Integrand size = 22, antiderivative size = 73 \[ \int x^3 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=-\frac {2 a (b c-a d) \left (a+b x^2\right )^{7/4}}{7 b^3}+\frac {2 (b c-2 a d) \left (a+b x^2\right )^{11/4}}{11 b^3}+\frac {2 d \left (a+b x^2\right )^{15/4}}{15 b^3} \] Output:
-2/7*a*(-a*d+b*c)*(b*x^2+a)^(7/4)/b^3+2/11*(-2*a*d+b*c)*(b*x^2+a)^(11/4)/b ^3+2/15*d*(b*x^2+a)^(15/4)/b^3
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int x^3 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {2 \left (a+b x^2\right )^{7/4} \left (-60 a b c+32 a^2 d+105 b^2 c x^2-56 a b d x^2+77 b^2 d x^4\right )}{1155 b^3} \] Input:
Integrate[x^3*(a + b*x^2)^(3/4)*(c + d*x^2),x]
Output:
(2*(a + b*x^2)^(7/4)*(-60*a*b*c + 32*a^2*d + 105*b^2*c*x^2 - 56*a*b*d*x^2 + 77*b^2*d*x^4))/(1155*b^3)
Time = 0.21 (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.
\(\displaystyle \int x^3 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int x^2 \left (b x^2+a\right )^{3/4} \left (d x^2+c\right )dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {d \left (b x^2+a\right )^{11/4}}{b^2}+\frac {(b c-2 a d) \left (b x^2+a\right )^{7/4}}{b^2}+\frac {a (a d-b c) \left (b x^2+a\right )^{3/4}}{b^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {4 \left (a+b x^2\right )^{11/4} (b c-2 a d)}{11 b^3}-\frac {4 a \left (a+b x^2\right )^{7/4} (b c-a d)}{7 b^3}+\frac {4 d \left (a+b x^2\right )^{15/4}}{15 b^3}\right )\) |
Input:
Int[x^3*(a + b*x^2)^(3/4)*(c + d*x^2),x]
Output:
((-4*a*(b*c - a*d)*(a + b*x^2)^(7/4))/(7*b^3) + (4*(b*c - 2*a*d)*(a + b*x^ 2)^(11/4))/(11*b^3) + (4*d*(a + b*x^2)^(15/4))/(15*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 = 0.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(\frac {64 \left (\frac {105 x^{2} \left (\frac {11 x^{2} d}{15}+c \right ) b^{2}}{32}-\frac {15 a \left (\frac {14 x^{2} d}{15}+c \right ) b}{8}+d \,a^{2}\right ) \left (b \,x^{2}+a \right )^{\frac {7}{4}}}{1155 b^{3}}\) | \(49\) |
gosper | \(\frac {2 \left (b \,x^{2}+a \right )^{\frac {7}{4}} \left (77 d \,b^{2} x^{4}-56 a b d \,x^{2}+105 b^{2} c \,x^{2}+32 d \,a^{2}-60 a b c \right )}{1155 b^{3}}\) | \(53\) |
orering | \(\frac {2 \left (b \,x^{2}+a \right )^{\frac {7}{4}} \left (77 d \,b^{2} x^{4}-56 a b d \,x^{2}+105 b^{2} c \,x^{2}+32 d \,a^{2}-60 a b c \right )}{1155 b^{3}}\) | \(53\) |
trager | \(\frac {2 \left (77 d \,x^{6} b^{3}+21 a \,b^{2} d \,x^{4}+105 b^{3} c \,x^{4}-24 a^{2} b d \,x^{2}+45 a \,b^{2} c \,x^{2}+32 a^{3} d -60 a^{2} b c \right ) \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{1155 b^{3}}\) | \(77\) |
risch | \(\frac {2 \left (77 d \,x^{6} b^{3}+21 a \,b^{2} d \,x^{4}+105 b^{3} c \,x^{4}-24 a^{2} b d \,x^{2}+45 a \,b^{2} c \,x^{2}+32 a^{3} d -60 a^{2} b c \right ) \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{1155 b^{3}}\) | \(77\) |
Input:
int(x^3*(b*x^2+a)^(3/4)*(d*x^2+c),x,method=_RETURNVERBOSE)
Output:
64/1155*(105/32*x^2*(11/15*x^2*d+c)*b^2-15/8*a*(14/15*x^2*d+c)*b+d*a^2)*(b *x^2+a)^(7/4)/b^3
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int x^3 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {2 \, {\left (77 \, b^{3} d x^{6} + 21 \, {\left (5 \, b^{3} c + a b^{2} d\right )} x^{4} - 60 \, a^{2} b c + 32 \, a^{3} d + 3 \, {\left (15 \, a b^{2} c - 8 \, a^{2} b d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}}}{1155 \, b^{3}} \] Input:
integrate(x^3*(b*x^2+a)^(3/4)*(d*x^2+c),x, algorithm="fricas")
Output:
2/1155*(77*b^3*d*x^6 + 21*(5*b^3*c + a*b^2*d)*x^4 - 60*a^2*b*c + 32*a^3*d + 3*(15*a*b^2*c - 8*a^2*b*d)*x^2)*(b*x^2 + a)^(3/4)/b^3
Time = 0.79 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.47 \[ \int x^3 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {c \left (\begin {cases} \frac {4 \left (- \frac {a \left (a + b x^{2}\right )^{\frac {7}{4}}}{7} + \frac {\left (a + b x^{2}\right )^{\frac {11}{4}}}{11}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {a^{\frac {3}{4}} x^{4}}{2} & \text {otherwise} \end {cases}\right )}{2} + \frac {d \left (\begin {cases} \frac {4 \left (\frac {a^{2} \left (a + b x^{2}\right )^{\frac {7}{4}}}{7} - \frac {2 a \left (a + b x^{2}\right )^{\frac {11}{4}}}{11} + \frac {\left (a + b x^{2}\right )^{\frac {15}{4}}}{15}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\frac {a^{\frac {3}{4}} x^{6}}{3} & \text {otherwise} \end {cases}\right )}{2} \] Input:
integrate(x**3*(b*x**2+a)**(3/4)*(d*x**2+c),x)
Output:
c*Piecewise((4*(-a*(a + b*x**2)**(7/4)/7 + (a + b*x**2)**(11/4)/11)/b**2, Ne(b, 0)), (a**(3/4)*x**4/2, True))/2 + d*Piecewise((4*(a**2*(a + b*x**2)* *(7/4)/7 - 2*a*(a + b*x**2)**(11/4)/11 + (a + b*x**2)**(15/4)/15)/b**3, Ne (b, 0)), (a**(3/4)*x**6/3, True))/2
Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int x^3 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {2}{1155} \, d {\left (\frac {77 \, {\left (b x^{2} + a\right )}^{\frac {15}{4}}}{b^{3}} - \frac {210 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} a}{b^{3}} + \frac {165 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a^{2}}{b^{3}}\right )} + \frac {2}{77} \, c {\left (\frac {7 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}}}{b^{2}} - \frac {11 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a}{b^{2}}\right )} \] Input:
integrate(x^3*(b*x^2+a)^(3/4)*(d*x^2+c),x, algorithm="maxima")
Output:
2/1155*d*(77*(b*x^2 + a)^(15/4)/b^3 - 210*(b*x^2 + a)^(11/4)*a/b^3 + 165*( b*x^2 + a)^(7/4)*a^2/b^3) + 2/77*c*(7*(b*x^2 + a)^(11/4)/b^2 - 11*(b*x^2 + a)^(7/4)*a/b^2)
Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (61) = 122\).
Time = 0.12 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.51 \[ \int x^3 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {2 \, {\left (\frac {55 \, {\left (3 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} - 7 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a\right )} a c}{b} + \frac {5 \, {\left (21 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} - 66 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a + 77 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{2}\right )} c}{b} + \frac {5 \, {\left (21 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} - 66 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a + 77 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{2}\right )} a d}{b^{2}} + \frac {{\left (77 \, {\left (b x^{2} + a\right )}^{\frac {15}{4}} - 315 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} a + 495 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a^{2} - 385 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{3}\right )} d}{b^{2}}\right )}}{1155 \, b} \] Input:
integrate(x^3*(b*x^2+a)^(3/4)*(d*x^2+c),x, algorithm="giac")
Output:
2/1155*(55*(3*(b*x^2 + a)^(7/4) - 7*(b*x^2 + a)^(3/4)*a)*a*c/b + 5*(21*(b* x^2 + a)^(11/4) - 66*(b*x^2 + a)^(7/4)*a + 77*(b*x^2 + a)^(3/4)*a^2)*c/b + 5*(21*(b*x^2 + a)^(11/4) - 66*(b*x^2 + a)^(7/4)*a + 77*(b*x^2 + a)^(3/4)* a^2)*a*d/b^2 + (77*(b*x^2 + a)^(15/4) - 315*(b*x^2 + a)^(11/4)*a + 495*(b* x^2 + a)^(7/4)*a^2 - 385*(b*x^2 + a)^(3/4)*a^3)*d/b^2)/b
Time = 0.44 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int x^3 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx={\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {2\,d\,x^6}{15}+\frac {64\,a^3\,d-120\,a^2\,b\,c}{1155\,b^3}+\frac {x^4\,\left (210\,c\,b^3+42\,a\,d\,b^2\right )}{1155\,b^3}-\frac {2\,a\,x^2\,\left (8\,a\,d-15\,b\,c\right )}{385\,b^2}\right ) \] Input:
int(x^3*(a + b*x^2)^(3/4)*(c + d*x^2),x)
Output:
(a + b*x^2)^(3/4)*((2*d*x^6)/15 + (64*a^3*d - 120*a^2*b*c)/(1155*b^3) + (x ^4*(210*b^3*c + 42*a*b^2*d))/(1155*b^3) - (2*a*x^2*(8*a*d - 15*b*c))/(385* b^2))
Time = 0.23 (sec) , antiderivative size = 261, normalized size of antiderivative = 3.58 \[ \int x^3 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) \, dx=\frac {2 \sqrt {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +a +b \,x^{2}}\, \sqrt {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}\, \left (-32 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a^{3} d x +60 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a^{2} b c x +24 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{3}-45 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{3}-21 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{5}-105 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{5}-77 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{7}+32 a^{4} d -60 a^{3} b c +8 a^{3} b d \,x^{2}-15 a^{2} b^{2} c \,x^{2}-3 a^{2} b^{2} d \,x^{4}+150 a \,b^{3} c \,x^{4}+98 a \,b^{3} d \,x^{6}+105 b^{4} c \,x^{6}+77 b^{4} d \,x^{8}\right )}{1155 a \,b^{3}} \] Input:
int(x^3*(b*x^2+a)^(3/4)*(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(b)*sqrt(a + b*x**2)*a**3*d*x + 60*sqrt(b)*sqrt(a + b *x**2)*a**2*b*c*x + 24*sqrt(b)*sqrt(a + b*x**2)*a**2*b*d*x**3 - 45*sqrt(b) *sqrt(a + b*x**2)*a*b**2*c*x**3 - 21*sqrt(b)*sqrt(a + b*x**2)*a*b**2*d*x** 5 - 105*sqrt(b)*sqrt(a + b*x**2)*b**3*c*x**5 - 77*sqrt(b)*sqrt(a + b*x**2) *b**3*d*x**7 + 32*a**4*d - 60*a**3*b*c + 8*a**3*b*d*x**2 - 15*a**2*b**2*c* x**2 - 3*a**2*b**2*d*x**4 + 150*a*b**3*c*x**4 + 98*a*b**3*d*x**6 + 105*b** 4*c*x**6 + 77*b**4*d*x**8))/(1155*a*b**3)