\(\int (a+b x)^m (c+d x)^{2-m} \, dx\) [3126]

Optimal result

Integrand size = 19, antiderivative size = 84 \[ \int (a+b x)^m (c+d x)^{2-m} \, dx=\frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-2+m,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{b^3 (1+m)} \]

[Out]

(-a*d+b*c)^2*(b*x+a)^(1+m)*(b*(d*x+c)/(-a*d+b*c))^m*hypergeom([1+m, -2+m],[2+m],-d*(b*x+a)/(-a*d+b*c))/b^3/(1+
m)/((d*x+c)^m)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {72, 71} \[ \int (a+b x)^m (c+d x)^{2-m} \, dx=\frac {(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-2,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right )}{b^3 (m+1)} \]

[In]

Int[(a + b*x)^m*(c + d*x)^(2 - m),x]

[Out]

((b*c - a*d)^2*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-2 + m, 1 + m, 2 + m, -((d*(a
+ b*x))/(b*c - a*d))])/(b^3*(1 + m)*(c + d*x)^m)

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}& = \frac {\left ((b c-a d)^2 (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{2-m} \, dx}{b^2} \\ & = \frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-2+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{b^3 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99 \[ \int (a+b x)^m (c+d x)^{2-m} \, dx=\frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-2+m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )}{b^3 (1+m)} \]

[In]

Integrate[(a + b*x)^m*(c + d*x)^(2 - m),x]

[Out]

((b*c - a*d)^2*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-2 + m, 1 + m, 2 + m, (d*(a +
b*x))/(-(b*c) + a*d)])/(b^3*(1 + m)*(c + d*x)^m)

Maple [F]

\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{2-m}d x\]

[In]

int((b*x+a)^m*(d*x+c)^(2-m),x)

[Out]

int((b*x+a)^m*(d*x+c)^(2-m),x)

Fricas [F]

\[ \int (a+b x)^m (c+d x)^{2-m} \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(2-m),x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m + 2), x)

Sympy [F(-2)]

Exception generated. \[ \int (a+b x)^m (c+d x)^{2-m} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate((b*x+a)**m*(d*x+c)**(2-m),x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int (a+b x)^m (c+d x)^{2-m} \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(2-m),x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 2), x)

Giac [F]

\[ \int (a+b x)^m (c+d x)^{2-m} \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(2-m),x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 2), x)

Mupad [F(-1)]

Timed out. \[ \int (a+b x)^m (c+d x)^{2-m} \, dx=\int {\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m} \,d x \]

[In]

int((a + b*x)^m*(c + d*x)^(2 - m),x)

[Out]

int((a + b*x)^m*(c + d*x)^(2 - m), x)