\(\int \frac {x^6 (c+d x^4)}{(a+b x^4)^{5/2}} \, dx\) [52]

Optimal result

Integrand size = 22, antiderivative size = 303 \[ \int \frac {x^6 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=-\frac {(b c-a d) x^3}{6 b^2 \left (a+b x^4\right )^{3/2}}+\frac {(b c-3 a d) x^3}{4 a b^2 \sqrt {a+b x^4}}-\frac {(b c-7 a d) x \sqrt {a+b x^4}}{4 a b^{5/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {(b c-7 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{3/4} b^{11/4} \sqrt {a+b x^4}}-\frac {(b c-7 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{8 a^{3/4} b^{11/4} \sqrt {a+b x^4}} \] Output:

-1/6*(-a*d+b*c)*x^3/b^2/(b*x^4+a)^(3/2)+1/4*(-3*a*d+b*c)*x^3/a/b^2/(b*x^4+ 
a)^(1/2)-1/4*(-7*a*d+b*c)*x*(b*x^4+a)^(1/2)/a/b^(5/2)/(a^(1/2)+b^(1/2)*x^2 
)+1/4*(-7*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)*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))/a^(3/4)/b 
^(11/4)/(b*x^4+a)^(1/2)-1/8*(-7*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^(3/4)/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 = 10.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.30 \[ \int \frac {x^6 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {x^3 \left (a \left (-b c+7 a d+3 b d x^4\right )+(b c-7 a d) \left (a+b x^4\right ) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{3 a b^2 \left (a+b x^4\right )^{3/2}} \] Input:

Integrate[(x^6*(c + d*x^4))/(a + b*x^4)^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 293, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {957, 817, 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.

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

\(\Big \downarrow \) 957

\(\displaystyle \frac {x^7 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}-\frac {(b c-7 a d) \int \frac {x^6}{\left (b x^4+a\right )^{3/2}}dx}{6 a b}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {x^7 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}-\frac {(b c-7 a d) \left (\frac {3 \int \frac {x^2}{\sqrt {b x^4+a}}dx}{2 b}-\frac {x^3}{2 b \sqrt {a+b x^4}}\right )}{6 a b}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {x^7 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}-\frac {(b c-7 a d) \left (\frac {3 \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {a} \sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{2 b}-\frac {x^3}{2 b \sqrt {a+b x^4}}\right )}{6 a b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^7 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}-\frac {(b c-7 a d) \left (\frac {3 \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{2 b}-\frac {x^3}{2 b \sqrt {a+b x^4}}\right )}{6 a b}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {x^7 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}-\frac {(b c-7 a d) \left (\frac {3 \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{2 b}-\frac {x^3}{2 b \sqrt {a+b x^4}}\right )}{6 a b}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {x^7 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}-\frac {(b c-7 a d) \left (\frac {3 \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^4}}-\frac {x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}}{\sqrt {b}}\right )}{2 b}-\frac {x^3}{2 b \sqrt {a+b x^4}}\right )}{6 a b}\)

Input:

Int[(x^6*(c + d*x^4))/(a + b*x^4)^(5/2),x]
 

Output:

((b*c - a*d)*x^7)/(6*a*b*(a + b*x^4)^(3/2)) - ((b*c - 7*a*d)*(-1/2*x^3/(b* 
Sqrt[a + b*x^4]) + (3*(-((-((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]*EllipticE[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])))/(2*b)))/(6*a*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 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 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 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)]))
 

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 = 1.87 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.63

method	result	size
elliptic	\(\frac {x^{3} \left (a d -c b \right ) \sqrt {b \,x^{4}+a}}{6 b^{4} \left (x^{4}+\frac {a}{b}\right )^{2}}-\frac {x^{3} \left (3 a d -c b \right )}{4 b^{2} a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {i \left (\frac {d}{b^{2}}+\frac {3 a d -c b}{4 a \,b^{2}}\right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\)	\(190\)
default	\(c \left (-\frac {x^{3} \sqrt {b \,x^{4}+a}}{6 b^{3} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x^{3}}{4 b a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{4 b^{\frac {3}{2}} \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (\frac {a \,x^{3} \sqrt {b \,x^{4}+a}}{6 b^{4} \left (x^{4}+\frac {a}{b}\right )^{2}}-\frac {3 x^{3}}{4 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {7 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{4 b^{\frac {5}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(302\)

Input:

int(x^6*(d*x^4+c)/(b*x^4+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*x^3/b^4*(a*d-b*c)*(b*x^4+a)^(1/2)/(x^4+a/b)^2-1/4/b^2/a*x^3*(3*a*d-b*c 
)/((x^4+a/b)*b)^(1/2)+I*(d/b^2+1/4*(3*a*d-b*c)/a/b^2)*a^(1/2)/(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)/b^(1/2)*(EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-Elli 
pticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.82 \[ \int \frac {x^6 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=-\frac {3 \, {\left ({\left (b^{3} c - 7 \, a b^{2} d\right )} x^{9} + 2 \, {\left (a b^{2} c - 7 \, a^{2} b d\right )} x^{5} + {\left (a^{2} b c - 7 \, a^{3} d\right )} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 3 \, {\left ({\left (b^{3} c - 7 \, a b^{2} d\right )} x^{9} + 2 \, {\left (a b^{2} c - 7 \, a^{2} b d\right )} x^{5} + {\left (a^{2} b c - 7 \, a^{3} d\right )} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (12 \, a b^{2} d x^{8} - 5 \, {\left (a b^{2} c - 7 \, a^{2} b d\right )} x^{4} - 3 \, a^{2} b c + 21 \, a^{3} d\right )} \sqrt {b x^{4} + a}}{12 \, {\left (a b^{5} x^{9} + 2 \, a^{2} b^{4} x^{5} + a^{3} b^{3} x\right )}} \] Input:

integrate(x^6*(d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="fricas")
 

Output:

-1/12*(3*((b^3*c - 7*a*b^2*d)*x^9 + 2*(a*b^2*c - 7*a^2*b*d)*x^5 + (a^2*b*c 
 - 7*a^3*d)*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) 
 - 3*((b^3*c - 7*a*b^2*d)*x^9 + 2*(a*b^2*c - 7*a^2*b*d)*x^5 + (a^2*b*c - 7 
*a^3*d)*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) - ( 
12*a*b^2*d*x^8 - 5*(a*b^2*c - 7*a^2*b*d)*x^4 - 3*a^2*b*c + 21*a^3*d)*sqrt( 
b*x^4 + a))/(a*b^5*x^9 + 2*a^2*b^4*x^5 + a^3*b^3*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 31.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.26 \[ \int \frac {x^6 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {c x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {5}{2} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {11}{4}\right )} + \frac {d x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {15}{4}\right )} \] Input:

integrate(x**6*(d*x**4+c)/(b*x**4+a)**(5/2),x)
 

Output:

c*x**7*gamma(7/4)*hyper((7/4, 5/2), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4* 
a**(5/2)*gamma(11/4)) + d*x**11*gamma(11/4)*hyper((5/2, 11/4), (15/4,), b* 
x**4*exp_polar(I*pi)/a)/(4*a**(5/2)*gamma(15/4))
 

Maxima [F]

\[ \int \frac {x^6 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{6}}{{\left (b x^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^6*(d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate((d*x^4 + c)*x^6/(b*x^4 + a)^(5/2), x)
 

Giac [F]

\[ \int \frac {x^6 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{6}}{{\left (b x^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^6*(d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^6/(b*x^4 + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\int \frac {x^6\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{5/2}} \,d x \] Input:

int((x^6*(c + d*x^4))/(a + b*x^4)^(5/2),x)
 

Output:

int((x^6*(c + d*x^4))/(a + b*x^4)^(5/2), x)
 

Reduce [F]

\[ \int \frac {x^6 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {7 \sqrt {b \,x^{4}+a}\, a d \,x^{3}-\sqrt {b \,x^{4}+a}\, b c \,x^{3}+3 \sqrt {b \,x^{4}+a}\, b d \,x^{7}-21 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{4} d +3 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{3} b c -42 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{3} b d \,x^{4}+6 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{2} b^{2} c \,x^{4}-21 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{2} b^{2} d \,x^{8}+3 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a \,b^{3} c \,x^{8}}{3 b^{2} \left (b^{2} x^{8}+2 a b \,x^{4}+a^{2}\right )} \] Input:

int(x^6*(d*x^4+c)/(b*x^4+a)^(5/2),x)
 

Output:

(7*sqrt(a + b*x**4)*a*d*x**3 - sqrt(a + b*x**4)*b*c*x**3 + 3*sqrt(a + b*x* 
*4)*b*d*x**7 - 21*int((sqrt(a + b*x**4)*x**2)/(a**3 + 3*a**2*b*x**4 + 3*a* 
b**2*x**8 + b**3*x**12),x)*a**4*d + 3*int((sqrt(a + b*x**4)*x**2)/(a**3 + 
3*a**2*b*x**4 + 3*a*b**2*x**8 + b**3*x**12),x)*a**3*b*c - 42*int((sqrt(a + 
 b*x**4)*x**2)/(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 + 6*int((sqrt(a + b*x**4)*x**2)/(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 - 21*int((sqrt(a + b*x**4)*x**2)/( 
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 + 3 
*int((sqrt(a + b*x**4)*x**2)/(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)/(3*b**2*(a**2 + 2*a*b*x**4 + b**2*x**8))