Integrand size = 21, antiderivative size = 166 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {(b c-a d)^2 x}{2 a b^2 \sqrt {a+b x^4}}+\frac {d^2 x \sqrt {a+b x^4}}{3 b^2}+\frac {\left (3 b^2 c^2+6 a b c d-5 a^2 d^2\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{12 a^{5/4} b^{9/4} \sqrt {a+b x^4}} \] Output:
1/2*(-a*d+b*c)^2*x/a/b^2/(b*x^4+a)^(1/2)+1/3*d^2*x*(b*x^4+a)^(1/2)/b^2+1/1 2*(-5*a^2*d^2+6*a*b*c*d+3*b^2*c^2)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/ 2)+b^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2 ^(1/2))/a^(5/4)/b^(9/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 14.33 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.97 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x \sqrt {1+\frac {b x^4}{a}} \left (13 a \left (45 c^2+18 c d x^4+5 d^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{2},\frac {13}{4},-\frac {b x^4}{a}\right )-12 b x^4 \left (7 c^2+10 c d x^4+3 d^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {5}{2},\frac {17}{4},-\frac {b x^4}{a}\right )-24 b x^4 \left (c+d x^4\right )^2 \, _3F_2\left (\frac {5}{4},2,\frac {5}{2};1,\frac {17}{4};-\frac {b x^4}{a}\right )\right )}{585 a^2 \sqrt {a+b x^4}} \] Input:
Integrate[(c + d*x^4)^2/(a + b*x^4)^(3/2),x]
Output:
(x*Sqrt[1 + (b*x^4)/a]*(13*a*(45*c^2 + 18*c*d*x^4 + 5*d^2*x^8)*Hypergeomet ric2F1[1/4, 3/2, 13/4, -((b*x^4)/a)] - 12*b*x^4*(7*c^2 + 10*c*d*x^4 + 3*d^ 2*x^8)*Hypergeometric2F1[5/4, 5/2, 17/4, -((b*x^4)/a)] - 24*b*x^4*(c + d*x ^4)^2*HypergeometricPFQ[{5/4, 2, 5/2}, {1, 17/4}, -((b*x^4)/a)]))/(585*a^2 *Sqrt[a + b*x^4])
Time = 0.47 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {930, 913, 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.
| \(\displaystyle \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 930 |
| \(\displaystyle \frac {\int \frac {c (b c+a d)-d (3 b c-5 a d) x^4}{\sqrt {b x^4+a}}dx}{2 a b}+\frac {x \left (c+d x^4\right ) (b c-a d)}{2 a b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 913 |
| \(\displaystyle \frac {\frac {\left (-5 a^2 d^2+6 a b c d+3 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^4+a}}dx}{3 b}-\frac {d x \sqrt {a+b x^4} (3 b c-5 a d)}{3 b}}{2 a b}+\frac {x \left (c+d x^4\right ) (b c-a d)}{2 a b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (-5 a^2 d^2+6 a b c d+3 b^2 c^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 \sqrt [4]{a} b^{5/4} \sqrt {a+b x^4}}-\frac {d x \sqrt {a+b x^4} (3 b c-5 a d)}{3 b}}{2 a b}+\frac {x \left (c+d x^4\right ) (b c-a d)}{2 a b \sqrt {a+b x^4}}\) |
Input:
Int[(c + d*x^4)^2/(a + b*x^4)^(3/2),x]
Output:
((b*c - a*d)*x*(c + d*x^4))/(2*a*b*Sqrt[a + b*x^4]) + (-1/3*(d*(3*b*c - 5* a*d)*x*Sqrt[a + b*x^4])/b + ((3*b^2*c^2 + 6*a*b*c*d - 5*a^2*d^2)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*Arc Tan[(b^(1/4)*x)/a^(1/4)], 1/2])/(6*a^(1/4)*b^(5/4)*Sqrt[a + b*x^4]))/(2*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[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( p + 1) + 1))/(b*(n*(p + 1) + 1)) Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b , c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a*d - c*b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 1))), x] - Simp[1/(a*b*n*(p + 1)) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*( p + q) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q, x]
Result contains complex when optimal does not.
Time = 5.88 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.11
| method | result | size |
| elliptic | \(\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x}{2 b^{2} a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {d^{2} x \sqrt {b \,x^{4}+a}}{3 b^{2}}+\frac {\left (-\frac {d \left (a d -2 b c \right )}{b^{2}}+\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{2 a \,b^{2}}-\frac {a \,d^{2}}{3 b^{2}}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(185\) |
| default | \(c^{2} \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d^{2} \left (\frac {x a}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {5 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+2 c d \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(310\) |
| risch | \(\frac {d^{2} x \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {a^{2} d^{2} \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+2 b d \left (2 a d -3 b c \right ) \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-3 b^{2} c^{2} \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )}{3 b^{2}}\) | \(333\) |
Input:
int((d*x^4+c)^2/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2/b^2*(a^2*d^2-2*a*b*c*d+b^2*c^2)/a*x/((x^4+a/b)*b)^(1/2)+1/3*d^2*x*(b*x ^4+a)^(1/2)/b^2+(-d*(a*d-2*b*c)/b^2+1/2*(a^2*d^2-2*a*b*c*d+b^2*c^2)/a/b^2- 1/3*a/b^2*d^2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I*b^(1/2)*x^2/a^(1/2))^(1/2)*( 1+I*b^(1/2)*x^2/a^(1/2))^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1 /2))^(1/2),I)
Time = 0.11 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.95 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {{\left (3 \, a b^{2} c^{2} + 6 \, a^{2} b c d - 5 \, a^{3} d^{2} + {\left (3 \, b^{3} c^{2} + 6 \, a b^{2} c d - 5 \, a^{2} b d^{2}\right )} x^{4}\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (2 \, a^{2} b d^{2} x^{5} + {\left (3 \, a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2}\right )} x\right )} \sqrt {b x^{4} + a}}{6 \, {\left (a^{2} b^{3} x^{4} + a^{3} b^{2}\right )}} \] Input:
integrate((d*x^4+c)^2/(b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
1/6*((3*a*b^2*c^2 + 6*a^2*b*c*d - 5*a^3*d^2 + (3*b^3*c^2 + 6*a*b^2*c*d - 5 *a^2*b*d^2)*x^4)*sqrt(b)*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), - 1) + (2*a^2*b*d^2*x^5 + (3*a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2)*x)*sqrt(b* x^4 + a))/(a^2*b^3*x^4 + a^3*b^2)
\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {\left (c + d x^{4}\right )^{2}}{\left (a + b x^{4}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((d*x**4+c)**2/(b*x**4+a)**(3/2),x)
Output:
Integral((c + d*x**4)**2/(a + b*x**4)**(3/2), x)
\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )}^{2}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((d*x^4+c)^2/(b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)^2/(b*x^4 + a)^(3/2), x)
\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )}^{2}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((d*x^4+c)^2/(b*x^4+a)^(3/2),x, algorithm="giac")
Output:
integrate((d*x^4 + c)^2/(b*x^4 + a)^(3/2), x)
Timed out. \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {{\left (d\,x^4+c\right )}^2}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \] Input:
int((c + d*x^4)^2/(a + b*x^4)^(3/2),x)
Output:
int((c + d*x^4)^2/(a + b*x^4)^(3/2), x)
\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {5 \sqrt {b \,x^{4}+a}\, a \,d^{2} x -6 \sqrt {b \,x^{4}+a}\, b c d x +\sqrt {b \,x^{4}+a}\, b \,d^{2} x^{5}-5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{3} d^{2}+6 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{2} b c d -5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{2} b \,d^{2} x^{4}+3 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a \,b^{2} c^{2}+6 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a \,b^{2} c d \,x^{4}+3 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) b^{3} c^{2} x^{4}}{3 b^{2} \left (b \,x^{4}+a \right )} \] Input:
int((d*x^4+c)^2/(b*x^4+a)^(3/2),x)
Output:
(5*sqrt(a + b*x**4)*a*d**2*x - 6*sqrt(a + b*x**4)*b*c*d*x + sqrt(a + b*x** 4)*b*d**2*x**5 - 5*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x) *a**3*d**2 + 6*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a** 2*b*c*d - 5*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**2*b *d**2*x**4 + 3*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a*b **2*c**2 + 6*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a*b** 2*c*d*x**4 + 3*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*b** 3*c**2*x**4)/(3*b**2*(a + b*x**4))