\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{2/3}} \, dx\) [134]

Optimal result

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

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

Mathematica [A] (verified)

Time = 15.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.83 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {x \left (-d \left (a+b x^3\right ) \left (2 a d-b \left (5 c+d x^3\right )\right )+\left (5 b^2 c^2-5 a b c d+2 a^2 d^2\right ) \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )\right )}{5 b^2 \left (a+b x^3\right )^{2/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {933, 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.

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

\(\Big \downarrow \) 933

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

\(\Big \downarrow \) 913

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

\(\Big \downarrow \) 779

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

\(\Big \downarrow \) 778

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

Input:

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

Output:

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

Defintions of rubi rules used

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^2*x^6 + 2*c*d*x^3 + c^2)/(b*x^3 + a)^(2/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.53 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {2 c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {7}{3}\right )} + \frac {d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {10}{3}\right )} \] Input:

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

Output:

c**2*x*gamma(1/3)*hyper((1/3, 2/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a 
**(2/3)*gamma(4/3)) + 2*c*d*x**4*gamma(4/3)*hyper((2/3, 4/3), (7/3,), b*x* 
*3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(7/3)) + d**2*x**7*gamma(7/3)*hyper 
((2/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(10/3))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) d^{2}+2 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) c^{2} \] Input:

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

Output:

int(x**6/(a + b*x**3)**(2/3),x)*d**2 + 2*int(x**3/(a + b*x**3)**(2/3),x)*c 
*d + int(1/(a + b*x**3)**(2/3),x)*c**2