3.53 \(\int \frac {(a-b x^3)^2}{(a+b x^3)^{11/3}} \, dx\)

Optimal. Leaf size=77 \[ \frac {3 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {8}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{4 a \left (a+b x^3\right )^{2/3}}+\frac {x \left (a-b x^3\right )}{4 \left (a+b x^3\right )^{8/3}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {413, 12, 246, 245} \[ \frac {3 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {8}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{4 a \left (a+b x^3\right )^{2/3}}+\frac {x \left (a-b x^3\right )}{4 \left (a+b x^3\right )^{8/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^3)^2/(a + b*x^3)^(11/3),x]

[Out]

(x*(a - b*x^3))/(4*(a + b*x^3)^(8/3)) + (3*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 8/3, 4/3, -((b*x^3)/
a)])/(4*a*(a + b*x^3)^(2/3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 245

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[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 413

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*} \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{11/3}} \, dx &=\frac {x \left (a-b x^3\right )}{4 \left (a+b x^3\right )^{8/3}}+\frac {\int \frac {6 a^2 b}{\left (a+b x^3\right )^{8/3}} \, dx}{8 a b}\\ &=\frac {x \left (a-b x^3\right )}{4 \left (a+b x^3\right )^{8/3}}+\frac {1}{4} (3 a) \int \frac {1}{\left (a+b x^3\right )^{8/3}} \, dx\\ &=\frac {x \left (a-b x^3\right )}{4 \left (a+b x^3\right )^{8/3}}+\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 )^{8/3}} \, dx}{4 a \left (a+b x^3\right )^{2/3}}\\ &=\frac {x \left (a-b x^3\right )}{4 \left (a+b x^3\right )^{8/3}}+\frac {3 x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {8}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{4 a \left (a+b x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 85, normalized size = 1.10 \[ \frac {7 a^2 x+3 x \left (a+b x^3\right )^2 \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )+5 a b x^4+3 b^2 x^7}{10 a \left (a+b x^3\right )^{8/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^3)^2/(a + b*x^3)^(11/3),x]

[Out]

(7*a^2*x + 5*a*b*x^4 + 3*b^2*x^7 + 3*x*(a + b*x^3)^2*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -(
(b*x^3)/a)])/(10*a*(a + b*x^3)^(8/3))

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{6} - 2 \, a b x^{3} + a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{b^{4} x^{12} + 4 \, a b^{3} x^{9} + 6 \, a^{2} b^{2} x^{6} + 4 \, a^{3} b x^{3} + a^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2/(b*x^3+a)^(11/3),x, algorithm="fricas")

[Out]

integral((b^2*x^6 - 2*a*b*x^3 + a^2)*(b*x^3 + a)^(1/3)/(b^4*x^12 + 4*a*b^3*x^9 + 6*a^2*b^2*x^6 + 4*a^3*b*x^3 +
a^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {11}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2/(b*x^3+a)^(11/3),x, algorithm="giac")

[Out]

integrate((b*x^3 - a)^2/(b*x^3 + a)^(11/3), x)

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maple [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \int \frac {\left (-b \,x^{3}+a \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {11}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^3+a)^2/(b*x^3+a)^(11/3),x)

[Out]

int((-b*x^3+a)^2/(b*x^3+a)^(11/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {11}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2/(b*x^3+a)^(11/3),x, algorithm="maxima")

[Out]

integrate((b*x^3 - a)^2/(b*x^3 + a)^(11/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a-b\,x^3\right )}^2}{{\left (b\,x^3+a\right )}^{11/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^3)^2/(a + b*x^3)^(11/3),x)

[Out]

int((a - b*x^3)^2/(a + b*x^3)^(11/3), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x**3+a)**2/(b*x**3+a)**(11/3),x)

[Out]

Timed out

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