\(\int \frac {(a-b x^3)^2}{(a+b x^3)^{8/3}} \, dx\) [52]

Optimal result

Integrand size = 22, antiderivative size = 74 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{8/3}} \, dx=\frac {2 x \left (a-b x^3\right )}{5 \left (a+b x^3\right )^{5/3}}+\frac {3 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}} \]

[Out]

2/5*x*(-b*x^3+a)/(b*x^3+a)^(5/3)+3/5*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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {424, 21, 252, 251} \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{8/3}} \, dx=\frac {3 x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}+\frac {2 x \left (a-b x^3\right )}{5 \left (a+b x^3\right )^{5/3}} \]

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 424

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x \left (a-b x^3\right )}{5 \left (a+b x^3\right )^{5/3}}+\frac {\int \frac {3 a^2 b+3 a b^2 x^3}{\left (a+b x^3\right )^{5/3}} \, dx}{5 a b} \\ & = \frac {2 x \left (a-b x^3\right )}{5 \left (a+b x^3\right )^{5/3}}+\frac {3}{5} \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx \\ & = \frac {2 x \left (a-b x^3\right )}{5 \left (a+b x^3\right )^{5/3}}+\frac {\left (3 \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx}{5 \left (a+b x^3\right )^{2/3}} \\ & = \frac {2 x \left (a-b x^3\right )}{5 \left (a+b x^3\right )^{5/3}}+\frac {3 x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{8/3}} \, dx=\frac {2 x \left (a-b x^3\right )+3 x \left (a+b x^3\right ) \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{5/3}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((b^2*x^6 - 2*a*b*x^3 + a^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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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