Integrand size = 22, antiderivative size = 71 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=-\frac {a (b c-a d) \sqrt [4]{a+b x^4}}{b^3}+\frac {(b c-2 a d) \left (a+b x^4\right )^{5/4}}{5 b^3}+\frac {d \left (a+b x^4\right )^{9/4}}{9 b^3} \] Output:
-a*(-a*d+b*c)*(b*x^4+a)^(1/4)/b^3+1/5*(-2*a*d+b*c)*(b*x^4+a)^(5/4)/b^3+1/9 *d*(b*x^4+a)^(9/4)/b^3
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{a+b x^4} \left (-36 a b c+32 a^2 d+9 b^2 c x^4-8 a b d x^4+5 b^2 d x^8\right )}{45 b^3} \] Input:
Integrate[(x^7*(c + d*x^4))/(a + b*x^4)^(3/4),x]
Output:
((a + b*x^4)^(1/4)*(-36*a*b*c + 32*a^2*d + 9*b^2*c*x^4 - 8*a*b*d*x^4 + 5*b ^2*d*x^8))/(45*b^3)
Time = 0.36 (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 = {948, 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.
\(\displaystyle \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {x^4 \left (d x^4+c\right )}{\left (b x^4+a\right )^{3/4}}dx^4\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{4} \int \left (\frac {d \left (b x^4+a\right )^{5/4}}{b^2}+\frac {(b c-2 a d) \sqrt [4]{b x^4+a}}{b^2}+\frac {a (a d-b c)}{b^2 \left (b x^4+a\right )^{3/4}}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {4 \left (a+b x^4\right )^{5/4} (b c-2 a d)}{5 b^3}-\frac {4 a \sqrt [4]{a+b x^4} (b c-a d)}{b^3}+\frac {4 d \left (a+b x^4\right )^{9/4}}{9 b^3}\right )\) |
Input:
Int[(x^7*(c + d*x^4))/(a + b*x^4)^(3/4),x]
Output:
((-4*a*(b*c - a*d)*(a + b*x^4)^(1/4))/b^3 + (4*(b*c - 2*a*d)*(a + b*x^4)^( 5/4))/(5*b^3) + (4*d*(a + b*x^4)^(9/4))/(9*b^3))/4
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_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(\frac {32 \left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (\frac {9 x^{4} \left (\frac {5 d \,x^{4}}{9}+c \right ) b^{2}}{32}-\frac {9 \left (\frac {2 d \,x^{4}}{9}+c \right ) a b}{8}+a^{2} d \right )}{45 b^{3}}\) | \(49\) |
gosper | \(\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (5 d \,b^{2} x^{8}-8 a b d \,x^{4}+9 b^{2} c \,x^{4}+32 a^{2} d -36 a b c \right )}{45 b^{3}}\) | \(53\) |
trager | \(\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (5 d \,b^{2} x^{8}-8 a b d \,x^{4}+9 b^{2} c \,x^{4}+32 a^{2} d -36 a b c \right )}{45 b^{3}}\) | \(53\) |
risch | \(\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (5 d \,b^{2} x^{8}-8 a b d \,x^{4}+9 b^{2} c \,x^{4}+32 a^{2} d -36 a b c \right )}{45 b^{3}}\) | \(53\) |
orering | \(\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (5 d \,b^{2} x^{8}-8 a b d \,x^{4}+9 b^{2} c \,x^{4}+32 a^{2} d -36 a b c \right )}{45 b^{3}}\) | \(53\) |
Input:
int(x^7*(d*x^4+c)/(b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)
Output:
32/45*(b*x^4+a)^(1/4)*(9/32*x^4*(5/9*d*x^4+c)*b^2-9/8*(2/9*d*x^4+c)*a*b+a^ 2*d)/b^3
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.72 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {{\left (5 \, b^{2} d x^{8} + {\left (9 \, b^{2} c - 8 \, a b d\right )} x^{4} - 36 \, a b c + 32 \, a^{2} d\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{45 \, b^{3}} \] Input:
integrate(x^7*(d*x^4+c)/(b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
1/45*(5*b^2*d*x^8 + (9*b^2*c - 8*a*b*d)*x^4 - 36*a*b*c + 32*a^2*d)*(b*x^4 + a)^(1/4)/b^3
Time = 0.95 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.70 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\begin {cases} \frac {32 a^{2} d \sqrt [4]{a + b x^{4}}}{45 b^{3}} - \frac {4 a c \sqrt [4]{a + b x^{4}}}{5 b^{2}} - \frac {8 a d x^{4} \sqrt [4]{a + b x^{4}}}{45 b^{2}} + \frac {c x^{4} \sqrt [4]{a + b x^{4}}}{5 b} + \frac {d x^{8} \sqrt [4]{a + b x^{4}}}{9 b} & \text {for}\: b \neq 0 \\\frac {\frac {c x^{8}}{8} + \frac {d x^{12}}{12}}{a^{\frac {3}{4}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**7*(d*x**4+c)/(b*x**4+a)**(3/4),x)
Output:
Piecewise((32*a**2*d*(a + b*x**4)**(1/4)/(45*b**3) - 4*a*c*(a + b*x**4)**( 1/4)/(5*b**2) - 8*a*d*x**4*(a + b*x**4)**(1/4)/(45*b**2) + c*x**4*(a + b*x **4)**(1/4)/(5*b) + d*x**8*(a + b*x**4)**(1/4)/(9*b), Ne(b, 0)), ((c*x**8/ 8 + d*x**12/12)/a**(3/4), True))
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.17 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {1}{45} \, d {\left (\frac {5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}}}{b^{3}} - \frac {18 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a}{b^{3}} + \frac {45 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}}{b^{3}}\right )} + \frac {1}{5} \, c {\left (\frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{b^{2}} - \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{b^{2}}\right )} \] Input:
integrate(x^7*(d*x^4+c)/(b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
1/45*d*(5*(b*x^4 + a)^(9/4)/b^3 - 18*(b*x^4 + a)^(5/4)*a/b^3 + 45*(b*x^4 + a)^(1/4)*a^2/b^3) + 1/5*c*((b*x^4 + a)^(5/4)/b^2 - 5*(b*x^4 + a)^(1/4)*a/ b^2)
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (a b c - a^{2} d\right )}}{b^{3}} + \frac {9 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b c + 5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} d - 18 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a d}{45 \, b^{3}} \] Input:
integrate(x^7*(d*x^4+c)/(b*x^4+a)^(3/4),x, algorithm="giac")
Output:
-(b*x^4 + a)^(1/4)*(a*b*c - a^2*d)/b^3 + 1/45*(9*(b*x^4 + a)^(5/4)*b*c + 5 *(b*x^4 + a)^(9/4)*d - 18*(b*x^4 + a)^(5/4)*a*d)/b^3
Time = 3.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.80 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx={\left (b\,x^4+a\right )}^{1/4}\,\left (\frac {32\,a^2\,d-36\,a\,b\,c}{45\,b^3}+\frac {d\,x^8}{9\,b}+\frac {x^4\,\left (9\,b^2\,c-8\,a\,b\,d\right )}{45\,b^3}\right ) \] Input:
int((x^7*(c + d*x^4))/(a + b*x^4)^(3/4),x)
Output:
(a + b*x^4)^(1/4)*((32*a^2*d - 36*a*b*c)/(45*b^3) + (d*x^8)/(9*b) + (x^4*( 9*b^2*c - 8*a*b*d))/(45*b^3))
\[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\left (\int \frac {x^{11}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) d +\left (\int \frac {x^{7}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) c \] Input:
int(x^7*(d*x^4+c)/(b*x^4+a)^(3/4),x)
Output:
int(x**11/(a + b*x**4)**(3/4),x)*d + int(x**7/(a + b*x**4)**(3/4),x)*c