\(\int \frac {(c+d x^4)^3}{(a+b x^4)^{3/2}} \, dx\) [36]

Optimal result

Integrand size = 21, antiderivative size = 213 \[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {(b c-a d)^3 x}{2 a b^3 \sqrt {a+b x^4}}+\frac {d^2 (7 b c-4 a d) x \sqrt {a+b x^4}}{7 b^3}+\frac {d^3 x^5 \sqrt {a+b x^4}}{7 b^2}+\frac {\left (7 b^3 c^3+21 a b^2 c^2 d-35 a^2 b c d^2+15 a^3 d^3\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{28 a^{5/4} b^{13/4} \sqrt {a+b x^4}} \] Output:

1/2*(-a*d+b*c)^3*x/a/b^3/(b*x^4+a)^(1/2)+1/7*d^2*(-4*a*d+7*b*c)*x*(b*x^4+a 
)^(1/2)/b^3+1/7*d^3*x^5*(b*x^4+a)^(1/2)/b^2+1/28*(15*a^3*d^3-35*a^2*b*c*d^ 
2+21*a*b^2*c^2*d+7*b^3*c^3)*(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^(13/4)/(b*x^4+a)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 15.22 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.92 \[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} \left (7 (b c-a d)^3 x+2 a d^2 (7 b c-4 a d) x \left (a+b x^4\right )+2 a b d^3 x^5 \left (a+b x^4\right )\right )-i \left (7 b^3 c^3+21 a b^2 c^2 d-35 a^2 b c d^2+15 a^3 d^3\right ) \sqrt {1+\frac {b x^4}{a}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x\right ),-1\right )}{14 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} b^3 \sqrt {a+b x^4}} \] Input:

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

Output:

(Sqrt[(I*Sqrt[b])/Sqrt[a]]*(7*(b*c - a*d)^3*x + 2*a*d^2*(7*b*c - 4*a*d)*x* 
(a + b*x^4) + 2*a*b*d^3*x^5*(a + b*x^4)) - I*(7*b^3*c^3 + 21*a*b^2*c^2*d - 
 35*a^2*b*c*d^2 + 15*a^3*d^3)*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt 
[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(14*a*Sqrt[(I*Sqrt[b])/Sqrt[a]]*b^3*Sqrt[a 
+ b*x^4])
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {930, 1025, 913, 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.

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

\(\Big \downarrow \) 930

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

\(\Big \downarrow \) 1025

\(\displaystyle \frac {\frac {\int \frac {c \left (7 b^2 c^2+14 a b d c-9 a^2 d^2\right )-3 d (7 b c-5 a d) (b c-3 a d) x^4}{\sqrt {b x^4+a}}dx}{7 b}-\frac {d x \sqrt {a+b x^4} \left (c+d x^4\right ) (7 b c-9 a d)}{7 b}}{2 a b}+\frac {x \left (c+d x^4\right )^2 (b c-a d)}{2 a b \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {\frac {\left (15 a^3 d^3-35 a^2 b c d^2+21 a b^2 c^2 d+7 b^3 c^3\right ) \int \frac {1}{\sqrt {b x^4+a}}dx}{b}-\frac {d x \sqrt {a+b x^4} (7 b c-5 a d) (b c-3 a d)}{b}}{7 b}-\frac {d x \sqrt {a+b x^4} \left (c+d x^4\right ) (7 b c-9 a d)}{7 b}}{2 a b}+\frac {x \left (c+d x^4\right )^2 (b c-a d)}{2 a b \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {\frac {\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (15 a^3 d^3-35 a^2 b c d^2+21 a b^2 c^2 d+7 b^3 c^3\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} b^{5/4} \sqrt {a+b x^4}}-\frac {d x \sqrt {a+b x^4} (7 b c-5 a d) (b c-3 a d)}{b}}{7 b}-\frac {d x \sqrt {a+b x^4} \left (c+d x^4\right ) (7 b c-9 a d)}{7 b}}{2 a b}+\frac {x \left (c+d x^4\right )^2 (b c-a d)}{2 a b \sqrt {a+b x^4}}\)

Input:

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

Output:

((b*c - a*d)*x*(c + d*x^4)^2)/(2*a*b*Sqrt[a + b*x^4]) + (-1/7*(d*(7*b*c - 
9*a*d)*x*Sqrt[a + b*x^4]*(c + d*x^4))/b + (-((d*(7*b*c - 5*a*d)*(b*c - 3*a 
*d)*x*Sqrt[a + b*x^4])/b) + ((7*b^3*c^3 + 21*a*b^2*c^2*d - 35*a^2*b*c*d^2 
+ 15*a^3*d^3)*(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*a^(1/4)*b^(5/4)* 
Sqrt[a + b*x^4]))/(7*b))/(2*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 913

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

rule 930

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[(a*d - c*b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 
1))), x] - Simp[1/(a*b*n*(p + 1))   Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q 
- 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*( 
p + q) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q, x]
 

rule 1025

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 8.70 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.38

method	result	size
elliptic	\(-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{2 b^{3} a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {d^{3} x^{5} \sqrt {b \,x^{4}+a}}{7 b^{2}}+\frac {\left (-\frac {d^{2} \left (a d -3 b c \right )}{b^{2}}-\frac {5 a \,d^{3}}{7 b^{2}}\right ) x \sqrt {b \,x^{4}+a}}{3 b}+\frac {\left (\frac {d \left (a^{2} d^{2}-3 a b c d +3 b^{2} c^{2}\right )}{b^{3}}-\frac {a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}}{2 a \,b^{3}}-\frac {\left (-\frac {d^{2} \left (a d -3 b c \right )}{b^{2}}-\frac {5 a \,d^{3}}{7 b^{2}}\right ) a}{3 b}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(294\)
default	\(c^{3} \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d^{3} \left (-\frac {a^{2} x}{2 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x^{5} \sqrt {b \,x^{4}+a}}{7 b^{2}}-\frac {4 a x \sqrt {b \,x^{4}+a}}{7 b^{3}}+\frac {15 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{14 b^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+3 c \,d^{2} \left (\frac {x a}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {5 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+3 c^{2} d \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(450\)
risch	\(-\frac {d^{2} x \left (-b d \,x^{4}+4 a d -7 b c \right ) \sqrt {b \,x^{4}+a}}{7 b^{3}}+\frac {b d \left (11 a^{2} d^{2}-28 a b c d +21 b^{2} c^{2}\right ) \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+4 a^{3} d^{3} \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+7 b^{3} c^{3} \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-7 a^{2} b c \,d^{2} \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )}{7 b^{3}}\)	\(466\)

Input:

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

Output:

-1/2/b^3*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/a*x/((x^4+a/b)*b)^( 
1/2)+1/7*d^3*x^5*(b*x^4+a)^(1/2)/b^2+1/3*(-1/b^2*d^2*(a*d-3*b*c)-5/7*a/b^2 
*d^3)/b*x*(b*x^4+a)^(1/2)+(d*(a^2*d^2-3*a*b*c*d+3*b^2*c^2)/b^3-1/2*(a^3*d^ 
3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/a/b^3-1/3*(-1/b^2*d^2*(a*d-3*b*c)-5 
/7*a/b^2*d^3)/b*a)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I*b^(1/2)*x^2/a^(1/2))^(1/ 
2)*(1+I*b^(1/2)*x^2/a^(1/2))^(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 = 0.11 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.07 \[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {{\left (7 \, a b^{3} c^{3} + 21 \, a^{2} b^{2} c^{2} d - 35 \, a^{3} b c d^{2} + 15 \, a^{4} d^{3} + {\left (7 \, b^{4} c^{3} + 21 \, a b^{3} c^{2} d - 35 \, a^{2} b^{2} c d^{2} + 15 \, a^{3} b d^{3}\right )} x^{4}\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (2 \, a^{2} b^{2} d^{3} x^{9} + 2 \, {\left (7 \, a^{2} b^{2} c d^{2} - 3 \, a^{3} b d^{3}\right )} x^{5} + {\left (7 \, a b^{3} c^{3} - 21 \, a^{2} b^{2} c^{2} d + 35 \, a^{3} b c d^{2} - 15 \, a^{4} d^{3}\right )} x\right )} \sqrt {b x^{4} + a}}{14 \, {\left (a^{2} b^{4} x^{4} + a^{3} b^{3}\right )}} \] Input:

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

Output:

1/14*((7*a*b^3*c^3 + 21*a^2*b^2*c^2*d - 35*a^3*b*c*d^2 + 15*a^4*d^3 + (7*b 
^4*c^3 + 21*a*b^3*c^2*d - 35*a^2*b^2*c*d^2 + 15*a^3*b*d^3)*x^4)*sqrt(b)*(- 
a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + (2*a^2*b^2*d^3*x^9 + 2 
*(7*a^2*b^2*c*d^2 - 3*a^3*b*d^3)*x^5 + (7*a*b^3*c^3 - 21*a^2*b^2*c^2*d + 3 
5*a^3*b*c*d^2 - 15*a^4*d^3)*x)*sqrt(b*x^4 + a))/(a^2*b^4*x^4 + a^3*b^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {-15 \sqrt {b \,x^{4}+a}\, a^{2} d^{3} x +35 \sqrt {b \,x^{4}+a}\, a b c \,d^{2} x -3 \sqrt {b \,x^{4}+a}\, a b \,d^{3} x^{5}-21 \sqrt {b \,x^{4}+a}\, b^{2} c^{2} d x +7 \sqrt {b \,x^{4}+a}\, b^{2} c \,d^{2} x^{5}+\sqrt {b \,x^{4}+a}\, b^{2} d^{3} x^{9}+15 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{4} d^{3}-35 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{3} b c \,d^{2}+15 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{3} b \,d^{3} x^{4}+21 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{2} b^{2} c^{2} d -35 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{2} b^{2} c \,d^{2} x^{4}+7 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a \,b^{3} c^{3}+21 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a \,b^{3} c^{2} d \,x^{4}+7 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) b^{4} c^{3} x^{4}}{7 b^{3} \left (b \,x^{4}+a \right )} \] Input:

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

Output:

( - 15*sqrt(a + b*x**4)*a**2*d**3*x + 35*sqrt(a + b*x**4)*a*b*c*d**2*x - 3 
*sqrt(a + b*x**4)*a*b*d**3*x**5 - 21*sqrt(a + b*x**4)*b**2*c**2*d*x + 7*sq 
rt(a + b*x**4)*b**2*c*d**2*x**5 + sqrt(a + b*x**4)*b**2*d**3*x**9 + 15*int 
(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**4*d**3 - 35*int(sq 
rt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**3*b*c*d**2 + 15*int(s 
qrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**3*b*d**3*x**4 + 21*i 
nt(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**2*b**2*c**2*d - 
35*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**2*b**2*c*d** 
2*x**4 + 7*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a*b**3* 
c**3 + 21*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a*b**3*c 
**2*d*x**4 + 7*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*b** 
4*c**3*x**4)/(7*b**3*(a + b*x**4))