\(\int \frac {(c+d x^4)^2}{\sqrt [3]{a+b x^4}} \, dx\) [67]

Optimal result

Integrand size = 21, antiderivative size = 125 \[ \int \frac {\left (c+d x^4\right )^2}{\sqrt [3]{a+b x^4}} \, dx=\frac {3 d (46 b c-15 a d) x \left (a+b x^4\right )^{2/3}}{253 b^2}+\frac {3 d^2 x^5 \left (a+b x^4\right )^{2/3}}{23 b}+\frac {\left (253 c^2-\frac {3 a d (46 b c-15 a d)}{b^2}\right ) x \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{253 \sqrt [3]{a+b x^4}} \] Output:

3/253*d*(-15*a*d+46*b*c)*x*(b*x^4+a)^(2/3)/b^2+3/23*d^2*x^5*(b*x^4+a)^(2/3 
)/b+1/253*(253*c^2-3*a*d*(-15*a*d+46*b*c)/b^2)*x*(1+b*x^4/a)^(1/3)*hyperge 
om([1/4, 1/3],[5/4],-b*x^4/a)/(b*x^4+a)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 14.44 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.29 \[ \int \frac {\left (c+d x^4\right )^2}{\sqrt [3]{a+b x^4}} \, dx=\frac {x \sqrt [3]{1+\frac {b x^4}{a}} \left (39 a \left (45 c^2+18 c d x^4+5 d^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {13}{4},-\frac {b x^4}{a}\right )-8 b x^4 \left (7 c^2+10 c d x^4+3 d^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {4}{3},\frac {17}{4},-\frac {b x^4}{a}\right )-16 b x^4 \left (c+d x^4\right )^2 \, _3F_2\left (\frac {5}{4},\frac {4}{3},2;1,\frac {17}{4};-\frac {b x^4}{a}\right )\right )}{1755 a \sqrt [3]{a+b x^4}} \] Input:

Integrate[(c + d*x^4)^2/(a + b*x^4)^(1/3),x]
 

Output:

(x*(1 + (b*x^4)/a)^(1/3)*(39*a*(45*c^2 + 18*c*d*x^4 + 5*d^2*x^8)*Hypergeom 
etric2F1[1/4, 1/3, 13/4, -((b*x^4)/a)] - 8*b*x^4*(7*c^2 + 10*c*d*x^4 + 3*d 
^2*x^8)*Hypergeometric2F1[5/4, 4/3, 17/4, -((b*x^4)/a)] - 16*b*x^4*(c + d* 
x^4)^2*HypergeometricPFQ[{5/4, 4/3, 2}, {1, 17/4}, -((b*x^4)/a)]))/(1755*a 
*(a + b*x^4)^(1/3))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {933, 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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {3 \int \frac {5 d (7 b c-3 a d) x^4+c (23 b c-3 a d)}{3 \sqrt [3]{b x^4+a}}dx}{23 b}+\frac {3 d x \left (a+b x^4\right )^{2/3} \left (c+d x^4\right )}{23 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {5 d (7 b c-3 a d) x^4+c (23 b c-3 a d)}{\sqrt [3]{b x^4+a}}dx}{23 b}+\frac {3 d x \left (a+b x^4\right )^{2/3} \left (c+d x^4\right )}{23 b}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {\left (45 a^2 d^2-138 a b c d+253 b^2 c^2\right ) \int \frac {1}{\sqrt [3]{b x^4+a}}dx}{11 b}+\frac {15 d x \left (a+b x^4\right )^{2/3} (7 b c-3 a d)}{11 b}}{23 b}+\frac {3 d x \left (a+b x^4\right )^{2/3} \left (c+d x^4\right )}{23 b}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {\frac {\sqrt [3]{\frac {b x^4}{a}+1} \left (45 a^2 d^2-138 a b c d+253 b^2 c^2\right ) \int \frac {1}{\sqrt [3]{\frac {b x^4}{a}+1}}dx}{11 b \sqrt [3]{a+b x^4}}+\frac {15 d x \left (a+b x^4\right )^{2/3} (7 b c-3 a d)}{11 b}}{23 b}+\frac {3 d x \left (a+b x^4\right )^{2/3} \left (c+d x^4\right )}{23 b}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {\frac {x \sqrt [3]{\frac {b x^4}{a}+1} \left (45 a^2 d^2-138 a b c d+253 b^2 c^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{11 b \sqrt [3]{a+b x^4}}+\frac {15 d x \left (a+b x^4\right )^{2/3} (7 b c-3 a d)}{11 b}}{23 b}+\frac {3 d x \left (a+b x^4\right )^{2/3} \left (c+d x^4\right )}{23 b}\)

Input:

Int[(c + d*x^4)^2/(a + b*x^4)^(1/3),x]
 

Output:

(3*d*x*(a + b*x^4)^(2/3)*(c + d*x^4))/(23*b) + ((15*d*(7*b*c - 3*a*d)*x*(a 
 + b*x^4)^(2/3))/(11*b) + ((253*b^2*c^2 - 138*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)))/(23*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 933

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

Maple [F]

\[\int \frac {\left (d \,x^{4}+c \right )^{2}}{\left (b \,x^{4}+a \right )^{\frac {1}{3}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((d*x^4+c)^2/(b*x^4+a)^(1/3),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.58 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.99 \[ \int \frac {\left (c+d x^4\right )^2}{\sqrt [3]{a+b x^4}} \, dx=\frac {c^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{3} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} \Gamma \left (\frac {5}{4}\right )} + \frac {c d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt [3]{a} \Gamma \left (\frac {9}{4}\right )} + \frac {d^{2} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} \Gamma \left (\frac {13}{4}\right )} \] Input:

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

Output:

c**2*x*gamma(1/4)*hyper((1/4, 1/3), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a 
**(1/3)*gamma(5/4)) + c*d*x**5*gamma(5/4)*hyper((1/3, 5/4), (9/4,), b*x**4 
*exp_polar(I*pi)/a)/(2*a**(1/3)*gamma(9/4)) + d**2*x**9*gamma(9/4)*hyper(( 
1/3, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(1/3)*gamma(13/4))
 

Maxima [F]

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

integrate((d*x^4+c)^2/(b*x^4+a)^(1/3),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)^2/(b*x^4 + a)^(1/3), x)
 

Giac [F]

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

integrate((d*x^4+c)^2/(b*x^4+a)^(1/3),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((d*x^4 + c)^2/(b*x^4 + a)^(1/3), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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