\(\int \frac {(c+d x^4)^2}{(a+b x^4)^{4/3}} \, dx\) [71]

Optimal result

Integrand size = 21, antiderivative size = 133 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{4/3}} \, dx=\frac {3 (b c-a d)^2 x}{4 a b^2 \sqrt [3]{a+b x^4}}+\frac {3 d^2 x \left (a+b x^4\right )^{2/3}}{11 b^2}+\frac {\left (11 b^2 c^2+66 a b c d-45 a^2 d^2\right ) x \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{44 a b^2 \sqrt [3]{a+b x^4}} \] Output:

3/4*(-a*d+b*c)^2*x/a/b^2/(b*x^4+a)^(1/3)+3/11*d^2*x*(b*x^4+a)^(2/3)/b^2+1/ 
44*(-45*a^2*d^2+66*a*b*c*d+11*b^2*c^2)*x*(1+b*x^4/a)^(1/3)*hypergeom([1/4, 
 1/3],[5/4],-b*x^4/a)/a/b^2/(b*x^4+a)^(1/3)
 

Mathematica [A] (verified)

Time = 15.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.83 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{4/3}} \, dx=\frac {x \left (33 b^2 c^2+45 a^2 d^2+6 a b d \left (-11 c+2 d x^4\right )+\left (11 b^2 c^2+66 a b c d-45 a^2 d^2\right ) \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{44 a b^2 \sqrt [3]{a+b x^4}} \] Input:

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

Output:

(x*(33*b^2*c^2 + 45*a^2*d^2 + 6*a*b*d*(-11*c + 2*d*x^4) + (11*b^2*c^2 + 66 
*a*b*c*d - 45*a^2*d^2)*(1 + (b*x^4)/a)^(1/3)*Hypergeometric2F1[1/4, 1/3, 5 
/4, -((b*x^4)/a)]))/(44*a*b^2*(a + b*x^4)^(1/3))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {930, 27, 913, 779, 778}

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.

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

\(\Big \downarrow \) 930

\(\displaystyle \frac {3 \int \frac {c (b c+3 a d)-d (11 b c-15 a d) x^4}{3 \sqrt [3]{b x^4+a}}dx}{4 a b}+\frac {3 x \left (c+d x^4\right ) (b c-a d)}{4 a b \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {c (b c+3 a d)-d (11 b c-15 a d) x^4}{\sqrt [3]{b x^4+a}}dx}{4 a b}+\frac {3 x \left (c+d x^4\right ) (b c-a d)}{4 a b \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {\left (-45 a^2 d^2+66 a b c d+11 b^2 c^2\right ) \int \frac {1}{\sqrt [3]{b x^4+a}}dx}{11 b}-\frac {3 d x \left (a+b x^4\right )^{2/3} (11 b c-15 a d)}{11 b}}{4 a b}+\frac {3 x \left (c+d x^4\right ) (b c-a d)}{4 a b \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {\frac {\sqrt [3]{\frac {b x^4}{a}+1} \left (-45 a^2 d^2+66 a b c d+11 b^2 c^2\right ) \int \frac {1}{\sqrt [3]{\frac {b x^4}{a}+1}}dx}{11 b \sqrt [3]{a+b x^4}}-\frac {3 d x \left (a+b x^4\right )^{2/3} (11 b c-15 a d)}{11 b}}{4 a b}+\frac {3 x \left (c+d x^4\right ) (b c-a d)}{4 a b \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {\frac {x \sqrt [3]{\frac {b x^4}{a}+1} \left (-45 a^2 d^2+66 a b c d+11 b^2 c^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{11 b \sqrt [3]{a+b x^4}}-\frac {3 d x \left (a+b x^4\right )^{2/3} (11 b c-15 a d)}{11 b}}{4 a b}+\frac {3 x \left (c+d x^4\right ) (b c-a d)}{4 a b \sqrt [3]{a+b x^4}}\)

Input:

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

Output:

(3*(b*c - a*d)*x*(c + d*x^4))/(4*a*b*(a + b*x^4)^(1/3)) + ((-3*d*(11*b*c - 
 15*a*d)*x*(a + b*x^4)^(2/3))/(11*b) + ((11*b^2*c^2 + 66*a*b*c*d - 45*a^2* 
d^2)*x*(1 + (b*x^4)/a)^(1/3)*Hypergeometric2F1[1/4, 1/3, 5/4, -((b*x^4)/a) 
])/(11*b*(a + b*x^4)^(1/3)))/(4*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 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 779

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x 
^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(1 + b*(x^n/a))^p, x], x 
] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Si 
mplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^2*x^8 + 2*c*d*x^4 + c^2)*(b*x^4 + a)^(2/3)/(b^2*x^8 + 2*a*b*x^ 
4 + a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{4/3}} \, dx=\left (\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} a +\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,x^{4}}d x \right ) d^{2}+2 \left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} a +\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,x^{4}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} a +\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,x^{4}}d x \right ) c^{2} \] Input:

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

Output:

int(x**8/((a + b*x**4)**(1/3)*a + (a + b*x**4)**(1/3)*b*x**4),x)*d**2 + 2* 
int(x**4/((a + b*x**4)**(1/3)*a + (a + b*x**4)**(1/3)*b*x**4),x)*c*d + int 
(1/((a + b*x**4)**(1/3)*a + (a + b*x**4)**(1/3)*b*x**4),x)*c**2