3.32 \(\int \frac{a-b x^3}{(a+b x^3)^{13/3}} \, dx\)

Optimal. Leaf size=74 \[ \frac{18 x}{35 a^3 \sqrt [3]{a+b x^3}}+\frac{6 x}{35 a^2 \left (a+b x^3\right )^{4/3}}+\frac{4 x}{35 a \left (a+b x^3\right )^{7/3}}+\frac{x}{5 \left (a+b x^3\right )^{10/3}} \]

[Out]

x/(5*(a + b*x^3)^(10/3)) + (4*x)/(35*a*(a + b*x^3)^(7/3)) + (6*x)/(35*a^2*(a + b*x^3)^(4/3)) + (18*x)/(35*a^3*
(a + b*x^3)^(1/3))

________________________________________________________________________________________

Rubi [A]  time = 0.0189451, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {385, 192, 191} \[ \frac{18 x}{35 a^3 \sqrt [3]{a+b x^3}}+\frac{6 x}{35 a^2 \left (a+b x^3\right )^{4/3}}+\frac{4 x}{35 a \left (a+b x^3\right )^{7/3}}+\frac{x}{5 \left (a+b x^3\right )^{10/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^3)/(a + b*x^3)^(13/3),x]

[Out]

x/(5*(a + b*x^3)^(10/3)) + (4*x)/(35*a*(a + b*x^3)^(7/3)) + (6*x)/(35*a^2*(a + b*x^3)^(4/3)) + (18*x)/(35*a^3*
(a + b*x^3)^(1/3))

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{a-b x^3}{\left (a+b x^3\right )^{13/3}} \, dx &=\frac{x}{5 \left (a+b x^3\right )^{10/3}}+\frac{4}{5} \int \frac{1}{\left (a+b x^3\right )^{10/3}} \, dx\\ &=\frac{x}{5 \left (a+b x^3\right )^{10/3}}+\frac{4 x}{35 a \left (a+b x^3\right )^{7/3}}+\frac{24 \int \frac{1}{\left (a+b x^3\right )^{7/3}} \, dx}{35 a}\\ &=\frac{x}{5 \left (a+b x^3\right )^{10/3}}+\frac{4 x}{35 a \left (a+b x^3\right )^{7/3}}+\frac{6 x}{35 a^2 \left (a+b x^3\right )^{4/3}}+\frac{18 \int \frac{1}{\left (a+b x^3\right )^{4/3}} \, dx}{35 a^2}\\ &=\frac{x}{5 \left (a+b x^3\right )^{10/3}}+\frac{4 x}{35 a \left (a+b x^3\right )^{7/3}}+\frac{6 x}{35 a^2 \left (a+b x^3\right )^{4/3}}+\frac{18 x}{35 a^3 \sqrt [3]{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0210671, size = 51, normalized size = 0.69 \[ \frac{x \left (70 a^2 b x^3+35 a^3+60 a b^2 x^6+18 b^3 x^9\right )}{35 a^3 \left (a+b x^3\right )^{10/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^3)/(a + b*x^3)^(13/3),x]

[Out]

(x*(35*a^3 + 70*a^2*b*x^3 + 60*a*b^2*x^6 + 18*b^3*x^9))/(35*a^3*(a + b*x^3)^(10/3))

________________________________________________________________________________________

Maple [A]  time = 0.003, size = 48, normalized size = 0.7 \begin{align*}{\frac{x \left ( 18\,{b}^{3}{x}^{9}+60\,{b}^{2}{x}^{6}a+70\,b{x}^{3}{a}^{2}+35\,{a}^{3} \right ) }{35\,{a}^{3}} \left ( b{x}^{3}+a \right ) ^{-{\frac{10}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^3+a)/(b*x^3+a)^(13/3),x)

[Out]

1/35*x*(18*b^3*x^9+60*a*b^2*x^6+70*a^2*b*x^3+35*a^3)/(b*x^3+a)^(10/3)/a^3

________________________________________________________________________________________

Maxima [B]  time = 1.1912, size = 161, normalized size = 2.18 \begin{align*} -\frac{{\left (14 \, b^{2} - \frac{40 \,{\left (b x^{3} + a\right )} b}{x^{3}} + \frac{35 \,{\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} b x^{10}}{140 \,{\left (b x^{3} + a\right )}^{\frac{10}{3}} a^{3}} - \frac{{\left (14 \, b^{3} - \frac{60 \,{\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac{105 \,{\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac{140 \,{\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} x^{10}}{140 \,{\left (b x^{3} + a\right )}^{\frac{10}{3}} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)/(b*x^3+a)^(13/3),x, algorithm="maxima")

[Out]

-1/140*(14*b^2 - 40*(b*x^3 + a)*b/x^3 + 35*(b*x^3 + a)^2/x^6)*b*x^10/((b*x^3 + a)^(10/3)*a^3) - 1/140*(14*b^3
- 60*(b*x^3 + a)*b^2/x^3 + 105*(b*x^3 + a)^2*b/x^6 - 140*(b*x^3 + a)^3/x^9)*x^10/((b*x^3 + a)^(10/3)*a^3)

________________________________________________________________________________________

Fricas [A]  time = 1.53024, size = 197, normalized size = 2.66 \begin{align*} \frac{{\left (18 \, b^{3} x^{10} + 60 \, a b^{2} x^{7} + 70 \, a^{2} b x^{4} + 35 \, a^{3} x\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{35 \,{\left (a^{3} b^{4} x^{12} + 4 \, a^{4} b^{3} x^{9} + 6 \, a^{5} b^{2} x^{6} + 4 \, a^{6} b x^{3} + a^{7}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)/(b*x^3+a)^(13/3),x, algorithm="fricas")

[Out]

1/35*(18*b^3*x^10 + 60*a*b^2*x^7 + 70*a^2*b*x^4 + 35*a^3*x)*(b*x^3 + a)^(2/3)/(a^3*b^4*x^12 + 4*a^4*b^3*x^9 +
6*a^5*b^2*x^6 + 4*a^6*b*x^3 + a^7)

________________________________________________________________________________________

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)/(b*x**3+a)**(13/3),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{b x^{3} - a}{{\left (b x^{3} + a\right )}^{\frac{13}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)/(b*x^3+a)^(13/3),x, algorithm="giac")

[Out]

integrate(-(b*x^3 - a)/(b*x^3 + a)^(13/3), x)