3.1.26 \(\int (c+d x) \sqrt [3]{a+b x^3} \, dx\) [26]

Optimal. Leaf size=155 \[ \frac {1}{6} \left (3 c x+2 d x^2\right ) \sqrt [3]{a+b x^3}-\frac {a d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {a c x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}-\frac {a d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3}} \]

[Out]

1/6*(2*d*x^2+3*c*x)*(b*x^3+a)^(1/3)+1/2*a*c*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x^3/a)/(b*x^3+a)
^(2/3)-1/6*a*d*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)-1/9*a*d*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2
))/b^(2/3)*3^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1867, 1907, 252, 251, 337} \begin {gather*} -\frac {a d \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}-\frac {a d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3}}+\frac {1}{6} \sqrt [3]{a+b x^3} \left (3 c x+2 d x^2\right )+\frac {a c x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*(a + b*x^3)^(1/3),x]

[Out]

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

Rule 251

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

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 337

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

Rule 1867

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[(a + b*x^n)^p*Sum[Co
eff[Pq, x, i]*(x^(i + 1)/(n*p + i + 1)), {i, 0, q}], x] + Dist[a*n*p, Int[(a + b*x^n)^(p - 1)*Sum[Coeff[Pq, x,
i]*(x^i/(n*p + i + 1)), {i, 0, q}], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[(n - 1)/2, 0] && GtQ[
p, 0]

Rule 1907

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])

Rubi steps

\begin {align*} \int (c+d x) \sqrt [3]{a+b x^3} \, dx &=\frac {1}{6} \left (3 c x+2 d x^2\right ) \sqrt [3]{a+b x^3}+a \int \frac {\frac {c}{2}+\frac {d x}{3}}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac {1}{6} \left (3 c x+2 d x^2\right ) \sqrt [3]{a+b x^3}+a \int \left (\frac {c}{2 \left (a+b x^3\right )^{2/3}}+\frac {d x}{3 \left (a+b x^3\right )^{2/3}}\right ) \, dx\\ &=\frac {1}{6} \left (3 c x+2 d x^2\right ) \sqrt [3]{a+b x^3}+\frac {1}{2} (a c) \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx+\frac {1}{3} (a d) \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac {1}{6} \left (3 c x+2 d x^2\right ) \sqrt [3]{a+b x^3}+\frac {1}{3} (a d) \text {Subst}\left (\int \frac {x}{1-b x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )+\frac {\left (a c \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx}{2 \left (a+b x^3\right )^{2/3}}\\ &=\frac {1}{6} \left (3 c x+2 d x^2\right ) \sqrt [3]{a+b x^3}+\frac {a c x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}+\frac {(a d) \text {Subst}\left (\int \frac {1}{1-\sqrt [3]{b} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 \sqrt [3]{b}}-\frac {(a d) \text {Subst}\left (\int \frac {1-\sqrt [3]{b} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 \sqrt [3]{b}}\\ &=\frac {1}{6} \left (3 c x+2 d x^2\right ) \sqrt [3]{a+b x^3}+\frac {a c x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}-\frac {a d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {(a d) \text {Subst}\left (\int \frac {\sqrt [3]{b}+2 b^{2/3} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{18 b^{2/3}}-\frac {(a d) \text {Subst}\left (\int \frac {1}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{6 \sqrt [3]{b}}\\ &=\frac {1}{6} \left (3 c x+2 d x^2\right ) \sqrt [3]{a+b x^3}+\frac {a c x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}-\frac {a d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {a d \log \left (1+\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}}\right )}{18 b^{2/3}}+\frac {(a d) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}\\ &=\frac {1}{6} \left (3 c x+2 d x^2\right ) \sqrt [3]{a+b x^3}-\frac {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^{2/3}}+\frac {a c x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}-\frac {a d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {a d \log \left (1+\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}}\right )}{18 b^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 6.20, size = 75, normalized size = 0.48 \begin {gather*} \frac {x \sqrt [3]{a+b x^3} \left (2 c \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )+d x \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )\right )}{2 \sqrt [3]{1+\frac {b x^3}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \left (d x +c \right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}\, dx\]

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]

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

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Sympy [C] Result contains complex when optimal does not.
time = 1.32, size = 82, normalized size = 0.53 \begin {gather*} \frac {\sqrt [3]{a} 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 \Gamma \left (\frac {4}{3}\right )} + \frac {\sqrt [3]{a} 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 \Gamma \left (\frac {5}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(1/3)*c*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) + a**(1/3)*d*x**2*
gamma(2/3)*hyper((-1/3, 2/3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(5/3))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((b*x^3 + a)^(1/3)*(d*x + c), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,x^3+a\right )}^{1/3}\,\left (c+d\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/3)*(c + d*x),x)

[Out]

int((a + b*x^3)^(1/3)*(c + d*x), x)

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