Integrand size = 22, antiderivative size = 70 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {a (b c-a d)}{3 b^3 \left (a+b x^4\right )^{3/4}}+\frac {(b c-2 a d) \sqrt [4]{a+b x^4}}{b^3}+\frac {d \left (a+b x^4\right )^{5/4}}{5 b^3} \] Output:
1/3*a*(-a*d+b*c)/b^3/(b*x^4+a)^(3/4)+(-2*a*d+b*c)*(b*x^4+a)^(1/4)/b^3+1/5* d*(b*x^4+a)^(5/4)/b^3
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.80 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {20 a b c-32 a^2 d+15 b^2 c x^4-24 a b d x^4+3 b^2 d x^8}{15 b^3 \left (a+b x^4\right )^{3/4}} \] Input:
Integrate[(x^7*(c + d*x^4))/(a + b*x^4)^(7/4),x]
Output:
(20*a*b*c - 32*a^2*d + 15*b^2*c*x^4 - 24*a*b*d*x^4 + 3*b^2*d*x^8)/(15*b^3* (a + b*x^4)^(3/4))
Time = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.07, 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.
\(\displaystyle \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {x^4 \left (d x^4+c\right )}{\left (b x^4+a\right )^{7/4}}dx^4\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{4} \int \left (\frac {\sqrt [4]{b x^4+a} d}{b^2}+\frac {b c-2 a d}{b^2 \left (b x^4+a\right )^{3/4}}+\frac {a (a d-b c)}{b^2 \left (b x^4+a\right )^{7/4}}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {4 \sqrt [4]{a+b x^4} (b c-2 a d)}{b^3}+\frac {4 a (b c-a d)}{3 b^3 \left (a+b x^4\right )^{3/4}}+\frac {4 d \left (a+b x^4\right )^{5/4}}{5 b^3}\right )\) |
Input:
Int[(x^7*(c + d*x^4))/(a + b*x^4)^(7/4),x]
Output:
((4*a*(b*c - a*d))/(3*b^3*(a + b*x^4)^(3/4)) + (4*(b*c - 2*a*d)*(a + b*x^4 )^(1/4))/b^3 + (4*d*(a + b*x^4)^(5/4))/(5*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 = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.70
method | result | size |
pseudoelliptic | \(-\frac {32 \left (-\frac {15 \left (\frac {d \,x^{4}}{5}+c \right ) x^{4} b^{2}}{32}-\frac {5 \left (-\frac {6 d \,x^{4}}{5}+c \right ) a b}{8}+a^{2} d \right )}{15 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{3}}\) | \(49\) |
gosper | \(-\frac {-3 d \,b^{2} x^{8}+24 a b d \,x^{4}-15 b^{2} c \,x^{4}+32 a^{2} d -20 a b c}{15 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{3}}\) | \(53\) |
trager | \(-\frac {-3 d \,b^{2} x^{8}+24 a b d \,x^{4}-15 b^{2} c \,x^{4}+32 a^{2} d -20 a b c}{15 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{3}}\) | \(53\) |
orering | \(-\frac {-3 d \,b^{2} x^{8}+24 a b d \,x^{4}-15 b^{2} c \,x^{4}+32 a^{2} d -20 a b c}{15 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{3}}\) | \(53\) |
risch | \(-\frac {\left (-d b \,x^{4}+9 a d -5 c b \right ) \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{5 b^{3}}-\frac {a \left (a d -c b \right )}{3 b^{3} \left (b \,x^{4}+a \right )^{\frac {3}{4}}}\) | \(55\) |
Input:
int(x^7*(d*x^4+c)/(b*x^4+a)^(7/4),x,method=_RETURNVERBOSE)
Output:
-32/15/(b*x^4+a)^(3/4)*(-15/32*(1/5*d*x^4+c)*x^4*b^2-5/8*(-6/5*d*x^4+c)*a* b+a^2*d)/b^3
Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.91 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {{\left (3 \, b^{2} d x^{8} + 3 \, {\left (5 \, b^{2} c - 8 \, a b d\right )} x^{4} + 20 \, a b c - 32 \, a^{2} d\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{15 \, {\left (b^{4} x^{4} + a b^{3}\right )}} \] Input:
integrate(x^7*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="fricas")
Output:
1/15*(3*b^2*d*x^8 + 3*(5*b^2*c - 8*a*b*d)*x^4 + 20*a*b*c - 32*a^2*d)*(b*x^ 4 + a)^(1/4)/(b^4*x^4 + a*b^3)
Time = 1.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.70 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\begin {cases} - \frac {32 a^{2} d}{15 b^{3} \left (a + b x^{4}\right )^{\frac {3}{4}}} + \frac {4 a c}{3 b^{2} \left (a + b x^{4}\right )^{\frac {3}{4}}} - \frac {8 a d x^{4}}{5 b^{2} \left (a + b x^{4}\right )^{\frac {3}{4}}} + \frac {c x^{4}}{b \left (a + b x^{4}\right )^{\frac {3}{4}}} + \frac {d x^{8}}{5 b \left (a + b x^{4}\right )^{\frac {3}{4}}} & \text {for}\: b \neq 0 \\\frac {\frac {c x^{8}}{8} + \frac {d x^{12}}{12}}{a^{\frac {7}{4}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**7*(d*x**4+c)/(b*x**4+a)**(7/4),x)
Output:
Piecewise((-32*a**2*d/(15*b**3*(a + b*x**4)**(3/4)) + 4*a*c/(3*b**2*(a + b *x**4)**(3/4)) - 8*a*d*x**4/(5*b**2*(a + b*x**4)**(3/4)) + c*x**4/(b*(a + b*x**4)**(3/4)) + d*x**8/(5*b*(a + b*x**4)**(3/4)), Ne(b, 0)), ((c*x**8/8 + d*x**12/12)/a**(7/4), True))
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.19 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {1}{15} \, d {\left (\frac {3 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}}}{b^{3}} - \frac {30 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{b^{3}} - \frac {5 \, a^{2}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} b^{3}}\right )} + \frac {1}{3} \, c {\left (\frac {3 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{2}} + \frac {a}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} b^{2}}\right )} \] Input:
integrate(x^7*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="maxima")
Output:
1/15*d*(3*(b*x^4 + a)^(5/4)/b^3 - 30*(b*x^4 + a)^(1/4)*a/b^3 - 5*a^2/((b*x ^4 + a)^(3/4)*b^3)) + 1/3*c*(3*(b*x^4 + a)^(1/4)/b^2 + a/((b*x^4 + a)^(3/4 )*b^2))
Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.10 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {a b c - a^{2} d}{3 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b^{3}} + \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{13} c + {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{12} d - 10 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a b^{12} d}{5 \, b^{15}} \] Input:
integrate(x^7*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="giac")
Output:
1/3*(a*b*c - a^2*d)/((b*x^4 + a)^(3/4)*b^3) + 1/5*(5*(b*x^4 + a)^(1/4)*b^1 3*c + (b*x^4 + a)^(5/4)*b^12*d - 10*(b*x^4 + a)^(1/4)*a*b^12*d)/b^15
Time = 3.57 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.86 \[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {3\,d\,{\left (b\,x^4+a\right )}^2-5\,a^2\,d-30\,a\,d\,\left (b\,x^4+a\right )+15\,b\,c\,\left (b\,x^4+a\right )+5\,a\,b\,c}{15\,b^3\,{\left (b\,x^4+a\right )}^{3/4}} \] Input:
int((x^7*(c + d*x^4))/(a + b*x^4)^(7/4),x)
Output:
(3*d*(a + b*x^4)^2 - 5*a^2*d - 30*a*d*(a + b*x^4) + 15*b*c*(a + b*x^4) + 5 *a*b*c)/(15*b^3*(a + b*x^4)^(3/4))
\[ \int \frac {x^7 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\left (\int \frac {x^{11}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {x^{7}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) c \] Input:
int(x^7*(d*x^4+c)/(b*x^4+a)^(7/4),x)
Output:
int(x**11/((a + b*x**4)**(3/4)*a + (a + b*x**4)**(3/4)*b*x**4),x)*d + int( x**7/((a + b*x**4)**(3/4)*a + (a + b*x**4)**(3/4)*b*x**4),x)*c