Integrand size = 20, antiderivative size = 92 \[ \int x^3 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {d x^4 \left (a+b x^3\right )^{1+p}}{b (7+3 p)}+\frac {1}{4} \left (c-\frac {4 a d}{7 b+3 b p}\right ) x^4 \left (a+b x^3\right )^p \left (1+\frac {b x^3}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},-p,\frac {7}{3},-\frac {b x^3}{a}\right ) \] Output:
d*x^4*(b*x^3+a)^(p+1)/b/(7+3*p)+1/4*(c-4*a*d/(3*b*p+7*b))*x^4*(b*x^3+a)^p* hypergeom([4/3, -p],[7/3],-b*x^3/a)/((1+b*x^3/a)^p)
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.85 \[ \int x^3 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {1}{28} x^4 \left (a+b x^3\right )^p \left (1+\frac {b x^3}{a}\right )^{-p} \left (7 c \operatorname {Hypergeometric2F1}\left (\frac {4}{3},-p,\frac {7}{3},-\frac {b x^3}{a}\right )+4 d x^3 \operatorname {Hypergeometric2F1}\left (\frac {7}{3},-p,\frac {10}{3},-\frac {b x^3}{a}\right )\right ) \] Input:
Integrate[x^3*(a + b*x^3)^p*(c + d*x^3),x]
Output:
(x^4*(a + b*x^3)^p*(7*c*Hypergeometric2F1[4/3, -p, 7/3, -((b*x^3)/a)] + 4* d*x^3*Hypergeometric2F1[7/3, -p, 10/3, -((b*x^3)/a)]))/(28*(1 + (b*x^3)/a) ^p)
Time = 0.40 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {959, 889, 888}
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.
| \(\displaystyle \int x^3 \left (c+d x^3\right ) \left (a+b x^3\right )^p \, dx\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \left (c-\frac {4 a d}{3 b p+7 b}\right ) \int x^3 \left (b x^3+a\right )^pdx+\frac {d x^4 \left (a+b x^3\right )^{p+1}}{b (3 p+7)}\) |
| \(\Big \downarrow \) 889 |
| \(\displaystyle \left (a+b x^3\right )^p \left (\frac {b x^3}{a}+1\right )^{-p} \left (c-\frac {4 a d}{3 b p+7 b}\right ) \int x^3 \left (\frac {b x^3}{a}+1\right )^pdx+\frac {d x^4 \left (a+b x^3\right )^{p+1}}{b (3 p+7)}\) |
| \(\Big \downarrow \) 888 |
| \(\displaystyle \frac {1}{4} x^4 \left (a+b x^3\right )^p \left (\frac {b x^3}{a}+1\right )^{-p} \left (c-\frac {4 a d}{3 b p+7 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {4}{3},-p,\frac {7}{3},-\frac {b x^3}{a}\right )+\frac {d x^4 \left (a+b x^3\right )^{p+1}}{b (3 p+7)}\) |
Input:
Int[x^3*(a + b*x^3)^p*(c + d*x^3),x]
Output:
(d*x^4*(a + b*x^3)^(1 + p))/(b*(7 + 3*p)) + ((c - (4*a*d)/(7*b + 3*b*p))*x ^4*(a + b*x^3)^p*Hypergeometric2F1[4/3, -p, 7/3, -((b*x^3)/a)])/(4*(1 + (b *x^3)/a)^p)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[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])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
\[\int x^{3} \left (b \,x^{3}+a \right )^{p} \left (d \,x^{3}+c \right )d x\]
Input:
int(x^3*(b*x^3+a)^p*(d*x^3+c),x)
Output:
int(x^3*(b*x^3+a)^p*(d*x^3+c),x)
\[ \int x^3 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\int { {\left (d x^{3} + c\right )} {\left (b x^{3} + a\right )}^{p} x^{3} \,d x } \] Input:
integrate(x^3*(b*x^3+a)^p*(d*x^3+c),x, algorithm="fricas")
Output:
integral((d*x^6 + c*x^3)*(b*x^3 + a)^p, x)
Result contains complex when optimal does not.
Time = 96.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.83 \[ \int x^3 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {a^{p} c x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, - p \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{p} d x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{3}, - p \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \] Input:
integrate(x**3*(b*x**3+a)**p*(d*x**3+c),x)
Output:
a**p*c*x**4*gamma(4/3)*hyper((4/3, -p), (7/3,), b*x**3*exp_polar(I*pi)/a)/ (3*gamma(7/3)) + a**p*d*x**7*gamma(7/3)*hyper((7/3, -p), (10/3,), b*x**3*e xp_polar(I*pi)/a)/(3*gamma(10/3))
\[ \int x^3 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\int { {\left (d x^{3} + c\right )} {\left (b x^{3} + a\right )}^{p} x^{3} \,d x } \] Input:
integrate(x^3*(b*x^3+a)^p*(d*x^3+c),x, algorithm="maxima")
Output:
integrate((d*x^3 + c)*(b*x^3 + a)^p*x^3, x)
\[ \int x^3 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\int { {\left (d x^{3} + c\right )} {\left (b x^{3} + a\right )}^{p} x^{3} \,d x } \] Input:
integrate(x^3*(b*x^3+a)^p*(d*x^3+c),x, algorithm="giac")
Output:
integrate((d*x^3 + c)*(b*x^3 + a)^p*x^3, x)
Timed out. \[ \int x^3 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\int x^3\,{\left (b\,x^3+a\right )}^p\,\left (d\,x^3+c\right ) \,d x \] Input:
int(x^3*(a + b*x^3)^p*(c + d*x^3),x)
Output:
int(x^3*(a + b*x^3)^p*(c + d*x^3), x)
\[ \int x^3 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx =\text {Too large to display} \] Input:
int(x^3*(b*x^3+a)^p*(d*x^3+c),x)
Output:
( - 12*(a + b*x**3)**p*a**2*d*p*x + 9*(a + b*x**3)**p*a*b*c*p**2*x + 21*(a + b*x**3)**p*a*b*c*p*x + 9*(a + b*x**3)**p*a*b*d*p**2*x**4 + 3*(a + b*x** 3)**p*a*b*d*p*x**4 + 9*(a + b*x**3)**p*b**2*c*p**2*x**4 + 24*(a + b*x**3)* *p*b**2*c*p*x**4 + 7*(a + b*x**3)**p*b**2*c*x**4 + 9*(a + b*x**3)**p*b**2* d*p**2*x**7 + 15*(a + b*x**3)**p*b**2*d*p*x**7 + 4*(a + b*x**3)**p*b**2*d* x**7 + 324*int((a + b*x**3)**p/(27*a*p**3 + 108*a*p**2 + 117*a*p + 28*a + 27*b*p**3*x**3 + 108*b*p**2*x**3 + 117*b*p*x**3 + 28*b*x**3),x)*a**3*d*p** 4 + 1296*int((a + b*x**3)**p/(27*a*p**3 + 108*a*p**2 + 117*a*p + 28*a + 27 *b*p**3*x**3 + 108*b*p**2*x**3 + 117*b*p*x**3 + 28*b*x**3),x)*a**3*d*p**3 + 1404*int((a + b*x**3)**p/(27*a*p**3 + 108*a*p**2 + 117*a*p + 28*a + 27*b *p**3*x**3 + 108*b*p**2*x**3 + 117*b*p*x**3 + 28*b*x**3),x)*a**3*d*p**2 + 336*int((a + b*x**3)**p/(27*a*p**3 + 108*a*p**2 + 117*a*p + 28*a + 27*b*p* *3*x**3 + 108*b*p**2*x**3 + 117*b*p*x**3 + 28*b*x**3),x)*a**3*d*p - 243*in t((a + b*x**3)**p/(27*a*p**3 + 108*a*p**2 + 117*a*p + 28*a + 27*b*p**3*x** 3 + 108*b*p**2*x**3 + 117*b*p*x**3 + 28*b*x**3),x)*a**2*b*c*p**5 - 1539*in t((a + b*x**3)**p/(27*a*p**3 + 108*a*p**2 + 117*a*p + 28*a + 27*b*p**3*x** 3 + 108*b*p**2*x**3 + 117*b*p*x**3 + 28*b*x**3),x)*a**2*b*c*p**4 - 3321*in t((a + b*x**3)**p/(27*a*p**3 + 108*a*p**2 + 117*a*p + 28*a + 27*b*p**3*x** 3 + 108*b*p**2*x**3 + 117*b*p*x**3 + 28*b*x**3),x)*a**2*b*c*p**3 - 2709*in t((a + b*x**3)**p/(27*a*p**3 + 108*a*p**2 + 117*a*p + 28*a + 27*b*p**3*...