Integrand size = 20, antiderivative size = 100 \[ \int x^2 \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\frac {d x^3 \left (a+b x^n\right )^{1+p}}{b (3+n+n p)}+\frac {1}{3} \left (c-\frac {3 a d}{b (3+n+n p)}\right ) x^3 \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b x^n}{a}\right ) \] Output:
d*x^3*(a+b*x^n)^(p+1)/b/(n*p+n+3)+1/3*(c-3*a*d/b/(n*p+n+3))*x^3*(a+b*x^n)^ p*hypergeom([-p, 3/n],[(3+n)/n],-b*x^n/a)/((1+b*x^n/a)^p)
Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99 \[ \int x^2 \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\frac {x^3 \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \left (c (3+n) \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b x^n}{a}\right )+3 d x^n \operatorname {Hypergeometric2F1}\left (\frac {3+n}{n},-p,2+\frac {3}{n},-\frac {b x^n}{a}\right )\right )}{3 (3+n)} \] Input:
Integrate[x^2*(a + b*x^n)^p*(c + d*x^n),x]
Output:
(x^3*(a + b*x^n)^p*(c*(3 + n)*Hypergeometric2F1[3/n, -p, (3 + n)/n, -((b*x ^n)/a)] + 3*d*x^n*Hypergeometric2F1[(3 + n)/n, -p, 2 + 3/n, -((b*x^n)/a)]) )/(3*(3 + n)*(1 + (b*x^n)/a)^p)
Time = 0.40 (sec) , antiderivative size = 100, 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.
\(\displaystyle \int x^2 \left (c+d x^n\right ) \left (a+b x^n\right )^p \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \left (c-\frac {3 a d}{b (n p+n+3)}\right ) \int x^2 \left (b x^n+a\right )^pdx+\frac {d x^3 \left (a+b x^n\right )^{p+1}}{b (n p+n+3)}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \left (c-\frac {3 a d}{b (n p+n+3)}\right ) \int x^2 \left (\frac {b x^n}{a}+1\right )^pdx+\frac {d x^3 \left (a+b x^n\right )^{p+1}}{b (n p+n+3)}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \left (c-\frac {3 a d}{b (n p+n+3)}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {n+3}{n},-\frac {b x^n}{a}\right )+\frac {d x^3 \left (a+b x^n\right )^{p+1}}{b (n p+n+3)}\) |
Input:
Int[x^2*(a + b*x^n)^p*(c + d*x^n),x]
Output:
(d*x^3*(a + b*x^n)^(1 + p))/(b*(3 + n + n*p)) + ((c - (3*a*d)/(b*(3 + n + n*p)))*x^3*(a + b*x^n)^p*Hypergeometric2F1[3/n, -p, (3 + n)/n, -((b*x^n)/a )])/(3*(1 + (b*x^n)/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]
\[\int x^{2} \left (a +b \,x^{n}\right )^{p} \left (c +d \,x^{n}\right )d x\]
Input:
int(x^2*(a+b*x^n)^p*(c+d*x^n),x)
Output:
int(x^2*(a+b*x^n)^p*(c+d*x^n),x)
\[ \int x^2 \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\int { {\left (d x^{n} + c\right )} {\left (b x^{n} + a\right )}^{p} x^{2} \,d x } \] Input:
integrate(x^2*(a+b*x^n)^p*(c+d*x^n),x, algorithm="fricas")
Output:
integral((d*x^2*x^n + c*x^2)*(b*x^n + a)^p, x)
Result contains complex when optimal does not.
Time = 6.73 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.09 \[ \int x^2 \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\frac {a^{\frac {3}{n}} a^{p - \frac {3}{n}} c x^{3} \Gamma \left (\frac {3}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{n}, - p \\ 1 + \frac {3}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (1 + \frac {3}{n}\right )} + \frac {a^{1 + \frac {3}{n}} a^{p - 1 - \frac {3}{n}} d x^{n + 3} \Gamma \left (1 + \frac {3}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 + \frac {3}{n} \\ 2 + \frac {3}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (2 + \frac {3}{n}\right )} \] Input:
integrate(x**2*(a+b*x**n)**p*(c+d*x**n),x)
Output:
a**(3/n)*a**(p - 3/n)*c*x**3*gamma(3/n)*hyper((3/n, -p), (1 + 3/n,), b*x** n*exp_polar(I*pi)/a)/(n*gamma(1 + 3/n)) + a**(1 + 3/n)*a**(p - 1 - 3/n)*d* x**(n + 3)*gamma(1 + 3/n)*hyper((-p, 1 + 3/n), (2 + 3/n,), b*x**n*exp_pola r(I*pi)/a)/(n*gamma(2 + 3/n))
\[ \int x^2 \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\int { {\left (d x^{n} + c\right )} {\left (b x^{n} + a\right )}^{p} x^{2} \,d x } \] Input:
integrate(x^2*(a+b*x^n)^p*(c+d*x^n),x, algorithm="maxima")
Output:
integrate((d*x^n + c)*(b*x^n + a)^p*x^2, x)
\[ \int x^2 \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\int { {\left (d x^{n} + c\right )} {\left (b x^{n} + a\right )}^{p} x^{2} \,d x } \] Input:
integrate(x^2*(a+b*x^n)^p*(c+d*x^n),x, algorithm="giac")
Output:
integrate((d*x^n + c)*(b*x^n + a)^p*x^2, x)
Timed out. \[ \int x^2 \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\int x^2\,{\left (a+b\,x^n\right )}^p\,\left (c+d\,x^n\right ) \,d x \] Input:
int(x^2*(a + b*x^n)^p*(c + d*x^n),x)
Output:
int(x^2*(a + b*x^n)^p*(c + d*x^n), x)
\[ \int x^2 \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx =\text {Too large to display} \] Input:
int(x^2*(a+b*x^n)^p*(c+d*x^n),x)
Output:
(x**n*(x**n*b + a)**p*b*d*n*p*x**3 + 3*x**n*(x**n*b + a)**p*b*d*x**3 + (x* *n*b + a)**p*a*d*n*p*x**3 + (x**n*b + a)**p*b*c*n*p*x**3 + (x**n*b + a)**p *b*c*n*x**3 + 3*(x**n*b + a)**p*b*c*x**3 - 3*int(((x**n*b + a)**p*x**2)/(x **n*b*n**2*p**2 + x**n*b*n**2*p + 6*x**n*b*n*p + 3*x**n*b*n + 9*x**n*b + a *n**2*p**2 + a*n**2*p + 6*a*n*p + 3*a*n + 9*a),x)*a**2*d*n**3*p**3 - 3*int (((x**n*b + a)**p*x**2)/(x**n*b*n**2*p**2 + x**n*b*n**2*p + 6*x**n*b*n*p + 3*x**n*b*n + 9*x**n*b + a*n**2*p**2 + a*n**2*p + 6*a*n*p + 3*a*n + 9*a),x )*a**2*d*n**3*p**2 - 18*int(((x**n*b + a)**p*x**2)/(x**n*b*n**2*p**2 + x** n*b*n**2*p + 6*x**n*b*n*p + 3*x**n*b*n + 9*x**n*b + a*n**2*p**2 + a*n**2*p + 6*a*n*p + 3*a*n + 9*a),x)*a**2*d*n**2*p**2 - 9*int(((x**n*b + a)**p*x** 2)/(x**n*b*n**2*p**2 + x**n*b*n**2*p + 6*x**n*b*n*p + 3*x**n*b*n + 9*x**n* b + a*n**2*p**2 + a*n**2*p + 6*a*n*p + 3*a*n + 9*a),x)*a**2*d*n**2*p - 27* int(((x**n*b + a)**p*x**2)/(x**n*b*n**2*p**2 + x**n*b*n**2*p + 6*x**n*b*n* p + 3*x**n*b*n + 9*x**n*b + a*n**2*p**2 + a*n**2*p + 6*a*n*p + 3*a*n + 9*a ),x)*a**2*d*n*p + int(((x**n*b + a)**p*x**2)/(x**n*b*n**2*p**2 + x**n*b*n* *2*p + 6*x**n*b*n*p + 3*x**n*b*n + 9*x**n*b + a*n**2*p**2 + a*n**2*p + 6*a *n*p + 3*a*n + 9*a),x)*a*b*c*n**4*p**4 + 2*int(((x**n*b + a)**p*x**2)/(x** n*b*n**2*p**2 + x**n*b*n**2*p + 6*x**n*b*n*p + 3*x**n*b*n + 9*x**n*b + a*n **2*p**2 + a*n**2*p + 6*a*n*p + 3*a*n + 9*a),x)*a*b*c*n**4*p**3 + int(((x* *n*b + a)**p*x**2)/(x**n*b*n**2*p**2 + x**n*b*n**2*p + 6*x**n*b*n*p + 3...