\(\int \frac {x^4 (c+d x^4)}{(a+b x^4)^{5/2}} \, dx\) [47]

Optimal result

Integrand size = 22, antiderivative size = 154 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=-\frac {(b c-a d) x}{6 b^2 \left (a+b x^4\right )^{3/2}}+\frac {(b c-7 a d) x}{12 a b^2 \sqrt {a+b x^4}}+\frac {(b c+5 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{24 a^{5/4} b^{9/4} \sqrt {a+b x^4}} \] Output:

-1/6*(-a*d+b*c)*x/b^2/(b*x^4+a)^(3/2)+1/12*(-7*a*d+b*c)*x/a/b^2/(b*x^4+a)^ 
(1/2)+1/24*(5*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^(9/4)/(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 = 10.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.65 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {-5 a^2 d x+b^2 c x^5-a b x \left (c+7 d x^4\right )+(b c+5 a d) x \left (a+b x^4\right ) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{12 a b^2 \left (a+b x^4\right )^{3/2}} \] Input:

Integrate[(x^4*(c + d*x^4))/(a + b*x^4)^(5/2),x]
 

Output:

(-5*a^2*d*x + b^2*c*x^5 - a*b*x*(c + 7*d*x^4) + (b*c + 5*a*d)*x*(a + b*x^4 
)*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(12* 
a*b^2*(a + b*x^4)^(3/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {957, 817, 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.

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

\(\Big \downarrow \) 957

\(\displaystyle \frac {(5 a d+b c) \int \frac {x^4}{\left (b x^4+a\right )^{3/2}}dx}{6 a b}+\frac {x^5 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {(5 a d+b c) \left (\frac {\int \frac {1}{\sqrt {b x^4+a}}dx}{2 b}-\frac {x}{2 b \sqrt {a+b x^4}}\right )}{6 a b}+\frac {x^5 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(5 a d+b c) \left (\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{5/4} \sqrt {a+b x^4}}-\frac {x}{2 b \sqrt {a+b x^4}}\right )}{6 a b}+\frac {x^5 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\)

Input:

Int[(x^4*(c + d*x^4))/(a + b*x^4)^(5/2),x]
 

Output:

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

Defintions of rubi rules used

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 817

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( 
n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.51 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.03

method	result	size
elliptic	\(\frac {x \left (a d -c b \right ) \sqrt {b \,x^{4}+a}}{6 b^{4} \left (x^{4}+\frac {a}{b}\right )^{2}}-\frac {x \left (7 a d -c b \right )}{12 b^{2} a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\left (\frac {d}{b^{2}}-\frac {7 a d -c b}{12 b^{2} a}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(159\)
default	\(c \left (-\frac {x \sqrt {b \,x^{4}+a}}{6 b^{3} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x}{12 b a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 b a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (\frac {a x \sqrt {b \,x^{4}+a}}{6 b^{4} \left (x^{4}+\frac {a}{b}\right )^{2}}-\frac {7 x}{12 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {5 \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(251\)

Input:

int(x^4*(d*x^4+c)/(b*x^4+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*x/b^4*(a*d-b*c)*(b*x^4+a)^(1/2)/(x^4+a/b)^2-1/12/b^2/a*x*(7*a*d-b*c)/( 
(x^4+a/b)*b)^(1/2)+(d/b^2-1/12/b^2/a*(7*a*d-b*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 = 0.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.99 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=-\frac {{\left ({\left (b^{3} c + 5 \, a b^{2} d\right )} x^{8} + 2 \, {\left (a b^{2} c + 5 \, a^{2} b d\right )} x^{4} + a^{2} b c + 5 \, a^{3} d\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left ({\left (b^{3} c - 7 \, a b^{2} d\right )} x^{5} - {\left (a b^{2} c + 5 \, a^{2} b d\right )} x\right )} \sqrt {b x^{4} + a}}{12 \, {\left (a b^{5} x^{8} + 2 \, a^{2} b^{4} x^{4} + a^{3} b^{3}\right )}} \] Input:

integrate(x^4*(d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="fricas")
 

Output:

-1/12*(((b^3*c + 5*a*b^2*d)*x^8 + 2*(a*b^2*c + 5*a^2*b*d)*x^4 + a^2*b*c + 
5*a^3*d)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/4)), -1) - ((b 
^3*c - 7*a*b^2*d)*x^5 - (a*b^2*c + 5*a^2*b*d)*x)*sqrt(b*x^4 + a))/(a*b^5*x 
^8 + 2*a^2*b^4*x^4 + a^3*b^3)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 24.50 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.52 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {c x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {d x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {5}{2} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate(x**4*(d*x**4+c)/(b*x**4+a)**(5/2),x)
 

Output:

c*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)) + d*x**9*gamma(9/4)*hyper((9/4, 5/2), (13/4,), b*x**4* 
exp_polar(I*pi)/a)/(4*a**(5/2)*gamma(13/4))
 

Maxima [F]

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

integrate(x^4*(d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)*x^4/(b*x^4 + a)^(5/2), x)
 

Giac [F]

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

integrate(x^4*(d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^4/(b*x^4 + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\int \frac {x^4\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{5/2}} \,d x \] Input:

int((x^4*(c + d*x^4))/(a + b*x^4)^(5/2),x)
 

Output:

int((x^4*(c + d*x^4))/(a + b*x^4)^(5/2), x)
 

Reduce [F]

\[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {-5 \sqrt {b \,x^{4}+a}\, a d x -\sqrt {b \,x^{4}+a}\, b c x -5 \sqrt {b \,x^{4}+a}\, b d \,x^{5}+5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{4} d +\left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{3} b c +10 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{3} b d \,x^{4}+2 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{2} b^{2} c \,x^{4}+5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{2} b^{2} d \,x^{8}+\left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a \,b^{3} c \,x^{8}}{5 b^{2} \left (b^{2} x^{8}+2 a b \,x^{4}+a^{2}\right )} \] Input:

int(x^4*(d*x^4+c)/(b*x^4+a)^(5/2),x)
 

Output:

( - 5*sqrt(a + b*x**4)*a*d*x - sqrt(a + b*x**4)*b*c*x - 5*sqrt(a + b*x**4) 
*b*d*x**5 + 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**4*d + 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*b*c + 10*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*b*d*x**4 + 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**2*c 
*x**4 + 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**2*d*x**8 + 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**3*c*x**8)/(5*b**2*(a**2 + 2*a*b*x**4 
 + b**2*x**8))