Integrand size = 19, antiderivative size = 157 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {(b c-a d) x}{6 a b \left (a+b x^4\right )^{3/2}}+\frac {(5 b c+a d) x}{12 a^2 b \sqrt {a+b x^4}}+\frac {(5 b c+a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{24 a^{9/4} b^{5/4} \sqrt {a+b x^4}} \] Output:
1/6*(-a*d+b*c)*x/a/b/(b*x^4+a)^(3/2)+1/12*(a*d+5*b*c)*x/a^2/b/(b*x^4+a)^(1 /2)+1/24*(a*d+5*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)*InverseJacobiAM(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(9/4 )/b^(5/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.64 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {-a^2 d x+5 b^2 c x^5+a b x \left (7 c+d x^4\right )+(5 b c+a d) x \left (a+b x^4\right ) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{12 a^2 b \left (a+b x^4\right )^{3/2}} \] Input:
Integrate[(c + d*x^4)/(a + b*x^4)^(5/2),x]
Output:
(-(a^2*d*x) + 5*b^2*c*x^5 + a*b*x*(7*c + d*x^4) + (5*b*c + a*d)*x*(a + b*x ^4)*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(1 2*a^2*b*(a + b*x^4)^(3/2))
Time = 0.39 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {910, 749, 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.
| \(\displaystyle \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 910 |
| \(\displaystyle \frac {(a d+5 b c) \int \frac {1}{\left (b x^4+a\right )^{3/2}}dx}{6 a b}+\frac {x (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\) |
| \(\Big \downarrow \) 749 |
| \(\displaystyle \frac {(a d+5 b c) \left (\frac {\int \frac {1}{\sqrt {b x^4+a}}dx}{2 a}+\frac {x}{2 a \sqrt {a+b x^4}}\right )}{6 a b}+\frac {x (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {(a d+5 b c) \left (\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 a^{5/4} \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {x}{2 a \sqrt {a+b x^4}}\right )}{6 a b}+\frac {x (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\) |
Input:
Int[(c + d*x^4)/(a + b*x^4)^(5/2),x]
Output:
((b*c - a*d)*x)/(6*a*b*(a + b*x^4)^(3/2)) + ((5*b*c + a*d)*(x/(2*a*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^(5/4)*b^(1/4)*S qrt[a + b*x^4])))/(6*a*b)
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
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[(-(b*c - a*d))*x*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] - Simp[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)) Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/ n + p, 0])
Result contains complex when optimal does not.
Time = 0.59 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.97
| method | result | size |
| elliptic | \(-\frac {x \left (a d -c b \right ) \sqrt {b \,x^{4}+a}}{6 a \,b^{3} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x \left (a d +5 c b \right )}{12 b \,a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\left (a d +5 c b \right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 a^{2} b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(153\) |
| default | \(c \left (\frac {x \sqrt {b \,x^{4}+a}}{6 a \,b^{2} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {5 x}{12 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {5 \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (-\frac {x \sqrt {b \,x^{4}+a}}{6 b^{3} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x}{12 b a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 b a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(253\) |
Input:
int((d*x^4+c)/(b*x^4+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/6/a*x/b^3*(a*d-b*c)*(b*x^4+a)^(1/2)/(x^4+a/b)^2+1/12/b/a^2*x*(a*d+5*b*c )/((x^4+a/b)*b)^(1/2)+1/12*(a*d+5*b*c)/a^2/b/(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)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)
Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/2}} \, dx=-\frac {{\left ({\left (5 \, b^{3} c + a b^{2} d\right )} x^{8} + 2 \, {\left (5 \, a b^{2} c + a^{2} b d\right )} x^{4} + 5 \, a^{2} b c + a^{3} d\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left ({\left (5 \, b^{3} c + a b^{2} d\right )} x^{5} + {\left (7 \, a b^{2} c - a^{2} b d\right )} x\right )} \sqrt {b x^{4} + a}}{12 \, {\left (a^{2} b^{4} x^{8} + 2 \, a^{3} b^{3} x^{4} + a^{4} b^{2}\right )}} \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="fricas")
Output:
-1/12*(((5*b^3*c + a*b^2*d)*x^8 + 2*(5*a*b^2*c + a^2*b*d)*x^4 + 5*a^2*b*c + a^3*d)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/4)), -1) - ((5 *b^3*c + a*b^2*d)*x^5 + (7*a*b^2*c - a^2*b*d)*x)*sqrt(b*x^4 + a))/(a^2*b^4 *x^8 + 2*a^3*b^3*x^4 + a^4*b^2)
Result contains complex when optimal does not.
Time = 18.73 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.50 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate((d*x**4+c)/(b*x**4+a)**(5/2),x)
Output:
c*x*gamma(1/4)*hyper((1/4, 5/2), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**( 5/2)*gamma(5/4)) + d*x**5*gamma(5/4)*hyper((5/4, 5/2), (9/4,), b*x**4*exp_ polar(I*pi)/a)/(4*a**(5/2)*gamma(9/4))
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/2}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(5/2), x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/2}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(5/2), x)
Timed out. \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/2}} \, dx=\int \frac {d\,x^4+c}{{\left (b\,x^4+a\right )}^{5/2}} \,d x \] Input:
int((c + d*x^4)/(a + b*x^4)^(5/2),x)
Output:
int((c + d*x^4)/(a + b*x^4)^(5/2), x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {-\sqrt {b \,x^{4}+a}\, d x +\left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{3} d +5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{2} b c +2 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{2} b d \,x^{4}+10 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a \,b^{2} c \,x^{4}+\left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a \,b^{2} d \,x^{8}+5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) b^{3} c \,x^{8}}{5 b \left (b^{2} x^{8}+2 a b \,x^{4}+a^{2}\right )} \] Input:
int((d*x^4+c)/(b*x^4+a)^(5/2),x)
Output:
( - sqrt(a + b*x**4)*d*x + int(sqrt(a + b*x**4)/(a**3 + 3*a**2*b*x**4 + 3* a*b**2*x**8 + b**3*x**12),x)*a**3*d + 5*int(sqrt(a + b*x**4)/(a**3 + 3*a** 2*b*x**4 + 3*a*b**2*x**8 + b**3*x**12),x)*a**2*b*c + 2*int(sqrt(a + b*x**4 )/(a**3 + 3*a**2*b*x**4 + 3*a*b**2*x**8 + b**3*x**12),x)*a**2*b*d*x**4 + 1 0*int(sqrt(a + b*x**4)/(a**3 + 3*a**2*b*x**4 + 3*a*b**2*x**8 + b**3*x**12) ,x)*a*b**2*c*x**4 + int(sqrt(a + b*x**4)/(a**3 + 3*a**2*b*x**4 + 3*a*b**2* x**8 + b**3*x**12),x)*a*b**2*d*x**8 + 5*int(sqrt(a + b*x**4)/(a**3 + 3*a** 2*b*x**4 + 3*a*b**2*x**8 + b**3*x**12),x)*b**3*c*x**8)/(5*b*(a**2 + 2*a*b* x**4 + b**2*x**8))