Integrand size = 19, antiderivative size = 82 \[ \int \frac {A+B x^3}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {B x \sqrt [3]{a+b x^3}}{2 b}+\frac {(2 A b-a B) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 b \left (a+b x^3\right )^{2/3}} \] Output:
1/2*B*x*(b*x^3+a)^(1/3)/b+1/2*(2*A*b-B*a)*x*(1+b*x^3/a)^(2/3)*hypergeom([1 /3, 2/3],[4/3],-b*x^3/a)/b/(b*x^3+a)^(2/3)
Time = 10.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x^3}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {B x \left (a+b x^3\right )+(2 A b-a B) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 b \left (a+b x^3\right )^{2/3}} \] Input:
Integrate[(A + B*x^3)/(a + b*x^3)^(2/3),x]
Output:
(B*x*(a + b*x^3) + (2*A*b - a*B)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1 [1/3, 2/3, 4/3, -((b*x^3)/a)])/(2*b*(a + b*x^3)^(2/3))
Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.
\(\displaystyle \int \frac {A+B x^3}{\left (a+b x^3\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 913 |
\(\displaystyle \frac {(2 A b-a B) \int \frac {1}{\left (b x^3+a\right )^{2/3}}dx}{2 b}+\frac {B x \sqrt [3]{a+b x^3}}{2 b}\) |
\(\Big \downarrow \) 779 |
\(\displaystyle \frac {\left (\frac {b x^3}{a}+1\right )^{2/3} (2 A b-a B) \int \frac {1}{\left (\frac {b x^3}{a}+1\right )^{2/3}}dx}{2 b \left (a+b x^3\right )^{2/3}}+\frac {B x \sqrt [3]{a+b x^3}}{2 b}\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} (2 A b-a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 b \left (a+b x^3\right )^{2/3}}+\frac {B x \sqrt [3]{a+b x^3}}{2 b}\) |
Input:
Int[(A + B*x^3)/(a + b*x^3)^(2/3),x]
Output:
(B*x*(a + b*x^3)^(1/3))/(2*b) + ((2*A*b - a*B)*x*(1 + (b*x^3)/a)^(2/3)*Hyp ergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(2*b*(a + b*x^3)^(2/3))
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])
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])
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]
\[\int \frac {B \,x^{3}+A}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x\]
Input:
int((B*x^3+A)/(b*x^3+a)^(2/3),x)
Output:
int((B*x^3+A)/(b*x^3+a)^(2/3),x)
\[ \int \frac {A+B x^3}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="fricas")
Output:
integral((B*x^3 + A)/(b*x^3 + a)^(2/3), x)
Result contains complex when optimal does not.
Time = 1.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x^3}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {A x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {B x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {7}{3}\right )} \] Input:
integrate((B*x**3+A)/(b*x**3+a)**(2/3),x)
Output:
A*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)) + B*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))
\[ \int \frac {A+B x^3}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="maxima")
Output:
integrate((B*x^3 + A)/(b*x^3 + a)^(2/3), x)
\[ \int \frac {A+B x^3}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="giac")
Output:
integrate((B*x^3 + A)/(b*x^3 + a)^(2/3), x)
Timed out. \[ \int \frac {A+B x^3}{\left (a+b x^3\right )^{2/3}} \, dx=\int \frac {B\,x^3+A}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \] Input:
int((A + B*x^3)/(a + b*x^3)^(2/3),x)
Output:
int((A + B*x^3)/(a + b*x^3)^(2/3), x)
\[ \int \frac {A+B x^3}{\left (a+b x^3\right )^{2/3}} \, dx=\left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) b +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a \] Input:
int((B*x^3+A)/(b*x^3+a)^(2/3),x)
Output:
int(x**3/(a + b*x**3)**(2/3),x)*b + int(1/(a + b*x**3)**(2/3),x)*a