\(\int (a+b x)^m (c+d x)^n \, dx\) [1845]

Optimal result

Integrand size = 15, antiderivative size = 61 \[ \int (a+b x)^m (c+d x)^n \, dx=-\frac {(a+b x)^{1+m} (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,2+m+n,2+n,\frac {b (c+d x)}{b c-a d}\right )}{(b c-a d) (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.21, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {72, 71} \[ \int (a+b x)^m (c+d x)^n \, dx=\frac {(a+b x)^{m+1} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (m+1,-n,m+2,-\frac {d (a+b x)}{b c-a d}\right )}{b (m+1)} \]

[In]

Int[(a + b*x)^m*(c + d*x)^n,x]

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int (a+b x)^m (c+d x)^n \, dx=\frac {(a+b x)^{1+m} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,\frac {d (a+b x)}{-b c+a d}\right )}{b (1+m)} \]

[In]

Integrate[(a + b*x)^m*(c + d*x)^n,x]

[Out]

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

Maple [F]

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

[In]

int((b*x+a)^m*(d*x+c)^n,x)

[Out]

int((b*x+a)^m*(d*x+c)^n,x)

Fricas [F]

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

[In]

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

[Out]

integral((b*x + a)^m*(d*x + c)^n, x)

Sympy [F(-2)]

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

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[In]

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

[Out]

integrate((b*x + a)^m*(d*x + c)^n, x)

Giac [F]

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

[In]

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

[Out]

integrate((b*x + a)^m*(d*x + c)^n, x)

Mupad [F(-1)]

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

[In]

int((a + b*x)^m*(c + d*x)^n,x)

[Out]

int((a + b*x)^m*(c + d*x)^n, x)