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

Optimal. Leaf size=387 \[ \frac {a^2 d^4 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{5/3}}+\frac {a^2 d^4 \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{5/3}}-\frac {2 a c^3 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{2/3}}-\frac {4 a c^3 d \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {a c^4 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}+\frac {3 a c^2 d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )+\frac {a c d^3 x^4 \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}+\frac {a d^4 x^2 \sqrt [3]{a+b x^3}}{18 b} \]

[Out]

3/2*a*c^2*d^2*(b*x^3+a)^(1/3)/b+1/18*a*d^4*x^2*(b*x^3+a)^(1/3)/b+1/30*(b*x^3+a)^(1/3)*(5*d^4*x^5+24*c*d^3*x^4+
45*c^2*d^2*x^3+40*c^3*d*x^2+15*c^4*x)+1/2*a*c^4*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)+1/5*a*c*d^3*x^4*(1+b*x^3/a)^(2/3)*hypergeom([2/3, 4/3],[7/3],-b*x^3/a)/(b*x^3+a)^(2/3)-2/3*a*c^3*d*
ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)+1/18*a^2*d^4*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(5/3)-4/9*a*c^3*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)+1/27*a^2*d^4*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1
/3))*3^(1/2))/b^(5/3)*3^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.40, antiderivative size = 498, normalized size of antiderivative = 1.29, number of steps used = 23, number of rules used = 15, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.790, Rules used = {1853, 1893, 246, 245, 331, 292, 31, 634, 617, 204, 628, 261, 365, 364, 321} \[ \frac {a^2 d^4 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{5/3}}-\frac {a^2 d^4 \log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{54 b^{5/3}}+\frac {a^2 d^4 \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{5/3}}-\frac {4 a c^3 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {2 a c^3 d \log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{9 b^{2/3}}-\frac {4 a c^3 d \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {3 a c^2 d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {1}{30} \sqrt [3]{a+b x^3} \left (45 c^2 d^2 x^3+40 c^3 d x^2+15 c^4 x+24 c d^3 x^4+5 d^4 x^5\right )+\frac {a c^4 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}+\frac {a c d^3 x^4 \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}+\frac {a d^4 x^2 \sqrt [3]{a+b x^3}}{18 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^4*(a + b*x^3)^(1/3),x]

[Out]

(3*a*c^2*d^2*(a + b*x^3)^(1/3))/(2*b) + (a*d^4*x^2*(a + b*x^3)^(1/3))/(18*b) + ((a + b*x^3)^(1/3)*(15*c^4*x +
40*c^3*d*x^2 + 45*c^2*d^2*x^3 + 24*c*d^3*x^4 + 5*d^4*x^5))/30 - (4*a*c^3*d*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^
3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]*b^(2/3)) + (a^2*d^4*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(9*S
qrt[3]*b^(5/3)) + (a*c^4*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*c*d^3*x^4*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x^3)/a)])/(5*(a + b*x^3)^(
2/3)) - (4*a*c^3*d*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(9*b^(2/3)) + (a^2*d^4*Log[1 - (b^(1/3)*x)/(a + b*x
^3)^(1/3)])/(27*b^(5/3)) + (2*a*c^3*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)) - (a^2*d^4*Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(54*b^(5/3)
)

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 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]

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 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 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 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 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)^4 \sqrt [3]{a+b x^3} \, dx &=\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )+a \int \frac {\frac {c^4}{2}+\frac {4}{3} c^3 d x+\frac {3}{2} c^2 d^2 x^2+\frac {4}{5} c d^3 x^3+\frac {d^4 x^4}{6}}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )+a \int \left (\frac {c^4}{2 \left (a+b x^3\right )^{2/3}}+\frac {4 c^3 d x}{3 \left (a+b x^3\right )^{2/3}}+\frac {3 c^2 d^2 x^2}{2 \left (a+b x^3\right )^{2/3}}+\frac {4 c d^3 x^3}{5 \left (a+b x^3\right )^{2/3}}+\frac {d^4 x^4}{6 \left (a+b x^3\right )^{2/3}}\right ) \, dx\\ &=\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )+\frac {1}{2} \left (a c^4\right ) \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx+\frac {1}{3} \left (4 a c^3 d\right ) \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx+\frac {1}{2} \left (3 a c^2 d^2\right ) \int \frac {x^2}{\left (a+b x^3\right )^{2/3}} \, dx+\frac {1}{5} \left (4 a c d^3\right ) \int \frac {x^3}{\left (a+b x^3\right )^{2/3}} \, dx+\frac {1}{6} \left (a d^4\right ) \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac {3 a c^2 d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {a d^4 x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )+\frac {1}{3} \left (4 a c^3 d\right ) \operatorname {Subst}\left (\int \frac {x}{1-b x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )-\frac {\left (a^2 d^4\right ) \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx}{9 b}+\frac {\left (a c^4 \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 {\left (4 a c d^3 \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {x^3}{\left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx}{5 \left (a+b x^3\right )^{2/3}}\\ &=\frac {3 a c^2 d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {a d^4 x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )+\frac {a c^4 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 c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}+\frac {\left (4 a c^3 d\right ) \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 {\left (4 a c^3 d\right ) \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 {\left (a^2 d^4\right ) \operatorname {Subst}\left (\int \frac {x}{1-b x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 b}\\ &=\frac {3 a c^2 d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {a d^4 x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )+\frac {a c^4 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 c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}-\frac {4 a c^3 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {\left (2 a c^3 d\right ) \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 {\left (2 a c^3 d\right ) \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 {\left (a^2 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt [3]{b} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{4/3}}+\frac {\left (a^2 d^4\right ) \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 )}{27 b^{4/3}}\\ &=\frac {3 a c^2 d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {a d^4 x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )+\frac {a c^4 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 c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}-\frac {4 a c^3 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {a^2 d^4 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{5/3}}+\frac {2 a c^3 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 {\left (4 a c^3 d\right ) \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 {\left (a^2 d^4\right ) \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 )}{54 b^{5/3}}+\frac {\left (a^2 d^4\right ) \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 )}{18 b^{4/3}}\\ &=\frac {3 a c^2 d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {a d^4 x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )-\frac {4 a c^3 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^4 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 c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}-\frac {4 a c^3 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {a^2 d^4 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{5/3}}+\frac {2 a c^3 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 {a^2 d^4 \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 )}{54 b^{5/3}}-\frac {\left (a^2 d^4\right ) \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 )}{9 b^{5/3}}\\ &=\frac {3 a c^2 d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {a d^4 x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )-\frac {4 a c^3 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^2 d^4 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{5/3}}+\frac {a c^4 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 c d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}-\frac {4 a c^3 d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {a^2 d^4 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{5/3}}+\frac {2 a c^3 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 {a^2 d^4 \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 )}{54 b^{5/3}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 163, normalized size = 0.42 \[ \frac {\sqrt [3]{a+b x^3} \left (6 b c^4 x \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )+d x^2 \left (12 b c^3-a d^3\right ) \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )+d^2 \left (\left (a+b x^3\right ) \sqrt [3]{\frac {b x^3}{a}+1} \left (9 c^2+d^2 x^2\right )+6 b c d x^4 \, _2F_1\left (-\frac {1}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^3}{a}\right )\right )\right )}{6 b \sqrt [3]{\frac {b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^(1/3)*(6*b*c^4*x*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/a)] + d*(12*b*c^3 - a*d^3)*x^2*Hyper
geometric2F1[-1/3, 2/3, 5/3, -((b*x^3)/a)] + d^2*((9*c^2 + d^2*x^2)*(a + b*x^3)*(1 + (b*x^3)/a)^(1/3) + 6*b*c*
d*x^4*Hypergeometric2F1[-1/3, 4/3, 7/3, -((b*x^3)/a)])))/(6*b*(1 + (b*x^3)/a)^(1/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{4}\,{d x} \]

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((b*x^3 + a)^(1/3)*(d*x + c)^4, x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{4} \left (b \,x^{3}+a \right )^{\frac {1}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{4}\,{d x} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (b\,x^3+a\right )}^{1/3}\,{\left (c+d\,x\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/3)*(c + d*x)^4,x)

[Out]

int((a + b*x^3)^(1/3)*(c + d*x)^4, x)

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sympy [A]  time = 4.95, size = 212, normalized size = 0.55 \[ \frac {\sqrt [3]{a} c^{4} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {4 \sqrt [3]{a} c^{3} d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + \frac {4 \sqrt [3]{a} c d^{3} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {\sqrt [3]{a} d^{4} x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {8}{3}\right )} + 6 c^{2} d^{2} \left (\begin {cases} \frac {\sqrt [3]{a} x^{3}}{3} & \text {for}\: b = 0 \\\frac {\left (a + b x^{3}\right )^{\frac {4}{3}}}{4 b} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(1/3)*c**4*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) + 4*a**(1/3)*c*
*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)) + 4*a**(1/3)*c*d**3*x
**4*gamma(4/3)*hyper((-1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) + a**(1/3)*d**4*x**5*gamma(
5/3)*hyper((-1/3, 5/3), (8/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(8/3)) + 6*c**2*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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