3.31 \(\int \frac{(c+d x)^2}{\sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=147 \[ -\frac{c^2 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac{c^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}+\frac{c d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}+\frac{d^2 \left (a+b x^3\right )^{2/3}}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.101936, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {1886, 261, 1893, 239, 365, 364} \[ -\frac{c^2 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac{c^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}+\frac{c d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}+\frac{d^2 \left (a+b x^3\right )^{2/3}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2/(a + b*x^3)^(1/3),x]

[Out]

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

Rule 1886

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[Coeff[Pq, x, n - 1], Int[x^(n - 1)*(a + b*x^n)^p, x
], x] + Int[ExpandToSum[Pq - Coeff[Pq, x, n - 1]*x^(n - 1), x]*(a + b*x^n)^p, x] /; FreeQ[{a, b, p}, x] && Pol
yQ[Pq, x] && IGtQ[n, 0] && Expon[Pq, x] == n - 1

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

Mathematica [A]  time = 0.165305, size = 201, normalized size = 1.37 \[ \frac{c^2 \log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{6 \sqrt [3]{b}}-\frac{c^2 \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{b}}+\frac{c^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}+\frac{c d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}+\frac{d^2 \left (a+b x^3\right )^{2/3}}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2/(a + b*x^3)^(1/3),x]

[Out]

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

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx+c \right ) ^{2}{\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)^2/(b*x^3+a)^(1/3),x)

[Out]

int((d*x+c)^2/(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)^2/(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)^2/(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [A]  time = 2.60015, size = 110, normalized size = 0.75 \begin{align*} 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^{2} 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{2 c 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)**2/(b*x**3+a)**(1/3),x)

[Out]

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

[Out]

integrate((d*x + c)^2/(b*x^3 + a)^(1/3), x)