\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{8/3}} \, dx\) [136]

Optimal result

Integrand size = 21, antiderivative size = 130 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{8/3}} \, dx=\frac {(b c-a d)^2 x}{5 a b^2 \left (a+b x^3\right )^{5/3}}-\frac {d^2 x}{b^2 \left (a+b x^3\right )^{2/3}}+\frac {2 \left (2 b^2 c^2+a b c d+2 a^2 d^2\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 a^2 b^2 \left (a+b x^3\right )^{2/3}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 15.05 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.32 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{8/3}} \, dx=\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Gamma}\left (\frac {2}{3}\right ) \left (5 a \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {8}{3},\frac {10}{3},-\frac {b x^3}{a}\right )-2 b x^3 \left (11 c^2+16 c d x^3+5 d^2 x^6\right ) \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {11}{3},\frac {13}{3},-\frac {b x^3}{a}\right )-6 b x^3 \left (c+d x^3\right )^2 \, _3F_2\left (\frac {4}{3},2,\frac {11}{3};1,\frac {13}{3};-\frac {b x^3}{a}\right )\right )}{63 a^3 \left (a+b x^3\right )^{2/3} \operatorname {Gamma}\left (\frac {8}{3}\right )} \] Input:

Integrate[(c + d*x^3)^2/(a + b*x^3)^(8/3),x]
 

Output:

(x*(1 + (b*x^3)/a)^(2/3)*Gamma[2/3]*(5*a*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6)* 
Hypergeometric2F1[1/3, 8/3, 10/3, -((b*x^3)/a)] - 2*b*x^3*(11*c^2 + 16*c*d 
*x^3 + 5*d^2*x^6)*Hypergeometric2F1[4/3, 11/3, 13/3, -((b*x^3)/a)] - 6*b*x 
^3*(c + d*x^3)^2*HypergeometricPFQ[{4/3, 2, 11/3}, {1, 13/3}, -((b*x^3)/a) 
]))/(63*a^3*(a + b*x^3)^(2/3)*Gamma[8/3])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {930, 910, 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.

\(\displaystyle \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{8/3}} \, dx\)

\(\Big \downarrow \) 930

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

\(\Big \downarrow \) 910

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

\(\Big \downarrow \) 779

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

\(\Big \downarrow \) 778

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

Input:

Int[(c + d*x^3)^2/(a + b*x^3)^(8/3),x]
 

Output:

((b*c - a*d)*x*(c + d*x^3))/(5*a*b*(a + b*x^3)^(5/3)) + ((2*((b*c^2)/a - ( 
a*d^2)/b)*x)/(a + b*x^3)^(2/3) + (((2*b*c^2)/a + c*d + (2*a*d^2)/b)*x*(1 + 
 (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(a + b*x 
^3)^(2/3))/(5*a*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 910

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

\[\int \frac {\left (d \,x^{3}+c \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {8}{3}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((d*x^3+c)^2/(b*x^3+a)^(8/3),x, algorithm="fricas")
 

Output:

integral((d^2*x^6 + 2*c*d*x^3 + c^2)*(b*x^3 + a)^(1/3)/(b^3*x^9 + 3*a*b^2* 
x^6 + 3*a^2*b*x^3 + a^3), x)
 

Sympy [F]

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

integrate((d*x**3+c)**2/(b*x**3+a)**(8/3),x)
 

Output:

Integral((c + d*x**3)**2/(a + b*x**3)**(8/3), x)
 

Maxima [F]

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

integrate((d*x^3+c)^2/(b*x^3+a)^(8/3),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((d*x^3+c)^2/(b*x^3+a)^(8/3),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{8/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2}+2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} x^{6}}d x \right ) d^{2}+2 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2}+2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} x^{6}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2}+2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} x^{6}}d x \right ) c^{2} \] Input:

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

Output:

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