Integrand size = 22, antiderivative size = 73 \[ \int \frac {x^7 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=-\frac {a (b c-a d) \left (a+b x^4\right )^{3/4}}{3 b^3}+\frac {(b c-2 a d) \left (a+b x^4\right )^{7/4}}{7 b^3}+\frac {d \left (a+b x^4\right )^{11/4}}{11 b^3} \] Output:
-1/3*a*(-a*d+b*c)*(b*x^4+a)^(3/4)/b^3+1/7*(-2*a*d+b*c)*(b*x^4+a)^(7/4)/b^3 +1/11*d*(b*x^4+a)^(11/4)/b^3
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int \frac {x^7 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {\left (a+b x^4\right )^{3/4} \left (-44 a b c+32 a^2 d+33 b^2 c x^4-24 a b d x^4+21 b^2 d x^8\right )}{231 b^3} \] Input:
Integrate[(x^7*(c + d*x^4))/(a + b*x^4)^(1/4),x]
Output:
((a + b*x^4)^(3/4)*(-44*a*b*c + 32*a^2*d + 33*b^2*c*x^4 - 24*a*b*d*x^4 + 2 1*b^2*d*x^8))/(231*b^3)
Time = 0.37 (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 = {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.
\(\displaystyle \int \frac {x^7 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {x^4 \left (d x^4+c\right )}{\sqrt [4]{b x^4+a}}dx^4\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{4} \int \left (\frac {d \left (b x^4+a\right )^{7/4}}{b^2}+\frac {(b c-2 a d) \left (b x^4+a\right )^{3/4}}{b^2}+\frac {a (a d-b c)}{b^2 \sqrt [4]{b x^4+a}}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {4 \left (a+b x^4\right )^{7/4} (b c-2 a d)}{7 b^3}-\frac {4 a \left (a+b x^4\right )^{3/4} (b c-a d)}{3 b^3}+\frac {4 d \left (a+b x^4\right )^{11/4}}{11 b^3}\right )\) |
Input:
Int[(x^7*(c + d*x^4))/(a + b*x^4)^(1/4),x]
Output:
((-4*a*(b*c - a*d)*(a + b*x^4)^(3/4))/(3*b^3) + (4*(b*c - 2*a*d)*(a + b*x^ 4)^(7/4))/(7*b^3) + (4*d*(a + b*x^4)^(11/4))/(11*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 = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(\frac {32 \left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (\frac {33 \left (\frac {7 d \,x^{4}}{11}+c \right ) x^{4} b^{2}}{32}-\frac {11 \left (\frac {6 d \,x^{4}}{11}+c \right ) a b}{8}+a^{2} d \right )}{231 b^{3}}\) | \(49\) |
gosper | \(\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (21 d \,b^{2} x^{8}-24 a b d \,x^{4}+33 b^{2} c \,x^{4}+32 a^{2} d -44 a b c \right )}{231 b^{3}}\) | \(53\) |
trager | \(\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (21 d \,b^{2} x^{8}-24 a b d \,x^{4}+33 b^{2} c \,x^{4}+32 a^{2} d -44 a b c \right )}{231 b^{3}}\) | \(53\) |
risch | \(\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (21 d \,b^{2} x^{8}-24 a b d \,x^{4}+33 b^{2} c \,x^{4}+32 a^{2} d -44 a b c \right )}{231 b^{3}}\) | \(53\) |
orering | \(\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (21 d \,b^{2} x^{8}-24 a b d \,x^{4}+33 b^{2} c \,x^{4}+32 a^{2} d -44 a b c \right )}{231 b^{3}}\) | \(53\) |
Input:
int(x^7*(d*x^4+c)/(b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
32/231*(b*x^4+a)^(3/4)*(33/32*(7/11*d*x^4+c)*x^4*b^2-11/8*(6/11*d*x^4+c)*a *b+a^2*d)/b^3
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71 \[ \int \frac {x^7 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {{\left (21 \, b^{2} d x^{8} + 3 \, {\left (11 \, b^{2} c - 8 \, a b d\right )} x^{4} - 44 \, a b c + 32 \, a^{2} d\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{231 \, b^{3}} \] Input:
integrate(x^7*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
1/231*(21*b^2*d*x^8 + 3*(11*b^2*c - 8*a*b*d)*x^4 - 44*a*b*c + 32*a^2*d)*(b *x^4 + a)^(3/4)/b^3
Time = 0.54 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.66 \[ \int \frac {x^7 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\begin {cases} \frac {32 a^{2} d \left (a + b x^{4}\right )^{\frac {3}{4}}}{231 b^{3}} - \frac {4 a c \left (a + b x^{4}\right )^{\frac {3}{4}}}{21 b^{2}} - \frac {8 a d x^{4} \left (a + b x^{4}\right )^{\frac {3}{4}}}{77 b^{2}} + \frac {c x^{4} \left (a + b x^{4}\right )^{\frac {3}{4}}}{7 b} + \frac {d x^{8} \left (a + b x^{4}\right )^{\frac {3}{4}}}{11 b} & \text {for}\: b \neq 0 \\\frac {\frac {c x^{8}}{8} + \frac {d x^{12}}{12}}{\sqrt [4]{a}} & \text {otherwise} \end {cases} \] Input:
integrate(x**7*(d*x**4+c)/(b*x**4+a)**(1/4),x)
Output:
Piecewise((32*a**2*d*(a + b*x**4)**(3/4)/(231*b**3) - 4*a*c*(a + b*x**4)** (3/4)/(21*b**2) - 8*a*d*x**4*(a + b*x**4)**(3/4)/(77*b**2) + c*x**4*(a + b *x**4)**(3/4)/(7*b) + d*x**8*(a + b*x**4)**(3/4)/(11*b), Ne(b, 0)), ((c*x* *8/8 + d*x**12/12)/a**(1/4), True))
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int \frac {x^7 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {1}{231} \, d {\left (\frac {21 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}}}{b^{3}} - \frac {66 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a}{b^{3}} + \frac {77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2}}{b^{3}}\right )} + \frac {1}{21} \, c {\left (\frac {3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}}}{b^{2}} - \frac {7 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a}{b^{2}}\right )} \] Input:
integrate(x^7*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
1/231*d*(21*(b*x^4 + a)^(11/4)/b^3 - 66*(b*x^4 + a)^(7/4)*a/b^3 + 77*(b*x^ 4 + a)^(3/4)*a^2/b^3) + 1/21*c*(3*(b*x^4 + a)^(7/4)/b^2 - 7*(b*x^4 + a)^(3 /4)*a/b^2)
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08 \[ \int \frac {x^7 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {\frac {11 \, {\left (3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} - 7 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a\right )} c}{b} + \frac {{\left (21 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} - 66 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a + 77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2}\right )} d}{b^{2}}}{231 \, b} \] Input:
integrate(x^7*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
1/231*(11*(3*(b*x^4 + a)^(7/4) - 7*(b*x^4 + a)^(3/4)*a)*c/b + (21*(b*x^4 + a)^(11/4) - 66*(b*x^4 + a)^(7/4)*a + 77*(b*x^4 + a)^(3/4)*a^2)*d/b^2)/b
Time = 3.52 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78 \[ \int \frac {x^7 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx={\left (b\,x^4+a\right )}^{3/4}\,\left (\frac {32\,a^2\,d-44\,a\,b\,c}{231\,b^3}+\frac {d\,x^8}{11\,b}+\frac {x^4\,\left (33\,b^2\,c-24\,a\,b\,d\right )}{231\,b^3}\right ) \] Input:
int((x^7*(c + d*x^4))/(a + b*x^4)^(1/4),x)
Output:
(a + b*x^4)^(3/4)*((32*a^2*d - 44*a*b*c)/(231*b^3) + (d*x^8)/(11*b) + (x^4 *(33*b^2*c - 24*a*b*d))/(231*b^3))
\[ \int \frac {x^7 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\left (\int \frac {x^{11}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) d +\left (\int \frac {x^{7}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) c \] Input:
int(x^7*(d*x^4+c)/(b*x^4+a)^(1/4),x)
Output:
int(x**11/(a + b*x**4)**(1/4),x)*d + int(x**7/(a + b*x**4)**(1/4),x)*c