Integrand size = 22, antiderivative size = 101 \[ \int \frac {x^5 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 a^2 (b c-a d)}{5 b^4 \left (a+b x^2\right )^{5/4}}+\frac {2 a (2 b c-3 a d)}{b^4 \sqrt [4]{a+b x^2}}+\frac {2 (b c-3 a d) \left (a+b x^2\right )^{3/4}}{3 b^4}+\frac {2 d \left (a+b x^2\right )^{7/4}}{7 b^4} \] Output:
-2/5*a^2*(-a*d+b*c)/b^4/(b*x^2+a)^(5/4)+2*a*(-3*a*d+2*b*c)/b^4/(b*x^2+a)^( 1/4)+2/3*(-3*a*d+b*c)*(b*x^2+a)^(3/4)/b^4+2/7*d*(b*x^2+a)^(7/4)/b^4
Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79 \[ \int \frac {x^5 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 \left (224 a^2 b c-384 a^3 d+280 a b^2 c x^2-480 a^2 b d x^2+35 b^3 c x^4-60 a b^2 d x^4+15 b^3 d x^6\right )}{105 b^4 \left (a+b x^2\right )^{5/4}} \] Input:
Integrate[(x^5*(c + d*x^2))/(a + b*x^2)^(9/4),x]
Output:
(2*(224*a^2*b*c - 384*a^3*d + 280*a*b^2*c*x^2 - 480*a^2*b*d*x^2 + 35*b^3*c *x^4 - 60*a*b^2*d*x^4 + 15*b^3*d*x^6))/(105*b^4*(a + b*x^2)^(5/4))
Time = 0.23 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, 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.
\(\displaystyle \int \frac {x^5 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^4 \left (d x^2+c\right )}{\left (b x^2+a\right )^{9/4}}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {(a d-b c) a^2}{b^3 \left (b x^2+a\right )^{9/4}}+\frac {(3 a d-2 b c) a}{b^3 \left (b x^2+a\right )^{5/4}}+\frac {d \left (b x^2+a\right )^{3/4}}{b^3}+\frac {b c-3 a d}{b^3 \sqrt [4]{b x^2+a}}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {4 a^2 (b c-a d)}{5 b^4 \left (a+b x^2\right )^{5/4}}+\frac {4 a (2 b c-3 a d)}{b^4 \sqrt [4]{a+b x^2}}+\frac {4 \left (a+b x^2\right )^{3/4} (b c-3 a d)}{3 b^4}+\frac {4 d \left (a+b x^2\right )^{7/4}}{7 b^4}\right )\) |
Input:
Int[(x^5*(c + d*x^2))/(a + b*x^2)^(9/4),x]
Output:
((-4*a^2*(b*c - a*d))/(5*b^4*(a + b*x^2)^(5/4)) + (4*a*(2*b*c - 3*a*d))/(b ^4*(a + b*x^2)^(1/4)) + (4*(b*c - 3*a*d)*(a + b*x^2)^(3/4))/(3*b^4) + (4*d *(a + b*x^2)^(7/4))/(7*b^4))/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 = 0.42 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(-\frac {256 \left (-\frac {35 x^{4} \left (\frac {3 x^{2} d}{7}+c \right ) b^{3}}{384}-\frac {35 a \left (-\frac {3 x^{2} d}{14}+c \right ) x^{2} b^{2}}{48}-\frac {7 a^{2} \left (-\frac {15 x^{2} d}{7}+c \right ) b}{12}+a^{3} d \right )}{35 \left (b \,x^{2}+a \right )^{\frac {5}{4}} b^{4}}\) | \(68\) |
risch | \(-\frac {2 \left (-3 b d \,x^{2}+18 a d -7 b c \right ) \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{21 b^{4}}-\frac {2 a \left (15 a b d \,x^{2}-10 b^{2} c \,x^{2}+14 d \,a^{2}-9 a b c \right )}{5 b^{4} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}\) | \(76\) |
gosper | \(-\frac {2 \left (-15 d \,x^{6} b^{3}+60 a \,b^{2} d \,x^{4}-35 b^{3} c \,x^{4}+480 a^{2} b d \,x^{2}-280 a \,b^{2} c \,x^{2}+384 a^{3} d -224 a^{2} b c \right )}{105 \left (b \,x^{2}+a \right )^{\frac {5}{4}} b^{4}}\) | \(77\) |
trager | \(-\frac {2 \left (-15 d \,x^{6} b^{3}+60 a \,b^{2} d \,x^{4}-35 b^{3} c \,x^{4}+480 a^{2} b d \,x^{2}-280 a \,b^{2} c \,x^{2}+384 a^{3} d -224 a^{2} b c \right )}{105 \left (b \,x^{2}+a \right )^{\frac {5}{4}} b^{4}}\) | \(77\) |
orering | \(-\frac {2 \left (-15 d \,x^{6} b^{3}+60 a \,b^{2} d \,x^{4}-35 b^{3} c \,x^{4}+480 a^{2} b d \,x^{2}-280 a \,b^{2} c \,x^{2}+384 a^{3} d -224 a^{2} b c \right )}{105 \left (b \,x^{2}+a \right )^{\frac {5}{4}} b^{4}}\) | \(77\) |
Input:
int(x^5*(d*x^2+c)/(b*x^2+a)^(9/4),x,method=_RETURNVERBOSE)
Output:
-256/35/(b*x^2+a)^(5/4)*(-35/384*x^4*(3/7*x^2*d+c)*b^3-35/48*a*(-3/14*x^2* d+c)*x^2*b^2-7/12*a^2*(-15/7*x^2*d+c)*b+a^3*d)/b^4
Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98 \[ \int \frac {x^5 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 \, {\left (15 \, b^{3} d x^{6} + 5 \, {\left (7 \, b^{3} c - 12 \, a b^{2} d\right )} x^{4} + 224 \, a^{2} b c - 384 \, a^{3} d + 40 \, {\left (7 \, a b^{2} c - 12 \, a^{2} b d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}}}{105 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} \] Input:
integrate(x^5*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="fricas")
Output:
2/105*(15*b^3*d*x^6 + 5*(7*b^3*c - 12*a*b^2*d)*x^4 + 224*a^2*b*c - 384*a^3 *d + 40*(7*a*b^2*c - 12*a^2*b*d)*x^2)*(b*x^2 + a)^(3/4)/(b^6*x^4 + 2*a*b^5 *x^2 + a^2*b^4)
Time = 5.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.18 \[ \int \frac {x^5 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\begin {cases} \frac {4 \left (\frac {a^{2} \left (a d - b c\right )}{10 b^{3} \left (a + b x^{2}\right )^{\frac {5}{4}}} - \frac {a \left (3 a d - 2 b c\right )}{2 b^{3} \sqrt [4]{a + b x^{2}}} + \frac {d \left (a + b x^{2}\right )^{\frac {7}{4}}}{14 b^{3}} + \frac {\left (a + b x^{2}\right )^{\frac {3}{4}} \left (- 3 a d + b c\right )}{6 b^{3}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {\frac {c x^{6}}{3} + \frac {d x^{8}}{4}}{2 a^{\frac {9}{4}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**5*(d*x**2+c)/(b*x**2+a)**(9/4),x)
Output:
Piecewise((4*(a**2*(a*d - b*c)/(10*b**3*(a + b*x**2)**(5/4)) - a*(3*a*d - 2*b*c)/(2*b**3*(a + b*x**2)**(1/4)) + d*(a + b*x**2)**(7/4)/(14*b**3) + (a + b*x**2)**(3/4)*(-3*a*d + b*c)/(6*b**3))/b, Ne(b, 0)), ((c*x**6/3 + d*x* *8/4)/(2*a**(9/4)), True))
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.17 \[ \int \frac {x^5 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2}{35} \, d {\left (\frac {5 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}}}{b^{4}} - \frac {35 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a}{b^{4}} - \frac {105 \, a^{2}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} b^{4}} + \frac {7 \, a^{3}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} b^{4}}\right )} + \frac {2}{15} \, c {\left (\frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}}}{b^{3}} + \frac {30 \, a}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} b^{3}} - \frac {3 \, a^{2}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} b^{3}}\right )} \] Input:
integrate(x^5*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="maxima")
Output:
2/35*d*(5*(b*x^2 + a)^(7/4)/b^4 - 35*(b*x^2 + a)^(3/4)*a/b^4 - 105*a^2/((b *x^2 + a)^(1/4)*b^4) + 7*a^3/((b*x^2 + a)^(5/4)*b^4)) + 2/15*c*(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))
Time = 0.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04 \[ \int \frac {x^5 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 \, {\left (10 \, {\left (b x^{2} + a\right )} a b c - a^{2} b c - 15 \, {\left (b x^{2} + a\right )} a^{2} d + a^{3} d\right )}}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} b^{4}} + \frac {2 \, {\left (7 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} b^{25} c + 3 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} b^{24} d - 21 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a b^{24} d\right )}}{21 \, b^{28}} \] Input:
integrate(x^5*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="giac")
Output:
2/5*(10*(b*x^2 + a)*a*b*c - a^2*b*c - 15*(b*x^2 + a)*a^2*d + a^3*d)/((b*x^ 2 + a)^(5/4)*b^4) + 2/21*(7*(b*x^2 + a)^(3/4)*b^25*c + 3*(b*x^2 + a)^(7/4) *b^24*d - 21*(b*x^2 + a)^(3/4)*a*b^24*d)/b^28
Time = 0.69 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.94 \[ \int \frac {x^5 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {42\,a^3\,d-180\,a\,d\,{\left (b\,x^2+a\right )}^2+70\,b\,c\,{\left (b\,x^2+a\right )}^2-630\,a^2\,d\,\left (b\,x^2+a\right )-42\,a^2\,b\,c+30\,b\,d\,x^2\,{\left (b\,x^2+a\right )}^2+420\,a\,b\,c\,\left (b\,x^2+a\right )}{105\,b^4\,{\left (b\,x^2+a\right )}^{5/4}} \] Input:
int((x^5*(c + d*x^2))/(a + b*x^2)^(9/4),x)
Output:
(42*a^3*d - 180*a*d*(a + b*x^2)^2 + 70*b*c*(a + b*x^2)^2 - 630*a^2*d*(a + b*x^2) - 42*a^2*b*c + 30*b*d*x^2*(a + b*x^2)^2 + 420*a*b*c*(a + b*x^2))/(1 05*b^4*(a + b*x^2)^(5/4))
Time = 0.22 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.78 \[ \int \frac {x^5 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 \sqrt {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +a +b \,x^{2}}\, \sqrt {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}\, \left (384 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a^{3} d x -224 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a^{2} b c x +480 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{3}-280 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{3}+60 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{5}-35 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{5}-15 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{7}-384 a^{4} d +224 a^{3} b c -864 a^{3} b d \,x^{2}+504 a^{2} b^{2} c \,x^{2}-540 a^{2} b^{2} d \,x^{4}+315 a \,b^{3} c \,x^{4}-45 a \,b^{3} d \,x^{6}+35 b^{4} c \,x^{6}+15 b^{4} d \,x^{8}\right )}{105 a \,b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^5*(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)*(384*sqrt(b)*sqrt(a + b*x**2)*a**3*d*x - 224*sqrt(b)*sqrt(a + b* x**2)*a**2*b*c*x + 480*sqrt(b)*sqrt(a + b*x**2)*a**2*b*d*x**3 - 280*sqrt(b )*sqrt(a + b*x**2)*a*b**2*c*x**3 + 60*sqrt(b)*sqrt(a + b*x**2)*a*b**2*d*x* *5 - 35*sqrt(b)*sqrt(a + b*x**2)*b**3*c*x**5 - 15*sqrt(b)*sqrt(a + b*x**2) *b**3*d*x**7 - 384*a**4*d + 224*a**3*b*c - 864*a**3*b*d*x**2 + 504*a**2*b* *2*c*x**2 - 540*a**2*b**2*d*x**4 + 315*a*b**3*c*x**4 - 45*a*b**3*d*x**6 + 35*b**4*c*x**6 + 15*b**4*d*x**8))/(105*a*b**4*(a**2 + 2*a*b*x**2 + b**2*x* *4))