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

Optimal. Leaf size=242 \[ -\frac{a c^2 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}}-\frac{a c^2 d \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} b^{2/3}}+\frac{a x \left (\frac{b x^3}{a}+1\right )^{2/3} \left (5 b c^3-a d^3\right ) \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{10 b \left (a+b x^3\right )^{2/3}}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (20 c^2 d x^2+10 c^3 x+15 c d^2 x^3+4 d^3 x^4\right )+\frac{3 a c d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b} \]

[Out]

(3*a*c*d^2*(a + b*x^3)^(1/3))/(4*b) + (a*d^3*x*(a + b*x^3)^(1/3))/(10*b) + ((a + b*x^3)^(1/3)*(10*c^3*x + 20*c
^2*d*x^2 + 15*c*d^2*x^3 + 4*d^3*x^4))/20 - (a*c^2*d*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sq
rt[3]*b^(2/3)) + (a*(5*b*c^3 - a*d^3)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/
(10*b*(a + b*x^3)^(2/3)) - (a*c^2*d*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(2*b^(2/3))

________________________________________________________________________________________

Rubi [A]  time = 0.305283, antiderivative size = 297, normalized size of antiderivative = 1.23, number of steps used = 15, number of rules used = 14, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.737, Rules used = {1853, 1888, 1886, 261, 1893, 246, 245, 331, 292, 31, 634, 617, 204, 628} \[ -\frac{a c^2 d \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac{a c^2 d \log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{6 b^{2/3}}-\frac{a c^2 d \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} b^{2/3}}+\frac{a x \left (\frac{b x^3}{a}+1\right )^{2/3} \left (5 b c^3-a d^3\right ) \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{10 b \left (a+b x^3\right )^{2/3}}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (20 c^2 d x^2+10 c^3 x+15 c d^2 x^3+4 d^3 x^4\right )+\frac{3 a c d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*(a + b*x^3)^(1/3),x]

[Out]

(3*a*c*d^2*(a + b*x^3)^(1/3))/(4*b) + (a*d^3*x*(a + b*x^3)^(1/3))/(10*b) + ((a + b*x^3)^(1/3)*(10*c^3*x + 20*c
^2*d*x^2 + 15*c*d^2*x^3 + 4*d^3*x^4))/20 - (a*c^2*d*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sq
rt[3]*b^(2/3)) + (a*(5*b*c^3 - a*d^3)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/
(10*b*(a + b*x^3)^(2/3)) - (a*c^2*d*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(3*b^(2/3)) + (a*c^2*d*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^(2/3))

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 1888

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, With[{Pqq = Coeff[Pq, x, q]}, D
ist[1/(b*(q + n*p + 1)), Int[ExpandToSum[b*(q + n*p + 1)*(Pq - Pqq*x^q) - a*Pqq*(q - n + 1)*x^(q - n), x]*(a +
b*x^n)^p, x], x] + Simp[(Pqq*x^(q - n + 1)*(a + b*x^n)^(p + 1))/(b*(q + n*p + 1)), x]] /; NeQ[q + n*p + 1, 0]
&& q - n >= 0 && (IntegerQ[2*p] || IntegerQ[p + (q + 1)/(2*n)])] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IG
tQ[n, 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 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 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 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 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 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 31

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

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 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 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 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]

Rubi steps

\begin{align*} \int (c+d x)^3 \sqrt [3]{a+b x^3} \, dx &=\frac{1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )+a \int \frac{\frac{c^3}{2}+c^2 d x+\frac{3}{4} c d^2 x^2+\frac{d^3 x^3}{5}}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )+\frac{a \int \frac{\frac{1}{5} \left (5 b c^3-a d^3\right )+2 b c^2 d x+\frac{3}{2} b c d^2 x^2}{\left (a+b x^3\right )^{2/3}} \, dx}{2 b}\\ &=\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )+\frac{a \int \frac{\frac{1}{5} \left (5 b c^3-a d^3\right )+2 b c^2 d x}{\left (a+b x^3\right )^{2/3}} \, dx}{2 b}+\frac{1}{4} \left (3 a c d^2\right ) \int \frac{x^2}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac{3 a c d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )+\frac{a \int \left (\frac{5 b c^3-a d^3}{5 \left (a+b x^3\right )^{2/3}}+\frac{2 b c^2 d x}{\left (a+b x^3\right )^{2/3}}\right ) \, dx}{2 b}\\ &=\frac{3 a c d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )+\left (a c^2 d\right ) \int \frac{x}{\left (a+b x^3\right )^{2/3}} \, dx+\frac{\left (a \left (5 b c^3-a d^3\right )\right ) \int \frac{1}{\left (a+b x^3\right )^{2/3}} \, dx}{10 b}\\ &=\frac{3 a c d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )+\left (a c^2 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 \left (5 b c^3-a d^3\right ) \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}{10 b \left (a+b x^3\right )^{2/3}}\\ &=\frac{3 a c d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )+\frac{a \left (5 b c^3-a d^3\right ) 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 )}{10 b \left (a+b x^3\right )^{2/3}}+\frac{\left (a c^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt [3]{b} x} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{b}}-\frac{\left (a c^2 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 )}{3 \sqrt [3]{b}}\\ &=\frac{3 a c d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )+\frac{a \left (5 b c^3-a d^3\right ) 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 )}{10 b \left (a+b x^3\right )^{2/3}}-\frac{a c^2 d \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac{\left (a c^2 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 )}{6 b^{2/3}}-\frac{\left (a c^2 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 )}{2 \sqrt [3]{b}}\\ &=\frac{3 a c d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )+\frac{a \left (5 b c^3-a d^3\right ) 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 )}{10 b \left (a+b x^3\right )^{2/3}}-\frac{a c^2 d \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac{a c^2 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 )}{6 b^{2/3}}+\frac{\left (a c^2 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 )}{b^{2/3}}\\ &=\frac{3 a c d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac{a d^3 x \sqrt [3]{a+b x^3}}{10 b}+\frac{1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )-\frac{a c^2 d \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt{3}}\right )}{\sqrt{3} b^{2/3}}+\frac{a \left (5 b c^3-a d^3\right ) 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 )}{10 b \left (a+b x^3\right )^{2/3}}-\frac{a c^2 d \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac{a c^2 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 )}{6 b^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.132894, size = 142, normalized size = 0.59 \[ \frac{\sqrt [3]{a+b x^3} \left (d \left (6 b c^2 x^2 \, _2F_1\left (-\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )+d \left (3 c \left (a+b x^3\right ) \sqrt [3]{\frac{b x^3}{a}+1}+b d x^4 \, _2F_1\left (-\frac{1}{3},\frac{4}{3};\frac{7}{3};-\frac{b x^3}{a}\right )\right )\right )+4 b c^3 x \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{4}{3};-\frac{b x^3}{a}\right )\right )}{4 b \sqrt [3]{\frac{b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^(1/3)*(4*b*c^3*x*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/a)] + d*(6*b*c^2*x^2*Hypergeometric2
F1[-1/3, 2/3, 5/3, -((b*x^3)/a)] + d*(3*c*(a + b*x^3)*(1 + (b*x^3)/a)^(1/3) + b*d*x^4*Hypergeometric2F1[-1/3,
4/3, 7/3, -((b*x^3)/a)]))))/(4*b*(1 + (b*x^3)/a)^(1/3))

________________________________________________________________________________________

Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{3}\sqrt [3]{b{x}^{3}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*(b*x^3+a)^(1/3),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [A]  time = 3.29069, size = 160, normalized size = 0.66 \begin{align*} \frac{\sqrt [3]{a} c^{3} 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} c^{2} 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 )}}{\Gamma \left (\frac{5}{3}\right )} + \frac{\sqrt [3]{a} 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 )} + 3 c 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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

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