\(\int (a+b x^n)^p (c+d x^n)^2 \, dx\) [314]

Optimal result

Integrand size = 19, antiderivative size = 202 \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^2 \, dx=-\frac {d (a d (1+n)-b c (1+n (3+p))) x \left (a+b x^n\right )^{1+p}}{b^2 (1+n+n p) (1+n (2+p))}+\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{b (1+2 n+n p)}-\frac {(b c (1+n+n p) (a d-b c (1+n (2+p)))-a d (a d (1+n)-b c (1+n (3+p)))) x \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b^2 (1+n+n p) (1+n (2+p))} \]

[Out]

-d*(a*d*(1+n)-b*c*(1+n*(3+p)))*x*(a+b*x^n)^(p+1)/b^2/(n*p+n+1)/(1+n*(2+p))+d*x*(a+b*x^n)^(p+1)*(c+d*x^n)/b/(n*
p+2*n+1)-(b*c*(n*p+n+1)*(a*d-b*c*(1+n*(2+p)))-a*d*(a*d*(1+n)-b*c*(1+n*(3+p))))*x*(a+b*x^n)^p*hypergeom([1/n, -
p],[1+1/n],-b*x^n/a)/b^2/(n*p+n+1)/(1+n*(2+p))/((1+b*x^n/a)^p)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {427, 396, 252, 251} \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^2 \, dx=-\frac {d x \left (a+b x^n\right )^{p+1} (a d (n+1)-b (c n (p+3)+c))}{b^2 (n p+n+1) (n (p+2)+1)}-\frac {x \left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \left (c (a d-b (c n (p+2)+c))-\frac {a d (a d (n+1)-b (c n (p+3)+c))}{b (n p+n+1)}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b (n (p+2)+1)}+\frac {d x \left (c+d x^n\right ) \left (a+b x^n\right )^{p+1}}{b (n (p+2)+1)} \]

[In]

Int[(a + b*x^n)^p*(c + d*x^n)^2,x]

[Out]

-((d*(a*d*(1 + n) - b*(c + c*n*(3 + p)))*x*(a + b*x^n)^(1 + p))/(b^2*(1 + n + n*p)*(1 + n*(2 + p)))) + (d*x*(a
+ b*x^n)^(1 + p)*(c + d*x^n))/(b*(1 + n*(2 + p))) - ((c*(a*d - b*(c + c*n*(2 + p))) - (a*d*(a*d*(1 + n) - b*(
c + c*n*(3 + p))))/(b*(1 + n + n*p)))*x*(a + b*x^n)^p*Hypergeometric2F1[n^(-1), -p, 1 + n^(-1), -((b*x^n)/a)])
/(b*(1 + n*(2 + p))*(1 + (b*x^n)/a)^p)

Rule 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps \begin{align*} \text {integral}& = \frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{b (1+n (2+p))}+\frac {\int \left (a+b x^n\right )^p \left (-c (a d-b (c+c n (2+p)))-d (a d (1+n)-b (c+c n (3+p))) x^n\right ) \, dx}{b (1+n (2+p))} \\ & = -\frac {d (a d (1+n)-b (c+c n (3+p))) x \left (a+b x^n\right )^{1+p}}{b^2 (1+n+n p) (1+n (2+p))}+\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{b (1+n (2+p))}-\frac {\left (c (a d-b (c+c n (2+p)))-\frac {a d (a d (1+n)-b (c+c n (3+p)))}{b (1+n+n p)}\right ) \int \left (a+b x^n\right )^p \, dx}{b (1+n (2+p))} \\ & = -\frac {d (a d (1+n)-b (c+c n (3+p))) x \left (a+b x^n\right )^{1+p}}{b^2 (1+n+n p) (1+n (2+p))}+\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{b (1+n (2+p))}-\frac {\left (\left (c (a d-b (c+c n (2+p)))-\frac {a d (a d (1+n)-b (c+c n (3+p)))}{b (1+n+n p)}\right ) \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^n}{a}\right )^p \, dx}{b (1+n (2+p))} \\ & = -\frac {d (a d (1+n)-b (c+c n (3+p))) x \left (a+b x^n\right )^{1+p}}{b^2 (1+n+n p) (1+n (2+p))}+\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{b (1+n (2+p))}-\frac {\left (c (a d-b (c+c n (2+p)))-\frac {a d (a d (1+n)-b (c+c n (3+p)))}{b (1+n+n p)}\right ) x \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b x^n}{a}\right )}{b (1+n (2+p))} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.20 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.69 \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^2 \, dx=\frac {x \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \left (2 c d (1+2 n) x^n \operatorname {Hypergeometric2F1}\left (1+\frac {1}{n},-p,2+\frac {1}{n},-\frac {b x^n}{a}\right )+(1+n) \left (d^2 x^{2 n} \operatorname {Hypergeometric2F1}\left (2+\frac {1}{n},-p,3+\frac {1}{n},-\frac {b x^n}{a}\right )+c^2 (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b x^n}{a}\right )\right )\right )}{(1+n) (1+2 n)} \]

[In]

Integrate[(a + b*x^n)^p*(c + d*x^n)^2,x]

[Out]

(x*(a + b*x^n)^p*(2*c*d*(1 + 2*n)*x^n*Hypergeometric2F1[1 + n^(-1), -p, 2 + n^(-1), -((b*x^n)/a)] + (1 + n)*(d
^2*x^(2*n)*Hypergeometric2F1[2 + n^(-1), -p, 3 + n^(-1), -((b*x^n)/a)] + c^2*(1 + 2*n)*Hypergeometric2F1[n^(-1
), -p, 1 + n^(-1), -((b*x^n)/a)])))/((1 + n)*(1 + 2*n)*(1 + (b*x^n)/a)^p)

Maple [F]

\[\int \left (a +b \,x^{n}\right )^{p} \left (c +d \,x^{n}\right )^{2}d x\]

[In]

int((a+b*x^n)^p*(c+d*x^n)^2,x)

[Out]

int((a+b*x^n)^p*(c+d*x^n)^2,x)

Fricas [F]

\[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^2 \, dx=\int { {\left (d x^{n} + c\right )}^{2} {\left (b x^{n} + a\right )}^{p} \,d x } \]

[In]

integrate((a+b*x^n)^p*(c+d*x^n)^2,x, algorithm="fricas")

[Out]

integral((d^2*x^(2*n) + 2*c*d*x^n + c^2)*(b*x^n + a)^p, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 17.30 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.87 \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^2 \, dx=\frac {a^{\frac {1}{n}} a^{p - \frac {1}{n}} c^{2} x \Gamma \left (\frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{n}, - p \\ 1 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (1 + \frac {1}{n}\right )} + \frac {2 a^{1 + \frac {1}{n}} a^{p - 1 - \frac {1}{n}} c d x^{n + 1} \Gamma \left (1 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 + \frac {1}{n} \\ 2 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (2 + \frac {1}{n}\right )} + \frac {a^{2 + \frac {1}{n}} a^{p - 2 - \frac {1}{n}} d^{2} x^{2 n + 1} \Gamma \left (2 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 2 + \frac {1}{n} \\ 3 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (3 + \frac {1}{n}\right )} \]

[In]

integrate((a+b*x**n)**p*(c+d*x**n)**2,x)

[Out]

a**(1/n)*a**(p - 1/n)*c**2*x*gamma(1/n)*hyper((1/n, -p), (1 + 1/n,), b*x**n*exp_polar(I*pi)/a)/(n*gamma(1 + 1/
n)) + 2*a**(1 + 1/n)*a**(p - 1 - 1/n)*c*d*x**(n + 1)*gamma(1 + 1/n)*hyper((-p, 1 + 1/n), (2 + 1/n,), b*x**n*ex
p_polar(I*pi)/a)/(n*gamma(2 + 1/n)) + a**(2 + 1/n)*a**(p - 2 - 1/n)*d**2*x**(2*n + 1)*gamma(2 + 1/n)*hyper((-p
, 2 + 1/n), (3 + 1/n,), b*x**n*exp_polar(I*pi)/a)/(n*gamma(3 + 1/n))

Maxima [F]

\[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^2 \, dx=\int { {\left (d x^{n} + c\right )}^{2} {\left (b x^{n} + a\right )}^{p} \,d x } \]

[In]

integrate((a+b*x^n)^p*(c+d*x^n)^2,x, algorithm="maxima")

[Out]

integrate((d*x^n + c)^2*(b*x^n + a)^p, x)

Giac [F(-2)]

Exception generated. \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^2 \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a+b*x^n)^p*(c+d*x^n)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-1,[1,0,4,3,1,3,3,2,0]%%%}+%%%{-3,[1,0,4,3,1,2,3,2,0]%%%
}+%%%{-3,[1

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^2 \, dx=\int {\left (a+b\,x^n\right )}^p\,{\left (c+d\,x^n\right )}^2 \,d x \]

[In]

int((a + b*x^n)^p*(c + d*x^n)^2,x)

[Out]

int((a + b*x^n)^p*(c + d*x^n)^2, x)