Integrand size = 22, antiderivative size = 73 \[ \int x^3 \sqrt [4]{a+b x^2} \left (c+d x^2\right ) \, dx=-\frac {2 a (b c-a d) \left (a+b x^2\right )^{5/4}}{5 b^3}+\frac {2 (b c-2 a d) \left (a+b x^2\right )^{9/4}}{9 b^3}+\frac {2 d \left (a+b x^2\right )^{13/4}}{13 b^3} \] Output:
-2/5*a*(-a*d+b*c)*(b*x^2+a)^(5/4)/b^3+2/9*(-2*a*d+b*c)*(b*x^2+a)^(9/4)/b^3 +2/13*d*(b*x^2+a)^(13/4)/b^3
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int x^3 \sqrt [4]{a+b x^2} \left (c+d x^2\right ) \, dx=\frac {2 \left (a+b x^2\right )^{5/4} \left (-52 a b c+32 a^2 d+65 b^2 c x^2-40 a b d x^2+45 b^2 d x^4\right )}{585 b^3} \] Input:
Integrate[x^3*(a + b*x^2)^(1/4)*(c + d*x^2),x]
Output:
(2*(a + b*x^2)^(5/4)*(-52*a*b*c + 32*a^2*d + 65*b^2*c*x^2 - 40*a*b*d*x^2 + 45*b^2*d*x^4))/(585*b^3)
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05, 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.
\(\displaystyle \int x^3 \sqrt [4]{a+b x^2} \left (c+d x^2\right ) \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int x^2 \sqrt [4]{b x^2+a} \left (d x^2+c\right )dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {d \left (b x^2+a\right )^{9/4}}{b^2}+\frac {(b c-2 a d) \left (b x^2+a\right )^{5/4}}{b^2}+\frac {a (a d-b c) \sqrt [4]{b x^2+a}}{b^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {4 \left (a+b x^2\right )^{9/4} (b c-2 a d)}{9 b^3}-\frac {4 a \left (a+b x^2\right )^{5/4} (b c-a d)}{5 b^3}+\frac {4 d \left (a+b x^2\right )^{13/4}}{13 b^3}\right )\) |
Input:
Int[x^3*(a + b*x^2)^(1/4)*(c + d*x^2),x]
Output:
((-4*a*(b*c - a*d)*(a + b*x^2)^(5/4))/(5*b^3) + (4*(b*c - 2*a*d)*(a + b*x^ 2)^(9/4))/(9*b^3) + (4*d*(a + b*x^2)^(13/4))/(13*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 = 0.41 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(\frac {64 \left (\frac {65 \left (\frac {9 x^{2} d}{13}+c \right ) x^{2} b^{2}}{32}-\frac {13 a \left (\frac {10 x^{2} d}{13}+c \right ) b}{8}+d \,a^{2}\right ) \left (b \,x^{2}+a \right )^{\frac {5}{4}}}{585 b^{3}}\) | \(49\) |
gosper | \(\frac {2 \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (45 d \,b^{2} x^{4}-40 a b d \,x^{2}+65 b^{2} c \,x^{2}+32 d \,a^{2}-52 a b c \right )}{585 b^{3}}\) | \(53\) |
orering | \(\frac {2 \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (45 d \,b^{2} x^{4}-40 a b d \,x^{2}+65 b^{2} c \,x^{2}+32 d \,a^{2}-52 a b c \right )}{585 b^{3}}\) | \(53\) |
trager | \(\frac {2 \left (45 d \,x^{6} b^{3}+5 a \,b^{2} d \,x^{4}+65 b^{3} c \,x^{4}-8 a^{2} b d \,x^{2}+13 a \,b^{2} c \,x^{2}+32 a^{3} d -52 a^{2} b c \right ) \left (b \,x^{2}+a \right )^{\frac {1}{4}}}{585 b^{3}}\) | \(77\) |
risch | \(\frac {2 \left (45 d \,x^{6} b^{3}+5 a \,b^{2} d \,x^{4}+65 b^{3} c \,x^{4}-8 a^{2} b d \,x^{2}+13 a \,b^{2} c \,x^{2}+32 a^{3} d -52 a^{2} b c \right ) \left (b \,x^{2}+a \right )^{\frac {1}{4}}}{585 b^{3}}\) | \(77\) |
Input:
int(x^3*(b*x^2+a)^(1/4)*(d*x^2+c),x,method=_RETURNVERBOSE)
Output:
64/585*(65/32*(9/13*x^2*d+c)*x^2*b^2-13/8*a*(10/13*x^2*d+c)*b+d*a^2)*(b*x^ 2+a)^(5/4)/b^3
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01 \[ \int x^3 \sqrt [4]{a+b x^2} \left (c+d x^2\right ) \, dx=\frac {2 \, {\left (45 \, b^{3} d x^{6} + 5 \, {\left (13 \, b^{3} c + a b^{2} d\right )} x^{4} - 52 \, a^{2} b c + 32 \, a^{3} d + {\left (13 \, a b^{2} c - 8 \, a^{2} b d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}}}{585 \, b^{3}} \] Input:
integrate(x^3*(b*x^2+a)^(1/4)*(d*x^2+c),x, algorithm="fricas")
Output:
2/585*(45*b^3*d*x^6 + 5*(13*b^3*c + a*b^2*d)*x^4 - 52*a^2*b*c + 32*a^3*d + (13*a*b^2*c - 8*a^2*b*d)*x^2)*(b*x^2 + a)^(1/4)/b^3
Time = 0.70 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.47 \[ \int x^3 \sqrt [4]{a+b x^2} \left (c+d x^2\right ) \, dx=\frac {c \left (\begin {cases} \frac {4 \left (- \frac {a \left (a + b x^{2}\right )^{\frac {5}{4}}}{5} + \frac {\left (a + b x^{2}\right )^{\frac {9}{4}}}{9}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {\sqrt [4]{a} x^{4}}{2} & \text {otherwise} \end {cases}\right )}{2} + \frac {d \left (\begin {cases} \frac {4 \left (\frac {a^{2} \left (a + b x^{2}\right )^{\frac {5}{4}}}{5} - \frac {2 a \left (a + b x^{2}\right )^{\frac {9}{4}}}{9} + \frac {\left (a + b x^{2}\right )^{\frac {13}{4}}}{13}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\frac {\sqrt [4]{a} x^{6}}{3} & \text {otherwise} \end {cases}\right )}{2} \] Input:
integrate(x**3*(b*x**2+a)**(1/4)*(d*x**2+c),x)
Output:
c*Piecewise((4*(-a*(a + b*x**2)**(5/4)/5 + (a + b*x**2)**(9/4)/9)/b**2, Ne (b, 0)), (a**(1/4)*x**4/2, True))/2 + d*Piecewise((4*(a**2*(a + b*x**2)**( 5/4)/5 - 2*a*(a + b*x**2)**(9/4)/9 + (a + b*x**2)**(13/4)/13)/b**3, Ne(b, 0)), (a**(1/4)*x**6/3, True))/2
Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int x^3 \sqrt [4]{a+b x^2} \left (c+d x^2\right ) \, dx=\frac {2}{585} \, d {\left (\frac {45 \, {\left (b x^{2} + a\right )}^{\frac {13}{4}}}{b^{3}} - \frac {130 \, {\left (b x^{2} + a\right )}^{\frac {9}{4}} a}{b^{3}} + \frac {117 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} a^{2}}{b^{3}}\right )} + \frac {2}{45} \, c {\left (\frac {5 \, {\left (b x^{2} + a\right )}^{\frac {9}{4}}}{b^{2}} - \frac {9 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} a}{b^{2}}\right )} \] Input:
integrate(x^3*(b*x^2+a)^(1/4)*(d*x^2+c),x, algorithm="maxima")
Output:
2/585*d*(45*(b*x^2 + a)^(13/4)/b^3 - 130*(b*x^2 + a)^(9/4)*a/b^3 + 117*(b* x^2 + a)^(5/4)*a^2/b^3) + 2/45*c*(5*(b*x^2 + a)^(9/4)/b^2 - 9*(b*x^2 + a)^ (5/4)*a/b^2)
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int x^3 \sqrt [4]{a+b x^2} \left (c+d x^2\right ) \, dx=\frac {2 \, {\left (65 \, {\left (b x^{2} + a\right )}^{\frac {9}{4}} b c - 117 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} a b c + 45 \, {\left (b x^{2} + a\right )}^{\frac {13}{4}} d - 130 \, {\left (b x^{2} + a\right )}^{\frac {9}{4}} a d + 117 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} a^{2} d\right )}}{585 \, b^{3}} \] Input:
integrate(x^3*(b*x^2+a)^(1/4)*(d*x^2+c),x, algorithm="giac")
Output:
2/585*(65*(b*x^2 + a)^(9/4)*b*c - 117*(b*x^2 + a)^(5/4)*a*b*c + 45*(b*x^2 + a)^(13/4)*d - 130*(b*x^2 + a)^(9/4)*a*d + 117*(b*x^2 + a)^(5/4)*a^2*d)/b ^3
Time = 0.44 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int x^3 \sqrt [4]{a+b x^2} \left (c+d x^2\right ) \, dx={\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2\,d\,x^6}{13}+\frac {64\,a^3\,d-104\,a^2\,b\,c}{585\,b^3}+\frac {x^4\,\left (130\,c\,b^3+10\,a\,d\,b^2\right )}{585\,b^3}-\frac {2\,a\,x^2\,\left (8\,a\,d-13\,b\,c\right )}{585\,b^2}\right ) \] Input:
int(x^3*(a + b*x^2)^(1/4)*(c + d*x^2),x)
Output:
(a + b*x^2)^(1/4)*((2*d*x^6)/13 + (64*a^3*d - 104*a^2*b*c)/(585*b^3) + (x^ 4*(130*b^3*c + 10*a*b^2*d))/(585*b^3) - (2*a*x^2*(8*a*d - 13*b*c))/(585*b^ 2))
Time = 0.23 (sec) , antiderivative size = 237, normalized size of antiderivative = 3.25 \[ \int x^3 \sqrt [4]{a+b x^2} \left (c+d x^2\right ) \, dx=\frac {2 \sqrt {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +a +b \,x^{2}}\, \sqrt {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}\, \left (32 \sqrt {b \,x^{2}+a}\, a^{3} d -52 \sqrt {b \,x^{2}+a}\, a^{2} b c -8 \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{2}+13 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{2}+5 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{4}+65 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{4}+45 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{6}-32 \sqrt {b}\, a^{3} d x +52 \sqrt {b}\, a^{2} b c x +8 \sqrt {b}\, a^{2} b d \,x^{3}-13 \sqrt {b}\, a \,b^{2} c \,x^{3}-5 \sqrt {b}\, a \,b^{2} d \,x^{5}-65 \sqrt {b}\, b^{3} c \,x^{5}-45 \sqrt {b}\, b^{3} d \,x^{7}\right )}{585 a \,b^{3}} \] Input:
int(x^3*(b*x^2+a)^(1/4)*(d*x^2+c),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(a + b*x**2)*a**3*d - 52*sqrt(a + b*x**2)*a**2*b*c - 8*s qrt(a + b*x**2)*a**2*b*d*x**2 + 13*sqrt(a + b*x**2)*a*b**2*c*x**2 + 5*sqrt (a + b*x**2)*a*b**2*d*x**4 + 65*sqrt(a + b*x**2)*b**3*c*x**4 + 45*sqrt(a + b*x**2)*b**3*d*x**6 - 32*sqrt(b)*a**3*d*x + 52*sqrt(b)*a**2*b*c*x + 8*sqr t(b)*a**2*b*d*x**3 - 13*sqrt(b)*a*b**2*c*x**3 - 5*sqrt(b)*a*b**2*d*x**5 - 65*sqrt(b)*b**3*c*x**5 - 45*sqrt(b)*b**3*d*x**7))/(585*a*b**3)