Integrand size = 22, antiderivative size = 71 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 a (b c-a d)}{5 b^3 \left (a+b x^2\right )^{5/4}}-\frac {2 (b c-2 a d)}{b^3 \sqrt [4]{a+b x^2}}+\frac {2 d \left (a+b x^2\right )^{3/4}}{3 b^3} \] Output:
2/5*a*(-a*d+b*c)/b^3/(b*x^2+a)^(5/4)-2*(-2*a*d+b*c)/b^3/(b*x^2+a)^(1/4)+2/ 3*d*(b*x^2+a)^(3/4)/b^3
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 \left (-12 a b c+32 a^2 d-15 b^2 c x^2+40 a b d x^2+5 b^2 d x^4\right )}{15 b^3 \left (a+b x^2\right )^{5/4}} \] Input:
Integrate[(x^3*(c + d*x^2))/(a + b*x^2)^(9/4),x]
Output:
(2*(-12*a*b*c + 32*a^2*d - 15*b^2*c*x^2 + 40*a*b*d*x^2 + 5*b^2*d*x^4))/(15 *b^3*(a + b*x^2)^(5/4))
Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {354, 86, 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.
\(\displaystyle \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^2 \left (d x^2+c\right )}{\left (b x^2+a\right )^{9/4}}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {d}{b^2 \sqrt [4]{b x^2+a}}+\frac {b c-2 a d}{b^2 \left (b x^2+a\right )^{5/4}}+\frac {a (a d-b c)}{b^2 \left (b x^2+a\right )^{9/4}}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {4 (b c-2 a d)}{b^3 \sqrt [4]{a+b x^2}}+\frac {4 a (b c-a d)}{5 b^3 \left (a+b x^2\right )^{5/4}}+\frac {4 d \left (a+b x^2\right )^{3/4}}{3 b^3}\right )\) |
Input:
Int[(x^3*(c + d*x^2))/(a + b*x^2)^(9/4),x]
Output:
((4*a*(b*c - a*d))/(5*b^3*(a + b*x^2)^(5/4)) - (4*(b*c - 2*a*d))/(b^3*(a + b*x^2)^(1/4)) + (4*d*(a + b*x^2)^(3/4))/(3*b^3))/2
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 0.43 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(\frac {-2 x^{2} \left (-\frac {x^{2} d}{3}+c \right ) b^{2}-\frac {8 a \left (-\frac {10 x^{2} d}{3}+c \right ) b}{5}+\frac {64 d \,a^{2}}{15}}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} b^{3}}\) | \(49\) |
gosper | \(\frac {\frac {2}{3} d \,b^{2} x^{4}+\frac {16}{3} a b d \,x^{2}-2 b^{2} c \,x^{2}+\frac {64}{15} d \,a^{2}-\frac {8}{5} a b c}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} b^{3}}\) | \(53\) |
trager | \(\frac {\frac {2}{3} d \,b^{2} x^{4}+\frac {16}{3} a b d \,x^{2}-2 b^{2} c \,x^{2}+\frac {64}{15} d \,a^{2}-\frac {8}{5} a b c}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} b^{3}}\) | \(53\) |
orering | \(\frac {\frac {2}{3} d \,b^{2} x^{4}+\frac {16}{3} a b d \,x^{2}-2 b^{2} c \,x^{2}+\frac {64}{15} d \,a^{2}-\frac {8}{5} a b c}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} b^{3}}\) | \(53\) |
risch | \(\frac {2 d \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{3 b^{3}}+\frac {4 a b d \,x^{2}-2 b^{2} c \,x^{2}+\frac {18}{5} d \,a^{2}-\frac {8}{5} a b c}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} b^{3}}\) | \(60\) |
Input:
int(x^3*(d*x^2+c)/(b*x^2+a)^(9/4),x,method=_RETURNVERBOSE)
Output:
64/15*(-15/32*x^2*(-1/3*x^2*d+c)*b^2-3/8*a*(-10/3*x^2*d+c)*b+d*a^2)/(b*x^2 +a)^(5/4)/b^3
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 \, {\left (5 \, b^{2} d x^{4} - 12 \, a b c + 32 \, a^{2} d - 5 \, {\left (3 \, b^{2} c - 8 \, a b d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}}}{15 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} \] Input:
integrate(x^3*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="fricas")
Output:
2/15*(5*b^2*d*x^4 - 12*a*b*c + 32*a^2*d - 5*(3*b^2*c - 8*a*b*d)*x^2)*(b*x^ 2 + a)^(3/4)/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)
Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (66) = 132\).
Time = 0.53 (sec) , antiderivative size = 240, normalized size of antiderivative = 3.38 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\begin {cases} \frac {64 a^{2} d}{15 a b^{3} \sqrt [4]{a + b x^{2}} + 15 b^{4} x^{2} \sqrt [4]{a + b x^{2}}} - \frac {24 a b c}{15 a b^{3} \sqrt [4]{a + b x^{2}} + 15 b^{4} x^{2} \sqrt [4]{a + b x^{2}}} + \frac {80 a b d x^{2}}{15 a b^{3} \sqrt [4]{a + b x^{2}} + 15 b^{4} x^{2} \sqrt [4]{a + b x^{2}}} - \frac {30 b^{2} c x^{2}}{15 a b^{3} \sqrt [4]{a + b x^{2}} + 15 b^{4} x^{2} \sqrt [4]{a + b x^{2}}} + \frac {10 b^{2} d x^{4}}{15 a b^{3} \sqrt [4]{a + b x^{2}} + 15 b^{4} x^{2} \sqrt [4]{a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {\frac {c x^{4}}{4} + \frac {d x^{6}}{6}}{a^{\frac {9}{4}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(d*x**2+c)/(b*x**2+a)**(9/4),x)
Output:
Piecewise((64*a**2*d/(15*a*b**3*(a + b*x**2)**(1/4) + 15*b**4*x**2*(a + b* x**2)**(1/4)) - 24*a*b*c/(15*a*b**3*(a + b*x**2)**(1/4) + 15*b**4*x**2*(a + b*x**2)**(1/4)) + 80*a*b*d*x**2/(15*a*b**3*(a + b*x**2)**(1/4) + 15*b**4 *x**2*(a + b*x**2)**(1/4)) - 30*b**2*c*x**2/(15*a*b**3*(a + b*x**2)**(1/4) + 15*b**4*x**2*(a + b*x**2)**(1/4)) + 10*b**2*d*x**4/(15*a*b**3*(a + b*x* *2)**(1/4) + 15*b**4*x**2*(a + b*x**2)**(1/4)), Ne(b, 0)), ((c*x**4/4 + d* x**6/6)/a**(9/4), True))
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.18 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2}{15} \, d {\left (\frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}}}{b^{3}} + \frac {30 \, a}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} b^{3}} - \frac {3 \, a^{2}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} b^{3}}\right )} - \frac {2}{5} \, c {\left (\frac {5}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} b^{2}} - \frac {a}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} b^{2}}\right )} \] Input:
integrate(x^3*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="maxima")
Output:
2/15*d*(5*(b*x^2 + a)^(3/4)/b^3 + 30*a/((b*x^2 + a)^(1/4)*b^3) - 3*a^2/((b *x^2 + a)^(5/4)*b^3)) - 2/5*c*(5/((b*x^2 + a)^(1/4)*b^2) - a/((b*x^2 + a)^ (5/4)*b^2))
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.89 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} d}{3 \, b^{3}} - \frac {2 \, {\left (5 \, {\left (b x^{2} + a\right )} b c - a b c - 10 \, {\left (b x^{2} + a\right )} a d + a^{2} d\right )}}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} b^{3}} \] Input:
integrate(x^3*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="giac")
Output:
2/3*(b*x^2 + a)^(3/4)*d/b^3 - 2/5*(5*(b*x^2 + a)*b*c - a*b*c - 10*(b*x^2 + a)*a*d + a^2*d)/((b*x^2 + a)^(5/4)*b^3)
Time = 0.51 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.85 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {10\,d\,{\left (b\,x^2+a\right )}^2-6\,a^2\,d+60\,a\,d\,\left (b\,x^2+a\right )-30\,b\,c\,\left (b\,x^2+a\right )+6\,a\,b\,c}{15\,b^3\,{\left (b\,x^2+a\right )}^{5/4}} \] Input:
int((x^3*(c + d*x^2))/(a + b*x^2)^(9/4),x)
Output:
(10*d*(a + b*x^2)^2 - 6*a^2*d + 60*a*d*(a + b*x^2) - 30*b*c*(a + b*x^2) + 6*a*b*c)/(15*b^3*(a + b*x^2)^(5/4))
Time = 0.25 (sec) , antiderivative size = 213, normalized size of antiderivative = 3.00 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 \sqrt {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +a +b \,x^{2}}\, \sqrt {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}\, \left (-32 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a^{2} d x +12 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a b c x -40 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a b d \,x^{3}+15 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, b^{2} c \,x^{3}-5 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, b^{2} d \,x^{5}+32 a^{3} d -12 a^{2} b c +72 a^{2} b d \,x^{2}-27 a \,b^{2} c \,x^{2}+45 a \,b^{2} d \,x^{4}-15 b^{3} c \,x^{4}+5 b^{3} d \,x^{6}\right )}{15 a \,b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^3*(d*x^2+c)/(b*x^2+a)^(9/4),x)
Output:
(2*sqrt(sqrt(b)*sqrt(a + b*x**2)*x + a + b*x**2)*sqrt(sqrt(a + b*x**2) + s qrt(b)*x)*( - 32*sqrt(b)*sqrt(a + b*x**2)*a**2*d*x + 12*sqrt(b)*sqrt(a + b *x**2)*a*b*c*x - 40*sqrt(b)*sqrt(a + b*x**2)*a*b*d*x**3 + 15*sqrt(b)*sqrt( a + b*x**2)*b**2*c*x**3 - 5*sqrt(b)*sqrt(a + b*x**2)*b**2*d*x**5 + 32*a**3 *d - 12*a**2*b*c + 72*a**2*b*d*x**2 - 27*a*b**2*c*x**2 + 45*a*b**2*d*x**4 - 15*b**3*c*x**4 + 5*b**3*d*x**6))/(15*a*b**3*(a**2 + 2*a*b*x**2 + b**2*x* *4))