\(\int (b x)^m (c+d x)^n (e+f x) \, dx\) [953]

Optimal result

Integrand size = 18, antiderivative size = 108 \[ \int (b x)^m (c+d x)^n (e+f x) \, dx=\frac {f (b x)^{1+m} (c+d x)^{1+n}}{b d (2+m+n)}-\frac {(c f (1+m)-d e (2+m+n)) (b x)^{1+m} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,-\frac {d x}{c}\right )}{b d (1+m) (2+m+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {81, 68, 66} \[ \int (b x)^m (c+d x)^n (e+f x) \, dx=\frac {(b x)^{m+1} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (\frac {e}{m+1}-\frac {c f}{d (m+n+2)}\right ) \operatorname {Hypergeometric2F1}\left (m+1,-n,m+2,-\frac {d x}{c}\right )}{b}+\frac {f (b x)^{m+1} (c+d x)^{n+1}}{b d (m+n+2)} \]

[In]

Int[(b*x)^m*(c + d*x)^n*(e + f*x),x]

[Out]

(f*(b*x)^(1 + m)*(c + d*x)^(1 + n))/(b*d*(2 + m + n)) + ((e/(1 + m) - (c*f)/(d*(2 + m + n)))*(b*x)^(1 + m)*(c
+ d*x)^n*Hypergeometric2F1[1 + m, -n, 2 + m, -((d*x)/c)])/(b*(1 + (d*x)/c)^n)

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 68

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0])) |
|  !RationalQ[n])

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.75 \[ \int (b x)^m (c+d x)^n (e+f x) \, dx=\frac {x (b x)^m (c+d x)^n \left (f (c+d x)+\frac {(-c f (1+m)+d e (2+m+n)) \left (1+\frac {d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,-\frac {d x}{c}\right )}{1+m}\right )}{d (2+m+n)} \]

[In]

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

[Out]

(x*(b*x)^m*(c + d*x)^n*(f*(c + d*x) + ((-(c*f*(1 + m)) + d*e*(2 + m + n))*Hypergeometric2F1[1 + m, -n, 2 + m,
-((d*x)/c)])/((1 + m)*(1 + (d*x)/c)^n)))/(d*(2 + m + n))

Maple [F]

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

[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]

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

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.98 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.74 \[ \int (b x)^m (c+d x)^n (e+f x) \, dx=\frac {b^{m} c^{n} e x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - n, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 2\right )} + \frac {b^{m} c^{n} f x^{m + 2} \Gamma \left (m + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} - n, m + 2 \\ m + 3 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 3\right )} \]

[In]

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

[Out]

b**m*c**n*e*x**(m + 1)*gamma(m + 1)*hyper((-n, m + 1), (m + 2,), d*x*exp_polar(I*pi)/c)/gamma(m + 2) + b**m*c*
*n*f*x**(m + 2)*gamma(m + 2)*hyper((-n, m + 2), (m + 3,), d*x*exp_polar(I*pi)/c)/gamma(m + 3)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((e + f*x)*(b*x)^m*(c + d*x)^n,x)

[Out]

int((e + f*x)*(b*x)^m*(c + d*x)^n, x)