\(\int \frac {(b x)^m (c+d x)^n}{e+f x} \, dx\) [954]

Optimal result

Integrand size = 20, antiderivative size = 63 \[ \int \frac {(b x)^m (c+d x)^n}{e+f x} \, dx=\frac {(b x)^{1+m} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,1,2+m,-\frac {d x}{c},-\frac {f x}{e}\right )}{b e (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 63, 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} \[ \int \frac {(b x)^m (c+d x)^n}{e+f x} \, dx=\frac {(b x)^{m+1} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \operatorname {AppellF1}\left (m+1,-n,1,m+2,-\frac {d x}{c},-\frac {f x}{e}\right )}{b e (m+1)} \]

[In]

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

[Out]

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

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 ((c+d x)^n \left (1+\frac {d x}{c}\right )^{-n}\right ) \int \frac {(b x)^m \left (1+\frac {d x}{c}\right )^n}{e+f x} \, dx \\ & = \frac {(b x)^{1+m} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} F_1\left (1+m;-n,1;2+m;-\frac {d x}{c},-\frac {f x}{e}\right )}{b e (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int \frac {(b x)^m (c+d x)^n}{e+f x} \, dx=\frac {x (b x)^m (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,1,2+m,-\frac {d x}{c},-\frac {f x}{e}\right )}{e (1+m)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

int((b*x)^m*(d*x+c)^n/(f*x+e),x)

[Out]

int((b*x)^m*(d*x+c)^n/(f*x+e),x)

Fricas [F]

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

[In]

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

[Out]

integral((b*x)^m*(d*x + c)^n/(f*x + e), x)

Sympy [F(-2)]

Exception generated. \[ \int \frac {(b x)^m (c+d x)^n}{e+f x} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate((b*x)**m*(d*x+c)**n/(f*x+e),x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[In]

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

[Out]

integrate((b*x)^m*(d*x + c)^n/(f*x + e), x)

Giac [F]

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

[In]

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

[Out]

integrate((b*x)^m*(d*x + c)^n/(f*x + e), x)

Mupad [F(-1)]

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

[In]

int(((b*x)^m*(c + d*x)^n)/(e + f*x),x)

[Out]

int(((b*x)^m*(c + d*x)^n)/(e + f*x), x)