Integrand size = 22, antiderivative size = 84 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{4/3}} \, dx=-\frac {A}{a x \sqrt [3]{a+b x^3}}-\frac {(2 A b-a B) x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{2 a^2 \sqrt [3]{a+b x^3}} \] Output:
-A/a/x/(b*x^3+a)^(1/3)-1/2*(2*A*b-B*a)*x^2*(1+b*x^3/a)^(1/3)*hypergeom([2/ 3, 4/3],[5/3],-b*x^3/a)/a^2/(b*x^3+a)^(1/3)
Time = 10.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{4/3}} \, dx=\frac {-2 a A+(-2 A b+a B) x^3 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{2 a^2 x \sqrt [3]{a+b x^3}} \] Input:
Integrate[(A + B*x^3)/(x^2*(a + b*x^3)^(4/3)),x]
Output:
(-2*a*A + (-2*A*b + a*B)*x^3*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[2/3, 4/3, 5/3, -((b*x^3)/a)])/(2*a^2*x*(a + b*x^3)^(1/3))
Time = 0.34 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {955, 889, 888}
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.
\(\displaystyle \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{4/3}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(2 A b-a B) \int \frac {x}{\left (b x^3+a\right )^{4/3}}dx}{a}-\frac {A}{a x \sqrt [3]{a+b x^3}}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle -\frac {\sqrt [3]{\frac {b x^3}{a}+1} (2 A b-a B) \int \frac {x}{\left (\frac {b x^3}{a}+1\right )^{4/3}}dx}{a^2 \sqrt [3]{a+b x^3}}-\frac {A}{a x \sqrt [3]{a+b x^3}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle -\frac {x^2 \sqrt [3]{\frac {b x^3}{a}+1} (2 A b-a B) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{2 a^2 \sqrt [3]{a+b x^3}}-\frac {A}{a x \sqrt [3]{a+b x^3}}\) |
Input:
Int[(A + B*x^3)/(x^2*(a + b*x^3)^(4/3)),x]
Output:
-(A/(a*x*(a + b*x^3)^(1/3))) - ((2*A*b - a*B)*x^2*(1 + (b*x^3)/a)^(1/3)*Hy pergeometric2F1[2/3, 4/3, 5/3, -((b*x^3)/a)])/(2*a^2*(a + b*x^3)^(1/3))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(c*x) ^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0 ] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
\[\int \frac {B \,x^{3}+A}{x^{2} \left (b \,x^{3}+a \right )^{\frac {4}{3}}}d x\]
Input:
int((B*x^3+A)/x^2/(b*x^3+a)^(4/3),x)
Output:
int((B*x^3+A)/x^2/(b*x^3+a)^(4/3),x)
\[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{4/3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} x^{2}} \,d x } \] Input:
integrate((B*x^3+A)/x^2/(b*x^3+a)^(4/3),x, algorithm="fricas")
Output:
integral((B*x^3 + A)*(b*x^3 + a)^(2/3)/(b^2*x^8 + 2*a*b*x^5 + a^2*x^2), x)
Result contains complex when optimal does not.
Time = 6.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{4/3}} \, dx=\frac {A \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {4}{3}} x \Gamma \left (\frac {2}{3}\right )} + \frac {B x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {4}{3}} \Gamma \left (\frac {5}{3}\right )} \] Input:
integrate((B*x**3+A)/x**2/(b*x**3+a)**(4/3),x)
Output:
A*gamma(-1/3)*hyper((-1/3, 4/3), (2/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**( 4/3)*x*gamma(2/3)) + B*x**2*gamma(2/3)*hyper((2/3, 4/3), (5/3,), b*x**3*ex p_polar(I*pi)/a)/(3*a**(4/3)*gamma(5/3))
\[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{4/3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} x^{2}} \,d x } \] Input:
integrate((B*x^3+A)/x^2/(b*x^3+a)^(4/3),x, algorithm="maxima")
Output:
integrate((B*x^3 + A)/((b*x^3 + a)^(4/3)*x^2), x)
\[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{4/3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} x^{2}} \,d x } \] Input:
integrate((B*x^3+A)/x^2/(b*x^3+a)^(4/3),x, algorithm="giac")
Output:
integrate((B*x^3 + A)/((b*x^3 + a)^(4/3)*x^2), x)
Timed out. \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{4/3}} \, dx=\int \frac {B\,x^3+A}{x^2\,{\left (b\,x^3+a\right )}^{4/3}} \,d x \] Input:
int((A + B*x^3)/(x^2*(a + b*x^3)^(4/3)),x)
Output:
int((A + B*x^3)/(x^2*(a + b*x^3)^(4/3)), x)
\[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^{4/3}} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{2}}d x \] Input:
int((B*x^3+A)/x^2/(b*x^3+a)^(4/3),x)
Output:
int(1/((a + b*x**3)**(1/3)*x**2),x)