3.58 \(\int \frac{c+d x^3}{\sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=111 \[ -\frac{(3 b c-a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 b^{4/3}}+\frac{(3 b c-a d) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{4/3}}+\frac{d x \left (a+b x^3\right )^{2/3}}{3 b} \]

[Out]

(d*x*(a + b*x^3)^(2/3))/(3*b) + ((3*b*c - a*d)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[
3]*b^(4/3)) - ((3*b*c - a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(6*b^(4/3))

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Rubi [A]  time = 0.0303546, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {388, 239} \[ -\frac{(3 b c-a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 b^{4/3}}+\frac{(3 b c-a d) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{4/3}}+\frac{d x \left (a+b x^3\right )^{2/3}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)/(a + b*x^3)^(1/3),x]

[Out]

(d*x*(a + b*x^3)^(2/3))/(3*b) + ((3*b*c - a*d)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[
3]*b^(4/3)) - ((3*b*c - a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(6*b^(4/3))

Rule 388

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

Rule 239

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + (2*Rt[b, 3]*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{c+d x^3}{\sqrt [3]{a+b x^3}} \, dx &=\frac{d x \left (a+b x^3\right )^{2/3}}{3 b}-\frac{(-3 b c+a d) \int \frac{1}{\sqrt [3]{a+b x^3}} \, dx}{3 b}\\ &=\frac{d x \left (a+b x^3\right )^{2/3}}{3 b}+\frac{(3 b c-a d) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt{3}}\right )}{3 \sqrt{3} b^{4/3}}-\frac{(3 b c-a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 b^{4/3}}\\ \end{align*}

Mathematica [A]  time = 0.130079, size = 141, normalized size = 1.27 \[ \frac{\frac{(3 b c-a d) \left (\log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-2 \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )\right )}{6 \sqrt [3]{b}}+d x \left (a+b x^3\right )^{2/3}}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3)/(a + b*x^3)^(1/3),x]

[Out]

(d*x*(a + b*x^3)^(2/3) + ((3*b*c - a*d)*(2*Sqrt[3]*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]] - 2*L
og[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)] + Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3
)]))/(6*b^(1/3)))/(3*b)

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Maple [F]  time = 0.217, size = 0, normalized size = 0. \begin{align*} \int{(d{x}^{3}+c){\frac{1}{\sqrt [3]{b{x}^{3}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)/(b*x^3+a)^(1/3),x)

[Out]

int((d*x^3+c)/(b*x^3+a)^(1/3),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(1/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.97456, size = 965, normalized size = 8.69 \begin{align*} \left [\frac{6 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} b d x - 3 \, \sqrt{\frac{1}{3}}{\left (3 \, b^{2} c - a b d\right )} \sqrt{-\frac{1}{b^{\frac{2}{3}}}} \log \left (3 \, b x^{3} - 3 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} b^{\frac{2}{3}} x^{2} - 3 \, \sqrt{\frac{1}{3}}{\left (b^{\frac{4}{3}} x^{3} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} b x^{2} - 2 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} b^{\frac{2}{3}} x\right )} \sqrt{-\frac{1}{b^{\frac{2}{3}}}} + 2 \, a\right ) - 2 \,{\left (3 \, b c - a d\right )} b^{\frac{2}{3}} \log \left (-\frac{b^{\frac{1}{3}} x -{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{x}\right ) +{\left (3 \, b c - a d\right )} b^{\frac{2}{3}} \log \left (\frac{b^{\frac{2}{3}} x^{2} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} b^{\frac{1}{3}} x +{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{x^{2}}\right )}{18 \, b^{2}}, \frac{6 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} b d x - 2 \,{\left (3 \, b c - a d\right )} b^{\frac{2}{3}} \log \left (-\frac{b^{\frac{1}{3}} x -{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{x}\right ) +{\left (3 \, b c - a d\right )} b^{\frac{2}{3}} \log \left (\frac{b^{\frac{2}{3}} x^{2} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} b^{\frac{1}{3}} x +{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{x^{2}}\right ) - \frac{6 \, \sqrt{\frac{1}{3}}{\left (3 \, b^{2} c - a b d\right )} \arctan \left (\frac{\sqrt{\frac{1}{3}}{\left (b^{\frac{1}{3}} x + 2 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}}\right )}}{b^{\frac{1}{3}} x}\right )}{b^{\frac{1}{3}}}}{18 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)/(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

[1/18*(6*(b*x^3 + a)^(2/3)*b*d*x - 3*sqrt(1/3)*(3*b^2*c - a*b*d)*sqrt(-1/b^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^
(1/3)*b^(2/3)*x^2 - 3*sqrt(1/3)*(b^(4/3)*x^3 + (b*x^3 + a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^(2/3)*b^(2/3)*x)*sqrt(-
1/b^(2/3)) + 2*a) - 2*(3*b*c - a*d)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) + (3*b*c - a*d)*b^(2/3)*lo
g((b^(2/3)*x^2 + (b*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2))/b^2, 1/18*(6*(b*x^3 + a)^(2/3)*b*d*x -
2*(3*b*c - a*d)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) + (3*b*c - a*d)*b^(2/3)*log((b^(2/3)*x^2 + (b
*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) - 6*sqrt(1/3)*(3*b^2*c - a*b*d)*arctan(sqrt(1/3)*(b^(1/3)*
x + 2*(b*x^3 + a)^(1/3))/(b^(1/3)*x))/b^(1/3))/b^2]

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Sympy [C]  time = 2.31708, size = 78, normalized size = 0.7 \begin{align*} \frac{c x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac{4}{3}\right )} + \frac{d x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{4}{3} \\ \frac{7}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac{7}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)/(b*x**3+a)**(1/3),x)

[Out]

c*x*gamma(1/3)*hyper((1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(4/3)) + d*x**4*gamma(4/3)
*hyper((1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(7/3))

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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{1}{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)^(1/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)/(b*x^3 + a)^(1/3), x)