3.3146 \(\int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{1+n} \, dx\)

Optimal. Leaf size=139 \[ \frac{(b e-a f) (a+b x)^{m+1} (e+f x)^n (c+d x)^{-m-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-n} \left (\frac{b (c+d x)}{b c-a d}\right )^{m+n} F_1\left (m+1;m+n,-n-1;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^2 (m+1)} \]

[Out]

((b*e - a*f)*(a + b*x)^(1 + m)*(c + d*x)^(-m - n)*((b*(c + d*x))/(b*c - a*d))^(m + n)*(e + f*x)^n*AppellF1[1 +
m, m + n, -1 - n, 2 + m, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b^2*(1 + m)*((b*(e + f
*x))/(b*e - a*f))^n)

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Rubi [A]  time = 0.093506, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {140, 139, 138} \[ \frac{(b e-a f) (a+b x)^{m+1} (e+f x)^n (c+d x)^{-m-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-n} \left (\frac{b (c+d x)}{b c-a d}\right )^{m+n} F_1\left (m+1;m+n,-n-1;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*e - a*f)*(a + b*x)^(1 + m)*(c + d*x)^(-m - n)*((b*(c + d*x))/(b*c - a*d))^(m + n)*(e + f*x)^n*AppellF1[1 +
m, m + n, -1 - n, 2 + m, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b^2*(1 + m)*((b*(e + f
*x))/(b*e - a*f))^n)

Rule 140

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), 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*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !SimplerQ[e +
f*x, a + b*x]

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{1+n} \, dx &=\left ((c+d x)^{-m-n} \left (\frac{b (c+d x)}{b c-a d}\right )^{m+n}\right ) \int (a+b x)^m \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^{-m-n} (e+f x)^{1+n} \, dx\\ &=\frac{\left ((b e-a f) (c+d x)^{-m-n} \left (\frac{b (c+d x)}{b c-a d}\right )^{m+n} (e+f x)^n \left (\frac{b (e+f x)}{b e-a f}\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 )^{-m-n} \left (\frac{b e}{b e-a f}+\frac{b f x}{b e-a f}\right )^{1+n} \, dx}{b}\\ &=\frac{(b e-a f) (a+b x)^{1+m} (c+d x)^{-m-n} \left (\frac{b (c+d x)}{b c-a d}\right )^{m+n} (e+f x)^n \left (\frac{b (e+f x)}{b e-a f}\right )^{-n} F_1\left (1+m;m+n,-1-n;2+m;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^2 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.228209, size = 133, normalized size = 0.96 \[ \frac{(a+b x)^{m+1} (e+f x)^{n+1} (c+d x)^{-m-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-n-1} \left (\frac{b (c+d x)}{b c-a d}\right )^{m+n} F_1\left (m+1;m+n,-n-1;m+2;\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )}{b (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1 + m)*(c + d*x)^(-m - n)*((b*(c + d*x))/(b*c - a*d))^(m + n)*(e + f*x)^(1 + n)*((b*(e + f*x))/(b*
e - a*f))^(-1 - n)*AppellF1[1 + m, m + n, -1 - n, 2 + m, (d*(a + b*x))/(-(b*c) + a*d), (f*(a + b*x))/(-(b*e) +
a*f)])/(b*(1 + m))

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Maple [F]  time = 0.147, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-n-m} \left ( fx+e \right ) ^{1+n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-n-m)*(f*x+e)^(1+n),x)

[Out]

int((b*x+a)^m*(d*x+c)^(-n-m)*(f*x+e)^(1+n),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - n}{\left (f x + e\right )}^{n + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-m-n)*(f*x+e)^(1+n),x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - n)*(f*x + e)^(n + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - n}{\left (f x + e\right )}^{n + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-m-n)*(f*x+e)^(1+n),x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m - n)*(f*x + e)^(n + 1), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**m*(d*x+c)**(-m-n)*(f*x+e)**(1+n),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - n}{\left (f x + e\right )}^{n + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-m-n)*(f*x+e)^(1+n),x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - n)*(f*x + e)^(n + 1), x)