\(\int \frac {c+d x^3}{(a+b x^3)^{8/3}} \, dx\) [121]

Optimal result

Integrand size = 19, antiderivative size = 84 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{8/3}} \, dx=-\frac {d x}{4 b \left (a+b x^3\right )^{5/3}}+\frac {(4 b c+a d) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {8}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{4 a^2 b \left (a+b x^3\right )^{2/3}} \] Output:

-1/4*d*x/b/(b*x^3+a)^(5/3)+1/4*(a*d+4*b*c)*x*(1+b*x^3/a)^(2/3)*hypergeom([ 
1/3, 8/3],[4/3],-b*x^3/a)/a^2/b/(b*x^3+a)^(2/3)
 

Mathematica [A] (verified)

Time = 10.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{8/3}} \, dx=\frac {x \left (-d+\frac {(4 b c+a d) \left (a+b x^3\right ) \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {8}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{a^2}\right )}{4 b \left (a+b x^3\right )^{5/3}} \] Input:

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

Output:

(x*(-d + ((4*b*c + a*d)*(a + b*x^3)*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F 
1[1/3, 8/3, 4/3, -((b*x^3)/a)])/a^2))/(4*b*(a + b*x^3)^(5/3))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.

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

\(\Big \downarrow \) 910

\(\displaystyle \frac {(a d+4 b c) \int \frac {1}{\left (b x^3+a\right )^{5/3}}dx}{5 a b}+\frac {x (b c-a d)}{5 a b \left (a+b x^3\right )^{5/3}}\)

\(\Big \downarrow \) 779

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

\(\Big \downarrow \) 778

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

Input:

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

Output:

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

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])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 39.44 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{8/3}} \, dx=\frac {c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {8}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {8}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {8}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {8}{3}} \Gamma \left (\frac {7}{3}\right )} \] Input:

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

Output:

c*x*gamma(1/3)*hyper((1/3, 8/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**( 
8/3)*gamma(4/3)) + d*x**4*gamma(4/3)*hyper((4/3, 8/3), (7/3,), b*x**3*exp_ 
polar(I*pi)/a)/(3*a**(8/3)*gamma(7/3))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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)*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