\(\int (a+b x)^m (A+B x) (c+d x)^n \, dx\) [32]

Optimal result

Integrand size = 20, antiderivative size = 141 \[ \int (a+b x)^m (A+B x) (c+d x)^n \, dx=\frac {B (a+b x)^{1+m} (c+d x)^{1+n}}{b d (2+m+n)}+\frac {(A b d (2+m+n)-B (b c (1+m)+a d (1+n))) (a+b x)^{1+m} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{b^2 d (1+m) (2+m+n)} \]

[Out]

B*(b*x+a)^(1+m)*(d*x+c)^(1+n)/b/d/(2+m+n)+(A*b*d*(2+m+n)-B*(b*c*(1+m)+a*d*(1+n)))*(b*x+a)^(1+m)*(d*x+c)^n*hype
rgeom([-n, 1+m],[2+m],-d*(b*x+a)/(-a*d+b*c))/b^2/d/(1+m)/(2+m+n)/((b*(d*x+c)/(-a*d+b*c))^n)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {81, 72, 71} \[ \int (a+b x)^m (A+B x) (c+d x)^n \, dx=\frac {(a+b x)^{m+1} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (A b d (m+n+2)-B (a d (n+1)+b c (m+1))) \operatorname {Hypergeometric2F1}\left (m+1,-n,m+2,-\frac {d (a+b x)}{b c-a d}\right )}{b^2 d (m+1) (m+n+2)}+\frac {B (a+b x)^{m+1} (c+d x)^{n+1}}{b d (m+n+2)} \]

[In]

Int[(a + b*x)^m*(A + B*x)*(c + d*x)^n,x]

[Out]

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

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 {B (a+b x)^{1+m} (c+d x)^{1+n}}{b d (2+m+n)}+\left (A-\frac {B (b c (1+m)+a d (1+n))}{b d (2+m+n)}\right ) \int (a+b x)^m (c+d x)^n \, dx \\ & = \frac {B (a+b x)^{1+m} (c+d x)^{1+n}}{b d (2+m+n)}+\left (\left (A-\frac {B (b c (1+m)+a d (1+n))}{b d (2+m+n)}\right ) (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 {B (a+b x)^{1+m} (c+d x)^{1+n}}{b d (2+m+n)}+\frac {\left (A-\frac {B (b c (1+m)+a d (1+n))}{b d (2+m+n)}\right ) (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 = 0.10 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.83 \[ \int (a+b x)^m (A+B x) (c+d x)^n \, dx=\frac {(a+b x)^{1+m} (c+d x)^n \left (b B (c+d x)-\frac {(b B c (1+m)+a B d (1+n)-A b d (2+m+n)) \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,\frac {d (a+b x)}{-b c+a d}\right )}{1+m}\right )}{b^2 d (2+m+n)} \]

[In]

Integrate[(a + b*x)^m*(A + B*x)*(c + d*x)^n,x]

[Out]

((a + b*x)^(1 + m)*(c + d*x)^n*(b*B*(c + d*x) - ((b*B*c*(1 + m) + a*B*d*(1 + n) - A*b*d*(2 + m + n))*Hypergeom
etric2F1[1 + m, -n, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)])/((1 + m)*((b*(c + d*x))/(b*c - a*d))^n)))/(b^2*d*(2
+ m + n))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((B*x + A)*(b*x + a)^m*(d*x + c)^n, x)

Sympy [F(-2)]

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

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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