Integrand size = 22, antiderivative size = 119 \[ \int \frac {c+d x^4}{x^4 \sqrt {a+b x^4}} \, dx=-\frac {c \sqrt {a+b x^4}}{3 a x^3}-\frac {(b c-3 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 a^{5/4} \sqrt [4]{b} \sqrt {a+b x^4}} \] Output:
-1/3*c*(b*x^4+a)^(1/2)/a/x^3-1/6*(-3*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)*InverseJacobiAM(2*arctan(b^(1/4)*x/a^( 1/4)),1/2*2^(1/2))/a^(5/4)/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 = 10.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66 \[ \int \frac {c+d x^4}{x^4 \sqrt {a+b x^4}} \, dx=\frac {-c \left (a+b x^4\right )+(-b c+3 a d) x^4 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{3 a x^3 \sqrt {a+b x^4}} \] Input:
Integrate[(c + d*x^4)/(x^4*Sqrt[a + b*x^4]),x]
Output:
(-(c*(a + b*x^4)) + (-(b*c) + 3*a*d)*x^4*Sqrt[1 + (b*x^4)/a]*Hypergeometri c2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(3*a*x^3*Sqrt[a + b*x^4])
Time = 0.35 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {955, 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.
\(\displaystyle \int \frac {c+d x^4}{x^4 \sqrt {a+b x^4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(b c-3 a d) \int \frac {1}{\sqrt {b x^4+a}}dx}{3 a}-\frac {c \sqrt {a+b x^4}}{3 a x^3}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} (b c-3 a d) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 a^{5/4} \sqrt [4]{b} \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{3 a x^3}\) |
Input:
Int[(c + d*x^4)/(x^4*Sqrt[a + b*x^4]),x]
Output:
-1/3*(c*Sqrt[a + b*x^4])/(a*x^3) - ((b*c - 3*a*d)*(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])/(6*a^(5/4)*b^(1/4)*Sqrt[a + b*x^4])
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 = 0.68 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82
method | result | size |
elliptic | \(-\frac {c \sqrt {b \,x^{4}+a}}{3 a \,x^{3}}+\frac {\left (d -\frac {b c}{3 a}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(98\) |
risch | \(-\frac {c \sqrt {b \,x^{4}+a}}{3 a \,x^{3}}+\frac {\left (3 a d -c b \right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(102\) |
default | \(\frac {d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+c \left (-\frac {\sqrt {b \,x^{4}+a}}{3 a \,x^{3}}-\frac {b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(166\) |
Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/3*c*(b*x^4+a)^(1/2)/a/x^3+(d-1/3*b/a*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 = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.51 \[ \int \frac {c+d x^4}{x^4 \sqrt {a+b x^4}} \, dx=\frac {{\left (b c - 3 \, a d\right )} \sqrt {a} x^{3} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - \sqrt {b x^{4} + a} b c}{3 \, a b x^{3}} \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(1/2),x, algorithm="fricas")
Output:
1/3*((b*c - 3*a*d)*sqrt(a)*x^3*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/ 4)), -1) - sqrt(b*x^4 + a)*b*c)/(a*b*x^3)
Result contains complex when optimal does not.
Time = 1.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69 \[ \int \frac {c+d x^4}{x^4 \sqrt {a+b x^4}} \, dx=\frac {c \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x^{3} \Gamma \left (\frac {1}{4}\right )} + \frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((d*x**4+c)/x**4/(b*x**4+a)**(1/2),x)
Output:
c*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt (a)*x**3*gamma(1/4)) + d*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), b*x**4*exp _polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4))
\[ \int \frac {c+d x^4}{x^4 \sqrt {a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{\sqrt {b x^{4} + a} x^{4}} \,d x } \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(1/2),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)/(sqrt(b*x^4 + a)*x^4), x)
\[ \int \frac {c+d x^4}{x^4 \sqrt {a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{\sqrt {b x^{4} + a} x^{4}} \,d x } \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(1/2),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/(sqrt(b*x^4 + a)*x^4), x)
Timed out. \[ \int \frac {c+d x^4}{x^4 \sqrt {a+b x^4}} \, dx=\int \frac {d\,x^4+c}{x^4\,\sqrt {b\,x^4+a}} \,d x \] Input:
int((c + d*x^4)/(x^4*(a + b*x^4)^(1/2)),x)
Output:
int((c + d*x^4)/(x^4*(a + b*x^4)^(1/2)), x)
\[ \int \frac {c+d x^4}{x^4 \sqrt {a+b x^4}} \, dx=\frac {-\sqrt {b \,x^{4}+a}\, d -3 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{8}+a \,x^{4}}d x \right ) a d \,x^{3}+\left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{8}+a \,x^{4}}d x \right ) b c \,x^{3}}{b \,x^{3}} \] Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(1/2),x)
Output:
( - sqrt(a + b*x**4)*d - 3*int(sqrt(a + b*x**4)/(a*x**4 + b*x**8),x)*a*d*x **3 + int(sqrt(a + b*x**4)/(a*x**4 + b*x**8),x)*b*c*x**3)/(b*x**3)