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

Optimal. Leaf size=192 \[ -\frac {a c d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{2/3}}-\frac {2 a c d \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )+\frac {a c^2 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}+\frac {a d^2 \sqrt [3]{a+b x^3}}{4 b} \]

[Out]

1/4*a*d^2*(b*x^3+a)^(1/3)/b+1/12*(b*x^3+a)^(1/3)*(3*d^2*x^3+8*c*d*x^2+6*c^2*x)+1/2*a*c^2*x*(1+b*x^3/a)^(2/3)*h
ypergeom([1/3, 2/3],[4/3],-b*x^3/a)/(b*x^3+a)^(2/3)-1/3*a*c*d*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)-2/9*a*c*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 = 0.20, antiderivative size = 245, normalized size of antiderivative = 1.28, number of steps used = 14, number of rules used = 13, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {1853, 1886, 261, 1893, 246, 245, 331, 292, 31, 634, 617, 204, 628} \[ -\frac {2 a c d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {a c d \log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{9 b^{2/3}}-\frac {2 a c d \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )+\frac {a c^2 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}+\frac {a d^2 \sqrt [3]{a+b x^3}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*(a + b*x^3)^(1/3),x]

[Out]

(a*d^2*(a + b*x^3)^(1/3))/(4*b) + ((a + b*x^3)^(1/3)*(6*c^2*x + 8*c*d*x^2 + 3*d^2*x^3))/12 - (2*a*c*d*ArcTan[(
1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]*b^(2/3)) + (a*c^2*x*(1 + (b*x^3)/a)^(2/3)*Hypergeome
tric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(2*(a + b*x^3)^(2/3)) - (2*a*c*d*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)])
/(9*b^(2/3)) + (a*c*d*Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(9*b^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 245

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[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 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 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1853

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[(a + b*x^n)^p*Sum[(C
oeff[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 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 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])

Rubi steps

\begin {align*} \int (c+d x)^2 \sqrt [3]{a+b x^3} \, dx &=\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )+a \int \frac {\frac {c^2}{2}+\frac {2 c d x}{3}+\frac {d^2 x^2}{4}}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )+a \int \frac {\frac {c^2}{2}+\frac {2 c d x}{3}}{\left (a+b x^3\right )^{2/3}} \, dx+\frac {1}{4} \left (a d^2\right ) \int \frac {x^2}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac {a d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )+a \int \left (\frac {c^2}{2 \left (a+b x^3\right )^{2/3}}+\frac {2 c d x}{3 \left (a+b x^3\right )^{2/3}}\right ) \, dx\\ &=\frac {a d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )+\frac {1}{2} \left (a c^2\right ) \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx+\frac {1}{3} (2 a c d) \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac {a d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )+\frac {1}{3} (2 a c d) \operatorname {Subst}\left (\int \frac {x}{1-b x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )+\frac {\left (a c^2 \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 {a d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )+\frac {a c^2 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 {(2 a c d) \operatorname {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 {(2 a c d) \operatorname {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 {a d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )+\frac {a c^2 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 {2 a c d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {(a c d) \operatorname {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 )}{9 b^{2/3}}-\frac {(a c d) \operatorname {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 )}{3 \sqrt [3]{b}}\\ &=\frac {a d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )+\frac {a c^2 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 {2 a c d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {a c 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 )}{9 b^{2/3}}+\frac {(2 a c d) \operatorname {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 {a d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )-\frac {2 a c 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^2 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 {2 a c d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {a c 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 )}{9 b^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 111, normalized size = 0.58 \[ \frac {\sqrt [3]{a+b x^3} \left (4 b c^2 x \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )+d \left (4 b c x^2 \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )+d \left (a+b x^3\right ) \sqrt [3]{\frac {b x^3}{a}+1}\right )\right )}{4 b \sqrt [3]{\frac {b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 130.28, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}, x\right ) \]

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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{2}\,{d x} \]

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((b*x^3 + a)^(1/3)*(d*x + c)^2, x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{2}\,{d x} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,x^3+a\right )}^{1/3}\,{\left (c+d\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/3)*(c + d*x)^2,x)

[Out]

int((a + b*x^3)^(1/3)*(c + d*x)^2, x)

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sympy [A]  time = 3.24, size = 114, normalized size = 0.59 \[ \frac {\sqrt [3]{a} c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {2 \sqrt [3]{a} c d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + d^{2} \left (\begin {cases} \frac {\sqrt [3]{a} x^{3}}{3} & \text {for}\: b = 0 \\\frac {\left (a + b x^{3}\right )^{\frac {4}{3}}}{4 b} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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