Integrand size = 22, antiderivative size = 176 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{5/2}} \, dx=-\frac {c}{3 a x^3 \left (a+b x^4\right )^{3/2}}-\frac {(3 b c-a d) x}{6 a^2 \left (a+b x^4\right )^{3/2}}-\frac {5 (3 b c-a d) x}{12 a^3 \sqrt {a+b x^4}}-\frac {5 (3 b c-a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{24 a^{13/4} \sqrt [4]{b} \sqrt {a+b x^4}} \] Output:
-1/3*c/a/x^3/(b*x^4+a)^(3/2)-1/6*(-a*d+3*b*c)*x/a^2/(b*x^4+a)^(3/2)-5/12*( -a*d+3*b*c)*x/a^3/(b*x^4+a)^(1/2)-5/24*(-a*d+3*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^(13/4)/b^(1/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.06 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.65 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{5/2}} \, dx=\frac {-15 b^2 c x^8+a^2 \left (-4 c+7 d x^4\right )+a b \left (-21 c x^4+5 d x^8\right )+5 (-3 b c+a d) x^4 \left (a+b x^4\right ) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{12 a^3 x^3 \left (a+b x^4\right )^{3/2}} \] Input:
Integrate[(c + d*x^4)/(x^4*(a + b*x^4)^(5/2)),x]
Output:
(-15*b^2*c*x^8 + a^2*(-4*c + 7*d*x^4) + a*b*(-21*c*x^4 + 5*d*x^8) + 5*(-3* b*c + a*d)*x^4*(a + b*x^4)*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(12*a^3*x^3*(a + b*x^4)^(3/2))
Time = 0.41 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {955, 749, 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.
| \(\displaystyle \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 955 |
| \(\displaystyle -\frac {(3 b c-a d) \int \frac {1}{\left (b x^4+a\right )^{5/2}}dx}{a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{3/2}}\) |
| \(\Big \downarrow \) 749 |
| \(\displaystyle -\frac {(3 b c-a d) \left (\frac {5 \int \frac {1}{\left (b x^4+a\right )^{3/2}}dx}{6 a}+\frac {x}{6 a \left (a+b x^4\right )^{3/2}}\right )}{a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{3/2}}\) |
| \(\Big \downarrow \) 749 |
| \(\displaystyle -\frac {(3 b c-a d) \left (\frac {5 \left (\frac {\int \frac {1}{\sqrt {b x^4+a}}dx}{2 a}+\frac {x}{2 a \sqrt {a+b x^4}}\right )}{6 a}+\frac {x}{6 a \left (a+b x^4\right )^{3/2}}\right )}{a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{3/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle -\frac {(3 b c-a d) \left (\frac {5 \left (\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 a^{5/4} \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {x}{2 a \sqrt {a+b x^4}}\right )}{6 a}+\frac {x}{6 a \left (a+b x^4\right )^{3/2}}\right )}{a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{3/2}}\) |
Input:
Int[(c + d*x^4)/(x^4*(a + b*x^4)^(5/2)),x]
Output:
-1/3*c/(a*x^3*(a + b*x^4)^(3/2)) - ((3*b*c - a*d)*(x/(6*a*(a + b*x^4)^(3/2 )) + (5*(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)*Sqrt[a + b*x^4])))/(6*a)))/a
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Result contains complex when optimal does not.
Time = 3.32 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00
| method | result | size |
| elliptic | \(\frac {x \left (a d -c b \right ) \sqrt {b \,x^{4}+a}}{6 a^{2} b^{2} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x \left (5 a d -11 c b \right )}{12 a^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {c \sqrt {b \,x^{4}+a}}{3 a^{3} x^{3}}+\frac {\left (\frac {5 a d -11 c b}{12 a^{3}}-\frac {b c}{3 a^{3}}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(176\) |
| default | \(d \left (\frac {x \sqrt {b \,x^{4}+a}}{6 a \,b^{2} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {5 x}{12 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {5 \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+c \left (-\frac {x \sqrt {b \,x^{4}+a}}{6 a^{2} b \left (x^{4}+\frac {a}{b}\right )^{2}}-\frac {11 b x}{12 a^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {\sqrt {b \,x^{4}+a}}{3 a^{3} x^{3}}-\frac {5 b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{4 a^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(269\) |
| risch | \(-\frac {c \sqrt {b \,x^{4}+a}}{3 a^{3} x^{3}}-\frac {\frac {c b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-3 a^{2} \left (a d -c b \right ) \left (\frac {x \sqrt {b \,x^{4}+a}}{6 a \,b^{2} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {5 x}{12 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {5 \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+3 a b c \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )}{3 a^{3}}\) | \(330\) |
Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/6/a^2*x/b^2*(a*d-b*c)*(b*x^4+a)^(1/2)/(x^4+a/b)^2+1/12/a^3*x*(5*a*d-11*b *c)/((x^4+a/b)*b)^(1/2)-1/3/a^3*c*(b*x^4+a)^(1/2)/x^3+(1/12/a^3*(5*a*d-11* b*c)-1/3*b/a^3*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)*EllipticF(x*(I/a^(1/2)*b ^(1/2))^(1/2),I)
Time = 0.09 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{5/2}} \, dx=\frac {5 \, {\left ({\left (3 \, b^{3} c - a b^{2} d\right )} x^{11} + 2 \, {\left (3 \, a b^{2} c - a^{2} b d\right )} x^{7} + {\left (3 \, a^{2} b c - a^{3} d\right )} x^{3}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (5 \, {\left (3 \, b^{3} c - a b^{2} d\right )} x^{8} + 7 \, {\left (3 \, a b^{2} c - a^{2} b d\right )} x^{4} + 4 \, a^{2} b c\right )} \sqrt {b x^{4} + a}}{12 \, {\left (a^{3} b^{3} x^{11} + 2 \, a^{4} b^{2} x^{7} + a^{5} b x^{3}\right )}} \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(5/2),x, algorithm="fricas")
Output:
1/12*(5*((3*b^3*c - a*b^2*d)*x^11 + 2*(3*a*b^2*c - a^2*b*d)*x^7 + (3*a^2*b *c - a^3*d)*x^3)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/4)), - 1) - (5*(3*b^3*c - a*b^2*d)*x^8 + 7*(3*a*b^2*c - a^2*b*d)*x^4 + 4*a^2*b*c) *sqrt(b*x^4 + a))/(a^3*b^3*x^11 + 2*a^4*b^2*x^7 + a^5*b*x^3)
Result contains complex when optimal does not.
Time = 46.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.47 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{5/2}} \, dx=\frac {c \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} x^{3} \Gamma \left (\frac {1}{4}\right )} + \frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((d*x**4+c)/x**4/(b*x**4+a)**(5/2),x)
Output:
c*gamma(-3/4)*hyper((-3/4, 5/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**( 5/2)*x**3*gamma(1/4)) + d*x*gamma(1/4)*hyper((1/4, 5/2), (5/4,), b*x**4*ex p_polar(I*pi)/a)/(4*a**(5/2)*gamma(5/4))
\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{5/2}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{2}} x^{4}} \,d x } \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(5/2),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(5/2)*x^4), x)
\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{5/2}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{2}} x^{4}} \,d x } \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(5/2),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(5/2)*x^4), x)
Timed out. \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{5/2}} \, dx=\int \frac {d\,x^4+c}{x^4\,{\left (b\,x^4+a\right )}^{5/2}} \,d x \] Input:
int((c + d*x^4)/(x^4*(a + b*x^4)^(5/2)),x)
Output:
int((c + d*x^4)/(x^4*(a + b*x^4)^(5/2)), x)
\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{5/2}} \, dx=\frac {-\sqrt {b \,x^{4}+a}\, d -3 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{16}+3 a \,b^{2} x^{12}+3 a^{2} b \,x^{8}+a^{3} x^{4}}d x \right ) a^{3} d \,x^{3}+9 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{16}+3 a \,b^{2} x^{12}+3 a^{2} b \,x^{8}+a^{3} x^{4}}d x \right ) a^{2} b c \,x^{3}-6 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{16}+3 a \,b^{2} x^{12}+3 a^{2} b \,x^{8}+a^{3} x^{4}}d x \right ) a^{2} b d \,x^{7}+18 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{16}+3 a \,b^{2} x^{12}+3 a^{2} b \,x^{8}+a^{3} x^{4}}d x \right ) a \,b^{2} c \,x^{7}-3 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{16}+3 a \,b^{2} x^{12}+3 a^{2} b \,x^{8}+a^{3} x^{4}}d x \right ) a \,b^{2} d \,x^{11}+9 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{16}+3 a \,b^{2} x^{12}+3 a^{2} b \,x^{8}+a^{3} x^{4}}d x \right ) b^{3} c \,x^{11}}{9 b \,x^{3} \left (b^{2} x^{8}+2 a b \,x^{4}+a^{2}\right )} \] Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(5/2),x)
Output:
( - sqrt(a + b*x**4)*d - 3*int(sqrt(a + b*x**4)/(a**3*x**4 + 3*a**2*b*x**8 + 3*a*b**2*x**12 + b**3*x**16),x)*a**3*d*x**3 + 9*int(sqrt(a + b*x**4)/(a **3*x**4 + 3*a**2*b*x**8 + 3*a*b**2*x**12 + b**3*x**16),x)*a**2*b*c*x**3 - 6*int(sqrt(a + b*x**4)/(a**3*x**4 + 3*a**2*b*x**8 + 3*a*b**2*x**12 + b**3 *x**16),x)*a**2*b*d*x**7 + 18*int(sqrt(a + b*x**4)/(a**3*x**4 + 3*a**2*b*x **8 + 3*a*b**2*x**12 + b**3*x**16),x)*a*b**2*c*x**7 - 3*int(sqrt(a + b*x** 4)/(a**3*x**4 + 3*a**2*b*x**8 + 3*a*b**2*x**12 + b**3*x**16),x)*a*b**2*d*x **11 + 9*int(sqrt(a + b*x**4)/(a**3*x**4 + 3*a**2*b*x**8 + 3*a*b**2*x**12 + b**3*x**16),x)*b**3*c*x**11)/(9*b*x**3*(a**2 + 2*a*b*x**4 + b**2*x**8))