\(\int x^m (a+b x^n)^p (a (1+m+q) x^q+b (1+m+n (1+p)+q) x^{n+q}) \, dx\) [283]

Optimal result

Integrand size = 39, antiderivative size = 18 \[ \int x^m \left (a+b x^n\right )^p \left (a (1+m+q) x^q+b (1+m+n (1+p)+q) x^{n+q}\right ) \, dx=x^{1+m+q} \left (a+b x^n\right )^{1+p} \] Output:

x^(1+m+q)*(a+b*x^n)^(p+1)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.33 (sec) , antiderivative size = 116, normalized size of antiderivative = 6.44 \[ \int x^m \left (a+b x^n\right )^p \left (a (1+m+q) x^q+b (1+m+n (1+p)+q) x^{n+q}\right ) \, dx=x^{1+m+q} \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \left (a \operatorname {Hypergeometric2F1}\left (-p,\frac {1+m+q}{n},\frac {1+m+n+q}{n},-\frac {b x^n}{a}\right )+\frac {b (1+m+n+n p+q) x^n \operatorname {Hypergeometric2F1}\left (-p,\frac {1+m+n+q}{n},\frac {1+m+2 n+q}{n},-\frac {b x^n}{a}\right )}{1+m+n+q}\right ) \] Input:

Integrate[x^m*(a + b*x^n)^p*(a*(1 + m + q)*x^q + b*(1 + m + n*(1 + p) + q) 
*x^(n + q)),x]
 

Output:

(x^(1 + m + q)*(a + b*x^n)^p*(a*Hypergeometric2F1[-p, (1 + m + q)/n, (1 + 
m + n + q)/n, -((b*x^n)/a)] + (b*(1 + m + n + n*p + q)*x^n*Hypergeometric2 
F1[-p, (1 + m + n + q)/n, (1 + m + 2*n + q)/n, -((b*x^n)/a)])/(1 + m + n + 
 q)))/(1 + (b*x^n)/a)^p
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {10, 951}

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.

\(\displaystyle \int x^m \left (a+b x^n\right )^p \left (a (m+q+1) x^q+b x^{n+q} (m+n (p+1)+q+1)\right ) \, dx\)

\(\Big \downarrow \) 10

\(\displaystyle \int x^{m+q} \left (a+b x^n\right )^p \left (a (m+q+1)+b x^n (m+n p+n+q+1)\right )dx\)

\(\Big \downarrow \) 951

\(\displaystyle x^{m+q+1} \left (a+b x^n\right )^{p+1}\)

Input:

Int[x^m*(a + b*x^n)^p*(a*(1 + m + q)*x^q + b*(1 + m + n*(1 + p) + q)*x^(n 
+ q)),x]
 

Output:

x^(1 + m + q)*(a + b*x^n)^(1 + p)
 

Defintions of rubi rules used

rule 10

Int[(u_.)*((e_.)*(x_))^(m_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x 
_Symbol] :> Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*(a + b*x^(s - r))^p, x], 
 x] /; FreeQ[{a, b, e, m, r, s}, x] && IntegerQ[p] && (IntegerQ[p*r] || GtQ 
[e, 0]) && PosQ[s - r]
 

rule 951

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

Maple [F]

\[\int x^{m} \left (a +b \,x^{n}\right )^{p} \left (a \left (1+m +q \right ) x^{q}+b \left (1+m +n \left (p +1\right )+q \right ) x^{n +q}\right )d x\]

Input:

int(x^m*(a+b*x^n)^p*(a*(1+m+q)*x^q+b*(1+m+n*(p+1)+q)*x^(n+q)),x)
 

Output:

int(x^m*(a+b*x^n)^p*(a*(1+m+q)*x^q+b*(1+m+n*(p+1)+q)*x^(n+q)),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (18) = 36\).

Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.39 \[ \int x^m \left (a+b x^n\right )^p \left (a (1+m+q) x^q+b (1+m+n (1+p)+q) x^{n+q}\right ) \, dx={\left (b x x^{m} x^{n + q} + a x x^{m} x^{q}\right )} \left (\frac {b x^{n + q} + a x^{q}}{x^{q}}\right )^{p} \] Input:

integrate(x^m*(a+b*x^n)^p*(a*(1+m+q)*x^q+b*(1+m+n*(p+1)+q)*x^(n+q)),x, alg 
orithm="fricas")
 

Output:

(b*x*x^m*x^(n + q) + a*x*x^m*x^q)*((b*x^(n + q) + a*x^q)/x^q)^p
 

Sympy [F(-1)]

Timed out. \[ \int x^m \left (a+b x^n\right )^p \left (a (1+m+q) x^q+b (1+m+n (1+p)+q) x^{n+q}\right ) \, dx=\text {Timed out} \] Input:

integrate(x**m*(a+b*x**n)**p*(a*(1+m+q)*x**q+b*(1+m+n*(p+1)+q)*x**(n+q)),x 
)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.06 \[ \int x^m \left (a+b x^n\right )^p \left (a (1+m+q) x^q+b (1+m+n (1+p)+q) x^{n+q}\right ) \, dx={\left (a x x^{m} + b x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}\right )} e^{\left (p \log \left (b x^{n} + a\right ) + q \log \left (x\right )\right )} \] Input:

integrate(x^m*(a+b*x^n)^p*(a*(1+m+q)*x^q+b*(1+m+n*(p+1)+q)*x^(n+q)),x, alg 
orithm="maxima")
 

Output:

(a*x*x^m + b*x*e^(m*log(x) + n*log(x)))*e^(p*log(b*x^n + a) + q*log(x))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (18) = 36\).

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.67 \[ \int x^m \left (a+b x^n\right )^p \left (a (1+m+q) x^q+b (1+m+n (1+p)+q) x^{n+q}\right ) \, dx={\left (b x^{n} + a\right )}^{p} b x x^{n} e^{\left (m \log \left (x\right ) + q \log \left (x\right )\right )} + {\left (b x^{n} + a\right )}^{p} a x e^{\left (m \log \left (x\right ) + q \log \left (x\right )\right )} \] Input:

integrate(x^m*(a+b*x^n)^p*(a*(1+m+q)*x^q+b*(1+m+n*(p+1)+q)*x^(n+q)),x, alg 
orithm="giac")
 

Output:

(b*x^n + a)^p*b*x*x^n*e^(m*log(x) + q*log(x)) + (b*x^n + a)^p*a*x*e^(m*log 
(x) + q*log(x))
 

Mupad [F(-1)]

Timed out. \[ \int x^m \left (a+b x^n\right )^p \left (a (1+m+q) x^q+b (1+m+n (1+p)+q) x^{n+q}\right ) \, dx=\int x^m\,\left (a\,x^q\,\left (m+q+1\right )+b\,x^{n+q}\,\left (m+q+n\,\left (p+1\right )+1\right )\right )\,{\left (a+b\,x^n\right )}^p \,d x \] Input:

int(x^m*(a*x^q*(m + q + 1) + b*x^(n + q)*(m + q + n*(p + 1) + 1))*(a + b*x 
^n)^p,x)
 

Output:

int(x^m*(a*x^q*(m + q + 1) + b*x^(n + q)*(m + q + n*(p + 1) + 1))*(a + b*x 
^n)^p, x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int x^m \left (a+b x^n\right )^p \left (a (1+m+q) x^q+b (1+m+n (1+p)+q) x^{n+q}\right ) \, dx=x^{m +q} \left (x^{n} b +a \right )^{p} x \left (x^{n} b +a \right ) \] Input:

int(x^m*(a+b*x^n)^p*(a*(1+m+q)*x^q+b*(1+m+n*(p+1)+q)*x^(n+q)),x)
 

Output:

x**(m + q)*(x**n*b + a)**p*x*(x**n*b + a)