\(\int (b x)^m (\pi +d x)^n (e+f x)^p \, dx\) [967]

Optimal result

Integrand size = 20, antiderivative size = 65 \[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\frac {\pi ^n (b x)^{1+m} (e+f x)^p \left (1+\frac {f x}{e}\right )^{-p} \operatorname {AppellF1}\left (1+m,-n,-p,2+m,-\frac {d x}{\pi },-\frac {f x}{e}\right )}{b (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {140, 138} \[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\frac {\pi ^n (b x)^{m+1} (e+f x)^p \left (\frac {f x}{e}+1\right )^{-p} \operatorname {AppellF1}\left (m+1,-n,-p,m+2,-\frac {d x}{\pi },-\frac {f x}{e}\right )}{b (m+1)} \]

[In]

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

[Out]

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

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> 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] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 140

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[c^IntPart[n]*((c +
d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = \left ((e+f x)^p \left (1+\frac {f x}{e}\right )^{-p}\right ) \int (b x)^m (\pi +d x)^n \left (1+\frac {f x}{e}\right )^p \, dx \\ & = \frac {\pi ^n (b x)^{1+m} (e+f x)^p \left (1+\frac {f x}{e}\right )^{-p} F_1\left (1+m;-n,-p;2+m;-\frac {d x}{\pi },-\frac {f x}{e}\right )}{b (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\frac {\pi ^n x (b x)^m (e+f x)^p \left (\frac {e+f x}{e}\right )^{-p} \operatorname {AppellF1}\left (1+m,-n,-p,2+m,-\frac {d x}{\pi },-\frac {f x}{e}\right )}{1+m} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (b x)^m (\pi +d x)^n (e+f x)^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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