\(\int \frac {c+d x^4}{x^4 (a+b x^4)^{3/2}} \, dx\) [38]

Optimal result

Integrand size = 22, antiderivative size = 148 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {c}{3 a x^3 \sqrt {a+b x^4}}-\frac {(5 b c-3 a d) x}{6 a^2 \sqrt {a+b x^4}}-\frac {(5 b c-3 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{12 a^{9/4} \sqrt [4]{b} \sqrt {a+b x^4}} \] Output:

-1/3*c/a/x^3/(b*x^4+a)^(1/2)-1/6*(-3*a*d+5*b*c)*x/a^2/(b*x^4+a)^(1/2)-1/12 
*(-3*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^( 
1/4)/(b*x^4+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.58 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\frac {-2 a c-5 b c x^4+3 a d x^4+(-5 b c+3 a d) x^4 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{6 a^2 x^3 \sqrt {a+b x^4}} \] Input:

Integrate[(c + d*x^4)/(x^4*(a + b*x^4)^(3/2)),x]
 

Output:

(-2*a*c - 5*b*c*x^4 + 3*a*d*x^4 + (-5*b*c + 3*a*d)*x^4*Sqrt[1 + (b*x^4)/a] 
*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(6*a^2*x^3*Sqrt[a + b*x^4 
])
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {955, 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.

\(\displaystyle \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 955

\(\displaystyle -\frac {(5 b c-3 a d) \int \frac {1}{\left (b x^4+a\right )^{3/2}}dx}{3 a}-\frac {c}{3 a x^3 \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 749

\(\displaystyle -\frac {(5 b c-3 a d) \left (\frac {\int \frac {1}{\sqrt {b x^4+a}}dx}{2 a}+\frac {x}{2 a \sqrt {a+b x^4}}\right )}{3 a}-\frac {c}{3 a x^3 \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 761

\(\displaystyle -\frac {(5 b c-3 a d) \left (\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 a^{5/4} \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {x}{2 a \sqrt {a+b x^4}}\right )}{3 a}-\frac {c}{3 a x^3 \sqrt {a+b x^4}}\)

Input:

Int[(c + d*x^4)/(x^4*(a + b*x^4)^(3/2)),x]
 

Output:

-1/3*c/(a*x^3*Sqrt[a + b*x^4]) - ((5*b*c - 3*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]*El 
lipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(5/4)*b^(1/4)*Sqrt[a + b 
*x^4])))/(3*a)
 

Defintions of rubi rules used

rule 749

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])
 

rule 761

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]
 

rule 955

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.76 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93

method	result	size
elliptic	\(\frac {x \left (a d -c b \right )}{2 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {c \sqrt {b \,x^{4}+a}}{3 a^{2} x^{3}}+\frac {\left (\frac {a d -c b}{2 a^{2}}-\frac {b c}{3 a^{2}}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(137\)
default	\(d \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+c \left (-\frac {b x}{2 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {\sqrt {b \,x^{4}+a}}{3 a^{2} x^{3}}-\frac {5 b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(211\)
risch	\(-\frac {c \sqrt {b \,x^{4}+a}}{3 a^{2} x^{3}}-\frac {c \,b^{2} \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-3 a^{2} d \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+4 a b c \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )}{3 a^{2}}\)	\(321\)

Input:

int((d*x^4+c)/x^4/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/a^2*x*(a*d-b*c)/((x^4+a/b)*b)^(1/2)-1/3/a^2*c*(b*x^4+a)^(1/2)/x^3+(1/2 
/a^2*(a*d-b*c)-1/3*b/a^2*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)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.78 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\frac {{\left ({\left (5 \, b^{2} c - 3 \, a b d\right )} x^{7} + {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{3}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left ({\left (5 \, b^{2} c - 3 \, a b d\right )} x^{4} + 2 \, a b c\right )} \sqrt {b x^{4} + a}}{6 \, {\left (a^{2} b^{2} x^{7} + a^{3} b x^{3}\right )}} \] Input:

integrate((d*x^4+c)/x^4/(b*x^4+a)^(3/2),x, algorithm="fricas")
 

Output:

1/6*(((5*b^2*c - 3*a*b*d)*x^7 + (5*a*b*c - 3*a^2*d)*x^3)*sqrt(a)*(-b/a)^(3 
/4)*elliptic_f(arcsin(x*(-b/a)^(1/4)), -1) - ((5*b^2*c - 3*a*b*d)*x^4 + 2* 
a*b*c)*sqrt(b*x^4 + a))/(a^2*b^2*x^7 + a^3*b*x^3)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.55 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\frac {c \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} x^{3} \Gamma \left (\frac {1}{4}\right )} + \frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right )} \] Input:

integrate((d*x**4+c)/x**4/(b*x**4+a)**(3/2),x)
 

Output:

c*gamma(-3/4)*hyper((-3/4, 3/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**( 
3/2)*x**3*gamma(1/4)) + d*x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), b*x**4*ex 
p_polar(I*pi)/a)/(4*a**(3/2)*gamma(5/4))
 

Maxima [F]

\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{4}} \,d x } \] Input:

integrate((d*x^4+c)/x^4/(b*x^4+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(3/2)*x^4), x)
 

Giac [F]

\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{4}} \,d x } \] Input:

integrate((d*x^4+c)/x^4/(b*x^4+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(3/2)*x^4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\int \frac {d\,x^4+c}{x^4\,{\left (b\,x^4+a\right )}^{3/2}} \,d x \] Input:

int((c + d*x^4)/(x^4*(a + b*x^4)^(3/2)),x)
 

Output:

int((c + d*x^4)/(x^4*(a + b*x^4)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\frac {-\sqrt {b \,x^{4}+a}\, d -3 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{12}+2 a b \,x^{8}+a^{2} x^{4}}d x \right ) a^{2} d \,x^{3}+5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{12}+2 a b \,x^{8}+a^{2} x^{4}}d x \right ) a b c \,x^{3}-3 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{12}+2 a b \,x^{8}+a^{2} x^{4}}d x \right ) a b d \,x^{7}+5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{12}+2 a b \,x^{8}+a^{2} x^{4}}d x \right ) b^{2} c \,x^{7}}{5 b \,x^{3} \left (b \,x^{4}+a \right )} \] Input:

int((d*x^4+c)/x^4/(b*x^4+a)^(3/2),x)
 

Output:

( - sqrt(a + b*x**4)*d - 3*int(sqrt(a + b*x**4)/(a**2*x**4 + 2*a*b*x**8 + 
b**2*x**12),x)*a**2*d*x**3 + 5*int(sqrt(a + b*x**4)/(a**2*x**4 + 2*a*b*x** 
8 + b**2*x**12),x)*a*b*c*x**3 - 3*int(sqrt(a + b*x**4)/(a**2*x**4 + 2*a*b* 
x**8 + b**2*x**12),x)*a*b*d*x**7 + 5*int(sqrt(a + b*x**4)/(a**2*x**4 + 2*a 
*b*x**8 + b**2*x**12),x)*b**2*c*x**7)/(5*b*x**3*(a + b*x**4))