3.85 \(\int \frac{(a+b x^3)^3}{(c+d x^3)^{13/3}} \, dx\)

Optimal. Leaf size=109 \[ \frac{27 a^2 x \left (a+b x^3\right )}{140 c^3 \left (c+d x^3\right )^{4/3}}+\frac{81 a^3 x}{140 c^4 \sqrt [3]{c+d x^3}}+\frac{9 a x \left (a+b x^3\right )^2}{70 c^2 \left (c+d x^3\right )^{7/3}}+\frac{x \left (a+b x^3\right )^3}{10 c \left (c+d x^3\right )^{10/3}} \]

[Out]

(x*(a + b*x^3)^3)/(10*c*(c + d*x^3)^(10/3)) + (9*a*x*(a + b*x^3)^2)/(70*c^2*(c + d*x^3)^(7/3)) + (27*a^2*x*(a
+ b*x^3))/(140*c^3*(c + d*x^3)^(4/3)) + (81*a^3*x)/(140*c^4*(c + d*x^3)^(1/3))

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Rubi [A]  time = 0.0357897, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {378, 191} \[ \frac{27 a^2 x \left (a+b x^3\right )}{140 c^3 \left (c+d x^3\right )^{4/3}}+\frac{81 a^3 x}{140 c^4 \sqrt [3]{c+d x^3}}+\frac{9 a x \left (a+b x^3\right )^2}{70 c^2 \left (c+d x^3\right )^{7/3}}+\frac{x \left (a+b x^3\right )^3}{10 c \left (c+d x^3\right )^{10/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^3/(c + d*x^3)^(13/3),x]

[Out]

(x*(a + b*x^3)^3)/(10*c*(c + d*x^3)^(10/3)) + (9*a*x*(a + b*x^3)^2)/(70*c^2*(c + d*x^3)^(7/3)) + (27*a^2*x*(a
+ b*x^3))/(140*c^3*(c + d*x^3)^(4/3)) + (81*a^3*x)/(140*c^4*(c + d*x^3)^(1/3))

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 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 )^3}{\left (c+d x^3\right )^{13/3}} \, dx &=\frac{x \left (a+b x^3\right )^3}{10 c \left (c+d x^3\right )^{10/3}}+\frac{(9 a) \int \frac{\left (a+b x^3\right )^2}{\left (c+d x^3\right )^{10/3}} \, dx}{10 c}\\ &=\frac{x \left (a+b x^3\right )^3}{10 c \left (c+d x^3\right )^{10/3}}+\frac{9 a x \left (a+b x^3\right )^2}{70 c^2 \left (c+d x^3\right )^{7/3}}+\frac{\left (27 a^2\right ) \int \frac{a+b x^3}{\left (c+d x^3\right )^{7/3}} \, dx}{35 c^2}\\ &=\frac{x \left (a+b x^3\right )^3}{10 c \left (c+d x^3\right )^{10/3}}+\frac{9 a x \left (a+b x^3\right )^2}{70 c^2 \left (c+d x^3\right )^{7/3}}+\frac{27 a^2 x \left (a+b x^3\right )}{140 c^3 \left (c+d x^3\right )^{4/3}}+\frac{\left (81 a^3\right ) \int \frac{1}{\left (c+d x^3\right )^{4/3}} \, dx}{140 c^3}\\ &=\frac{x \left (a+b x^3\right )^3}{10 c \left (c+d x^3\right )^{10/3}}+\frac{9 a x \left (a+b x^3\right )^2}{70 c^2 \left (c+d x^3\right )^{7/3}}+\frac{27 a^2 x \left (a+b x^3\right )}{140 c^3 \left (c+d x^3\right )^{4/3}}+\frac{81 a^3 x}{140 c^4 \sqrt [3]{c+d x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0373065, size = 120, normalized size = 1.1 \[ \frac{x \left (3 a^2 b c x^3 \left (35 c^2+30 c d x^3+9 d^2 x^6\right )+a^3 \left (315 c^2 d x^3+140 c^3+270 c d^2 x^6+81 d^3 x^9\right )+6 a b^2 c^2 x^6 \left (10 c+3 d x^3\right )+14 b^3 c^3 x^9\right )}{140 c^4 \left (c+d x^3\right )^{10/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(14*b^3*c^3*x^9 + 6*a*b^2*c^2*x^6*(10*c + 3*d*x^3) + 3*a^2*b*c*x^3*(35*c^2 + 30*c*d*x^3 + 9*d^2*x^6) + a^3*
(140*c^3 + 315*c^2*d*x^3 + 270*c*d^2*x^6 + 81*d^3*x^9)))/(140*c^4*(c + d*x^3)^(10/3))

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Maple [A]  time = 0.005, size = 134, normalized size = 1.2 \begin{align*}{\frac{x \left ( 81\,{a}^{3}{d}^{3}{x}^{9}+27\,{a}^{2}bc{d}^{2}{x}^{9}+18\,a{b}^{2}{c}^{2}d{x}^{9}+14\,{b}^{3}{c}^{3}{x}^{9}+270\,{a}^{3}c{d}^{2}{x}^{6}+90\,{a}^{2}b{c}^{2}d{x}^{6}+60\,a{b}^{2}{c}^{3}{x}^{6}+315\,{a}^{3}{c}^{2}d{x}^{3}+105\,{a}^{2}b{c}^{3}{x}^{3}+140\,{a}^{3}{c}^{3} \right ) }{140\,{c}^{4}} \left ( d{x}^{3}+c \right ) ^{-{\frac{10}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/140*x*(81*a^3*d^3*x^9+27*a^2*b*c*d^2*x^9+18*a*b^2*c^2*d*x^9+14*b^3*c^3*x^9+270*a^3*c*d^2*x^6+90*a^2*b*c^2*d*
x^6+60*a*b^2*c^3*x^6+315*a^3*c^2*d*x^3+105*a^2*b*c^3*x^3+140*a^3*c^3)/(d*x^3+c)^(10/3)/c^4

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Maxima [A]  time = 0.97642, size = 246, normalized size = 2.26 \begin{align*} \frac{b^{3} x^{10}}{10 \,{\left (d x^{3} + c\right )}^{\frac{10}{3}} c} - \frac{3 \, a b^{2}{\left (7 \, d - \frac{10 \,{\left (d x^{3} + c\right )}}{x^{3}}\right )} x^{10}}{70 \,{\left (d x^{3} + c\right )}^{\frac{10}{3}} c^{2}} + \frac{3 \,{\left (14 \, d^{2} - \frac{40 \,{\left (d x^{3} + c\right )} d}{x^{3}} + \frac{35 \,{\left (d x^{3} + c\right )}^{2}}{x^{6}}\right )} a^{2} b x^{10}}{140 \,{\left (d x^{3} + c\right )}^{\frac{10}{3}} c^{3}} - \frac{{\left (14 \, d^{3} - \frac{60 \,{\left (d x^{3} + c\right )} d^{2}}{x^{3}} + \frac{105 \,{\left (d x^{3} + c\right )}^{2} d}{x^{6}} - \frac{140 \,{\left (d x^{3} + c\right )}^{3}}{x^{9}}\right )} a^{3} x^{10}}{140 \,{\left (d x^{3} + c\right )}^{\frac{10}{3}} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^3*x^10/((d*x^3 + c)^(10/3)*c) - 3/70*a*b^2*(7*d - 10*(d*x^3 + c)/x^3)*x^10/((d*x^3 + c)^(10/3)*c^2) + 3
/140*(14*d^2 - 40*(d*x^3 + c)*d/x^3 + 35*(d*x^3 + c)^2/x^6)*a^2*b*x^10/((d*x^3 + c)^(10/3)*c^3) - 1/140*(14*d^
3 - 60*(d*x^3 + c)*d^2/x^3 + 105*(d*x^3 + c)^2*d/x^6 - 140*(d*x^3 + c)^3/x^9)*a^3*x^10/((d*x^3 + c)^(10/3)*c^4
)

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Fricas [A]  time = 1.65728, size = 356, normalized size = 3.27 \begin{align*} \frac{{\left ({\left (14 \, b^{3} c^{3} + 18 \, a b^{2} c^{2} d + 27 \, a^{2} b c d^{2} + 81 \, a^{3} d^{3}\right )} x^{10} + 30 \,{\left (2 \, a b^{2} c^{3} + 3 \, a^{2} b c^{2} d + 9 \, a^{3} c d^{2}\right )} x^{7} + 140 \, a^{3} c^{3} x + 105 \,{\left (a^{2} b c^{3} + 3 \, a^{3} c^{2} d\right )} x^{4}\right )}{\left (d x^{3} + c\right )}^{\frac{2}{3}}}{140 \,{\left (c^{4} d^{4} x^{12} + 4 \, c^{5} d^{3} x^{9} + 6 \, c^{6} d^{2} x^{6} + 4 \, c^{7} d x^{3} + c^{8}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/140*((14*b^3*c^3 + 18*a*b^2*c^2*d + 27*a^2*b*c*d^2 + 81*a^3*d^3)*x^10 + 30*(2*a*b^2*c^3 + 3*a^2*b*c^2*d + 9*
a^3*c*d^2)*x^7 + 140*a^3*c^3*x + 105*(a^2*b*c^3 + 3*a^3*c^2*d)*x^4)*(d*x^3 + c)^(2/3)/(c^4*d^4*x^12 + 4*c^5*d^
3*x^9 + 6*c^6*d^2*x^6 + 4*c^7*d*x^3 + c^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)**3/(d*x**3+c)**(13/3),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 )}^{3}}{{\left (d x^{3} + c\right )}^{\frac{13}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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