3.45 \(\int \frac {(a-b x^3)^2}{(a+b x^3)^{13/3}} \, dx\)

Optimal. Leaf size=105 \[ \frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 x \left (a-b x^3\right )^2}{140 a^2 \left (a+b x^3\right )^{7/3}}+\frac {57 x \left (a-b x^3\right )}{280 a^2 \left (a+b x^3\right )^{4/3}}+\frac {171 x}{280 a^2 \sqrt [3]{a+b x^3}} \]

[Out]

1/20*x*(-b*x^3+a)^3/a^2/(b*x^3+a)^(10/3)+19/140*x*(-b*x^3+a)^2/a^2/(b*x^3+a)^(7/3)+57/280*x*(-b*x^3+a)/a^2/(b*
x^3+a)^(4/3)+171/280*x/a^2/(b*x^3+a)^(1/3)

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Rubi [A]  time = 0.04, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {382, 378, 191} \[ \frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 x \left (a-b x^3\right )^2}{140 a^2 \left (a+b x^3\right )^{7/3}}+\frac {57 x \left (a-b x^3\right )}{280 a^2 \left (a+b x^3\right )^{4/3}}+\frac {171 x}{280 a^2 \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^3)^2/(a + b*x^3)^(13/3),x]

[Out]

(x*(a - b*x^3)^3)/(20*a^2*(a + b*x^3)^(10/3)) + (19*x*(a - b*x^3)^2)/(140*a^2*(a + b*x^3)^(7/3)) + (57*x*(a -
b*x^3))/(280*a^2*(a + b*x^3)^(4/3)) + (171*x)/(280*a^2*(a + b*x^3)^(1/3))

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]

Rule 378

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] - Dist[(c*q)/(a*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx &=\frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx}{20 a}\\ &=\frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 x \left (a-b x^3\right )^2}{140 a^2 \left (a+b x^3\right )^{7/3}}+\frac {57 \int \frac {a-b x^3}{\left (a+b x^3\right )^{7/3}} \, dx}{70 a}\\ &=\frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 x \left (a-b x^3\right )^2}{140 a^2 \left (a+b x^3\right )^{7/3}}+\frac {57 x \left (a-b x^3\right )}{280 a^2 \left (a+b x^3\right )^{4/3}}+\frac {171 \int \frac {1}{\left (a+b x^3\right )^{4/3}} \, dx}{280 a}\\ &=\frac {x \left (a-b x^3\right )^3}{20 a^2 \left (a+b x^3\right )^{10/3}}+\frac {19 x \left (a-b x^3\right )^2}{140 a^2 \left (a+b x^3\right )^{7/3}}+\frac {57 x \left (a-b x^3\right )}{280 a^2 \left (a+b x^3\right )^{4/3}}+\frac {171 x}{280 a^2 \sqrt [3]{a+b x^3}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.49 \[ \frac {x \left (140 a^3+245 a^2 b x^3+230 a b^2 x^6+69 b^3 x^9\right )}{140 a^2 \left (a+b x^3\right )^{10/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^3)^2/(a + b*x^3)^(13/3),x]

[Out]

(x*(140*a^3 + 245*a^2*b*x^3 + 230*a*b^2*x^6 + 69*b^3*x^9))/(140*a^2*(a + b*x^3)^(10/3))

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fricas [A]  time = 0.91, size = 91, normalized size = 0.87 \[ \frac {{\left (69 \, b^{3} x^{10} + 230 \, a b^{2} x^{7} + 245 \, a^{2} b x^{4} + 140 \, a^{3} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{140 \, {\left (a^{2} b^{4} x^{12} + 4 \, a^{3} b^{3} x^{9} + 6 \, a^{4} b^{2} x^{6} + 4 \, a^{5} b x^{3} + a^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2/(b*x^3+a)^(13/3),x, algorithm="fricas")

[Out]

1/140*(69*b^3*x^10 + 230*a*b^2*x^7 + 245*a^2*b*x^4 + 140*a^3*x)*(b*x^3 + a)^(2/3)/(a^2*b^4*x^12 + 4*a^3*b^3*x^
9 + 6*a^4*b^2*x^6 + 4*a^5*b*x^3 + a^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {13}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2/(b*x^3+a)^(13/3),x, algorithm="giac")

[Out]

integrate((b*x^3 - a)^2/(b*x^3 + a)^(13/3), x)

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maple [A]  time = 0.05, size = 48, normalized size = 0.46 \[ \frac {\left (69 b^{3} x^{9}+230 a \,b^{2} x^{6}+245 a^{2} b \,x^{3}+140 a^{3}\right ) x}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^3+a)^2/(b*x^3+a)^(13/3),x)

[Out]

1/140*x*(69*b^3*x^9+230*a*b^2*x^6+245*a^2*b*x^3+140*a^3)/(b*x^3+a)^(10/3)/a^2

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maxima [A]  time = 0.59, size = 155, normalized size = 1.48 \[ -\frac {{\left (7 \, b - \frac {10 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} b^{2} x^{10}}{70 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{2}} - \frac {{\left (14 \, b^{2} - \frac {40 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {35 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} b x^{10}}{70 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{2}} - \frac {{\left (14 \, b^{3} - \frac {60 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {105 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {140 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} x^{10}}{140 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^3+a)^2/(b*x^3+a)^(13/3),x, algorithm="maxima")

[Out]

-1/70*(7*b - 10*(b*x^3 + a)/x^3)*b^2*x^10/((b*x^3 + a)^(10/3)*a^2) - 1/70*(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^2) - 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^2)

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mupad [B]  time = 1.39, size = 56, normalized size = 0.53 \[ \frac {69\,x}{140\,a^2\,{\left (b\,x^3+a\right )}^{1/3}}-\frac {2\,x}{35\,{\left (b\,x^3+a\right )}^{7/3}}+\frac {23\,x}{140\,a\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {2\,a\,x}{5\,{\left (b\,x^3+a\right )}^{10/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^3)^2/(a + b*x^3)^(13/3),x)

[Out]

(69*x)/(140*a^2*(a + b*x^3)^(1/3)) - (2*x)/(35*(a + b*x^3)^(7/3)) + (23*x)/(140*a*(a + b*x^3)^(4/3)) + (2*a*x)
/(5*(a + b*x^3)^(10/3))

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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)**(13/3),x)

[Out]

Timed out

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