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

Optimal. Leaf size=124 \[ -\frac{c \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac{c \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}+\frac{d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}} \]

[Out]

(c*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3)) + (d*x^2*(1 + (b*x^3)/a)^(1/3)*Hyp
ergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(2*(a + b*x^3)^(1/3)) - (c*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])
/(2*b^(1/3))

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Rubi [A]  time = 0.0663397, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {1893, 239, 365, 364} \[ -\frac{c \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac{c \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}+\frac{d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*x^3)^(1/3),x]

[Out]

(c*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3)) + (d*x^2*(1 + (b*x^3)/a)^(1/3)*Hyp
ergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(2*(a + b*x^3)^(1/3)) - (c*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])
/(2*b^(1/3))

Rule 1893

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n])

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]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{c+d x}{\sqrt [3]{a+b x^3}} \, dx &=\int \left (\frac{c}{\sqrt [3]{a+b x^3}}+\frac{d x}{\sqrt [3]{a+b x^3}}\right ) \, dx\\ &=c \int \frac{1}{\sqrt [3]{a+b x^3}} \, dx+d \int \frac{x}{\sqrt [3]{a+b x^3}} \, dx\\ &=\frac{c \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}-\frac{c \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}}+\frac{\left (d \sqrt [3]{1+\frac{b x^3}{a}}\right ) \int \frac{x}{\sqrt [3]{1+\frac{b x^3}{a}}} \, dx}{\sqrt [3]{a+b x^3}}\\ &=\frac{c \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}+\frac{d x^2 \sqrt [3]{1+\frac{b x^3}{a}} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}}-\frac{c \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}}\\ \end{align*}

Mathematica [A]  time = 0.114559, size = 163, normalized size = 1.31 \[ \frac{1}{6} \left (\frac{c \left (\log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-2 \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )\right )}{\sqrt [3]{b}}+\frac{3 d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*d*x^2*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(a + b*x^3)^(1/3) + (c*(2*Sqrt
[3]*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]] - 2*Log[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)]))/b^(1/3))/6

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{(dx+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+c)/(b*x^3+a)^(1/3),x)

[Out]

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

[Out]

Exception raised: ValueError

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

[Out]

Timed out

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Sympy [C]  time = 1.86781, size = 78, normalized size = 0.63 \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^{2} \Gamma \left (\frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{2}{3} \\ \frac{5}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac{5}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+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**2*gamma(2/3)
*hyper((1/3, 2/3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(5/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + 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+c)/(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

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