3.1880 \(\int (a+b x)^m (c+d x)^{1+2 n-2 (1+n)} \, dx\)

Optimal. Leaf size=51 \[ \frac{(a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0091459, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {7, 68} \[ \frac{(a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 7

Int[(u_.)*(Px_)^(p_), x_Symbol] :> Int[u*Px^Simplify[p], x] /; PolyQ[Px, x] &&  !RationalQ[p] && FreeQ[p, x] &
& RationalQ[Simplify[p]]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.0039276, size = 51, normalized size = 1. \[ -\frac{(a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )}{(m+1) (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m}}{dx+c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}}{d x + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^m/(d*x + c), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{m}}{d x + c}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x + a)^m/(d*x + c), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{m}}{c + d x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**m/(c + d*x), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}}{d x + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^m/(d*x + c), x)