\(\int (b x)^m (\pi +d x)^n (e+f x)^p \, dx\) [596]

Optimal result

Integrand size = 20, antiderivative size = 49 \[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\frac {e^p \pi ^n (b x)^{1+m} \operatorname {AppellF1}\left (1+m,-n,-p,2+m,-\frac {d x}{\pi },-\frac {f x}{e}\right )}{b (1+m)} \] Output:

exp(p)*Pi^n*(b*x)^(1+m)*AppellF1(1+m,-n,-p,2+m,-d*x/Pi,-f*x*exp(-1))/b/(1+ 
m)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.92 \[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\frac {e^p \pi ^n x (b x)^m \operatorname {AppellF1}\left (1+m,-n,-p,2+m,-\frac {d x}{\pi },-\frac {f x}{e}\right )}{1+m} \] Input:

Integrate[(b*x)^m*(Pi + d*x)^n*(E + f*x)^p,x]
 

Output:

(E^p*Pi^n*x*(b*x)^m*AppellF1[1 + m, -n, -p, 2 + m, -((d*x)/Pi), -((f*x)/E) 
])/(1 + m)
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {150}

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.

\(\displaystyle \int (b x)^m (d x+\pi )^n (f x+e)^p \, dx\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\pi ^n e^p (b x)^{m+1} \operatorname {AppellF1}\left (m+1,-n,-p,m+2,-\frac {d x}{\pi },-\frac {f x}{e}\right )}{b (m+1)}\)

Input:

Int[(b*x)^m*(Pi + d*x)^n*(E + f*x)^p,x]
 

Output:

(E^p*Pi^n*(b*x)^(1 + m)*AppellF1[1 + m, -n, -p, 2 + m, -((d*x)/Pi), -((f*x 
)/E)])/(b*(1 + m))
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

Maple [F]

\[\int \left (b x \right )^{m} \left (x d +\pi \right )^{n} \left (f x +{\mathrm e}\right )^{p}d x\]

Input:

int((b*x)^m*(d*x+Pi)^n*(f*x+exp(1))^p,x)
 

Output:

int((b*x)^m*(d*x+Pi)^n*(f*x+exp(1))^p,x)
 

Fricas [F]

\[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\int { {\left (\pi + d x\right )}^{n} \left (b x\right )^{m} {\left (f x + e\right )}^{p} \,d x } \] Input:

integrate((b*x)^m*(d*x+pi)^n*(f*x+exp(1))^p,x, algorithm="fricas")
 

Output:

integral((pi + d*x)^n*(b*x)^m*(f*x + e)^p, x)
 

Sympy [F]

\[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\int \left (b x\right )^{m} \left (d x + \pi \right )^{n} \left (f x + e\right )^{p}\, dx \] Input:

integrate((b*x)**m*(d*x+pi)**n*(f*x+exp(1))**p,x)
 

Output:

Integral((b*x)**m*(d*x + pi)**n*(f*x + E)**p, x)
 

Maxima [F]

\[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\int { {\left (\pi + d x\right )}^{n} \left (b x\right )^{m} {\left (f x + e\right )}^{p} \,d x } \] Input:

integrate((b*x)^m*(d*x+pi)^n*(f*x+exp(1))^p,x, algorithm="maxima")
 

Output:

integrate((pi + d*x)^n*(b*x)^m*(f*x + e)^p, x)
 

Giac [F]

\[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\int { {\left (\pi + d x\right )}^{n} \left (b x\right )^{m} {\left (f x + e\right )}^{p} \,d x } \] Input:

integrate((b*x)^m*(d*x+pi)^n*(f*x+exp(1))^p,x, algorithm="giac")
 

Output:

integrate((pi + d*x)^n*(b*x)^m*(f*x + e)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\int {\left (\mathrm {e}+f\,x\right )}^p\,{\left (b\,x\right )}^m\,{\left (\Pi +d\,x\right )}^n \,d x \] Input:

int((exp(1) + f*x)^p*(b*x)^m*(Pi + d*x)^n,x)
 

Output:

int((exp(1) + f*x)^p*(b*x)^m*(Pi + d*x)^n, x)
 

Reduce [F]

\[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\text {too large to display} \] Input:

int((b*x)^m*(d*x+Pi)^n*(f*x+exp(1))^p,x)
 

Output:

(b**m*(x**m*(e + f*x)**p*(d*x + pi)**n*d*e*m*x + x**m*(e + f*x)**p*(d*x + 
pi)**n*d*e*n*x + x**m*(e + f*x)**p*(d*x + pi)**n*e*n*pi + x**m*(e + f*x)** 
p*(d*x + pi)**n*e*p*pi + x**m*(e + f*x)**p*(d*x + pi)**n*f*m*pi*x + x**m*( 
e + f*x)**p*(d*x + pi)**n*f*p*pi*x + int((x**m*(e + f*x)**p*(d*x + pi)**n* 
x)/(d**2*e**2*m**2*x + 2*d**2*e**2*m*n*x + d**2*e**2*m*p*x + d**2*e**2*m*x 
 + d**2*e**2*n**2*x + d**2*e**2*n*p*x + d**2*e**2*n*x + d**2*e*f*m**2*x**2 
 + 2*d**2*e*f*m*n*x**2 + d**2*e*f*m*p*x**2 + d**2*e*f*m*x**2 + d**2*e*f*n* 
*2*x**2 + d**2*e*f*n*p*x**2 + d**2*e*f*n*x**2 + d*e**2*m**2*pi + 2*d*e**2* 
m*n*pi + d*e**2*m*p*pi + d*e**2*m*pi + d*e**2*n**2*pi + d*e**2*n*p*pi + d* 
e**2*n*pi + 2*d*e*f*m**2*pi*x + 3*d*e*f*m*n*pi*x + 3*d*e*f*m*p*pi*x + 2*d* 
e*f*m*pi*x + d*e*f*n**2*pi*x + 2*d*e*f*n*p*pi*x + d*e*f*n*pi*x + d*e*f*p** 
2*pi*x + d*e*f*p*pi*x + d*f**2*m**2*pi*x**2 + d*f**2*m*n*pi*x**2 + 2*d*f** 
2*m*p*pi*x**2 + d*f**2*m*pi*x**2 + d*f**2*n*p*pi*x**2 + d*f**2*p**2*pi*x** 
2 + d*f**2*p*pi*x**2 + e*f*m**2*pi**2 + e*f*m*n*pi**2 + 2*e*f*m*p*pi**2 + 
e*f*m*pi**2 + e*f*n*p*pi**2 + e*f*p**2*pi**2 + e*f*p*pi**2 + f**2*m**2*pi* 
*2*x + f**2*m*n*pi**2*x + 2*f**2*m*p*pi**2*x + f**2*m*pi**2*x + f**2*n*p*p 
i**2*x + f**2*p**2*pi**2*x + f**2*p*pi**2*x),x)*d**3*e**3*m**3*p + 3*int(( 
x**m*(e + f*x)**p*(d*x + pi)**n*x)/(d**2*e**2*m**2*x + 2*d**2*e**2*m*n*x + 
 d**2*e**2*m*p*x + d**2*e**2*m*x + d**2*e**2*n**2*x + d**2*e**2*n*p*x + d* 
*2*e**2*n*x + d**2*e*f*m**2*x**2 + 2*d**2*e*f*m*n*x**2 + d**2*e*f*m*p*x...