3.29 \(\int \frac{(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=310 \[ \frac{2 a c d^3 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{3 b^{4/3}}-\frac{4 a c d^3 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{4/3}}+\frac{3 c^2 d^2 \left (a+b x^3\right )^{2/3}}{b}+\frac{2 c^3 d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}-\frac{c^4 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac{c^4 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}+\frac{4 c d^3 x \left (a+b x^3\right )^{2/3}}{3 b}+\frac{d^4 x^5 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{5}{3};\frac{8}{3};-\frac{b x^3}{a}\right )}{5 \sqrt [3]{a+b x^3}} \]

[Out]

(3*c^2*d^2*(a + b*x^3)^(2/3))/b + (4*c*d^3*x*(a + b*x^3)^(2/3))/(3*b) + (c^4*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*
x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3)) - (4*a*c*d^3*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3
*Sqrt[3]*b^(4/3)) + (2*c^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) + (d^4*x^5*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 5/3, 8/3, -((b*x^3)/a)])/(5*(a + b*x^3)^(1/
3)) - (c^4*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(2*b^(1/3)) + (2*a*c*d^3*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3
)])/(3*b^(4/3))

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Rubi [A]  time = 0.17824, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ \frac{2 a c d^3 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{3 b^{4/3}}-\frac{4 a c d^3 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{4/3}}+\frac{3 c^2 d^2 \left (a+b x^3\right )^{2/3}}{b}+\frac{2 c^3 d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}-\frac{c^4 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac{c^4 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{b}}+\frac{4 c d^3 x \left (a+b x^3\right )^{2/3}}{3 b}+\frac{d^4 x^5 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{5}{3};\frac{8}{3};-\frac{b x^3}{a}\right )}{5 \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^4/(a + b*x^3)^(1/3),x]

[Out]

(3*c^2*d^2*(a + b*x^3)^(2/3))/b + (4*c*d^3*x*(a + b*x^3)^(2/3))/(3*b) + (c^4*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*
x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3)) - (4*a*c*d^3*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3
*Sqrt[3]*b^(4/3)) + (2*c^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) + (d^4*x^5*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 5/3, 8/3, -((b*x^3)/a)])/(5*(a + b*x^3)^(1/
3)) - (c^4*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(2*b^(1/3)) + (2*a*c*d^3*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3
)])/(3*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)^4}{\sqrt [3]{a+b x^3}} \, dx &=\int \left (\frac{c^4}{\sqrt [3]{a+b x^3}}+\frac{4 c^3 d x}{\sqrt [3]{a+b x^3}}+\frac{6 c^2 d^2 x^2}{\sqrt [3]{a+b x^3}}+\frac{4 c d^3 x^3}{\sqrt [3]{a+b x^3}}+\frac{d^4 x^4}{\sqrt [3]{a+b x^3}}\right ) \, dx\\ &=c^4 \int \frac{1}{\sqrt [3]{a+b x^3}} \, dx+\left (4 c^3 d\right ) \int \frac{x}{\sqrt [3]{a+b x^3}} \, dx+\left (6 c^2 d^2\right ) \int \frac{x^2}{\sqrt [3]{a+b x^3}} \, dx+\left (4 c d^3\right ) \int \frac{x^3}{\sqrt [3]{a+b x^3}} \, dx+d^4 \int \frac{x^4}{\sqrt [3]{a+b x^3}} \, dx\\ &=\frac{3 c^2 d^2 \left (a+b x^3\right )^{2/3}}{b}+\frac{4 c d^3 x \left (a+b x^3\right )^{2/3}}{3 b}+\frac{c^4 \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^4 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}}-\frac{\left (4 a c d^3\right ) \int \frac{1}{\sqrt [3]{a+b x^3}} \, dx}{3 b}+\frac{\left (4 c^3 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{\left (d^4 \sqrt [3]{1+\frac{b x^3}{a}}\right ) \int \frac{x^4}{\sqrt [3]{1+\frac{b x^3}{a}}} \, dx}{\sqrt [3]{a+b x^3}}\\ &=\frac{3 c^2 d^2 \left (a+b x^3\right )^{2/3}}{b}+\frac{4 c d^3 x \left (a+b x^3\right )^{2/3}}{3 b}+\frac{c^4 \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{4 a c 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{2 c^3 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{d^4 x^5 \sqrt [3]{1+\frac{b x^3}{a}} \, _2F_1\left (\frac{1}{3},\frac{5}{3};\frac{8}{3};-\frac{b x^3}{a}\right )}{5 \sqrt [3]{a+b x^3}}-\frac{c^4 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}}+\frac{2 a c d^3 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{3 b^{4/3}}\\ \end{align*}

Mathematica [A]  time = 0.464989, size = 392, normalized size = 1.26 \[ \frac{5 c \left (3 b c^3 \sqrt [3]{a+b x^3} \log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-4 a d^3 \sqrt [3]{a+b x^3} \log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )+2 \sqrt [3]{a+b x^3} \left (4 a d^3-3 b c^3\right ) \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt{3} \sqrt [3]{a+b x^3} \left (3 b c^3-4 a d^3\right ) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )+54 a \sqrt [3]{b} c d^2+24 a \sqrt [3]{b} d^3 x+54 b^{4/3} c d^2 x^3+24 b^{4/3} d^3 x^4\right )+180 b^{4/3} c^3 d x^2 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )+18 b^{4/3} d^4 x^5 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{5}{3};\frac{8}{3};-\frac{b x^3}{a}\right )}{90 b^{4/3} \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^4/(a + b*x^3)^(1/3),x]

[Out]

(180*b^(4/3)*c^3*d*x^2*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)] + 18*b^(4/3)*d^4*x
^5*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 5/3, 8/3, -((b*x^3)/a)] + 5*c*(54*a*b^(1/3)*c*d^2 + 24*a*b^(1/
3)*d^3*x + 54*b^(4/3)*c*d^2*x^3 + 24*b^(4/3)*d^3*x^4 + 2*Sqrt[3]*(3*b*c^3 - 4*a*d^3)*(a + b*x^3)^(1/3)*ArcTan[
(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]] + 2*(-3*b*c^3 + 4*a*d^3)*(a + b*x^3)^(1/3)*Log[1 - (b^(1/3)*x)/
(a + b*x^3)^(1/3)] + 3*b*c^3*(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)] - 4*a*d^3*(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)]
))/(90*b^(4/3)*(a + b*x^3)^(1/3))

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Maple [F]  time = 0.034, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx+c \right ) ^{4}{\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)^4/(b*x^3+a)^(1/3),x)

[Out]

int((d*x+c)^4/(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)^4/(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)^4/(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [A]  time = 4.22527, size = 206, normalized size = 0.66 \begin{align*} 6 c^{2} 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^{4} 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{4 c^{3} 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 )} + \frac{4 c 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 )} + \frac{d^{4} x^{5} \Gamma \left (\frac{5}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{5}{3} \\ \frac{8}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac{8}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**4/(b*x**3+a)**(1/3),x)

[Out]

6*c**2*d**2*Piecewise((x**3/(3*a**(1/3)), Eq(b, 0)), ((a + b*x**3)**(2/3)/(2*b), True)) + c**4*x*gamma(1/3)*hy
per((1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(4/3)) + 4*c**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)) + 4*c*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)) + d**4*x**5*gamma(5/3)*hyper((1/3, 5/3), (8/3,), b
*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(8/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{4}}{{\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)^4/(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

integrate((d*x + c)^4/(b*x^3 + a)^(1/3), x)