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

Optimal. Leaf size=93 \[ \frac{891 x}{1820 a^4 \sqrt [3]{a+b x^3}}+\frac{297 x}{1820 a^3 \left (a+b x^3\right )^{4/3}}+\frac{99 x}{910 a^2 \left (a+b x^3\right )^{7/3}}+\frac{11 x}{130 a \left (a+b x^3\right )^{10/3}}+\frac{2 x}{13 \left (a+b x^3\right )^{13/3}} \]

[Out]

(2*x)/(13*(a + b*x^3)^(13/3)) + (11*x)/(130*a*(a + b*x^3)^(10/3)) + (99*x)/(910*a^2*(a + b*x^3)^(7/3)) + (297*
x)/(1820*a^3*(a + b*x^3)^(4/3)) + (891*x)/(1820*a^4*(a + b*x^3)^(1/3))

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Rubi [A]  time = 0.0263103, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {385, 192, 191} \[ \frac{891 x}{1820 a^4 \sqrt [3]{a+b x^3}}+\frac{297 x}{1820 a^3 \left (a+b x^3\right )^{4/3}}+\frac{99 x}{910 a^2 \left (a+b x^3\right )^{7/3}}+\frac{11 x}{130 a \left (a+b x^3\right )^{10/3}}+\frac{2 x}{13 \left (a+b x^3\right )^{13/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^3)/(a + b*x^3)^(16/3),x]

[Out]

(2*x)/(13*(a + b*x^3)^(13/3)) + (11*x)/(130*a*(a + b*x^3)^(10/3)) + (99*x)/(910*a^2*(a + b*x^3)^(7/3)) + (297*
x)/(1820*a^3*(a + b*x^3)^(4/3)) + (891*x)/(1820*a^4*(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 )^{16/3}} \, dx &=\frac{2 x}{13 \left (a+b x^3\right )^{13/3}}+\frac{11}{13} \int \frac{1}{\left (a+b x^3\right )^{13/3}} \, dx\\ &=\frac{2 x}{13 \left (a+b x^3\right )^{13/3}}+\frac{11 x}{130 a \left (a+b x^3\right )^{10/3}}+\frac{99 \int \frac{1}{\left (a+b x^3\right )^{10/3}} \, dx}{130 a}\\ &=\frac{2 x}{13 \left (a+b x^3\right )^{13/3}}+\frac{11 x}{130 a \left (a+b x^3\right )^{10/3}}+\frac{99 x}{910 a^2 \left (a+b x^3\right )^{7/3}}+\frac{297 \int \frac{1}{\left (a+b x^3\right )^{7/3}} \, dx}{455 a^2}\\ &=\frac{2 x}{13 \left (a+b x^3\right )^{13/3}}+\frac{11 x}{130 a \left (a+b x^3\right )^{10/3}}+\frac{99 x}{910 a^2 \left (a+b x^3\right )^{7/3}}+\frac{297 x}{1820 a^3 \left (a+b x^3\right )^{4/3}}+\frac{891 \int \frac{1}{\left (a+b x^3\right )^{4/3}} \, dx}{1820 a^3}\\ &=\frac{2 x}{13 \left (a+b x^3\right )^{13/3}}+\frac{11 x}{130 a \left (a+b x^3\right )^{10/3}}+\frac{99 x}{910 a^2 \left (a+b x^3\right )^{7/3}}+\frac{297 x}{1820 a^3 \left (a+b x^3\right )^{4/3}}+\frac{891 x}{1820 a^4 \sqrt [3]{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0231695, size = 62, normalized size = 0.67 \[ \frac{x \left (6435 a^2 b^2 x^6+5005 a^3 b x^3+1820 a^4+3861 a b^3 x^9+891 b^4 x^{12}\right )}{1820 a^4 \left (a+b x^3\right )^{13/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(1820*a^4 + 5005*a^3*b*x^3 + 6435*a^2*b^2*x^6 + 3861*a*b^3*x^9 + 891*b^4*x^12))/(1820*a^4*(a + b*x^3)^(13/3
))

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Maple [A]  time = 0.004, size = 59, normalized size = 0.6 \begin{align*}{\frac{x \left ( 891\,{b}^{4}{x}^{12}+3861\,{b}^{3}{x}^{9}a+6435\,{b}^{2}{x}^{6}{a}^{2}+5005\,b{x}^{3}{a}^{3}+1820\,{a}^{4} \right ) }{1820\,{a}^{4}} \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)/(b*x^3+a)^(16/3),x)

[Out]

1/1820*x*(891*b^4*x^12+3861*a*b^3*x^9+6435*a^2*b^2*x^6+5005*a^3*b*x^3+1820*a^4)/(b*x^3+a)^(13/3)/a^4

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Maxima [B]  time = 1.38391, size = 207, normalized size = 2.23 \begin{align*} \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}}{1820 \,{\left (b x^{3} + a\right )}^{\frac{13}{3}} a^{4}} + \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^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1820*(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^4) + 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^4)

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Fricas [A]  time = 1.48221, size = 262, normalized size = 2.82 \begin{align*} \frac{{\left (891 \, b^{4} x^{13} + 3861 \, a b^{3} x^{10} + 6435 \, a^{2} b^{2} x^{7} + 5005 \, a^{3} b x^{4} + 1820 \, a^{4} x\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{1820 \,{\left (a^{4} b^{5} x^{15} + 5 \, a^{5} b^{4} x^{12} + 10 \, a^{6} b^{3} x^{9} + 10 \, a^{7} b^{2} x^{6} + 5 \, a^{8} b x^{3} + a^{9}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1820*(891*b^4*x^13 + 3861*a*b^3*x^10 + 6435*a^2*b^2*x^7 + 5005*a^3*b*x^4 + 1820*a^4*x)*(b*x^3 + a)^(2/3)/(a^
4*b^5*x^15 + 5*a^5*b^4*x^12 + 10*a^6*b^3*x^9 + 10*a^7*b^2*x^6 + 5*a^8*b*x^3 + a^9)

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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)/(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{b x^{3} - a}{{\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)/(b*x^3+a)^(16/3),x, algorithm="giac")

[Out]

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