3.46 \(\int \frac{(a-b x^3)^2}{(a+b x^3)^{16/3}} \, dx\)

Optimal. Leaf size=98 \[ \frac{423 x}{910 a^3 \sqrt [3]{a+b x^3}}+\frac{141 x}{910 a^2 \left (a+b x^3\right )^{4/3}}+\frac{47 x}{455 a \left (a+b x^3\right )^{7/3}}+\frac{8 x}{65 \left (a+b x^3\right )^{10/3}}+\frac{2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}} \]

[Out]

(2*x*(a - b*x^3))/(13*(a + b*x^3)^(13/3)) + (8*x)/(65*(a + b*x^3)^(10/3)) + (47*x)/(455*a*(a + b*x^3)^(7/3)) +
(141*x)/(910*a^2*(a + b*x^3)^(4/3)) + (423*x)/(910*a^3*(a + b*x^3)^(1/3))

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Rubi [A]  time = 0.0353347, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {413, 385, 192, 191} \[ \frac{423 x}{910 a^3 \sqrt [3]{a+b x^3}}+\frac{141 x}{910 a^2 \left (a+b x^3\right )^{4/3}}+\frac{47 x}{455 a \left (a+b x^3\right )^{7/3}}+\frac{8 x}{65 \left (a+b x^3\right )^{10/3}}+\frac{2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^3)^2/(a + b*x^3)^(16/3),x]

[Out]

(2*x*(a - b*x^3))/(13*(a + b*x^3)^(13/3)) + (8*x)/(65*(a + b*x^3)^(10/3)) + (47*x)/(455*a*(a + b*x^3)^(7/3)) +
(141*x)/(910*a^2*(a + b*x^3)^(4/3)) + (423*x)/(910*a^3*(a + b*x^3)^(1/3))

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]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(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])

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx &=\frac{2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}+\frac{\int \frac{11 a^2 b-5 a b^2 x^3}{\left (a+b x^3\right )^{13/3}} \, dx}{13 a b}\\ &=\frac{2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}+\frac{8 x}{65 \left (a+b x^3\right )^{10/3}}+\frac{47}{65} \int \frac{1}{\left (a+b x^3\right )^{10/3}} \, dx\\ &=\frac{2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}+\frac{8 x}{65 \left (a+b x^3\right )^{10/3}}+\frac{47 x}{455 a \left (a+b x^3\right )^{7/3}}+\frac{282 \int \frac{1}{\left (a+b x^3\right )^{7/3}} \, dx}{455 a}\\ &=\frac{2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}+\frac{8 x}{65 \left (a+b x^3\right )^{10/3}}+\frac{47 x}{455 a \left (a+b x^3\right )^{7/3}}+\frac{141 x}{910 a^2 \left (a+b x^3\right )^{4/3}}+\frac{423 \int \frac{1}{\left (a+b x^3\right )^{4/3}} \, dx}{910 a^2}\\ &=\frac{2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}+\frac{8 x}{65 \left (a+b x^3\right )^{10/3}}+\frac{47 x}{455 a \left (a+b x^3\right )^{7/3}}+\frac{141 x}{910 a^2 \left (a+b x^3\right )^{4/3}}+\frac{423 x}{910 a^3 \sqrt [3]{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0351586, size = 62, normalized size = 0.63 \[ \frac{x \left (3055 a^2 b^2 x^6+2275 a^3 b x^3+910 a^4+1833 a b^3 x^9+423 b^4 x^{12}\right )}{910 a^3 \left (a+b x^3\right )^{13/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^3)^2/(a + b*x^3)^(16/3),x]

[Out]

(x*(910*a^4 + 2275*a^3*b*x^3 + 3055*a^2*b^2*x^6 + 1833*a*b^3*x^9 + 423*b^4*x^12))/(910*a^3*(a + b*x^3)^(13/3))

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Maple [A]  time = 0.007, size = 59, normalized size = 0.6 \begin{align*}{\frac{x \left ( 423\,{b}^{4}{x}^{12}+1833\,{b}^{3}{x}^{9}a+3055\,{b}^{2}{x}^{6}{a}^{2}+2275\,b{x}^{3}{a}^{3}+910\,{a}^{4} \right ) }{910\,{a}^{3}} \left ( b{x}^{3}+a \right ) ^{-{\frac{13}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^3+a)^2/(b*x^3+a)^(16/3),x)

[Out]

1/910*x*(423*b^4*x^12+1833*a*b^3*x^9+3055*a^2*b^2*x^6+2275*a^3*b*x^3+910*a^4)/(b*x^3+a)^(13/3)/a^3

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Maxima [B]  time = 0.96415, size = 278, normalized size = 2.84 \begin{align*} \frac{{\left (35 \, b^{2} - \frac{91 \,{\left (b x^{3} + a\right )} b}{x^{3}} + \frac{65 \,{\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} b^{2} x^{13}}{455 \,{\left (b x^{3} + a\right )}^{\frac{13}{3}} a^{3}} + \frac{{\left (140 \, b^{3} - \frac{546 \,{\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac{780 \,{\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac{455 \,{\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} b x^{13}}{910 \,{\left (b x^{3} + a\right )}^{\frac{13}{3}} a^{3}} + \frac{{\left (35 \, b^{4} - \frac{182 \,{\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac{390 \,{\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}} - \frac{455 \,{\left (b x^{3} + a\right )}^{3} b}{x^{9}} + \frac{455 \,{\left (b x^{3} + a\right )}^{4}}{x^{12}}\right )} x^{13}}{455 \,{\left (b x^{3} + a\right )}^{\frac{13}{3}} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2/(b*x^3+a)^(16/3),x, algorithm="maxima")

[Out]

1/455*(35*b^2 - 91*(b*x^3 + a)*b/x^3 + 65*(b*x^3 + a)^2/x^6)*b^2*x^13/((b*x^3 + a)^(13/3)*a^3) + 1/910*(140*b^
3 - 546*(b*x^3 + a)*b^2/x^3 + 780*(b*x^3 + a)^2*b/x^6 - 455*(b*x^3 + a)^3/x^9)*b*x^13/((b*x^3 + a)^(13/3)*a^3)
+ 1/455*(35*b^4 - 182*(b*x^3 + a)*b^3/x^3 + 390*(b*x^3 + a)^2*b^2/x^6 - 455*(b*x^3 + a)^3*b/x^9 + 455*(b*x^3
+ a)^4/x^12)*x^13/((b*x^3 + a)^(13/3)*a^3)

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Fricas [A]  time = 2.04223, size = 259, normalized size = 2.64 \begin{align*} \frac{{\left (423 \, b^{4} x^{13} + 1833 \, a b^{3} x^{10} + 3055 \, a^{2} b^{2} x^{7} + 2275 \, a^{3} b x^{4} + 910 \, a^{4} x\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{910 \,{\left (a^{3} b^{5} x^{15} + 5 \, a^{4} b^{4} x^{12} + 10 \, a^{5} b^{3} x^{9} + 10 \, a^{6} b^{2} x^{6} + 5 \, a^{7} b x^{3} + a^{8}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2/(b*x^3+a)^(16/3),x, algorithm="fricas")

[Out]

1/910*(423*b^4*x^13 + 1833*a*b^3*x^10 + 3055*a^2*b^2*x^7 + 2275*a^3*b*x^4 + 910*a^4*x)*(b*x^3 + a)^(2/3)/(a^3*
b^5*x^15 + 5*a^4*b^4*x^12 + 10*a^5*b^3*x^9 + 10*a^6*b^2*x^6 + 5*a^7*b*x^3 + a^8)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x**3+a)**2/(b*x**3+a)**(16/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac{16}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2/(b*x^3+a)^(16/3),x, algorithm="giac")

[Out]

integrate((b*x^3 - a)^2/(b*x^3 + a)^(16/3), x)