3.63 \(\int \frac{c+d x^3}{(a+b x^3)^{16/3}} \, dx\)

Optimal. Leaf size=151 \[ \frac{81 x (a d+12 b c)}{1820 a^5 b \sqrt [3]{a+b x^3}}+\frac{27 x (a d+12 b c)}{1820 a^4 b \left (a+b x^3\right )^{4/3}}+\frac{9 x (a d+12 b c)}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac{x (a d+12 b c)}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac{x (b c-a d)}{13 a b \left (a+b x^3\right )^{13/3}} \]

[Out]

((b*c - a*d)*x)/(13*a*b*(a + b*x^3)^(13/3)) + ((12*b*c + a*d)*x)/(130*a^2*b*(a + b*x^3)^(10/3)) + (9*(12*b*c +
a*d)*x)/(910*a^3*b*(a + b*x^3)^(7/3)) + (27*(12*b*c + a*d)*x)/(1820*a^4*b*(a + b*x^3)^(4/3)) + (81*(12*b*c +
a*d)*x)/(1820*a^5*b*(a + b*x^3)^(1/3))

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Rubi [A]  time = 0.0469943, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {385, 192, 191} \[ \frac{81 x (a d+12 b c)}{1820 a^5 b \sqrt [3]{a+b x^3}}+\frac{27 x (a d+12 b c)}{1820 a^4 b \left (a+b x^3\right )^{4/3}}+\frac{9 x (a d+12 b c)}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac{x (a d+12 b c)}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac{x (b c-a d)}{13 a b \left (a+b x^3\right )^{13/3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)/(a + b*x^3)^(16/3),x]

[Out]

((b*c - a*d)*x)/(13*a*b*(a + b*x^3)^(13/3)) + ((12*b*c + a*d)*x)/(130*a^2*b*(a + b*x^3)^(10/3)) + (9*(12*b*c +
a*d)*x)/(910*a^3*b*(a + b*x^3)^(7/3)) + (27*(12*b*c + a*d)*x)/(1820*a^4*b*(a + b*x^3)^(4/3)) + (81*(12*b*c +
a*d)*x)/(1820*a^5*b*(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{c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx &=\frac{(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac{(12 b c+a d) \int \frac{1}{\left (a+b x^3\right )^{13/3}} \, dx}{13 a b}\\ &=\frac{(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac{(12 b c+a d) x}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac{(9 (12 b c+a d)) \int \frac{1}{\left (a+b x^3\right )^{10/3}} \, dx}{130 a^2 b}\\ &=\frac{(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac{(12 b c+a d) x}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac{9 (12 b c+a d) x}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac{(27 (12 b c+a d)) \int \frac{1}{\left (a+b x^3\right )^{7/3}} \, dx}{455 a^3 b}\\ &=\frac{(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac{(12 b c+a d) x}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac{9 (12 b c+a d) x}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac{27 (12 b c+a d) x}{1820 a^4 b \left (a+b x^3\right )^{4/3}}+\frac{(81 (12 b c+a d)) \int \frac{1}{\left (a+b x^3\right )^{4/3}} \, dx}{1820 a^4 b}\\ &=\frac{(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac{(12 b c+a d) x}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac{9 (12 b c+a d) x}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac{27 (12 b c+a d) x}{1820 a^4 b \left (a+b x^3\right )^{4/3}}+\frac{81 (12 b c+a d) x}{1820 a^5 b \sqrt [3]{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0375034, size = 100, normalized size = 0.66 \[ \frac{x \left (351 a^2 b^2 x^6 \left (20 c+d x^3\right )+195 a^3 b x^3 \left (28 c+3 d x^3\right )+455 a^4 \left (4 c+d x^3\right )+81 a b^3 x^9 \left (52 c+d x^3\right )+972 b^4 c x^{12}\right )}{1820 a^5 \left (a+b x^3\right )^{13/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(972*b^4*c*x^12 + 455*a^4*(4*c + d*x^3) + 351*a^2*b^2*x^6*(20*c + d*x^3) + 81*a*b^3*x^9*(52*c + d*x^3) + 19
5*a^3*b*x^3*(28*c + 3*d*x^3)))/(1820*a^5*(a + b*x^3)^(13/3))

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Maple [A]  time = 0.004, size = 105, normalized size = 0.7 \begin{align*}{\frac{x \left ( 81\,a{b}^{3}d{x}^{12}+972\,{b}^{4}c{x}^{12}+351\,{a}^{2}{b}^{2}d{x}^{9}+4212\,a{b}^{3}c{x}^{9}+585\,{a}^{3}bd{x}^{6}+7020\,{a}^{2}{b}^{2}c{x}^{6}+455\,{a}^{4}d{x}^{3}+5460\,{a}^{3}bc{x}^{3}+1820\,c{a}^{4} \right ) }{1820\,{a}^{5}} \left ( b{x}^{3}+a \right ) ^{-{\frac{13}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)/(b*x^3+a)^(16/3),x)

[Out]

1/1820*x*(81*a*b^3*d*x^12+972*b^4*c*x^12+351*a^2*b^2*d*x^9+4212*a*b^3*c*x^9+585*a^3*b*d*x^6+7020*a^2*b^2*c*x^6
+455*a^4*d*x^3+5460*a^3*b*c*x^3+1820*a^4*c)/(b*x^3+a)^(13/3)/a^5

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Maxima [A]  time = 0.973927, size = 208, normalized size = 1.38 \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 )} d 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 )} c x^{13}}{455 \,{\left (b x^{3} + a\right )}^{\frac{13}{3}} a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(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)*d*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)*c*x^13/((b*x^3 + a)^(13/3)*a^5)

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Fricas [A]  time = 1.78946, size = 352, normalized size = 2.33 \begin{align*} \frac{{\left (81 \,{\left (12 \, b^{4} c + a b^{3} d\right )} x^{13} + 351 \,{\left (12 \, a b^{3} c + a^{2} b^{2} d\right )} x^{10} + 585 \,{\left (12 \, a^{2} b^{2} c + a^{3} b d\right )} x^{7} + 1820 \, a^{4} c x + 455 \,{\left (12 \, a^{3} b c + a^{4} d\right )} x^{4}\right )}{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{1820 \,{\left (a^{5} b^{5} x^{15} + 5 \, a^{6} b^{4} x^{12} + 10 \, a^{7} b^{3} x^{9} + 10 \, a^{8} b^{2} x^{6} + 5 \, a^{9} b x^{3} + a^{10}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1820*(81*(12*b^4*c + a*b^3*d)*x^13 + 351*(12*a*b^3*c + a^2*b^2*d)*x^10 + 585*(12*a^2*b^2*c + a^3*b*d)*x^7 +
1820*a^4*c*x + 455*(12*a^3*b*c + a^4*d)*x^4)*(b*x^3 + a)^(2/3)/(a^5*b^5*x^15 + 5*a^6*b^4*x^12 + 10*a^7*b^3*x^9
+ 10*a^8*b^2*x^6 + 5*a^9*b*x^3 + a^10)

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

[Out]

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