)^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82 \[ \int \frac {x \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\sqrt {b \,x^{2}+a}\, \left (3 b^{3} d \,x^{6}-6 a \,b^{2} d \,x^{4}+5 b^{3} c \,x^{4}+24 a^{2} b d \,x^{2}-20 a \,b^{2} c \,x^{2}+15 b^{4} x^{2}+48 a^{3} d -40 a^{2} b c +15 a \,b^{3}\right )}{15 b^{4} \left (b \,x^{2}+a \right )} \] Input:
int(x*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/2),x)
Output:
(sqrt(a + b*x**2)*(48*a**3*d - 40*a**2*b*c + 24*a**2*b*d*x**2 + 15*a*b**3 - 20*a*b**2*c*x**2 - 6*a*b**2*d*x**4 + 15*b**4*x**2 + 5*b**3*c*x**4 + 3*b* *3*d*x**6))/(15*b**4*(a + b*x**2))