\(\int \frac {x^6 (c+d x^4)}{\sqrt {a+b x^4}} \, dx\) [31]

Optimal result

Integrand size = 22, antiderivative size = 295 \[ \int \frac {x^6 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {(9 b c-7 a d) x^3 \sqrt {a+b x^4}}{45 b^2}+\frac {d x^7 \sqrt {a+b x^4}}{9 b}-\frac {a (9 b c-7 a d) x \sqrt {a+b x^4}}{15 b^{5/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {a^{5/4} (9 b c-7 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^4}}-\frac {a^{5/4} (9 b c-7 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{30 b^{11/4} \sqrt {a+b x^4}} \] Output:

1/45*(-7*a*d+9*b*c)*x^3*(b*x^4+a)^(1/2)/b^2+1/9*d*x^7*(b*x^4+a)^(1/2)/b-1/ 
15*a*(-7*a*d+9*b*c)*x*(b*x^4+a)^(1/2)/b^(5/2)/(a^(1/2)+b^(1/2)*x^2)+1/15*a 
^(5/4)*(-7*a*d+9*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)*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))/b^(11/ 
4)/(b*x^4+a)^(1/2)-1/30*a^(5/4)*(-7*a*d+9*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))/b^(11/4)/(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 = 10.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.31 \[ \int \frac {x^6 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {x^3 \left (-\left (\left (a+b x^4\right ) \left (-9 b c+7 a d-5 b d x^4\right )\right )+a (-9 b c+7 a d) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{45 b^2 \sqrt {a+b x^4}} \] Input:

Integrate[(x^6*(c + d*x^4))/Sqrt[a + b*x^4],x]
 

Output:

(x^3*(-((a + b*x^4)*(-9*b*c + 7*a*d - 5*b*d*x^4)) + a*(-9*b*c + 7*a*d)*Sqr 
t[1 + (b*x^4)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((b*x^4)/a)]))/(45*b^2* 
Sqrt[a + b*x^4])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {959, 843, 834, 27, 761, 1510}

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.

\(\displaystyle \int \frac {x^6 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx\)

\(\Big \downarrow \) 959

\(\displaystyle \frac {(9 b c-7 a d) \int \frac {x^6}{\sqrt {b x^4+a}}dx}{9 b}+\frac {d x^7 \sqrt {a+b x^4}}{9 b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {(9 b c-7 a d) \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^4+a}}dx}{5 b}\right )}{9 b}+\frac {d x^7 \sqrt {a+b x^4}}{9 b}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {(9 b c-7 a d) \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {a} \sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{9 b}+\frac {d x^7 \sqrt {a+b x^4}}{9 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(9 b c-7 a d) \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{9 b}+\frac {d x^7 \sqrt {a+b x^4}}{9 b}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(9 b c-7 a d) \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{9 b}+\frac {d x^7 \sqrt {a+b x^4}}{9 b}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {(9 b c-7 a d) \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^4}}-\frac {x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}+\frac {d x^7 \sqrt {a+b x^4}}{9 b}\)

Input:

Int[(x^6*(c + d*x^4))/Sqrt[a + b*x^4],x]
 

Output:

(d*x^7*Sqrt[a + b*x^4])/(9*b) + ((9*b*c - 7*a*d)*((x^3*Sqrt[a + b*x^4])/(5 
*b) - (3*a*(-((-((x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2)) + (a^(1/4)*( 
Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Ellipti 
cE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(b^(1/4)*Sqrt[a + b*x^4]))/Sqrt[b] 
) + (a^(1/4)*(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])/(2*b^(3/4)*Sqrt[a + 
b*x^4])))/(5*b)))/(9*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 843

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*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 959

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p 
+ 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p 
 + 1) + 1))   Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, 
 n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.66 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.47

method	result	size
risch	\(-\frac {x^{3} \left (-5 d b \,x^{4}+7 a d -9 c b \right ) \sqrt {b \,x^{4}+a}}{45 b^{2}}+\frac {i a^{\frac {3}{2}} \left (7 a d -9 c b \right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 b^{\frac {5}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(140\)
elliptic	\(\frac {d \,x^{7} \sqrt {b \,x^{4}+a}}{9 b}+\frac {\left (c -\frac {7 a d}{9 b}\right ) x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {3 i \left (c -\frac {7 a d}{9 b}\right ) a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(151\)
default	\(c \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (\frac {x^{7} \sqrt {b \,x^{4}+a}}{9 b}-\frac {7 a \,x^{3} \sqrt {b \,x^{4}+a}}{45 b^{2}}+\frac {7 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 b^{\frac {5}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(252\)

Input:

int(x^6*(d*x^4+c)/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/45*x^3*(-5*b*d*x^4+7*a*d-9*b*c)/b^2*(b*x^4+a)^(1/2)+1/15*I*a^(3/2)*(7*a 
*d-9*b*c)/b^(5/2)/(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)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.47 \[ \int \frac {x^6 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=-\frac {3 \, {\left (9 \, a b c - 7 \, a^{2} d\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 3 \, {\left (9 \, a b c - 7 \, a^{2} d\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (5 \, b^{2} d x^{8} + {\left (9 \, b^{2} c - 7 \, a b d\right )} x^{4} - 27 \, a b c + 21 \, a^{2} d\right )} \sqrt {b x^{4} + a}}{45 \, b^{3} x} \] Input:

integrate(x^6*(d*x^4+c)/(b*x^4+a)^(1/2),x, algorithm="fricas")
 

Output:

-1/45*(3*(9*a*b*c - 7*a^2*d)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/ 
b)^(1/4)/x), -1) - 3*(9*a*b*c - 7*a^2*d)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f 
(arcsin((-a/b)^(1/4)/x), -1) - (5*b^2*d*x^8 + (9*b^2*c - 7*a*b*d)*x^4 - 27 
*a*b*c + 21*a^2*d)*sqrt(b*x^4 + a))/(b^3*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.53 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.27 \[ \int \frac {x^6 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {c x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} + \frac {d x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {15}{4}\right )} \] Input:

integrate(x**6*(d*x**4+c)/(b*x**4+a)**(1/2),x)
 

Output:

c*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4* 
sqrt(a)*gamma(11/4)) + d*x**11*gamma(11/4)*hyper((1/2, 11/4), (15/4,), b*x 
**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(15/4))
 

Maxima [F]

\[ \int \frac {x^6 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{6}}{\sqrt {b x^{4} + a}} \,d x } \] Input:

integrate(x^6*(d*x^4+c)/(b*x^4+a)^(1/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate((d*x^4 + c)*x^6/sqrt(b*x^4 + a), x)
 

Giac [F]

\[ \int \frac {x^6 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{6}}{\sqrt {b x^{4} + a}} \,d x } \] Input:

integrate(x^6*(d*x^4+c)/(b*x^4+a)^(1/2),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^6/sqrt(b*x^4 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\int \frac {x^6\,\left (d\,x^4+c\right )}{\sqrt {b\,x^4+a}} \,d x \] Input:

int((x^6*(c + d*x^4))/(a + b*x^4)^(1/2),x)
 

Output:

int((x^6*(c + d*x^4))/(a + b*x^4)^(1/2), x)
 

Reduce [F]

\[ \int \frac {x^6 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {-7 \sqrt {b \,x^{4}+a}\, a d \,x^{3}+9 \sqrt {b \,x^{4}+a}\, b c \,x^{3}+5 \sqrt {b \,x^{4}+a}\, b d \,x^{7}+21 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b \,x^{4}+a}d x \right ) a^{2} d -27 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b \,x^{4}+a}d x \right ) a b c}{45 b^{2}} \] Input:

int(x^6*(d*x^4+c)/(b*x^4+a)^(1/2),x)
 

Output:

( - 7*sqrt(a + b*x**4)*a*d*x**3 + 9*sqrt(a + b*x**4)*b*c*x**3 + 5*sqrt(a + 
 b*x**4)*b*d*x**7 + 21*int((sqrt(a + b*x**4)*x**2)/(a + b*x**4),x)*a**2*d 
- 27*int((sqrt(a + b*x**4)*x**2)/(a + b*x**4),x)*a*b*c)/(45*b**2)