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

Optimal. Leaf size=255 \[ \frac{a d^3 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 b^{4/3}}-\frac{a d^3 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{4/3}}+\frac{3 c^2 d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}}-\frac{c^3 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac{c^3 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}+\frac{3 c d^2 \left (a+b x^3\right )^{2/3}}{2 b}+\frac{d^3 x \left (a+b x^3\right )^{2/3}}{3 b} \]

[Out]

(3*c*d^2*(a + b*x^3)^(2/3))/(2*b) + (d^3*x*(a + b*x^3)^(2/3))/(3*b) + (c^3*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^
3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3)) - (a*d^3*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[
3]*b^(4/3)) + (3*c^2*d*x^2*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(2*(a + b*x^3
)^(1/3)) - (c^3*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(2*b^(1/3)) + (a*d^3*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/
3)])/(6*b^(4/3))

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Rubi [A]  time = 0.137653, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {1893, 239, 365, 364, 261, 321} \[ \frac{a d^3 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 b^{4/3}}-\frac{a d^3 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{4/3}}+\frac{3 c^2 d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}}-\frac{c^3 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac{c^3 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}+\frac{3 c d^2 \left (a+b x^3\right )^{2/3}}{2 b}+\frac{d^3 x \left (a+b x^3\right )^{2/3}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3/(a + b*x^3)^(1/3),x]

[Out]

(3*c*d^2*(a + b*x^3)^(2/3))/(2*b) + (d^3*x*(a + b*x^3)^(2/3))/(3*b) + (c^3*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^
3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3)) - (a*d^3*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[
3]*b^(4/3)) + (3*c^2*d*x^2*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(2*(a + b*x^3
)^(1/3)) - (c^3*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(2*b^(1/3)) + (a*d^3*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/
3)])/(6*b^(4/3))

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])

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 321

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

Rubi steps

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

Mathematica [A]  time = 0.377238, size = 287, normalized size = 1.13 \[ \frac{1}{18} \left (\frac{3 b c^3 \log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-a d^3 \log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )+\left (2 a d^3-6 b c^3\right ) \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt{3} \left (3 b c^3-a d^3\right ) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )+27 \sqrt [3]{b} c d^2 \left (a+b x^3\right )^{2/3}+6 \sqrt [3]{b} d^3 x \left (a+b x^3\right )^{2/3}}{b^{4/3}}+\frac{27 c^2 d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((27*c^2*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) + (27*b
^(1/3)*c*d^2*(a + b*x^3)^(2/3) + 6*b^(1/3)*d^3*x*(a + b*x^3)^(2/3) + 2*Sqrt[3]*(3*b*c^3 - a*d^3)*ArcTan[(1 + (
2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]] + (-6*b*c^3 + 2*a*d^3)*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)] + 3*b*c
^3*Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)] - a*d^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^(4/3))/18

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx+c \right ) ^{3}{\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)^3/(b*x^3+a)^(1/3),x)

[Out]

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

[Out]

Timed out

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Sympy [A]  time = 3.552, size = 155, normalized size = 0.61 \begin{align*} 3 c d^{2} \left (\begin{cases} \frac{x^{3}}{3 \sqrt [3]{a}} & \text{for}\: b = 0 \\\frac{\left (a + b x^{3}\right )^{\frac{2}{3}}}{2 b} & \text{otherwise} \end{cases}\right ) + \frac{c^{3} 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{c^{2} 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 )}}{\sqrt [3]{a} \Gamma \left (\frac{5}{3}\right )} + \frac{d^{3} 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+c)**3/(b*x**3+a)**(1/3),x)

[Out]

3*c*d**2*Piecewise((x**3/(3*a**(1/3)), Eq(b, 0)), ((a + b*x**3)**(2/3)/(2*b), True)) + c**3*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)) + c**2*d*x**2*gamma(2/3)*hyper((1/3, 2/
3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(a**(1/3)*gamma(5/3)) + d**3*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{{\left (d x + c\right )}^{3}}{{\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)^3/(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

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