3.1.24 \(\int (c+d x)^3 \sqrt [3]{a+b x^3} \, dx\) [24]

3.1.24.1 Optimal result

Integrand size = 19, antiderivative size = 242 \[ \int (c+d x)^3 \sqrt [3]{a+b x^3} \, dx=\frac {3 a c d^2 \sqrt [3]{a+b x^3}}{4 b}+\frac {a d^3 x \sqrt [3]{a+b x^3}}{10 b}+\frac {1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )-\frac {a c^2 d \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}+\frac {a \left (5 b c^3-a d^3\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{10 b \left (a+b x^3\right )^{2/3}}-\frac {a c^2 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}} \]

output

3/4*a*c*d^2*(b*x^3+a)^(1/3)/b+1/10*a*d^3*x*(b*x^3+a)^(1/3)/b+1/20*(b*x^3+a 
)^(1/3)*(4*d^3*x^4+15*c*d^2*x^3+20*c^2*d*x^2+10*c^3*x)+1/10*a*(-a*d^3+5*b* 
c^3)*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x^3/a)/b/(b*x^3+a)^ 
(2/3)-1/2*a*c^2*d*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)-1/3*a*c^2*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)
 

3.1.24.2 Mathematica [A] (verified)

Time = 6.49 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.59 \[ \int (c+d x)^3 \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt [3]{a+b x^3} \left (4 b c^3 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )+d \left (6 b c^2 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )+d \left (3 c \left (a+b x^3\right ) \sqrt [3]{1+\frac {b x^3}{a}}+b d x^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )\right )\right )\right )}{4 b \sqrt [3]{1+\frac {b x^3}{a}}} \]

input

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

output

((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*Hypergeometric2F1[-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))
 

3.1.24.3 Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {2392, 27, 2427, 27, 2425, 793, 2432, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2392

\(\displaystyle a \int \frac {10 c^3+20 d x c^2+15 d^2 x^2 c+4 d^3 x^3}{20 \left (b x^3+a\right )^{2/3}}dx+\frac {1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{20} a \int \frac {10 c^3+20 d x c^2+15 d^2 x^2 c+4 d^3 x^3}{\left (b x^3+a\right )^{2/3}}dx+\frac {1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )\)

\(\Big \downarrow \) 2427

\(\displaystyle \frac {1}{20} a \left (\frac {\int \frac {2 \left (20 b d x c^2+15 b d^2 x^2 c+2 \left (5 b c^3-a d^3\right )\right )}{\left (b x^3+a\right )^{2/3}}dx}{2 b}+\frac {2 d^3 x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{20} a \left (\frac {\int \frac {20 b d x c^2+15 b d^2 x^2 c+2 \left (5 b c^3-a d^3\right )}{\left (b x^3+a\right )^{2/3}}dx}{b}+\frac {2 d^3 x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )\)

\(\Big \downarrow \) 2425

\(\displaystyle \frac {1}{20} a \left (\frac {\int \frac {20 b d x c^2+2 \left (5 b c^3-a d^3\right )}{\left (b x^3+a\right )^{2/3}}dx+15 b c d^2 \int \frac {x^2}{\left (b x^3+a\right )^{2/3}}dx}{b}+\frac {2 d^3 x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )\)

\(\Big \downarrow \) 793

\(\displaystyle \frac {1}{20} a \left (\frac {\int \frac {20 b d x c^2+2 \left (5 b c^3-a d^3\right )}{\left (b x^3+a\right )^{2/3}}dx+15 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {2 d^3 x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )\)

\(\Big \downarrow \) 2432

\(\displaystyle \frac {1}{20} a \left (\frac {\int \left (\frac {20 b d x c^2}{\left (b x^3+a\right )^{2/3}}+\frac {2 \left (5 b c^3-a d^3\right )}{\left (b x^3+a\right )^{2/3}}\right )dx+15 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {2 d^3 x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{20} a \left (\frac {-\frac {20 \sqrt [3]{b} c^2 d \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {2 x \left (\frac {b x^3}{a}+1\right )^{2/3} \left (5 b c^3-a d^3\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-10 \sqrt [3]{b} c^2 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )+15 c d^2 \sqrt [3]{a+b x^3}}{b}+\frac {2 d^3 x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{20} \sqrt [3]{a+b x^3} \left (10 c^3 x+20 c^2 d x^2+15 c d^2 x^3+4 d^3 x^4\right )\)

input

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

output

((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))/2 
0 + (a*((2*d^3*x*(a + b*x^3)^(1/3))/b + (15*c*d^2*(a + b*x^3)^(1/3) - (20* 
b^(1/3)*c^2*d*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/Sqrt[ 
3] + (2*(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)])/(a + b*x^3)^(2/3) - 10*b^(1/3)*c^2*d*Log[b^(1/3)*x 
 - (a + b*x^3)^(1/3)])/b))/20
 

3.1.24.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 793

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2392

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Module[{q = Expon[Pq 
, x], i}, Simp[(a + b*x^n)^p*Sum[Coeff[Pq, x, i]*(x^(i + 1)/(n*p + i + 1)), 
 {i, 0, q}], x] + Simp[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 2425

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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] && PolyQ[P 
q, x] && IGtQ[n, 0] && Expon[Pq, x] == n - 1
 

rule 2427

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x 
]}, With[{Pqq = Coeff[Pq, x, q]}, Simp[Pqq*x^(q - n + 1)*((a + b*x^n)^(p + 
1)/(b*(q + n*p + 1))), x] + Simp[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]] /; 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] && IGtQ[n, 0]
 

rule 2432

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] || Poly 
Q[Pq, x^n])
 

3.1.24.4 Maple [F]

\[\int \left (d x +c \right )^{3} \left (b \,x^{3}+a \right )^{\frac {1}{3}}d x\]

input

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

output

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

3.1.24.5 Fricas [F]

\[ \int (c+d x)^3 \sqrt [3]{a+b x^3} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{3} \,d x } \]

input

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

output

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

3.1.24.6 Sympy [A] (verification not implemented)

Time = 1.75 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.66 \[ \int (c+d x)^3 \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt [3]{a} c^{3} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {\sqrt [3]{a} c^{2} d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{\Gamma \left (\frac {5}{3}\right )} + \frac {\sqrt [3]{a} d^{3} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + 3 c d^{2} \left (\begin {cases} \frac {\sqrt [3]{a} x^{3}}{3} & \text {for}\: b = 0 \\\frac {\left (a + b x^{3}\right )^{\frac {4}{3}}}{4 b} & \text {otherwise} \end {cases}\right ) \]

input

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

output

a**(1/3)*c**3*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*p 
i)/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))
 

3.1.24.7 Maxima [F]

\[ \int (c+d x)^3 \sqrt [3]{a+b x^3} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{3} \,d x } \]

input

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

output

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

3.1.24.8 Giac [F]

\[ \int (c+d x)^3 \sqrt [3]{a+b x^3} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{3} \,d x } \]

input

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

output

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

3.1.24.9 Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \sqrt [3]{a+b x^3} \, dx=\int {\left (b\,x^3+a\right )}^{1/3}\,{\left (c+d\,x\right )}^3 \,d x \]

input

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

output

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