Integrand size = 22, antiderivative size = 183 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\frac {a (b c-a d) x}{10 b^3 \left (a+b x^4\right )^{5/2}}-\frac {(11 b c-21 a d) x}{60 b^3 \left (a+b x^4\right )^{3/2}}+\frac {(b c-15 a d) x}{24 a b^3 \sqrt {a+b x^4}}+\frac {(b c+9 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{48 a^{5/4} b^{13/4} \sqrt {a+b x^4}} \] Output:
1/10*a*(-a*d+b*c)*x/b^3/(b*x^4+a)^(5/2)-1/60*(-21*a*d+11*b*c)*x/b^3/(b*x^4 +a)^(3/2)+1/24*(-15*a*d+b*c)*x/a/b^3/(b*x^4+a)^(1/2)+1/48*(9*a*d+b*c)*(a^( 1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*InverseJacobiA M(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(5/4)/b^(13/4)/(b*x^4+a)^(1/2 )
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.10 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.70 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\frac {-x \left (45 a^3 d-5 b^3 c x^8+3 a b^2 x^4 \left (4 c+25 d x^4\right )+a^2 b \left (5 c+108 d x^4\right )\right )+5 (b c+9 a d) x \left (a+b x^4\right )^2 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{120 a b^3 \left (a+b x^4\right )^{5/2}} \] Input:
Integrate[(x^8*(c + d*x^4))/(a + b*x^4)^(7/2),x]
Output:
(-(x*(45*a^3*d - 5*b^3*c*x^8 + 3*a*b^2*x^4*(4*c + 25*d*x^4) + a^2*b*(5*c + 108*d*x^4))) + 5*(b*c + 9*a*d)*x*(a + b*x^4)^2*Sqrt[1 + (b*x^4)/a]*Hyperg eometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(120*a*b^3*(a + b*x^4)^(5/2))
Time = 0.48 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {957, 817, 817, 761}
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.
| \(\displaystyle \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle \frac {(9 a d+b c) \int \frac {x^8}{\left (b x^4+a\right )^{5/2}}dx}{10 a b}+\frac {x^9 (b c-a d)}{10 a b \left (a+b x^4\right )^{5/2}}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {(9 a d+b c) \left (\frac {5 \int \frac {x^4}{\left (b x^4+a\right )^{3/2}}dx}{6 b}-\frac {x^5}{6 b \left (a+b x^4\right )^{3/2}}\right )}{10 a b}+\frac {x^9 (b c-a d)}{10 a b \left (a+b x^4\right )^{5/2}}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {(9 a d+b c) \left (\frac {5 \left (\frac {\int \frac {1}{\sqrt {b x^4+a}}dx}{2 b}-\frac {x}{2 b \sqrt {a+b x^4}}\right )}{6 b}-\frac {x^5}{6 b \left (a+b x^4\right )^{3/2}}\right )}{10 a b}+\frac {x^9 (b c-a d)}{10 a b \left (a+b x^4\right )^{5/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {(9 a d+b c) \left (\frac {5 \left (\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{5/4} \sqrt {a+b x^4}}-\frac {x}{2 b \sqrt {a+b x^4}}\right )}{6 b}-\frac {x^5}{6 b \left (a+b x^4\right )^{3/2}}\right )}{10 a b}+\frac {x^9 (b c-a d)}{10 a b \left (a+b x^4\right )^{5/2}}\) |
Input:
Int[(x^8*(c + d*x^4))/(a + b*x^4)^(7/2),x]
Output:
((b*c - a*d)*x^9)/(10*a*b*(a + b*x^4)^(5/2)) + ((b*c + 9*a*d)*(-1/6*x^5/(b *(a + b*x^4)^(3/2)) + (5*(-1/2*x/(b*Sqrt[a + b*x^4]) + ((Sqrt[a] + Sqrt[b] *x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1 /4)*x)/a^(1/4)], 1/2])/(4*a^(1/4)*b^(5/4)*Sqrt[a + b*x^4])))/(6*b)))/(10*a *b)
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
Result contains complex when optimal does not.
Time = 2.94 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.07
| method | result | size |
| elliptic | \(-\frac {a x \left (a d -c b \right ) \sqrt {b \,x^{4}+a}}{10 b^{6} \left (x^{4}+\frac {a}{b}\right )^{3}}+\frac {x \left (21 a d -11 c b \right ) \sqrt {b \,x^{4}+a}}{60 b^{5} \left (x^{4}+\frac {a}{b}\right )^{2}}-\frac {x \left (15 a d -c b \right )}{24 b^{3} a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\left (\frac {d}{b^{3}}-\frac {15 a d -c b}{24 b^{3} a}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(195\) |
| default | \(c \left (\frac {a x \sqrt {b \,x^{4}+a}}{10 b^{5} \left (x^{4}+\frac {a}{b}\right )^{3}}-\frac {11 x \sqrt {b \,x^{4}+a}}{60 b^{4} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x}{24 b^{2} a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{24 b^{2} a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (-\frac {a^{2} x \sqrt {b \,x^{4}+a}}{10 b^{6} \left (x^{4}+\frac {a}{b}\right )^{3}}+\frac {7 a x \sqrt {b \,x^{4}+a}}{20 b^{5} \left (x^{4}+\frac {a}{b}\right )^{2}}-\frac {5 x}{8 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{8 b^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(307\) |
Input:
int(x^8*(d*x^4+c)/(b*x^4+a)^(7/2),x,method=_RETURNVERBOSE)
Output:
-1/10*a*x/b^6*(a*d-b*c)*(b*x^4+a)^(1/2)/(x^4+a/b)^3+1/60*x*(21*a*d-11*b*c) /b^5*(b*x^4+a)^(1/2)/(x^4+a/b)^2-1/24/b^3/a*x*(15*a*d-b*c)/((x^4+a/b)*b)^( 1/2)+(d/b^3-1/24/b^3/a*(15*a*d-b*c))/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2 )*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*Ellip ticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)
Time = 0.11 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.15 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=-\frac {5 \, {\left ({\left (b^{4} c + 9 \, a b^{3} d\right )} x^{12} + 3 \, {\left (a b^{3} c + 9 \, a^{2} b^{2} d\right )} x^{8} + a^{3} b c + 9 \, a^{4} d + 3 \, {\left (a^{2} b^{2} c + 9 \, a^{3} b d\right )} x^{4}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (5 \, {\left (b^{4} c - 15 \, a b^{3} d\right )} x^{9} - 12 \, {\left (a b^{3} c + 9 \, a^{2} b^{2} d\right )} x^{5} - 5 \, {\left (a^{2} b^{2} c + 9 \, a^{3} b d\right )} x\right )} \sqrt {b x^{4} + a}}{120 \, {\left (a b^{7} x^{12} + 3 \, a^{2} b^{6} x^{8} + 3 \, a^{3} b^{5} x^{4} + a^{4} b^{4}\right )}} \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(7/2),x, algorithm="fricas")
Output:
-1/120*(5*((b^4*c + 9*a*b^3*d)*x^12 + 3*(a*b^3*c + 9*a^2*b^2*d)*x^8 + a^3* b*c + 9*a^4*d + 3*(a^2*b^2*c + 9*a^3*b*d)*x^4)*sqrt(a)*(-b/a)^(3/4)*ellipt ic_f(arcsin(x*(-b/a)^(1/4)), -1) - (5*(b^4*c - 15*a*b^3*d)*x^9 - 12*(a*b^3 *c + 9*a^2*b^2*d)*x^5 - 5*(a^2*b^2*c + 9*a^3*b*d)*x)*sqrt(b*x^4 + a))/(a*b ^7*x^12 + 3*a^2*b^6*x^8 + 3*a^3*b^5*x^4 + a^4*b^4)
Result contains complex when optimal does not.
Time = 128.00 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.44 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\frac {c x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {7}{2} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {7}{2}} \Gamma \left (\frac {13}{4}\right )} + \frac {d x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {13}{4}, \frac {7}{2} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {7}{2}} \Gamma \left (\frac {17}{4}\right )} \] Input:
integrate(x**8*(d*x**4+c)/(b*x**4+a)**(7/2),x)
Output:
c*x**9*gamma(9/4)*hyper((9/4, 7/2), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4* a**(7/2)*gamma(13/4)) + d*x**13*gamma(13/4)*hyper((13/4, 7/2), (17/4,), b* x**4*exp_polar(I*pi)/a)/(4*a**(7/2)*gamma(17/4))
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(7/2),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)*x^8/(b*x^4 + a)^(7/2), x)
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(7/2),x, algorithm="giac")
Output:
integrate((d*x^4 + c)*x^8/(b*x^4 + a)^(7/2), x)
Timed out. \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\int \frac {x^8\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{7/2}} \,d x \] Input:
int((x^8*(c + d*x^4))/(a + b*x^4)^(7/2),x)
Output:
int((x^8*(c + d*x^4))/(a + b*x^4)^(7/2), x)
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\frac {-45 \sqrt {b \,x^{4}+a}\, a^{2} d x -5 \sqrt {b \,x^{4}+a}\, a b c x -81 \sqrt {b \,x^{4}+a}\, a b d \,x^{5}-9 \sqrt {b \,x^{4}+a}\, b^{2} c \,x^{5}-45 \sqrt {b \,x^{4}+a}\, b^{2} d \,x^{9}+45 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{6} d +5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{5} b c +135 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{5} b d \,x^{4}+15 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{4} b^{2} c \,x^{4}+135 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{4} b^{2} d \,x^{8}+15 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{3} b^{3} c \,x^{8}+45 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{3} b^{3} d \,x^{12}+5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{2} b^{4} c \,x^{12}}{45 b^{3} \left (b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}\right )} \] Input:
int(x^8*(d*x^4+c)/(b*x^4+a)^(7/2),x)
Output:
( - 45*sqrt(a + b*x**4)*a**2*d*x - 5*sqrt(a + b*x**4)*a*b*c*x - 81*sqrt(a + b*x**4)*a*b*d*x**5 - 9*sqrt(a + b*x**4)*b**2*c*x**5 - 45*sqrt(a + b*x**4 )*b**2*d*x**9 + 45*int(sqrt(a + b*x**4)/(a**4 + 4*a**3*b*x**4 + 6*a**2*b** 2*x**8 + 4*a*b**3*x**12 + b**4*x**16),x)*a**6*d + 5*int(sqrt(a + b*x**4)/( a**4 + 4*a**3*b*x**4 + 6*a**2*b**2*x**8 + 4*a*b**3*x**12 + b**4*x**16),x)* a**5*b*c + 135*int(sqrt(a + b*x**4)/(a**4 + 4*a**3*b*x**4 + 6*a**2*b**2*x* *8 + 4*a*b**3*x**12 + b**4*x**16),x)*a**5*b*d*x**4 + 15*int(sqrt(a + b*x** 4)/(a**4 + 4*a**3*b*x**4 + 6*a**2*b**2*x**8 + 4*a*b**3*x**12 + b**4*x**16) ,x)*a**4*b**2*c*x**4 + 135*int(sqrt(a + b*x**4)/(a**4 + 4*a**3*b*x**4 + 6* a**2*b**2*x**8 + 4*a*b**3*x**12 + b**4*x**16),x)*a**4*b**2*d*x**8 + 15*int (sqrt(a + b*x**4)/(a**4 + 4*a**3*b*x**4 + 6*a**2*b**2*x**8 + 4*a*b**3*x**1 2 + b**4*x**16),x)*a**3*b**3*c*x**8 + 45*int(sqrt(a + b*x**4)/(a**4 + 4*a* *3*b*x**4 + 6*a**2*b**2*x**8 + 4*a*b**3*x**12 + b**4*x**16),x)*a**3*b**3*d *x**12 + 5*int(sqrt(a + b*x**4)/(a**4 + 4*a**3*b*x**4 + 6*a**2*b**2*x**8 + 4*a*b**3*x**12 + b**4*x**16),x)*a**2*b**4*c*x**12)/(45*b**3*(a**3 + 3*a** 2*b*x**4 + 3*a*b**2*x**8 + b**3*x**12))