3.76 \(\int \frac {(c+d x^3)^2}{(a+b x^3)^{13/3}} \, dx\)

Optimal. Leaf size=174 \[ \frac {9 c^2 x (9 b c-10 a d)}{140 a^4 \sqrt [3]{a+b x^3} (b c-a d)}+\frac {3 c x \left (c+d x^3\right ) (9 b c-10 a d)}{140 a^3 \left (a+b x^3\right )^{4/3} (b c-a d)}+\frac {x \left (c+d x^3\right )^2 (9 b c-10 a d)}{70 a^2 \left (a+b x^3\right )^{7/3} (b c-a d)}+\frac {b x \left (c+d x^3\right )^3}{10 a \left (a+b x^3\right )^{10/3} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {382, 378, 191} \[ \frac {9 c^2 x (9 b c-10 a d)}{140 a^4 \sqrt [3]{a+b x^3} (b c-a d)}+\frac {x \left (c+d x^3\right )^2 (9 b c-10 a d)}{70 a^2 \left (a+b x^3\right )^{7/3} (b c-a d)}+\frac {3 c x \left (c+d x^3\right ) (9 b c-10 a d)}{140 a^3 \left (a+b x^3\right )^{4/3} (b c-a d)}+\frac {b x \left (c+d x^3\right )^3}{10 a \left (a+b x^3\right )^{10/3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 5.11, size = 106, normalized size = 0.61 \[ \frac {x \left (10 a^3 \left (14 c^2+7 c d x^3+2 d^2 x^6\right )+3 a^2 b x^3 \left (105 c^2+20 c d x^3+2 d^2 x^6\right )+18 a b^2 c x^6 \left (15 c+d x^3\right )+81 b^3 c^2 x^9\right )}{140 a^4 \left (a+b x^3\right )^{10/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(81*b^3*c^2*x^9 + 18*a*b^2*c*x^6*(15*c + d*x^3) + 10*a^3*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6) + 3*a^2*b*x^3*(10
5*c^2 + 20*c*d*x^3 + 2*d^2*x^6)))/(140*a^4*(a + b*x^3)^(10/3))

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fricas [A]  time = 0.73, size = 152, normalized size = 0.87 \[ \frac {{\left (3 \, {\left (27 \, b^{3} c^{2} + 6 \, a b^{2} c d + 2 \, a^{2} b d^{2}\right )} x^{10} + 10 \, {\left (27 \, a b^{2} c^{2} + 6 \, a^{2} b c d + 2 \, a^{3} d^{2}\right )} x^{7} + 140 \, a^{3} c^{2} x + 35 \, {\left (9 \, a^{2} b c^{2} + 2 \, a^{3} c d\right )} x^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{140 \, {\left (a^{4} b^{4} x^{12} + 4 \, a^{5} b^{3} x^{9} + 6 \, a^{6} b^{2} x^{6} + 4 \, a^{7} b x^{3} + a^{8}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/140*(3*(27*b^3*c^2 + 6*a*b^2*c*d + 2*a^2*b*d^2)*x^10 + 10*(27*a*b^2*c^2 + 6*a^2*b*c*d + 2*a^3*d^2)*x^7 + 140
*a^3*c^2*x + 35*(9*a^2*b*c^2 + 2*a^3*c*d)*x^4)*(b*x^3 + a)^(2/3)/(a^4*b^4*x^12 + 4*a^5*b^3*x^9 + 6*a^6*b^2*x^6
+ 4*a^7*b*x^3 + a^8)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 115, normalized size = 0.66 \[ \frac {\left (6 a^{2} b \,d^{2} x^{9}+18 a \,b^{2} c d \,x^{9}+81 b^{3} c^{2} x^{9}+20 a^{3} d^{2} x^{6}+60 a^{2} b c d \,x^{6}+270 a \,b^{2} c^{2} x^{6}+70 a^{3} c d \,x^{3}+315 a^{2} b \,c^{2} x^{3}+140 c^{2} a^{3}\right ) x}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/140*x*(6*a^2*b*d^2*x^9+18*a*b^2*c*d*x^9+81*b^3*c^2*x^9+20*a^3*d^2*x^6+60*a^2*b*c*d*x^6+270*a*b^2*c^2*x^6+70*
a^3*c*d*x^3+315*a^2*b*c^2*x^3+140*a^3*c^2)/(b*x^3+a)^(10/3)/a^4

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maxima [A]  time = 0.71, size = 159, normalized size = 0.91 \[ -\frac {{\left (7 \, b - \frac {10 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} d^{2} x^{10}}{70 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{2}} + \frac {{\left (14 \, b^{2} - \frac {40 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {35 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} c d x^{10}}{70 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{3}} - \frac {{\left (14 \, b^{3} - \frac {60 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {105 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {140 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} c^{2} x^{10}}{140 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/70*(7*b - 10*(b*x^3 + a)/x^3)*d^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)*c*d*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)*c^2*x^10/((b*x^3 + a)^(10/3)*a^4)

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mupad [B]  time = 1.45, size = 176, normalized size = 1.01 \[ \frac {x\,\left (\frac {c^2}{10\,a}+\frac {a\,\left (\frac {d^2}{10\,b}-\frac {c\,d}{5\,a}\right )}{b}\right )}{{\left (b\,x^3+a\right )}^{10/3}}-\frac {x\,\left (\frac {d^2}{7\,b^2}-\frac {-a^2\,d^2+2\,a\,b\,c\,d+9\,b^2\,c^2}{70\,a^2\,b^2}\right )}{{\left (b\,x^3+a\right )}^{7/3}}+\frac {x\,\left (2\,a^2\,d^2+6\,a\,b\,c\,d+27\,b^2\,c^2\right )}{140\,a^3\,b^2\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {x\,\left (6\,a^2\,d^2+18\,a\,b\,c\,d+81\,b^2\,c^2\right )}{140\,a^4\,b^2\,{\left (b\,x^3+a\right )}^{1/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(c^2/(10*a) + (a*(d^2/(10*b) - (c*d)/(5*a)))/b))/(a + b*x^3)^(10/3) - (x*(d^2/(7*b^2) - (9*b^2*c^2 - a^2*d^
2 + 2*a*b*c*d)/(70*a^2*b^2)))/(a + b*x^3)^(7/3) + (x*(2*a^2*d^2 + 27*b^2*c^2 + 6*a*b*c*d))/(140*a^3*b^2*(a + b
*x^3)^(4/3)) + (x*(6*a^2*d^2 + 81*b^2*c^2 + 18*a*b*c*d))/(140*a^4*b^2*(a + b*x^3)^(1/3))

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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((d*x**3+c)**2/(b*x**3+a)**(13/3),x)

[Out]

Timed out

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